8086 PINS and SIGNALS
The 8086 pins and signals are shown below. Unless otherwise
indicated, all 8086 pins are TTL compatible. The 8086 can
operate in two modes. These are minimum mode (uniprocessor
system – single 8086) and maximum mode (multiprocessor
system system – more than one 8086). Pin Diagram for the 8086
is given below:
REFER DIAGRAM 3:1.1
Pin(s) Symbol Input/ Output
Description
1 GND - Ground 2-16 AD14-
ADO I/O-3 Output address during
the first part of the bus cycle and inputs or outputs data during the remaining part of the bus cycle.
17 NMI I Nonmaskable interrupt request level triggered
18 INTR I Maskable interrupt request level triggered
19 CLK I Generates clock signals that synchronize the operation of processor.
20 GND Ground 21 RESET I Terminates activity,
clears PSW, IP, DS,SS,ES, and the instruction queue, and
sets CS to FFFF; IP to 0000H; SS to 0000H; DS to 0000H; PSW to 0000H. Processing begins at FFFFO when signal is dropped. Signal must be 1 for at least 4 clock cycles.
22 READY I Acknowledgment from memory or I/O interface that cpu can complete the current bus cucle.
23 TEST I Used in conjuction with the WAIT instruction in multiprocessing environments. A WAIT instruction will cause the cpu to idle, except for processing interruptsm, until a 0 is applied to this pin see chp-11
24-31 Definition depends on mode
32 RD 0-3 Indicates a memory or I/O read is to be performed
33 MN/MX I Cpu is in minimum mode when strapped to +5 v and in maximum mode when grounded
34 BHE/s7 0-3 If o during first of bus cycle this pin indicates that at least one byte of
the current transfer is to be made on pins AD15-AD8 if 1 the transfer is made on AD7-AD0. Status s7 is output during the latter part of bus assigned a meaning
35-38 A19/s6- A16/s3
0-3 During the first part of the bus cycle the upper 4 bits of the address are output and during the remainder of the bus cycle status is output. S3 and S4 indicates the segment register being used as follows; S4 s3 Register 0 0 ES 0 1 SS
1 0 CS or more 1 1 DS s5 gives the current setting of IF. S6 is always 0.
39 AD15 I/0-3 Same as AD14-AD0
40 Vcc Supply voltage - +5 v 10%
1. MN/MX is an input pin used to select one of these modes.
When MN/MX is HIGH, the 8086 operates in the minimum
mode. When MN/MX is LOW, 8086 is configured to support
multiprocessor systems. In this case, the Intel 8288 bus
controller is added to the 8086 to provide bus controls.
2. Pins 2 through 16 and 39 (AD15 - AD0) are a 16-bit
multiplexed address/data bus. During the first clock cycle
AD0-AD15 are the low order 16 bits of address. The 8086
have a total of 20 address lines. The upper four lines (35 –
38)are multiplexed with the status signals for the 8086. These
are the A16/S3, A17/S4, A18/S5, and A19/S6. During the first
clock period of a bus cycle, the entire 20-bit address is
available on these lines. During all other clock cycles for
memory and I/O operations, AD15-AD0 contains the 16-bit
data, and S3, S4, S5, and S6 become status lines. S3 and S4
lines are decoded as follows:
A17/S4 A16/S3 Function
0 0 Extra segment
0 1 Stack segment
1 0 Code or no segment
1 1 Data Segment
Status bits S3 and S4 indicate the segment register that is
being used to generate the address the address and bit S5
reflects the contents of the IF flag. S6 is always held at 0 and
indicates that an 8086 is controlling the system bus.
3. Pins 1 and 20 are grounded. Pin 17 is NMI. This is the
Non Maskable Interrupt input activated by a leading edge. Pin
18 is INTR. INTR is the maskable interrupt input. Pin 19 is
CLK is for supplying the clock signal that synchronizes the
activity within the CPU.
4. Pin 21 (RESET) is the system reset input signal. When the
8086 detects the positive going edge of a pulse on RESET, it
stops all activities until the signal goes LOW. When the reset
is low, the 8086 initializes as follows:
8086 Component Content
Falgs Clear
IP 0000H
CS FFFFH
DS 0000H
SS 0000H
ES 0000H
Queue Empty
5. Pin 22 (READY) is for inputting an acknowledge from a
memory or I/O interface that input data will be put on the data
bus or output data will be accepted from the data bus within
the next clock cycle. In either case, the CPU and its bus
control logic can complete the current bus cycle after the next
clock cycle.
6. Pin 23 TEST is an input pin and is only used by the WAIT
instruction. The 8086 enter a wait state after execution of the
WAIT instruction until a LOW is seen on the TEST pin.
7. Pins 24 to 31 are mode dependent and are considered
later.
8. Pin 32 (RD) is LOW whenever the 8086 is reading data
from memory or an I/O location.
9. Pin 34 (BHE/S7) is used as BHE (BUS High Enable) during
the first clock cycle of an instruction execution. BHE can be
used in conjunction with AD0 to select memory banks. During
all other clock cycles BHE/S7 is used as S7.
Operation BHE AD0 Data pins used
Write/Read a word at an even
Address 0 0 AD15 – AD0
Write/Read a byte at an even
Address 1 0 AD7 – AD0
Write/Read a byte at an odd
Address 0 1 AD15 – AD8
Write/Read a word at an odd
Address 0 1 AD15 – AD8
1 0 AD7 - AD0
10. Pin 40 (VCC) receives the supply voltage, which must be
+5 V.
TIME, SPEED and DISTANCE - Worksheet
1. A person crosses a 600 m long street in 5 minutes. What is his
speed in km per hour?
(a).3.6 (b).7.2 (c).8.4 (d).10
2. An aeroplane covers a certain distance at a speed of 240 kmph
in 5 hours. To cover the same distance in 1 hours, it must travel at
a speed of:
(a).300 kmph (b).360 kmph (c).600 kmph (d).720
kmph
3. A train can travel 50% faster than a car. Both start from point A
at the same time and reach point B 75 kms away from A at the
same time. On the way, however, the train lost about 12.5 minutes
while stopping at the stations. The speed of the car is:
(a).100 kmph (b).110 kmph (c).120 kmph (d).130
kmph
4. A swimmer swims from a point A against a current for 5min and
then in favour of the current for next 5 min and comes to the point
B. if AB=100mts find the speed of the current?
(a)0.5 km/hr (b)0.6km/hr (c)0.7km/hr (d)0.8
km/hr
5. If the current flows at 2 mph and it takes me 3 hrs. to row 9
miles upstream, how long will it take to return?
(a) 1 2/7 hrs (b) 2 5/7 hrs (c)2 2/3 hrs (d)1 5/7
hrs
6. A train 110 metres long travels at 60 km/hr. In what time will it
pass a man walking at 6 km/hr?
(i). against it
(ii). in the same direction
(a) 6 secs and 7 secs (b) 6 secs and 7 1/3 secs
(c) 6 1/3 secs and 7 secs (d) 6 secs and 7.1 secs
7. The ratio between the speeds of two trains is 7: 8. If the second
train runs 400 kms in 4 hours, then the speed of the first train is:
(a).70 km/hr (b).75 km/hr (c).84 km/hr (d).87.5
km/hr
8. A farmer travelled a distance of 61 km in 9 hours. He travelled
partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The
distance travelled on foot is:
(a).14 km (b).15 km (c).16 km (d).17 km
9. Excluding stoppages, the speed of a bus is 54 kmph and
including stoppages, it is 45 kmph. For how many minutes does
the bus stop per hour?
(a).9mins (b).10mins (c).12mins (d).20mins
10. Robert is travelling on his cycle and has calculated to reach
point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12
noon if he travels at 15 kmph. At what speed must he travel to
reach A at 1 P.M.?
(a).8 kmph (b).11 kmph (c).12 kmph (d).14 kmph
11. It takes eight hours for a 600 km journey, if 120 km is done by
train and the rest by car. It takes 20 minutes more, if 200 km is
done by train and the rest by car. The ratio of the speed of the
train to that of the cars is:
(a).2 : 3 (b).3 : 2 (c).3 : 4 (d).4 : 3
12. There are two towns A and B. Anil goes from A to B at 40
kmph and comes back to the starting point at 60 kmph. What is his
average speed during the whole journey?
(a). 84 kmph (b). 4.8 kmph (c). 48 kmph (d). 8.4
kmph
13. A man is standing on a railway bridge which is 180 meter long;
he finds that a train crosses a bridge in 20 secs, but himself in 8
secs. Find the length of the train and its speed?
(a). 120m and 54 kmph (b). 120 kmph and 54m (c). 120m and
48 kmph
(d). 1.20m and 5.4 kmph
14. Two trains 137 meters and 163 meters in length are running
towards each other on parallel lines, one at the rate of 42 kmph
and another at 48 kmph. In what time will they be clear of each
other from the moment they meet?
(a). 12 mins (b). 1.2 mins (c). 12 secs (d). 24 hrs
15. A train 100 meters long takes 6 secs to cross a man walking at
5 kmph in a direction opposite to that of the train. Find the speed
of the train?
(a). 5 kmph (b). 5.5 kmph (c). 60 kmph (d). 55
kmph
16. A train running at 54 kmph takes 20 secs to pass a platform,
next it takes 12 secs to pass a man walking at 6 kmph in the same
direction in which the trains is going. Find the length of the train
and the length of the platform?
(a). 160m and 100m (b). 160m and 140m (c). 130m and 140m
(d). 160m and 130m
17. If a boy walks from house to school at the rate of 4 kmph, he
reaches the school 10 mins earlier than the scheduled time.
However if he walks at the rate of 3 kmph, he reaches 10 mins
late. The distance of the school from his house is
(a). 6 km (b). 4.5 km (c). 4 km (d). 3 km
18. A man drives 150 km from A to B in 3 hrs 20 mins and return
to A in 4 hrs 10 mins. Then the average speed from A to B
exceeds the average speed for the entire trip by
(a). 5 kmph (b). 4.5 kmph (c). 4 kmph (d). 2.5
kmph
19. The ratio between the speeds of A and B is 4:3 and therefore
A takes 10 mins more than the time taken by B to reach a
destination. If A had walked at double the speed he would have
covered the distance in?
(a). 30 mins (b). 25 mins (c). 20 mins (d). 15
mins
20. A car travelling with of its actual speed covers 42 km in 1 hr
40 min 48 sec. Find the actual speed of the car.
(a). 17 6/7 km/hr (b). 25 km/hr (c). 30 km/hr (d). 35
km/hr
SET THEORY and CUBES - WORKSHEET
SET THEORY
1). Which of the following is/are true?
1. A set of all natural numbers is a well defined set.
2. A null set contains an element ‘0’
3. S1 = ( a, b, c ) and S2 = ( a, b, c), S2 is a subset of S1.
4. S1 = (a, b, c, d, e) and S2 = (a, b, c, m). S2 is an improper
subset of S1.
A. 1 only B. 1and 3 only C. 2 and 4 only D. 1, 2 and 4
only
2). Which of the following is not an empty set?
1) A = ( x/x є I, x2 is not positive )
2) B = ( x/x є N, 2x + 1 is even )
3) C = ( x/x є N, x is odd and x2
is even )
4) D = ( x/x є R, x2 + 1 = 0 )
3). If A = ( x: x = 2n – 2, n ≤ 3, n є N ); B = (4n-1: n ≤ 5, n є N ),
find A ∩ B.
A. (0, 2, 4) B. Φ C. (3, 7) D. (2, 3)
4). DIRECTIONS for questions 1 and 2: choose the correct
alternative.
1. Set A, B and C are such that A = ( x2 – 5x + 6 = 0); B = ( y
2
– 8y + 15 = 0 )
and C = ( z2 – 7x + 10 = 0 ). Find ( A ∩ B∩ C ) ∩ ( A U B U C ).
A. ( ) B. (2, 5) C. ( 2,3,5) D. (2,5)
2. If A = (6x2 + x – 15 = 0 ), B = ( 2x
2 – 5X +3 = 0), and C =
(2x2 + x -3 = 0 ); find
A ∩ B∩ C.
A. ( 1 ) B. (3/2) C. ( 3/2, -5/3, 1) D. None of
these
5). In a class of 100, 64% of the students have taken politics and
56% of the students have taken history, how many students have
taken both subjects if all the students take atleast one of these
subjects?
A. 18 B. 20 C. 0 D. 25
6). After recent examination every candidate took French or Latin.
75.8% took French and 49.4% took latin. If the total number of
candidates was 2500, how many took both French and latin?
A. 460 B. 560 C. 1999 D. 640
7). Describe the shaded region?
8). In a class of 120 students numbered 1 to 120, all even
numbered students opt for Physics, whose numbers are divisible
by 5 opt for Chemistry and those whose numbers are divisible by
7 opt for Math. How many opt for none of the three subjects?
A. 19 B. 41 C. 21 D. 57 E. 26
9). Of the 200 candidates who were interviewed for a position at a
call center, 100 had a two-wheeler, 70 had a credit card and 140
had a mobile phone. 40 of them had both, a two-wheeler and a
credit card, 30 had both, a credit card and a mobile phone and 60
had both, a two wheeler and mobile phone and 10 had all three.
How many candidates had none of the three?
A. 0 B. 20 C. 10 D. 18 E. 25
10). In a class of 40 students, 12 enrolled for both English and
German. 22 enrolled for German. If the students of the class
enrolled for at least one of the two subjects, then how many
students enrolled for only English and not German?
A. 30 B. 10 C. 18 D. 28 E. 32
11). In a class 40% of the students enrolled for Math and 70%
enrolled for Economics. If 15% of the students enrolled for both
Math and Economics, what % of the students of the class did not
enroll for either of the two subjects?
A B
A. 5% B. 15% C. 0% D. 25% E. None of these
12).There are three different cable channels namely Ahead, Luck
and Bang. In a survey it was found that 850 viewers respond to
Bang, 200 to Luck, and 300 to Ahead. 200 viewers respond to
exactly two channels and 50 to none. What percentage of the
viewers responded to all three?
A. 10 B. 12 C. 14 D. None of these
13). Assuming 20% respond to Ahead and Bang, and 16%
respond to Bang and Luck, what is the percentage of viewers who
watch only Luck?
A. 20 B. 10 C. 16 D. None of these
14). In a survey about the popularity of magazines A, B and C
among 200 pupils, 78 pupils like A, 72 like B, 90 like C, 28 like A
and B only, 22 like B and C only, 10 like A and C only. Find the
maximum number of pupils who do not like any of the magazines?
A. 20 B. 24 C. 16 D. None of these
CUBES
Directions to Solve
The following questions are based on the information given below:
1. A cuboid shaped wooden block has 6 cm length, 4 cm breadth
and 1 cm height.
2. Two faces measuring 4 cm x 1 cm are coloured in black.
3. Two faces measuring 6 cm x 1 cm are coloured in red.
4. Two faces measuring 6 cm x 4 cm are coloured in green.
5. The block is divided into 6 equal cubes of side 1 cm (from 6 cm
side), 4 equal cubes of side 1 cm(from 4 cm side).
1. How many cubes having red, green and black colours on at
least one side of the cube will be formed?
A.16 B.12 C.10 D.4
2. How many small cubes will be formed?
A.6 B.12 C.16 D.24
3. How many cubes will have 4 coloured sides and two non-
coloured sides ?
A.8 B.4 C.16 D.10
4. How many cubes will have green colour on two sides and rest of
the four sides having no colour?
A.12 B.10 C.8 D.4
5. How many cubes will remain if the cubes having black and
green coloured are removed?
A.4 B.8 C.12 D.16
Directions to Solve
The following questions are based on the information given below:
1. All the faces of cubes are painted with red colour.
2. The cubes is cut into 64 equal small cubes.
1. How many small cubes have only one face coloured?
A.4 B.8 C.16 D.24
2. How many small cubes have no faces coloured ?
A.24 B.8 C.16 D.0
3. How many small cubes are there whose three faces are
coloured ?
A.4 B.8 C.16 D.24
4. How many small cubes are there whose two adjacent faces are
coloured red ?
A.0 B.8 C.16 D.24
Directions to Solve
All the six faces of a cube of a cube are coloured with six different
colours - black, brown, green, red, white and blue.
1. Red face is opposite to the black face.
2. Green face is between red and black faces.
3. Blue face is adjacent to white face.
4. Brown face is adjacent to blue face.
5. Red face is in the bottom.
1. The upper face is
A. White B. Black C. Brown D. None of these
2. The face opposite to brown is _________
A. Blue B. White C. Green D. Red
3. Which of the following is adjacent to green?
A. Black, white, brown, red B. Blue, black, red, white
C. Red, black, blue, white D. None of these
4. Which face is opposite to green?
A. Red B. White C. Blue D. Brown
Ratio Proportion and Mixtures Solutions
Work Sheet
Ratio Proportion
1. If a:b=3:7,find the value of (5a+b):(4a+5b)
(a)15:44 (b)22:35 (c)15:49 (d)22:47
2. Which of the following ratios is greatest?
(a) 7:15 (b)22:35 (c)15:49 (d)21:29
3. If 76 is divided in to four parts proportional to 7,5,3,4 the smallest part
is
(a) 12 (b) 15 (c)16 (d)19
4. Suppose x varies inversely as square of y and when y=3 and x=4. Find
x, when y=6
(a) 1 (b)2 (c)3 (d)4
5. Divide Rs. 1162 among A, B, C in the ratio 35:28:20
(a) 120 (b)280 (c)170 (d)210
6. A bag contains one rupee, 50 paisa and 25 paisa coins in the ratio
1:2:4.If the total amount is Rs. 75,then find the number of 50 paisa coins
in the bag
(a)25 (b)50 (c)75 (d)100
7. If a is 75% of b,b is 150% of c and d is 25% of c,then find a:d
(a)9:1 (b)9:2 (c)8:3 (d)8:1
8. The present ages of two persons are in the ratio 7:8.Twenty years ago
the ratio of their ages was 9:11.Find the present age of the older son
(a)64 years (b)72 years (c)56 years (d)40 years
9. A certain sum is divided among A,B and C in a manner that for every
rupee that A gets, B gets 75 paisa and for every rupee that B gets,C gets
50 paisae.If B's share in the total sum is Rs 840.Find the share of A
(a)Rs 2380 (b)Rs 2240 (c)Rs 1750 (d)Rs 1120
10.A certain amount of money is divided among A,B and C such that A
gets half of what B and C gets together. B gets one third of what A and C
together get.If A got Rs 500 more than B, then how much money was
divided?
(a)Rs4500 (b)Rs. 6000 (c)Rs. 8000 (d)none of these
11. Annual income of A and B are in the ratio 4:3 and their expenses, as
3:2. If each saves Rs 600 at the end of the year, find the annual income of
each
(a) 1200, 900 (b) 1500, 1125
(c) 2400, 1800 (d) 3600, 2700
12. The value of a silver coin varies directly as the square of its diameter
when the thickness remains the same and directly as its thickness while
the diameter is constant.
Two silver coins have their diameter as 4:3. Find the ratio of their
thickness if the value of the first is 4 times the value of the second.
(a) 4:3 (b) 3:2 (c) 2:1 (d) 9:4
Mixture Solutions
1.15 litres of mixture contains 20% alcohol and the rest water. If 3 litres
of water be mixed with it, the percentage of alcohol in the new mixture
would be
(a)15 b)16 2/3 (c)17 (d)none of these
2.85 kg of a mixture contains milk and water in the ratio 27:7.How much
more water is to be added to get a new mixture containing milk and water
in the ratio 3:1?
(a)5kg (b)6.5kg (c)7.25kg (d)8kg
3. Two liquids are mixed in the ratio 3:2 and the solution is sold at Rs.11
per liter at 10%
Profit. If the first liquid costs Rs.2 more than the second, what is the cost
price of the first liquid?
(a) Rs.8.80 (b) Rs.10.80 (c) Rs.9.80 (d) none
4. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third
variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 kg, the price
of the third variety per kg will be:
(a) Rs. 169.50 (b) Rs. 170 (c)Rs. 175.50 (d) Rs. 180
5. In what ratio must a person mix three kinds of tea costing Rs.60/kg,
Rs.75/kg and
Rs.100 /kg, so that the resultant mixture when sold at Rs.96/kg yields a
profit of 20%?
(a)1 : 2 : 4 (b)3 : 7 : 6 (c)1 : 4 : 2 (d)None of these
6. Two liquids A and B are in the ratio 5:1 in container 1 and in container
2, they are in the ratio 1:3. In what ratio should the contents of the two
containers be mixed so as to obtain a mixture of A and B in the ratio 1:1?
(a) 2 : 3 (b) 4 : 3 (c) 3 : 2 (d) 3 : 4
7. 85% kg of a mixture contains milk and water in the ratio 27:7. How
much more water is to be added to get a new mixture containing milk and
water in the ratio 3:1?
(a) 5 kg (b) 6.5 kg (c) 7.25 kg (d) 8 kg
8. Find the proportion in which 2 kinds of tea costing Rs.3 and Rs.3.30
per pound are mixed up to produce a mixture of cost Rs. 3.20 per pound?
(a) 2:1 (b) 1:2 (c) 3:2 (d) 2:3
9. A grocer sells one kind of tea at Rs.2.70 per kg and loss is 10% and
another at Rs.4.50 per kg and gains 12 ½ %. How should the two
quantities of tea be mixed so that the mixture may be sold at Rs.3.95 per
kg at the profit of 25%?
(a) 42:8 (b) 16:2 (c) 21:4 (d) 4:21
10. In a mixture 60 litres, the ratio of milk and water 2 : 1. If the this ratio
is to be 1 : 2, then the quanity of water to be further added is:
(a) 20 litres (b) 30 litres (c) 40 litres (d) 60 litres
PROFIT AND LOSS
WORK SHEET
1. By selling a book for Rs.115.20, a man losses 10%. At what price
should he sell it to gain?
a). Rs.117.50 b). Rs.134.40 c). Rs.120 d). Rs.5
2. Ashok purchased a radio set and sold it to Shyam at a profit of 25%
and Shyam sold it to Mohan at a loss of 10% and Mohan paid Rs.675 for
it. For how much did Ashok purchase it?
a). 599 b). 674 c). 600 d). 300
3. A man sold two houses for Rs.6,75,958 each. On one he gains 16%
while on the other he loses 16%. How much does he gain or lose in the
whole transaction?
a). 2.56% b). 25.6% c). 15.6% d). 1.56%
4. A grain dealer cheats to the extent of 10% while buying as well as
selling by using false weights. His total gain is
a). 200% b). 8.34% c). 21% d). 12%
5. A dishonest dealer professes to sell his goods at cost price but uses a
weight of 960 gms, for a kg weight. Find his gain percent.
a). 6 ¼% b). 8.5% c). 4 2/3% d). 4 1/6%
6. Find a single discount equivalent to series discount of 20%, 10% and
5%?
a). 31.6% b). 3.16% c). 1.8% d). 0.31%
7. If C.P of 21 oranges is equal to S.P of 18 oranges, then profit percent
is____?
a). 15 2/3% b). 16 2/3% c). 21/18% d). 3%
8. A man sold two horses for Rs.3000 each gaining 25% on the one and
losing 25% on the other. His loss percent is___?
a). 6.15% b). 5.25% c). 6.25% d). 25%
9. A man buys lemon at the rate of 9 for 80 paisa and sells them at 11 for
120 paisa. His gain per lemon is
a). 200/99 b). 100/99 c). 99/200 d). 99/100
10. A man sells an article at 5% above its C.P. If he had bought it at 5%
less than what he paid for it and sold it for Rs.2 less, he would have
gained 10%. Find the C.P of the article?
a). 401 b). 399 c). 400 d). 991
11. If a person makes a profit of 10% on one fourth of the quantity sold
and a loss of 20% on the rest, then what is his average percent profit or
loss?
a). 12.5% loss b). 12.5 % profit c). 11.9% profit d). 12% loss
12. An article was sold at a profit of 10%. Had it been sold at a loss of
20%, the S.P would have been Rs.90 less. Find the C.P of the article.
a). 299 b). 399 c). 328.5 d). 300
13. A person sold his watch for Rs.75 and got a % profit equal to C.P.
Find the C.P of the watch?
a). 50 b). 75 c). 25 d). 100
14. If goods be purchased for Rs.450 and one-third be sold at a loss of
10%, at what gain % should the remainder be sold as to gain 20% on the
whole transaction?
a). 28% b). 35% c). 14.8% d). 100%
15. A man loses Rs.20 by selling a toy at Rs.3 per piece and gains Rs. 30
by selling the same piece at Rs. 3.25 per piece. The number of pieces sold
by the man is?
a). 800 b). 5 c). 200 d). 199
16. A sold an article to B at a gain of 15%. B sold it to C at a gain of
20%. If C paid Rs.897, then A would have paid?
a). 6,500 b). 1,500 c). 65 d). 650
17. A man bought a horse and sold it at a gain of 10%. If he had bought
the horse at 20% less and sold it for Rs.100 more, he would have a profit
of 40%. The C.P of the horse is?
a). 5000 b). 50,000 c). 500 d). 499
18. A merchant blends two varieties of tea from two different tea gardens,
one costing Rs.45 per kg and other Rs.60 per kg in the ratio of 7:3
respectively. He sells the blended variety at Rs.54.45 per kg. His profit %
is?
a). no profit b). 1% c). 10% d). 12%
19. A shopkeeper professes to sell his articles on C.P. But he uses false
weight of 900 gm for 1 kg. His gain percentage is?
a). 11% b). 1/9% c). 9% d). 11 1/9%
20. A publisher sells book to retail dealer at Rs.5 a copy but allows 25
copies to count as 24. If the retailer sells each of the 25 copies at Rs.6, the
profit percent made by him is?
a). 0.25% b). 25% c). 2.5% d). 250%
21. After getting two successive discounts, a shirt with a list price of
Rs.150 is available at Rs.105. If the second discount is 12.5%, find the
first discount?
a). 18.7% b). 45% c). 20% d). 2%
22. A merchant sold two dresses for Rs 2400 each. He made a 25% profit
on one and lost 20% on the other. What was his net gain or loss on the
sale of two dresses?
a). 120 b). 100 c). 130 d). 150
PROFIT AND LOSS
WORK SHEET
1. By selling a book for Rs.115.20, a man losses 10%. At what price
should he sell it to gain?
a). Rs.117.50 b). Rs.134.40 c). Rs.120 d). Rs.5
2. Ashok purchased a radio set and sold it to Shyam at a profit of 25%
and Shyam sold it to Mohan at a loss of 10% and Mohan paid Rs.675 for
it. For how much did Ashok purchase it?
a). 599 b). 674 c). 600 d). 300
3. A man sold two houses for Rs.6,75,958 each. On one he gains 16%
while on the other he loses 16%. How much does he gain or lose in the
whole transaction?
a). 2.56% b). 25.6% c). 15.6% d). 1.56%
4. A grain dealer cheats to the extent of 10% while buying as well as
selling by using false weights. His total gain is
a). 200% b). 8.34% c). 21% d). 12%
5. A dishonest dealer professes to sell his goods at cost price but uses a
weight of 960 gms, for a kg weight. Find his gain percent.
a). 6 ¼% b). 8.5% c). 4 2/3% d). 4 1/6%
6. Find a single discount equivalent to series discount of 20%, 10% and
5%?
a). 31.6% b). 3.16% c). 1.8% d). 0.31%
7. If C.P of 21 oranges is equal to S.P of 18 oranges, then profit percent
is____?
a). 15 2/3% b). 16 2/3% c). 21/18% d). 3%
8. A man sold two horses for Rs.3000 each gaining 25% on the one and
losing 25% on the other. His loss percent is___?
a). 6.15% b). 5.25% c). 6.25% d). 25%
9. A man buys lemon at the rate of 9 for 80 paisa and sells them at 11 for
120 paisa. His gain per lemon is
a). 200/99 b). 100/99 c). 99/200 d). 99/100
10. A man sells an article at 5% above its C.P. If he had bought it at 5%
less than what he paid for it and sold it for Rs.2 less, he would have
gained 10%. Find the C.P of the article?
a). 401 b). 399 c). 400 d). 991
11. If a person makes a profit of 10% on one fourth of the quantity sold
and a loss of 20% on the rest, then what is his average percent profit or
loss?
a). 12.5% loss b). 12.5 % profit c). 11.9% profit d). 12% loss
12. An article was sold at a profit of 10%. Had it been sold at a loss of
20%, the S.P would have been Rs.90 less. Find the C.P of the article.
a). 299 b). 399 c). 328.5 d). 300
13. A person sold his watch for Rs.75 and got a % profit equal to C.P.
Find the C.P of the watch?
a). 50 b). 75 c). 25 d). 100
14. If goods be purchased for Rs.450 and one-third be sold at a loss of
10%, at what gain % should the remainder be sold as to gain 20% on the
whole transaction?
a). 28% b). 35% c). 14.8% d). 100%
15. A man loses Rs.20 by selling a toy at Rs.3 per piece and gains Rs. 30
by selling the same piece at Rs. 3.25 per piece. The number of pieces sold
by the man is?
a). 800 b). 5 c). 200 d). 199
16. A sold an article to B at a gain of 15%. B sold it to C at a gain of
20%. If C paid Rs.897, then A would have paid?
a). 6,500 b). 1,500 c). 65 d). 650
17. A man bought a horse and sold it at a gain of 10%. If he had bought
the horse at 20% less and sold it for Rs.100 more, he would have a profit
of 40%. The C.P of the horse is?
a). 5000 b). 50,000 c). 500 d). 499
18. A merchant blends two varieties of tea from two different tea gardens,
one costing Rs.45 per kg and other Rs.60 per kg in the ratio of 7:3
respectively. He sells the blended variety at Rs.54.45 per kg. His profit %
is?
a). no profit b). 1% c). 10% d). 12%
19. A shopkeeper professes to sell his articles on C.P. But he uses false
weight of 900 gm for 1 kg. His gain percentage is?
a). 11% b). 1/9% c). 9% d). 11 1/9%
20. A publisher sells book to retail dealer at Rs.5 a copy but allows 25
copies to count as 24. If the retailer sells each of the 25 copies at Rs.6, the
profit percent made by him is?
a). 0.25% b). 25% c). 2.5% d). 250%
21. After getting two successive discounts, a shirt with a list price of
Rs.150 is available at Rs.105. If the second discount is 12.5%, find the
first discount?
a). 18.7% b). 45% c). 20% d). 2%
22. A merchant sold two dresses for Rs 2400 each. He made a 25% profit
on one and lost 20% on the other. What was his net gain or loss on the
sale of two dresses?
a). 120 b). 100 c). 130 d). 150
PERCENTAGE
WORKSHEET I
1. The length and breadth of a rectangle are changed by +20% and by -10%
respectively. What is the percentage change in the area of the rectangle?
(a) 4% decrease (b) 4 % increase (c) 8 % decrease (d) 8%
increase
2. If A’s income is 33% more than that of B, then how much % is B’s
income is less than that of A?
(a) 17.8% (b) 24.8% (c) 23.8% (d) 21%
3. If the price of tea is increased by 20%, find by how much percent must a
house holder reduce her consumption of tea, so as not to increase the
expenditure?
(a) 4% (b) 13 3/2% (c) 16 2/3 % (d) 16%
4. The population of a town is 1, 76,400. If it increases at the rate of 5% per
annum, what will be its population 2 years hence? What was it 2 years ago?
(a) 1,94,481 and 1,60,000 (b) 2,00,000 and 1,28,500
(c)1,55,800 and 2,78,000 (d) 1,10,100 and 1,76,400
5. A man spent 20% of his monthly income on house rent. Out of the
remaining, he spent 70% on food. If he had balance of Rs.250 at month
end, the monthly income of the man is approximately:
(a) 1042 rupees (b) 2000 rupees (c) 1800 rupees (d) 900 rupees
6. A woman has a certain number of apples, of which 13% are bad. She
gives 75% of the remainder to charity, and then has 261 left. How many did
she have initially?
(a) 1300 (b) 1200 (c) 1250 (d) 610
7. In an election between 2 candidates, Bhiku gets 65% of the total valid
votes. If the total number of votes is 12000, what is the number of votes
that the other candidate Sourabh gets if 25% of the total votes polled were
declared invalid?
(a) 3000 (b) 4150 (c) 5150 (d) 3150
8. 80 litres of milk and water contains 5% water in it. How much more
water must be added to the mixture to make it 25% in the resulting mixture?
(a) 10 litres (b) 21.33 litres (c) 20 litres (d) 25 litres
This is the required quantity of water.
9. A’s salary is 20% lower than B’s salary which is 25% lower than C’s
salary. By what percent is C’s salary more than A’s salary?
(a) 60% (b)65% (c)66.67% (d)70%
10. A’s salary is first decreased by 30% and then increased by 40%. The
result
is the same as B’s salary being first increased by 30% and then decreased
by 40%. Find the ratio of B’s salary to A’s salary.
(a) 49:39 (b) 47:37 (c) 19:17 (d) 7:5
11. A water melon weighs 5000gms. 99% of its weight is water. It is kept in
a drying room and after some time it turns out that only 98% of its weight is
water. What is its weight now?
(a) 4500 (b) 2500 (c) 4950 (d) none
12. In an examination 65% of the students passed in Science, 68% passed in
Mathematics and 25% failed in both. Find the pass percentage?
(a) 55% (b) 58% (c) 60% (d) 75%
13. In a class of 52 students, 25% are rich and the others are poor. There
are 20 female members in the class, of whom 55% are poor. How many
rich male members are there in the class?
(a) 4 (b) 8 (c) 5 (d) 6
14. At King’s college, 60% of the students are boys and the rest are girls.
Further, 15% of the boys and 7.5% of the girls are getting a fee waiver. If
the number of those getting a fee waiver is 90, find the total number of
students getting 50% fee-concession, if it is given that 10% of those not
getting a fee-waiver are eligible for half-fee concession?
(a) 64 (b) 66 (c) 68 (d) 63
15. A person fails in an exam for want of 12 marks. Had he got 27 marks
more he would have passed by scoring a clear 50% in the exam. Find the
maximum marks given that the pass percentage is 40?
(a) 100 (b) 175 (c) 300 (d) 150
16. A pile of coins is to be divided amongst 6 people A,B,C,D,E,F in that
order such that each gets twice the other . If A gets the least and F the
highest, what is the % increase in number of coins received by (a)F over C
and (b)E over B?
(a) 700%, 700% (b) 600%, 500% (c) 400%, 300% (d) 500%, 400%
17. Anindo’s project report consists of 25 pages each of 60 lines with 75
characters on each line. In case, the number of lines is reduced to 50, but
the number of characters is increased to 100 per line, what is the percentage
change in the number of pages?
(a) 10% decrease (b) 12% decrease (c) 20% increase (d) 5% decrease
ASSIGNMENT SHEET –I
1. At an election where there are only two candidates. The candidate who
gets 32% of the votes is elected by a majority of 144 votes. Find the total
number of votes recorded assuming that no vote is void.
(a) 500 (b) 300 (c) 450 (d) 600
2. A pie-chart giving the marks distribution in a certain exam shows that
20% got 80%, 25%got 70%, 25%got 60% and the rest got 50%. Find the
average percentage of marks scored by the students.
(a) 65 (b) 60 (c) 61 (d) 63.5
3. A man buys a truck for Rs.2,50,000. The annual repair cost comes to
2.0% of the price of purchase. Besides, he has to pay an annual tax of
Rs.2000. At what monthly rent must he rent out the truck to get a return of
24% on his net annual investment?
(a) 4000 rupees (b) 3140 rupees (c) 5140 rupees (d) 4140 rupees
4. The value of a machine depreciates at 20 % per annum. If the present
value of the machine is Rs.10, 000, find its value after three years.
(a) 5000 rupees (b) 5120 rupees (c) 5400 rupees (d) 6000 rupees
5. In an exam, 20% failed in English and 15% failed in Maths. If 12%
failed in both the exams, find the pass percentage.
(a) 70% (b) 72% (c) 77% (d) 75%
6. The side of a square is increased by 10%. Find the percentage change in
its area.
(a) 21% (b) 30% (c) 10% (d) 100%
7. Vicky’s salary is 50% more than Rahul’s. Vicky got a raise of 40% on
his salary while Rahul got a 25% raise on his salary. By what percent is
Vicky’s new salary more than Rahul’s?
(a) 56% (b) 68% (c) 72% (d) 84%
8. An ore contains 25% of an alloy that has 80% iron. Other than this, in
the remaining 75% of the ore, there is no iron. How many kg of the ore are
needed to obtain 60 kgs of pure iron?
(a) 250 kg. (b) 275 kg. (c) 350 kg. (d) 300 kg.
9. Sandeep spends 20% of his salary on house rent, and 20% of the rest he
spends on his children’s education and 14% of his total salary he spends on
clothes. After his expenditure, he is left with Rs.7000. What is his total
salary?
(a) 15000 rupees (b) 14000 rupees (c) 9000 rupees (d) 12000
rupees
10. Fresh fruit contains 68% water and dry fruit contains 20% water. How
much
dry fruit can be obtained from 100 kgs of fresh fruits?
(a) 25 kg (b) 40 kg (c) 30 kg (d) 20 kg
11. If A’s height is 24% less than that of B, then how much percent is B’s
height more than that of A?
(a) 91.0% (b) 31.57% (c) 29.1% (d) 13.8%
12. If the price of sugar decreases by 20%, find by how much percent must
a house holder increases her consumption of sugar so as not to decrease the
expenditure?
(a) 27% (b) 26.5% (c) 25% (d) 25.5%
13. The value of the machine depreciates at the rate of 10% per annum, if
its present value is Rs. 16,200 what would be its worth after 2 years? What
was the value of the machine 2 years ago?
(a) 1,94,481 and 1,60,000 (b) 2,00,000 and 1,28,500
(c) 1,55,800 and 2,78,000 (d) 1,31,220 and 2,00,000
LOGICAL REASONING - Worksheet
CODING - DECODING
1. If a meaningful word can be formed by rearranging the letters
USCALA, the first letter so formed is the answer. If no such letter can be
formed, the answer is X.
(a) C (b) A (c) S (d) L
2. If Cushion is called Pillow, Pillow is called mat, mat is called bed
sheet, and bed sheet is called cover, which will be spread on the floor?
(a) Pillow (b) Mat (c) Bed sheet (d) Cover
3. If ROTARY is coded as 36 and ROTARIAN is coded as 64, how will
you code ROTARACT?
(a) 64 (b) 26 (c) 72 (d) 36
4. In a certain code language
‘tom na rod’ means ‘give me sweet’
‘jo ta rod’ means ‘you and me’
‘pot ta noc’ means ‘you are good’
‘jo mit noc’ means ‘good and bad’
Which of the following represents ‘bad’ in the language?
(a)mit (b)noc (c)jo (d)rod
5. In a certain code language LANDMINE is written as PYRBQGRC.
How will HOMEMADE be written in that code language?
(a)LMPCPZID (b)LMPCOYHC (c)LMQCQYHC (d)LMOCOZID
6. If in the word ‘DISTURBANCE’ the first letter is interchanged with
the last letter, the second letter is interchanged with the tenth letter and so
on, which letter would come after the letter “T” in the newly formed
word?
(a) I (b) N (c) S (d) U
7. In a certain code language-
‘743’ means ‘Mangoes are good’.
‘657’ means ‘Eat good food’
‘934’ means ‘Mangoes are ripe’.
Which digit means ‘ripe’ in that language ?
(a) 5 (b) 4 (c) 9 (d) 7
BLOOD RELATIONSHIP
8. James goes to the local church and sees a man sitting to his left. He
finds that the man is his relative. The man is the husband of the sister of
his mother. How is the man related to James?
(a) Brother (b) Uncle (c) Nephew (d) Father
9. Pointing to a photograph a man said “He is my brother’s mother’s
brother’s only sibling’s son.” Whose photograph was it?
(a) His own (b) Brother (c) Cousin (d) Cannot say
10. C is A’s father’s nephew. D is A’s cousin but not the brother of C.
How is D related to C?
(a) Cousin (b) Sister (c) Father (d) Cannot say
11. When Anil saw Mani, he recalled ‘He is the son of the father of the
mother of my daughter.’ How is Mani related to Anil?
(a) Brother-in-law (b) Brother (c) Cousin (d) Uncle
12. Pointing out to a lady, a girl said,” She is the daughter-in-law of the
grandmother of my father’s only son.” How is the lady related to the girl?
(a) Sister-in-law (b)Aunt (c) Mother (d) mother-in-law
SYLLOGISM
Directions Q 11 to Q 13
Each question consists of two statements followed by four options
consisting of two options put together in a specific order. Choose the
option which indicates a valid argument, that is, the third statement is a
conclusion drawn from the preceding two statements.
13. All shares are debentures. No debentures are deposits.
(a) No shares are deposits
(b) Some shares are deposits
(c) All shares are deposits
(d) None of the above
14. Some students are smart. All students are hardworking
(a) some hardworking students are smart
(b) some smart students are hardworking
(c) Both (a) and (b)
(d) None of the above
15.Some actors are singers. All the singers are dancers.
1. Some actors are dancers.
2. No singer is actor.
A. Only (1) conclusion follows
B. Only (2) conclusion follows
C. Either (1) or (2) follows
D. Neither (1) nor (2) follows
E. Both (1) and (2) follow
16. All scientists are fools. All fools are literates.
(a) All scientists are literates
(b) All literates are scientists
(c) No scientists are literates
(d) Both (a) and (b) are correct
17. Statements:
some roses are red
some red are black.
Conclusions:
(I) No black is a rose.
(II) No rose is a black
(a) I (b) II (c) Both (d) None of these
ODD ONE OUT:
19 (a) Paradeep (b) Kalpakkam (c) Kota (d) Tarapur
20 (a) Silk (b) Cotton (c) Nylon (d) wool
21 (a) Gold (b) Silver (c) Bronze (d) Iron
22 (a) Motorcar (b) Tractor (c) Bus (d) Tram
23 (a) triangle (b) Pentagon (c) square (d) tangent
24. Find the odd one out
(a) Cubic metre (b) cubic centimetre (c) Gallons (d) Square metre
VENN-DIAGRAM:
25. Read the following Venn-diagram carefully and answer the question
given below:
Which letters represent girls who are athletes but not singers?
(a) C (b) A (c) D (d) E
26. Study the following figure and answer the questions given below.
i). How many educated people are employed ?
A. 9 B. 18
C. 20 D. 15
ii). How many backward people are educated ?
A. 9 B. 28
C. 14 D. 6
iii). How many backward uneducated people are employed ?
A. 14 B. 5
C. 7 D. 11
iv). How many backward people are not educated ?
A. 3 B. 14 C. 22 D. 25
NUMBER SERIES
27. Find the next of the series. 83, 82, 81, 77, 69, 60, ?
(a) 54 (b) 33 (c) 35 (d) 52
28. Find the missing term of the series. 0, 2, 6, ?, 20, 30, 42.
(a) 8 (b) 10 © 12 (d) 14
29. Look at this series: 36, 34, 30, 28, 24, ... What number should come
next?
A.20 B.22 C.23 D.26
30. Look at this series: 22, 21, 23, 22, 24, 23, ... What number should
come next?
A.22 B.24 C.25 D.26
31. Look at this series: 53, 53, 40, 40, 27, 27, ... What number should
come next?
A.12 B.14 C.27 D.53
33. Raji’s house is to the right of Vel’s house at a distance of 20 metres in
the same row facing north. Shakthi’s house is in the north-east direction
of Vel’s house at a distance of 25 metres. Determine Raji’s house is in
which direction with respect to Shakthi’s house.
(a) South (b) South-east (c) South-west (d) None of these
34. In a row of boys, Rajan is tenth from the right and Suraj is tenth from
the left. When Rajan and Suraj interchange their positions, Suraj will be
twenty-seven from the left. Which of the following will be Rajan’s
position from the right?
(a) Twenty fifth (b) Twenty sixth (c) Twenty seven (d)
Twenty eight
1. CUBOID
Let length = l, breadth = b and height = h units. Then
i. Volume = (l x b x h) cubic units.
ii. Surface area = 2(lb + bh + lh) sq. units.
iii. Diagonal = l2 + b
2 + h
2 units.
2. CUBE
Let each edge of a cube be of length a. Then,
i. Volume = a3 cubic units.
ii. Surface area = 6a2 sq. units.
iii. Diagonal = 3a units.
3. CYLINDER
Let radius of base = r and Height (or length) = h. Then,
i. Volume = ( r2h) cubic units.
ii. Curved surface area = (2 rh) sq. units.
iii. Total surface area = 2 r(h + r) sq. units.
4. CONE
Let radius of base = r and Height = h. Then,
i. Slant height, l = h2 + r
2 units.
ii. Volume = r2h cubic units.
iii. Curved surface area = ( rl) sq. units.
iv. Total surface area = ( rl + r2) sq. units.
5. SPHERE
Let the radius of the sphere be r. Then,
i. Volume = r3
cubic units.
ii. Surface area = (4 r2) sq. units.
6. HEMISPHERE
Let the radius of a hemisphere be r. Then,
i. Volume = r3
cubic units.
ii. Curved surface area = (2 r2) sq. units.
iii. Total surface area = (3 r2) sq. units.
Note: 1 litre = 1000 cm3.
FUNDAMENTAL CONCEPTS
1. Results on Triangles:
i. Sum of the angles of a triangle is 180°.
ii. The sum of any two sides of a triangle is greater
than the third side.
iii. Pythagoras Theorem:
In a right-angled triangle, (Hypotenuse)2 =
(Base)2 + (Height)2.
iv. The line joining the mid-point of a side of a triangle
to the positive vertex is called the median.
v. The point where the three medians of a triangle
meet, is called centroid.The centroid divided each
of the medians in the ratio 2 : 1.
vi. In an isosceles triangle, the altitude from the vertex
bisects the base.
vii. The median of a triangle divides it into two
triangles of the same area.
viii. The area of the triangle formed by joining the mid-
points of the sides of a given triangle is one-fourth
of the area of the given triangle.
2. Results on Quadrilaterals:
i. The diagonals of a parallelogram bisect each other.
ii. Each diagonal of a parallelogram divides it into
triangles of the same area.
iii. The diagonals of a rectangle are equal and bisect
each other.
iv. The diagonals of a square are equal and bisect each
other at right angles.
v. The diagonals of a rhombus are unequal and bisect
each other at right angles.
vi. A parallelogram and a rectangle on the same base
and between the same parallels are equal in area.
vii. Of all the parallelogram of given sides, the
parallelogram which is a rectangle has the greatest
area.
IMPORTANT FORMULAE
I. 1. Area of a rectangle = (Length x Breadth).
Length = Area
and Breadth = Area
. Breadth Length
II. 2. Perimeter of a rectangle = 2(Length + Breadth)
III. Area of a square = (side)2 = (diagonal)2.
IV. Area of 4 walls of a room = 2 (Length + Breadth) x
Height.
V. 1. Area of a triangle = x Base x Height. (Or) x any
side x length of perpendicular dropped on that side
2. Area of a triangle = s(s-a)(s-b)(s-c)
where a, b, c are the sides of the triangle and s =
(a + b + c).
3. Area of an equilateral triangle = 3
x (side)2. 4
4. Radius of incircle of an equilateral triangle of
side a =
a .
23
5. Radius of circumcircle of an equilateral triangle of side a =
a .
3
6. Radius of incircle of a triangle of area and semi-
perimeter s =
s
In a right angled triangle, the radius of the incircle is given by the
following relation
.
VI. 1. Area of parallelogram = (Base x Height).
2. Area of a rhombus = x (Product of diagonals).
3. Area of a trapezium = x (sum of parallel sides) x
distance between them.
VII. 1. Area of a circle = R2, where R is the radius.
2. Circumference of a circle = 2 R.
3. Length of an arc =
2 R
, where is the central angle.
360
4. Area of a sector = 1 (arc x R) =
R2 .
2 360
VIII. 1. Circumference of a semi-circle = R.
2. Area of semi-circle = R2
. 2
MENSURATION – Worksheet
1). A sizes of a triangle are 25m, 39m and 66m
respectively. Find the perpendicular from the opposite
angles and the greatest side?
A. 13m B. 15m C. 9.5m D. 6m
2). The perimeter of an equilateral triangle is 15m. Find
the length of the perpendicular and the area of the
triangle?
3). The parallel sides of a trapezoid are 8m and 6m and
its altitude is 3m. Find the area of the trapezoid?
A. 21m2 B. 31m2 C. 12m2 D. 13m2
4). Find the area and the circumference of a circle whose
radius is 3.5m?
5). A piece of wire is bent in the shape of an equilateral
triangle of each side 6.6m, it is re-bent to form a circular
ring. Find the diameter of the ring?
A. 3.6m B. 6.6m C. 6.3m D. 6m
6). A right triangle with sides 3 cm, 4 cm and 5 cm is
rotated the side of 3 cm to form a cone. The volume of
the cone so formed is:
A. 12∏cm3 B. 15∏cm3 C. 16∏cm3 D. 20∏cm3
7). In a shower, 5 cm of rain falls. The volume of water
that falls on 1.5 hectares of ground is:
A. 75 cu. m B. 750 cu. m C. 7500 cu. m D.75000 cu.
m
8). The slant height of a right circular cone is 10 m and its
height is 8 m. Find the area of its curved surface.
A. 30∏m2 B. 40∏m2 C. 60∏m2 D. 80∏m2
9). The curved surface area of a cylindrical pillar is 264
m2 and its volume is 924 m3. Find the ratio of its
diameter to its height.
A. 3 : 7 B. 7 : 3 C. 6 : 7 D. 7 : 6
10). What is the total surface area of a right circular cone
of height 14 cm and base radius 7 cm?
A. 344.35 cm2 B. 462 cm2 C. 498.35 cm2 D.
None of these
11). A boat having a length 3 m and breadth 2 m is
floating on a lake. The boat sinks by 1 cm when a man
gets on it. The mass of the man is:
A. 12 kg B. 60 kg C. 72 kg D. 96 kg
12). 50 men took a dip in a water tank 40 m long and 20
m broad on a religious day. If the average displacement
of water by a man is 4 m3, then the rise in the water level
in the tank will be:
A. 20 cm B. 25 cm C. 35 cm D. 50 cm
13). How many bricks, each measuring 25 cm x 11.25 cm
x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5
cm?
A. 5600 B. 6000 C. 6400 D. 7200
14). A hall is 15 m long and 12 m broad. If the sum of the
areas of the floor and the ceiling is equal to the sum of
the areas of four walls, the volume of the hall is:
A. 720m3 B. 900m3 C. 1200m3 D. 1800m3
15). The ratio between the length and the breadth of a
rectangular park is 3 : 2. If a man cycling along the
boundary of the park at the speed of 12 km/hr completes
one round in 8 minutes, then the area of the park (in sq.
m) is:
A. 15360m2 B. 153600m2 C. 30720m2 D. 307200m2
16). The percentage increase in the area of a rectangle, if
each of its sides is increased by 20% is:
A. 40% B. 42% C. 44% D. 46%
17). A towel, when bleached, was found to have lost 20%
of its length and 10% of its breadth. The percentage of
decrease in area is:
A. 10% B. 10.08% C. 20% D. 28%
18). The diagonal of a rectangle is 41 cm and its area is
20 sq. cm. The perimeter of the rectangle must be:
A. 9 cm B. 18 cm C. 20 cm D. 41 cm
19). The difference between the length and bradth of a
rectangle is 23 m. If its perimeter is 206 m, then its area
is:
A. 1520 m2 B. 2420 m2 C. 2480 m2 D. 2520 m2
20). The length of a rectangle is halved, while its breadth
is tripled. What is the percentage change in area?
A. 25% increase B. 50% increase C. 50% decrease
D. 75% decrease
GEOMETRY– Worksheet
1. Vertices of a quadrilateral ABCD are A(0, 0), B(4, 5),
C(9, 9) and D(5, 4). What is the shape of the
quadrilateral?
A. Square B.Rectangle but not a square C.Rhombus
D.Parallelogram but not a rhombus
2. What is the radius of the incircle of the triangle whose
sides measure 5, 12 and 13 units?
A. 2 units B. 12 units C. 6.5 units D. 6 units
3. How many diagonals does a 63-sided convex polygon
have?
A. 3780 B. 1890 C. 3843 D. 3906
4. What is the area of the following square, if the length
of BD is ?
A. 1 B.2 C.3 D.4
5. In the figure below, what is the value of y ?
A. 40 B.60 C.100 D.120
6. The square ABCD touches the circle at 4 points. The
length of the side of the square is 2 cm. Find the area of
the shaded region.
A. π – 4 B. 2π – 4 C.3 π – 4 D. 4π – 4
7. Find the length of the hypotenuse of a right triangle if
the lengths of the other two sides are 6 inches and 8
inches.
8. Find the length of one side of a right triangle if the
length of the hypotenuse is 15 inches and the length of
the other side is 12 inches.
9. Find the length of the hypotenuse of a right triangle if
the lengths of the other two sides are both 3 inches.
10. How many triangles of perimeter 14 can be formed if
the sides measure integral values only?
A. 1 B.2 C.3 D.4
11. In the given figure, AD is the bisector of BAC , AB = 6 cm, AC = 5
cm and
BD = 3 cm. Find DC
A. 11.3 cm B. 2.5 cm C. 3.5 cm D. None of these
12. A toy has a hemisphere base and a conical top as
shown in the figure. The perpendicular height of the cone
is 10 cm and radius of the hemisphere is 4 cm. Find the
volume of the toy.
A. 302 cm3 B. 408 cm3 C. 364 cm3 D. None of these
13. A rectangular tank is 50 m long and 29 m deep. If
2000 m3 of water is drawn off the tank, the level of the
water in the tank goes down by 4m, how many cubic
metre of the water can the tank hold?
A. 18,000 m3 B. 12,500 m3 C. 13,000 m3
D. 14,500 m3
14. The base of a prism is a triangle of which the sides
are 17, 28 and 25 m respectively. Volume of the prism is
2,100 m3. Find its height.
A. 14 m B. 10 m C. 12 m D. 15 m
15. Two sides of a triangle are 22 and 18. Which of the
following cannot be the area of the triangle?
A. 196 B. 197 C. 198 D. 199
DATA SUFFICIENCY – WORKSHEET
Each of the questions given below consists of a statement and / or a
question and two statements numbered I and II given below it. You have
to decide whether the data provided in the statement(s) is / are sufficient
to answer the given question. Read the both statements and
Give answer (A) if the data in Statement I alone are sufficient to answer
the question, while the data in Statement II alone are not sufficient to
answer the question.
Give answer (B) if the data in Statement II alone are sufficient to answer
the question, while the data in Statement I alone are not sufficient to
answer the question.
Give answer (C) if the data either in Statement I or in Statement II alone
are sufficient to answer the question.
Give answer (D) if the data even in both Statements I and II together are
not sufficient to answer the question.
Give answer(E) if the data in both Statements I and II together are
necessary to answer the question.
1. What is the volume of 32 metre high cylindrical tank?
I. The area of its base is 154 m2.
II. The diameter of the base is 14 m.
A. I alone sufficient while II alone not sufficient to answer
B. II alone sufficient while I alone not sufficient to answer
C. Either I or II alone sufficient to answer
D. Both I and II are not sufficient to answer
E. Both I and II are necessary to answer
2. What is the capacity of a cylindrical tank?
I. Radius of the base is half of its height which is 28 metres.
II. Area of the base is 616 sq. metres and its height is 28 metres.
( Same options from first problem)
3. What is the capacity of the cylindrical tank?
I. The area of the base is 61,600 sq. cm.
II. The height of the tank is 1.5 times the radius
III. The circumference of base is 880 cm.
A. Only I and II
B. Only II and III
C. Only I and III
D. Any two of the three
E. Only II and either I or III
4. Two cars pass each other in opposite direction. How long would they
take to be 500 km apart?
I. The sum of their speeds is 135 km/hr.
II. The difference of their speed is 25 km/hr.
( Same options from first problem)
5.How much time did X take to reach the destination?
I. The ratio between the speed of X and Y is 3 : 4.
II. Y takes 36 minutes to reach the same destination.
( Same options from first problem)
6.The towns A, B and C are on a straight line. Town C is between A and
B. The distance from A to B is 100 km. How far is A from C?
I. The distance from A to B is 25% more than the distance from C to B.
II. The distance from A to C is 1/4 of the distance C to B.
( Same options from first problem)
Directions for 7 and 8
The problems consist of a question and two statements, labelled (I) and
(II), in which certain data are given. You have to decide whether the data
given in the statements are sufficient for answering the question.
Mark (a) if statement (I) alone is sufficient, but statement (II) alone is not
sufficient to answer the question;
Mark (b) if statement (II) alone is sufficient, but statement (I) alone is not
sufficient to answer the question;
Mark (c) if both statements (I) and (II) together are sufficient to answer
the question but neither statement alone is sufficient;
Mark (d) if each statement alone is sufficient to answer the question;
Mark if both statements (I) and (II) together are not sufficient to answer
the question and additional data specific to the problem are needed.
7.The distance from Madhavi’s office to her house is 45 miles. On
Monday Madhavi went to the office for a while but returned home early.
what was the total time spent in the travelling?
I. She travelled at an uniform rate of 40 mph both in the onward as
well as the return journey.
II. If she travelled 50 mph faster then she actually did it would have
taken her half the time.
8. What is the speed of the train?
I. A motorcycle takes an hour more then the train to cover the same
distance of 120 kms.
II. The train moves 6 km faster than the motorcycle.
9.By selling an article what is the profit percent gained?
I. 5% discount is given on list price.
II. If discount is not given, 20% profit is gained.
III. The cost price of the articles is Rs. 5000.
A. Only I and II B. Only II and II
C. Only I and III D. All I, II and III
E. None of these
10. A man mixes two types of rice (X and Y) and sells the mixture at the
rate of Rs. 17 per kg. Find his profit percentage.
I. The rate of X is Rs. 20 per kg.
II. The rate of Y is Rs. 13 per kg.
( Same options from first problem)
11. A shopkeeper sells some toys at Rs. 250 each. What percent profit
does he make? To find the answer, which of the following information
given in Statements I and II is/are necessary?
I. Number of toys sold.
II. Cost price of each toy.
A. Only I is necessary
B. Only II is necessary
C. Both I and II are necessary
D. Either I or II is necessary
E. None of these
12. A and B together can complete a task in 7 days. B alone can do it in
20 days. What part of the work was carried out by A?
I. A completed the job alone after A and B worked together for 5 days.
II. Part of the work done by A could have been done by B and C together
in 6 days.
( Same options from first problem)
13. How long will Machine Y, working alone, take to produce x candles?
I. Machine X produces x candles in 5 minutes.
II. Machine X and Machine Y working at the same time produce x
candles in 2 minutes.
( Same options from first problem)
14. How many workers are required for completing the construction work
in 10 days?
I. 20% of the work can be completed by 8 workers in 8 days.
II. 20 workers can complete the work in 16 days.
III. One-eighth of the work can be completed by 8 workers in 5 days.
A. I only
B. II and III only
C. III only
D. I and III only
E. Any one of the three
15. The area of playground is 1600 m2. What is the perimeter?
I. It is a perfect square playground.
II. It costs Rs. 3200 to put a fence around the playground at the rate of
Rs. 20 per metre.
( Same options from first problem)
16. What is the area of a right-angled triangle?
I. The perimeter of the triangle is 30 cm.
II. The ratio between the base and the height of the triangle is 5 : 12.
III. The area of the triangle is equal to the area of a rectangle of length 10
cm.
A. I and II only
B. II and III only
C. I and III only
D. III, and either I or II only
E. None of these
17. The area of a rectangle is equal to the area of right-angles triangle.
What is the length of the rectangle?
I. The base of the triangle is 40 cm.
II. The height of the triangle is 50 cm.
( Same options from first problem)
DATA INTERPRETATION - WORKSHEET
Problem 1:
Study the following table and answer the questions based on it
Expenditures of a Company (in Lakh Rupees) per Annum Over the given
Years.
Year
Item of Expenditure
Salary Fuel and
Transport Bonus
Interest on
Loans Taxes
1998 288 98 3.00 23.4 83
1999 342 112 2.52 32.5 108
2000 324 101 3.84 41.6 74
2001 336 133 3.68 36.4 88
2002 420 142 3.96 49.4 98
1. What is the average amount of interest per year which the company
had to pay during this period?
A. Rs. 32.43 lakhs B. Rs. 33.72 lakhs
C. Rs. 34.18 lakhs D. Rs. 36.66 lakhs
2. The total amount of bonus paid by the company during the given
period is approximately what percent of the total amount of salary paid
during this period?
A. 0.1% B. 0.5% C. 1% D. 1.25%
3. Total expenditure on all these items in 1998 was approximately what
percent of the total expenditure in 2002?
A. 62% B. 66% C. 69% D. 71%
4. The total expenditure of the company over these items during the year
2000 is?
A. Rs. 544.44 lakhs B. Rs. 501.11 lakhs C. Rs. 446.46 lakhs D. Rs.
478.87 lakhs
5. The ratio between the total expenditure on Taxes for all the years and
the total expenditure on Fuel and Transport for all the years respectively
is approximately?
A. 4:7 B. 10:13 C. 15:18 D. 5:8
Problem 2:
The bar graph given below shows the sales of books (in thousand
number) from six branches of a publishing company during two
consecutive years 2000 and 2001.
Sales of Books (in thousand numbers) from Six Branches - B1, B2,
B3, B4, B5 and B6 of a publishing Company in 2000 and 2001.
1. What is the ratio of the total sales of branch B2 for both years to the
total sales of branch B4 for both years?
A. 2:3 B. 3:5 C. 4:5 D. 7:9
2. Total sales of branch B6 for both the years is what percent of the total
sales of branches B3 for both the years?
A. 68.54% B. 71.11% C. 73.17% D. 75.55%
3. What percent of the average sales of branches B1, B2 and B3 in 2001
is the average sales of branches B1, B3 and B6 in 2000?
A. 75% B. 77.5% C. 82.5% D. 87.5%
4. What is the average sales of all the branches (in thousand numbers) for
the year 2000?
A. 73 B. 80 C. 83 D. 88
5. Total sales of branches B1, B3 and B5 together for both the years (in
thousand numbers) is?
A. 250 B. 310 C. 435 D. 560
Problem 3:
The following pie-chart shows the percentage distribution of the
expenditure incurred in publishing a book. Study the pie-chart and the
answer the questions based on it.
Various Expenditures (in percentage) Incurred in Publishing a Book
1. If for a certain quantity of books, the publisher has to pay Rs. 30,600
as printing cost, then what will be amount of royalty to be paid for these
books?
A. Rs. 19,450 B. Rs. 21,200 C. Rs. 22,950 D. Rs.
26,150
2. What is the central angle of the sector corresponding to the expenditure
incurred on Royalty?
A. 15º B. 24º C. 54º D. 48º
3. The price of the book is marked 20% above the C.P. If the marked
price of the book is Rs. 180, then what is the cost of the paper used in a
single copy of the book?
A. Rs. 36 B. Rs. 37.50 C.Rs. 42 D.Rs. 44.25
4. If 5500 copies are published and the transportation cost on them
amounts to Rs. 82500, then what should be the selling price of the book
so that the publisher can earn a profit of 25%?
A. Rs. 187.50 B. Rs. 191.50 C.Rs. 175 D.Rs. 180
5. Royalty on the book is less than the printing cost by:
A. 5% B.33 1/5% C.20% D.25%
Problem 4:
The following line graph gives the ratio of the amounts of imports by a
company to the amount of exports from that company over the period
from 1995 to 2001.
Ratio of Value of Imports to Exports by a Company Over the Years.
1. If the imports in 1998 was Rs. 250 crores and the total exports in the
years 1998 and 1999 together was Rs. 500 crores, then the imports in
1999 was ?
A. Rs. 250 crores B. Rs. 300 crores C. Rs. 357 crores D. Rs. 420
crores
2. The imports were minimum proportionate to the exports of the
company in the year ?
A. 1995 B. 1996 C. 1997 D.2000
3. What was the percentage increase in imports from 1997 to 1998 ?
A. 72 B. 56 C. 28 D. Data inadequate
4. If the imports of the company in 1996 was Rs. 272 crores, the exports
from the company in 1996 was?
A. Rs. 370 crores B. Rs. 320 crores C. Rs. 280 crores D. Rs. 275
crores
5. In how many of the given years were the exports more than the imports
?
A. 1 B. 2 C. 3 D. 4
CONTENTS
SL.NO. TOPIC PAGE
NO.
1 ACKNOWLEDGEMENT 3
2 BASICS OF COMPUTER 4-9
3 CPU 10-14
4 NETWORKING AND
INTERNET
15-17
5 COMPUTER DATA STORAGE 18-26
6 BASIC HARDWARE
COMPONENTS
27-29
7 COMPUTER NETWORKS 30
8 OSI LAYERS 30-36
9 INTERFACES 37
ACKNOWLEDGEMENT
We owe a great many thanks to a great many people who helped and
supported us during the writing of this report.
Our deep sense of gratitude to the trainees of HCL Technologies for their
support and guidance.
We would also thank our Institution and our faculty members without
whom this industrial training in HCL would have been a distant reality.
We also extend my heartfelt thanks to my family and well wishers.
BASICS OF COMPUTERS.
A computer is a programmable machine designed to sequentially
and automatically carry out a sequence of arithmetic or logical
operations. The particular sequence of operations can be changed
readily, allowing the computer to solve more than one kind of
problem. An important class of computer operations on some
computing platforms is the accepting of input from human
operators and the output of results formatted for human
consumption. The interface between the computer and the human
operator is known as the user interface.
Conventionally a computer consists of some form of memory for
data storage, at least one element that carries out arithmetic and
logic operations, and a sequencing and control element that can
change the order of operations based on the information that is
stored. Peripheral devices allow information to be entered from an
external source, and allow the results of operations to be sent out.
A computer's processing unit executes series of instructions that
make it read, manipulate and then store data. Conditional
instructions change the sequence of instructions as a function of
the current state of the machine or its environment.
The first electronic digital computers were developed in the mid-
20th century (1940–1945). Originally, they were the size of a large
room, consuming as much power as several hundred modern
personal computers (PCs). In this era mechanical analog
computers were used for military applications.
Modern computers based on integrated circuits are millions to
billions of times more capable than the early machines, and
occupy a fraction of the space. Simple computers are small
enough to fit into mobile devices, and mobile computers can be
powered by small batteries. Personal computers in their various
forms are icons of the Information Age and are what most people
think of as "computers". However, the embedded computers found
in many devices from mp3 players to fighter aircraft and
from toys to industrial robots are the most numerous.
First general-purpose computers
In 1801, Joseph Marie Jacquard made an improvement to
the textile loom by introducing a series of punched paper cards as
a template which allowed his loom to weave intricate patterns
automatically. The resulting Jacquard loom was an important step
in the development of computers because the use of punched
cards to define woven patterns can be viewed as an early, albeit
limited, form of programmability.
It was the fusion of automatic calculation with programmability that
produced the first recognizable computers. In 1837, Charles
Babbage was the first to conceptualize and design a fully
programmable mechanical computer, his analytical engine. Limited
finances and Babbage's inability to resist tinkering with the design
meant that the device was never completed ; nevertheless his son,
Henry Babbage, completed a simplified version of the analytical
engine's computing unit (the mill) in 1888. He gave a successful
demonstration of its use in computing tables in 1906. This machine
was given to the Science museum in South Kensington in 1910.
In the late 1880s, Herman Hollerith invented the recording of data
on a machine readable medium. Prior uses of machine readable
media, above, had been for control, not data. "After some initial
trials with paper tape, he settled on punched cards. To process
these punched cards he invented the tabulator, and
the keypunch machines. These three inventions were the
foundation of the modern information processing industry. Large-
scale automated data processing of punched cards was performed
for the 1890 United States Census by Hollerith's company, which
later became the core of IBM. By the end of the 19th century a
number of ideas and technologies, that would later prove useful in
the realization of practical computers, had begun to
appear: Boolean algebra, the vacuum tube (thermionic valve),
punched cards and tape, and the teleprinter.
During the first half of the 20th century, many
scientific computing needs were met by increasingly
sophisticated analog computers, which used a direct mechanical
or electrical model of the problem as a basis for computation.
However, these were not programmable and generally lacked the
versatility and accuracy of modern digital computers.
Atanasoff–Berry Computer (ABC) is the world's first electronic
digital computer Atanasoff is considered the father of the
computer . Conceived in 1937 by Iowa State College physics
professor John Atanasoff, and built with the assistance of graduate
student Clifford Berry the machine was not programmable, being
designed only to solve systems of linear equations. The computer
did employ parallel computation. A 1973 court ruling in a patent
dispute found that the patent for the 1946 ENIAC computer
derived from the Atanasoff–Berry Computer.
The inventor of the program-controlled computer was Konrad
Zuse, who built the first working computer in 1941 and later in
1955 the first computer based on magnetic storage
George Stibitz is internationally recognized as a father of the
modern digital computer. While working at Bell Labs in November
1937, Stibitz invented and built a relay-based calculator he dubbed
the "Model K" (for "kitchen table", on which he had assembled it),
which was the first to use binary circuits to perform an arithmetic
operation. Later models added greater sophistication including
complex arithmetic and programmability.
A succession of steadily more powerful and
flexible computing devices were constructed in the 1930s and
1940s, gradually adding the key features that are seen in modern
computers. The use of digital electronics (largely invented
by Claude Shannon in 1937) and more flexible programmability
were vitally important steps, but defining one point along this road
as "the first digital electronic computer" is difficult. Shannon
194 Notable achievements include.
Stored-program architecture
Several developers of ENIAC, recognizing its flaws, came up with
a far more flexible and elegant design, which came to be known as
the "stored program architecture" or von Neumann architecture.
This design was first formally described by John von Neumann in
the paper First Draft of a Report on the EDVAC, distributed in
1945. A number of projects to develop computers based on the
stored-program architecture commenced around this time, the first
of these being completed in Great Britain. The first working
prototype to be demonstrated was the Manchester Small-Scale
Experimental Machine (SSEM or "Baby") in 1948. The Electronic
Delay Storage Automatic Calculator (EDSAC), completed a year
after the SSEM at Cambridge University, was the first practical,
non-experimental implementation of the stored program design
and was put to use immediately for research work at the university.
Shortly thereafter, the machine originally described by von
Neumann's paper—EDVAC—was completed but did not see full-
time use for an additional two years.
Nearly all modern computers implement some form of the stored-
program architecture, making it the single trait by which the word
"computer" is now defined. While the technologies used in
computers have changed dramatically since the first electronic,
general-purpose computers of the 1940s, most still use the von
Neumann architecture.
Beginning in the 1950s, Soviet scientists Sergei
Sobolev and Nikolay Brusentsov conducted research on ternary
computers, devices that operated on a base three numbering
system of −1, 0, and 1 rather than the conventional binary
numbering system upon which most computers are based. They
designed theSetun, a functional ternary computer, at Moscow
State University. The device was put into limited production in the
Soviet Union, but supplanted by the more common binary
architecture.
Semiconductors and microprocessors
Computers using vacuum tubes as their electronic elements were
in use throughout the 1950s, but by the 1960s had been largely
replaced by transistor-based machines, which were smaller, faster,
cheaper to produce, required less power, and were more reliable.
The first transistorised computer was demonstrated at
the University of Manchester in 1953. In the 1970s, integrated
circuit technology and the subsequent creation
of microprocessors, such as the Intel 4004, further decreased size
and cost and further increased speed and reliability of computers.
By the late 1970s, many products such as video
recorders contained dedicated computers called microcontrollers,
and they started to appear as a replacement to mechanical
controls in domestic appliances such as washing machines. The
1980s witnessed home computers and the now
ubiquitous personal computer. With the evolution of the Internet,
personal computers are becoming as common as
the television and the telephone in the house.
Modern smartphones are fully programmable computers in their
own right, and as of 2009 may well be the most common form of
such computers in existence.
Programs
The defining feature of modern computers which distinguishes
them from all other machines is that they can be programmed.
That is to say that some type of instructions (the program) can be
given to the computer, and it will carry process them. While some
computers may have strange concepts "instructions" and "output"
(see quantum computing), modern computers based on the von
Neumann architecture often have machine code in the form of
an imperative programming language.
In practical terms, a computer program may be just a few
instructions or extend to many millions of instructions, as do the
programs for word processors and web browsers for example. A
typical modern computer can execute billions of instructions per
second (gigaflops) and rarely makes a mistake over many years of
operation. Large computer programs consisting of several million
instructions may take teams of programmers years to write, and
due to the complexity of the task almost certainly contain errors.
Stored program architecture
A 1970s punched card containing one line from a
FORTRAN program. The card reads: "Z(1) = Y + W(1)" and is
labelled "PROJ039" for identification purposes.
This section applies to most common RAM machine-based
computers.
In most cases, computer instructions are simple: add one number
to another, move some data from one location to another, send a
message to some external device, etc. These instructions are read
from the computer's memory and are generally carried out
(executed) in the order they were given. However, there are
usually specialized instructions to tell the computer to jump ahead
or backwards to some other place in the program and to carry on
executing from there. These are called "jump" instructions
(or branches). Furthermore, jump instructions may be made to
happen conditionally so that different sequences of instructions
may be used depending on the result of some previous calculation
or some external event. Many computers directly
support subroutines by providing a type of jump that "remembers"
the location it jumped from and another instruction to return to the
instruction following that jump instruction.
Program execution might be likened to reading a book. While a
person will normally read each word and line in sequence, they
may at times jump back to an earlier place in the text or skip
sections that are not of interest. Similarly, a computer may
sometimes go back and repeat the instructions in some section of
the program over and over again until some internal condition is
met. This is called the flow of control within the program and it is
what allows the computer to perform tasks repeatedly without
human intervention.
Comparatively, a person using a pocket calculator can perform a
basic arithmetic operation such as adding two numbers with just a
few button presses. But to add together all of the numbers from 1
to 1,000 would take thousands of button presses and a lot of
time—with a near certainty of making a mistake. On the other
hand, a computer may be programmed to do this with just a few
simple instructions. For example:
mov #0, sum ; set sum to 0
mov #1, num ; set num to 1
loop: add num, sum ; add num to sum
add #1, num ; add 1 to num
cmp num, #1000 ; compare num to 1000
ble loop ; if num <= 1000, go back
to 'loop'
halt ; end of program. stop
running
Once told to run this program, the computer will perform the
repetitive addition task without further human intervention. It will
almost never make a mistake and a modern PC can complete the
task in about a millionth of a second.
Machine code
In most computers, individual instructions are stored as machine
code with each instruction being given a unique number (its
operation code or opcode for short). The command to add two
numbers together would have one opcode, the command to
multiply them would have a different opcode and so on. The
simplest computers are able to perform any of a handful of
different instructions; the more complex computers have several
hundred to choose from—each with a unique numerical code.
Since the computer's memory is able to store numbers, it can also
store the instruction codes. This leads to the important fact that
entire programs (which are just lists of these instructions) can be
represented as lists of numbers and can themselves be
manipulated inside the computer in the same way as numeric
data. The fundamental concept of storing programs in the
computer's memory alongside the data they operate on is the crux
of the von Neumann, or stored program, architecture. In some
cases, a computer might store some or all of its program in
memory that is kept separate from the data it operates on. This is
called the Harvard architecture after theHarvard Mark I computer.
Modern von Neumann computers display some traits of the
Harvard architecture in their designs, such as in CPU caches.
While it is possible to write computer programs as long lists of
numbers (machine language) and while this technique was used
with many early computers, it is extremely tedious and potentially
error-prone to do so in practice, especially for complicated
programs. Instead, each basic instruction can be given a short
name that is indicative of its function and easy to remember—
a mnemonic such as ADD, SUB, MULT or JUMP. These
mnemonics are collectively known as a computer's assembly
language. Converting programs written in assembly language into
something the computer can actually understand (machine
language) is usually done by a computer program called an
assembler. Machine languages and the assembly languages that
represent them (collectively termed low-level programming
languages) tend to be unique to a particular type of computer. For
instance, an ARM architecture computer (such as may be found in
a PDA or a hand-held videogame) cannot understand the machine
language of an Intel Pentium or the AMD Athlon 64 computer that
might be in a PC
Higher-level languages and program design
Though considerably easier than in machine language, writing
long programs in assembly language is often difficult and is also
error prone. Therefore, most practical programs are written in
more abstract high-level programming languages that are able to
express the needs of the programmer more conveniently (and
thereby help reduce programmer error). High level languages are
usually "compiled" into machine language (or sometimes into
assembly language and then into machine language) using
another computer program called a compiler. High level languages
are less related to the workings of the target computer than
assembly language, and more related to the language and
structure of the problem(s) to be solved by the final program. It is
therefore often possible to use different compilers to translate the
same high level language program into the machine language of
many different types of computer. This is part of the means by
which software like video games may be made available for
different computer architectures such as personal computers and
various video game consoles.
The task of developing large software systems presents a
significant intellectual challenge. Producing software with an
acceptably high reliability within a predictable schedule and budget
has historically been difficult; the academic and professional
discipline of software engineering concentrates specifically on this
challenge.
Function
A general purpose computer has four main components:
the arithmetic logic unit (ALU), the control unit, the memory, and
the input and output devices (collectively termed I/O). These parts
are interconnected by busses, often made of groups of wires.
Inside each of these parts are thousands to trillions of
small electrical circuits which can be turned off or on by means of
an electronic switch. Each circuit represents a bit (binary digit) of
information so that when the circuit is on it represents a "1", and
when off it represents a "0" (in positive logic representation). The
circuits are arranged in logic gates so that one or more of the
circuits may control the state of one or more of the other circuits.
The control unit, ALU, registers, and basic I/O (and often other
hardware closely linked with these) are collectively known as
a central processing unit (CPU). Early CPUs were composed of
many separate components but since the mid-1970s CPUs have
typically been constructed on a single integrated circuit called
a microprocessor.
Control unit
Diagram showing how a particular MIPS architecture instruction
would be decoded by the control system.
The control unit (often called a control system or central controller)
manages the computer's various components; it reads and
interprets (decodes) the program instructions, transforming them
into a series of control signals which activate other parts of the
computer. Control systems in advanced computers may change
the order of some instructions so as to improve performance.
A key component common to all CPUs is the program counter, a
special memory cell (a register) that keeps track of which location
in memory the next instruction is to be read from.
Since the program counter is (conceptually) just another set of
memory cells, it can be changed by calculations done in the ALU.
Adding 100 to the program counter would cause the next
instruction to be read from a place 100 locations further down the
program. Instructions that modify the program counter are often
known as "jumps" and allow for loops (instructions that are
repeated by the computer) and often conditional instruction
execution (both examples of control flow).
It is noticeable that the sequence of operations that the control unit
goes through to process an instruction is in itself like a short
computer program—and indeed, in some more complex CPU
designs, there is another yet smaller computer called
a microsequencer that runs a microcode program that causes all
of these events to happen.
Arithmetic/logic unit (ALU)
The ALU is capable of performing two classes of operations:
arithmetic and logic.
The set of arithmetic operations that a particular ALU supports
may be limited to adding and subtracting or might include
multiplying or dividing, trigonometry functions (sine, cosine, etc.)
and square roots. Some can only operate on whole numbers
(integers) whilst others use floating point to represent real
numbers—albeit with limited precision. However, any computer
that is capable of performing just the simplest operations can be
programmed to break down the more complex operations into
simple steps that it can perform. Therefore, any computer can be
programmed to perform any arithmetic operation—although it will
take more time to do so if its ALU does not directly support the
operation. An ALU may also compare numbers and return boolean
truth values (true or false) depending on whether one is equal to,
greater than or less than the other ("is 64 greater than 65?").
Logic operations involve Boolean logic: AND, OR, XOR and NOT.
These can be useful both for creating complicated conditional
statements and processing boolean logic.
Superscalar computers may contain multiple ALUs so that they
can process several instructions at the same time. Graphics
processors and computers with SIMD and MIMD features often
provide ALUs that can perform arithmetic on vectors and matrices.
Memory
Magnetic core memory was the computer memory of choice
throughout the 1960s, until it was replaced by semiconductor
memory.
A computer's memory can be viewed as a list of cells into which
numbers can be placed or read. Each cell has a numbered
"address" and can store a single number. The computer can be
instructed to "put the number 123 into the cell numbered 1357" or
to "add the number that is in cell 1357 to the number that is in cell
2468 and put the answer into cell 1595". The information stored in
memory may represent practically anything. Letters, numbers,
even computer instructions can be placed into memory with equal
ease. Since the CPU does not differentiate between different types
of information, it is the software's responsibility to give significance
to what the memory sees as nothing but a series of numbers.
In almost all modern computers, each memory cell is set up to
store binary numbers in groups of eight bits (called a byte). Each
byte is able to represent 256 different numbers (2^8 = 256); either
from 0 to 255 or −128 to +127. To store larger numbers, several
consecutive bytes may be used (typically, two, four or eight).
When negative numbers are required, they are usually stored
in two's complement notation. Other arrangements are possible,
but are usually not seen outside of specialized applications or
historical contexts. A computer can store any kind of information in
memory if it can be represented numerically. Modern computers
have billions or even trillions of bytes of memory.
The CPU contains a special set of memory cells
called registers that can be read and written to much more rapidly
than the main memory area. There are typically between two and
one hundred registers depending on the type of CPU. Registers
are used for the most frequently needed data items to avoid
having to access main memory every time data is needed. As data
is constantly being worked on, reducing the need to access main
memory (which is often slow compared to the ALU and control
units) greatly increases the computer's speed.
Computer main memory comes in two principal varieties: random-
access memory or RAM and read-only memory or ROM. RAM can
be read and written to anytime the CPU commands it, but ROM is
pre-loaded with data and software that never changes, so the CPU
can only read from it. ROM is typically used to store the
computer's initial start-up instructions. In general, the contents of
RAM are erased when the power to the computer is turned off, but
ROM retains its data indefinitely. In a PC, the ROM contains a
specialized program called the BIOS that orchestrates loading the
computer'soperating system from the hard disk drive into RAM
whenever the computer is turned on or reset. In embedded
computers, which frequently do not have disk drives, all of the
required software may be stored in ROM. Software stored in ROM
is often called firmware, because it is notionally more like
hardware than software. Flash memory blurs the distinction
between ROM and RAM, as it retains its data when turned off but
is also rewritable. It is typically much slower than conventional
ROM and RAM however, so its use is restricted to applications
where high speed is unnecessary.
In more sophisticated computers there may be one or more
RAM cache memories which are slower than registers but faster
than main memory. Generally computers with this sort of cache
are designed to move frequently needed data into the cache
automatically, often without the need for any intervention on the
programmer's part.
Input/output (I/O)
Hard disk drives are common storage devices used with
computers.
I/O is the means by which a computer exchanges information with
the outside world. Devices that provide input or output to the
computer are called peripherals. On a typical personal computer,
peripherals include input devices like the keyboard and mouse,
and output devices such as the display and printer. Hard disk
drives, floppy disk drives and optical disc drives serve as both
input and output devices. Computer networking is another form of
I/O.
Often, I/O devices are complex computers in their own right with
their own CPU and memory. A graphics processing unit might
contain fifty or more tiny computers that perform the calculations
necessary to display 3Dgraphics. Modern desktop
computers contain many smaller computers that assist the main
CPU in performing I/O.
Multitasking
While a computer may be viewed as running one gigantic program
stored in its main memory, in some systems it is necessary to give
the appearance of running several programs simultaneously. This
is achieved by multitasking i.e. having the computer switch rapidly
between running each program in turn.
One means by which this is done is with a special signal called
an interrupt which can periodically cause the computer to stop
executing instructions where it was and do something else instead.
By remembering where it was executing prior to the interrupt, the
computer can return to that task later. If several programs are
running "at the same time", then the interrupt generator might be
causing several hundred interrupts per second, causing a program
switch each time. Since modern computers typically execute
instructions several orders of magnitude faster than human
perception, it may appear that many programs are running at the
same time even though only one is ever executing in any given
instant. This method of multitasking is sometimes termed "time-
sharing" since each program is allocated a "slice" of time in turn.
Before the era of cheap computers, the principal use for
multitasking was to allow many people to share the same
computer.
Seemingly, multitasking would cause a computer that is switching
between several programs to run more slowly — in direct
proportion to the number of programs it is running. However, most
programs spend much of their time waiting for slow input/output
devices to complete their tasks. If a program is waiting for the user
to click on the mouse or press a key on the keyboard, then it will
not take a "time slice" until the event it is waiting for has occurred.
This frees up time for other programs to execute so that many
programs may be run at the same time without unacceptable
speed loss.
Multiprocessing
Cray designed many supercomputers that used multiprocessing
heavily.
Some computers are designed to distribute their work across
several CPUs in a multiprocessing configuration, a technique once
employed only in large and powerful machines such
as supercomputers, mainframe computers and servers.
Multiprocessor and multi-core (multiple CPUs on a single
integrated circuit) personal and laptop computers are now widely
available, and are being increasingly used in lower-end markets as
a result.
Supercomputers in particular often have highly unique
architectures that differ significantly from the basic stored-program
architecture and from general purpose computers. They often
feature thousands of CPUs, customized high-speed interconnects,
and specialized computing hardware. Such designs tend to be
useful only for specialized tasks due to the large scale of program
organization required to successfully utilize most of the available
resources at once. Supercomputers usually see usage in large-
scale simulation, graphics rendering,
and cryptography applications, as well as with other so-called
"embarrassingly parallel" tasks.
Networking and the Internet
Computers have been used to coordinate information between
multiple locations since the 1950s. The U.S.
military's SAGE system was the first large-scale example of such a
system, which led to a number of special-purpose commercial
systems like Sabre.
In the 1970s, computer engineers at research institutions
throughout the United States began to link their computers
together using telecommunications technology. This effort was
funded by ARPA (now DARPA), and the computer network that it
produced was called the ARPANET. The technologies that made
the Arpanet possible spread and evolved.
In time, the network spread beyond academic and military
institutions and became known as the Internet. The emergence of
networking involved a redefinition of the nature and boundaries of
the computer. Computer operating systems and applications were
modified to include the ability to define and access the resources
of other computers on the network, such as peripheral devices,
stored information, and the like, as extensions of the resources of
an individual computer. Initially these facilities were available
primarily to people working in high-tech environments, but in the
1990s the spread of applications like e-mail and the World Wide
Web, combined with the development of cheap, fast networking
technologies like Ethernet and ADSL saw computer networking
become almost ubiquitous. In fact, the number of computers that
are networked is growing phenomenally. A very large proportion
of personal computers regularly connect to the Internet to
communicate and receive information. "Wireless" networking,
often utilizing mobile phone networks, has meant networking is
becoming increasingly ubiquitous even in mobile computing
environments.
Required technology
Computational systems as flexible as a personal computer can be
built out of almost anything. For example, a computer can be
made out of billiard balls (billiard ball computer); this is an
unintuitive and pedagogical example that a computer can be made
out of almost anything. More realistically, modern computers are
made out of transistors made
of photolithographed semiconductors.
Historically, computers evolved from mechanical computers and
eventually from vacuum tubes to transistors.
There is active research to make computers out of many promising
new types of technology, such as optical computing, DNA
computers, neural computers, and quantum computers. Some of
these can easily tackle problems that modern computers cannot
(such as how quantum computers can break some modern
encryption algorithms by quantum factoring).
Logic gates are a common abstraction which can apply to most of
the above digital or analog paradigms.
The ability to store and execute lists of instructions
called programs makes computers extremely versatile,
distinguishing them from calculators. The Church–Turing thesis is
a mathematical statement of this versatility: any computer with
a minimum capability (being Turing-complete) is, in principle,
capable of performing the same tasks that any other computer can
perform. Therefore any type of computer
(netbook, supercomputer, cellular automaton, etc.) is able to
perform the same computational tasks, given enough time and
storage capacity.
Limited-function computers
Conversely, a computer which is limited in function (one that is not
"Turing-complete") cannot simulate arbitrary things. For example,
simple four-function calculators cannot simulate a real computer
without human intervention. As a more complicated example,
without the ability to program a gaming console, it can never
accomplish what a programmable calculator from the 1990s could
(given enough time); the system as a whole is not Turing-
complete, even though it contains a Turing-complete component
(the microprocessor). Living organisms (the body, not the brain)
are also limited-function computers designed to make copies of
themselves; they cannot be reprogrammed without genetic
engineering.
Virtual computers
A "computer" is commonly considered to be a physical device.
However, one can create a computer program which describes
how to run a different computer, i.e. "simulating a computer in a
computer". Not only is this a constructive proof of the Church-
Turing thesis, but is also extremely common in all modern
computers. For example, some programming languages use
something called an interpreter, which is a simulated computer
built using software that runs on a real, physical computer; this
allows programmers to write code (computer input) in a different
language than the one understood by the base computer (the
alternative is to use a compiler). Additionally, virtual machines are
simulated computers which virtually replicate a physical computer
in software, and are very commonly used by IT. Virtual
machines are also a common technique used to create emulators,
such game console emulators.
Artificial intelligence
A computer will solve problems in exactly the way they are
programmed to, without regard to efficiency nor alternative
solutions nor possible shortcuts nor possible errors in the code.
Computer programs which learn and adapt are part of the
emerging field of artificial intelligence and machine learning.
Hardware
The term hardware covers all of those parts of a computer that
are tangible objects. Circuits, displays, power supplies, cables,
keyboards, printers and mice are all hardware.
Software
Software refers to parts of the computer which do not have a
material form, such as programs, data, protocols, etc. When
software is stored in hardware that cannot easily be modified (such
as BIOSROM in an IBM PC compatible), it is sometimes called
"firmware" to indicate that it falls into an uncertain area somewhere
between hardware and software.
Programming languages
Programming languages provide various ways of specifying
programs for computers to run. Unlike natural languages,
programming languages are designed to permit no ambiguity and
to be concise. They are purely written languages and are often
difficult to read aloud. They are generally either translated
into machine code by a compiler or an assembler before being
run, or translated directly at run time by an interpreter. Sometimes
programs are executed by a hybrid method of the two techniques.
There are thousands of different programming languages—some
intended to be general purpose, others useful only for highly
specialized applications.
Computer data storage
1 GB of SDRAM mounted in a personal computer. An example
of primary storage.
40 GB PATA hard disk drive (HDD); when connected to a
computer it serves assecondary storage.
160 GB SDLT tape cartridge, an example of off-line storage. When
used within a robotic tape library, it is classified as tertiarystorage
instead.
Computer data storage, often called storage or memory, refers
to computer components and recording media that retain
digital data. Data storage is one of the core functions and
fundamental components of computers.
In contemporary usage, memory usually refers
to semiconductor storage random-access memory,
typically DRAM (Dynamic-RAM). Memory can refer to other forms
of fast but temporary storage. Storage refers to storage devices
and their media not directly accessible by the CPU,
(secondary or tertiary storage) — typically hard disk drives, optical
disc drives, and other devices slower than RAM but more
permanent. Historically, memory has been calledmain
memory, real storage or internal memory while storage devices
have been referred to as secondary storage, external
memory or auxiliary/peripheral storage.
The distinctions are fundamental to the architecture of computers.
The distinctions also reflect an important and significant technical
difference between memory and mass storage devices, which has
been blurred by the historical usage of the term storage.
Nevertheless, this article uses the traditional nomenclature.
Many different forms of storage, based on various natural
phenomena, have been invented. So far, no practical universal
storage medium exists, and all forms of storage have some
drawbacks. Therefore a computer system usually contains several
kinds of storage, each with an individual purpose.
A digital computer represents data using the binary numeral
system. Text, numbers, pictures, audio, and nearly any other form
of information can be converted into a string of bits, or binary
digits, each of which has a value of 1 or 0. The most common unit
of storage is the byte, equal to 8 bits. A piece of information can be
handled by any computer whose storage space is large enough to
accommodate the binary representation of the piece of
information, or simply data. For example, using eight million bits,
or about one megabyte, a typical computer could store a short
novel.
Traditionally the most important part of every computer is
the central processing unit (CPU, or simply a processor), because
it actually operates on data, performs any calculations, and
controls all the other components. The CPU consists of two (2)
main parts: Control Unit and Arithmetic Logic Unit (ALU). The
former controls the flow of data between the CPU and memory
whilst the latter is used for performing arithmetic and logical
operations on data.
Without a significant amount of memory, a computer would merely
be able to perform fixed operations and immediately output the
result. It would have to be reconfigured to change its behavior.
This is acceptable for devices such as desk calculators or
simple digital signal processors. Von Neumann machines differ in
having a memory in which they store their
operating instructions and data. Such computers are more
versatile in that they do not need to have their hardware
reconfigured for each new program, but can simply
be reprogrammed with new in-memory instructions; they also tend
to be simpler to design, in that a relatively simple processor may
keep state between successive computations to build up complex
procedural results. Most modern computers are von Neumann
machines.
In practice, almost all computers use a variety of memory types,
organized in a storage hierarchy around the CPU, as a trade-off
between performance and cost. Generally, the lower a storage is
in the hierarchy, the lesser its bandwidth and the greater its
access latency is from the CPU. This traditional division of storage
to primary, secondary, tertiary and off-line storage is also guided
by cost per bit.
Hierarchy of storage
Various forms of storage, divided according to their distance from
the central processing unit. The fundamental components of a
general-purpose computer are arithmetic and logic unit, control
circuitry, storage space, and input/output devices. Technology and
capacity as in common home computers around 2005
Primary storage
Primary storage (or main memory or internal memory), often
referred to simply as memory, is the only one directly accessible
to the CPU. The CPU continuously reads instructions stored there
and executes them as required. Any data actively operated on is
also stored there in uniform manner.
Historically, early computers used delay lines, Williams tubes, or
rotating magnetic drums as primary storage. By 1954, those
unreliable methods were mostly replaced by magnetic core
memory. Core memory remained dominant until the 1970s, when
advances in integrated circuit technology allowed semiconductor
memory to become economically competitive.
This led to modern random-access memory (RAM). It is small-
sized, light, but quite expensive at the same time. (The particular
types of RAM used for primary storage are also volatile, i.e. they
lose the information when not powered).
As shown in the diagram, traditionally there are two more sub-
layers of the primary storage, besides main large-capacity RAM:
Processor registers are located inside the processor. Each register
typically holds a word of data (often 32 or 64 bits). CPU
instructions instruct the arithmetic and logic unit to perform various
calculations or other operations on this data (or with the help of it).
Registers are the fastest of all forms of computer data storage.
Processor cache is an intermediate stage between ultra-fast
registers and much slower main memory. It's introduced solely to
increase performance of the computer. Most actively used
information in the main memory is just duplicated in the cache
memory, which is faster, but of much lesser capacity. On the other
hand, main memory is much slower, but has a much greater
storage capacity than processor registers. Multi-level hierarchical
cache setup is also commonly used—primary cache being
smallest, fastest and located inside the processor; secondary
cache being somewhat larger and slower.
Main memory is directly or indirectly connected to the central
processing unit via a memory bus. It is actually two buses (not on
the diagram): an address bus and a data bus. The CPU firstly
sends a number through an address bus, a number called memory
address, that indicates the desired location of data. Then it reads
or writes the data itself using the data bus. Additionally, a memory
management unit (MMU) is a small device between CPU and RAM
recalculating the actual memory address, for example to provide
an abstraction of virtual memory or other tasks.
As the RAM types used for primary storage are volatile (cleared at
start up), a computer containing only such storage would not have
a source to read instructions from, in order to start the computer.
Hence, non-volatile primary storage containing a small startup
program (BIOS) is used to bootstrap the computer, that is, to read
a larger program from non-volatile secondary storage to RAM and
start to execute it. A non-volatile technology used for this purpose
is called ROM, for read-only memory (the terminology may be
somewhat confusing as most ROM types are also capable
of random access).
Many types of "ROM" are not literally read only, as updates are
possible; however it is slow and memory must be erased in large
portions before it can be re-written. Some embedded systems run
programs directly from ROM (or similar), because such programs
are rarely changed. Standard computers do not store non-
rudimentary programs in ROM, rather use large capacities of
secondary storage, which is non-volatile as well, and not as costly.
Secondary storage
A hard disk drive with protective cover removed.
Secondary storage (also known as external memory or auxiliary
storage), differs from primary storage in that it is not directly
accessible by the CPU. The computer usually uses
its input/output channels to access secondary storage and
transfers the desired data using intermediate area in primary
storage. Secondary storage does not lose the data when the
device is powered down—it is non-volatile. Per unit, it is typically
also two orders of magnitude less expensive than primary storage.
Consequently, modern computer systems typically have two
orders of magnitude more secondary storage than primary storage
and data is kept for a longer time there.
In modern computers, hard disk drives are usually used as
secondary storage. The time taken to access a given byte of
information stored on a hard disk is typically a few thousandths of
a second, or milliseconds. By contrast, the time taken to access a
given byte of information stored in random access memory is
measured in billionths of a second, or nanoseconds. This
illustrates the significant access-time difference which
distinguishes solid-state memory from rotating magnetic storage
devices: hard disks are typically about a million times slower than
memory. Rotating optical storage devices, such
as CD and DVDdrives, have even longer access times. With disk
drives, once the disk read/write head reaches the proper
placement and the data of interest rotates under it, subsequent
data on the track are very fast to access. As a result, in order to
hide the initial seek time and rotational latency, data is transferred
to and from disks in large contiguous blocks.
When data reside on disk, block access to hide latency offers a ray
of hope in designing efficient external memory algorithms.
Sequential or block access on disks is orders of magnitude faster
than random access, and many sophisticated paradigms have
been developed to design efficient algorithms based upon
sequential and block access. Another way to reduce the I/O
bottleneck is to use multiple disks in parallel in order to increase
the bandwidth between primary and secondary memory.
Some other examples of secondary storage technologies
are: flash memory (e.g. USB flash drives or keys), floppy
disks, magnetic tape, paper tape, punched cards, standalone RAM
disks, andIomega Zip drives.
The secondary storage is often formatted according to a file
system format, which provides the abstraction necessary to
organize data into files and directories, providing also additional
information (called metadata) describing the owner of a certain file,
the access time, the access permissions, and other information.
Most computer operating systems use the concept of virtual
memory, allowing utilization of more primary storage capacity than
is physically available in the system. As the primary memory fills
up, the system moves the least-used chunks (pages) to secondary
storage devices (to a swap file or page file), retrieving them later
when they are needed. As more of these retrievals from slower
secondary storage are necessary, the more the overall system
performance is degraded.
Tertiary storage
Large tape library. Tape cartridges placed on shelves in the front,
robotic arm moving in the back. Visible height of the library is
about 180 cm.
Tertiary storage or tertiary memory, provides a third level of
storage. Typically it involves a robotic mechanism which
will mount (insert) and dismountremovable mass storage media
into a storage device according to the system's demands; this data
is often copied to secondary storage before use. It is primarily
used for archiving rarely accessed information since it is much
slower than secondary storage (e.g. 5–60 seconds vs. 1–10
milliseconds). This is primarily useful for extraordinarily large data
stores, accessed without human operators. Typical examples
include tape libraries and optical jukeboxes.
When a computer needs to read information from the tertiary
storage, it will first consult a catalog database to determine which
tape or disc contains the information. Next, the computer will
instruct a robotic arm to fetch the medium and place it in a drive.
When the computer has finished reading the information, the
robotic arm will return the medium to its place in the library.
Off-line storage
Off-line storage is a computer data storage on a medium or a
device that is not under the control of a processing unit. The
medium is recorded, usually in a secondary or tertiary storage
device, and then physically removed or disconnected. It must be
inserted or connected by a human operator before a computer can
access it again. Unlike tertiary storage, it cannot be accessed
without human interaction.
Off-line storage is used to transfer information, since the detached
medium can be easily physically transported. Additionally, in case
a disaster, for example a fire, destroys the original data, a medium
in a remote location will probably be unaffected, enabling disaster
recovery. Off-line storage increases generalinformation security,
since it is physically inaccessible from a computer, and data
confidentiality or integrity cannot be affected by computer-based
attack techniques. Also, if the information stored for archival
purposes is accessed seldom or never, off-line storage is less
expensive than tertiary storage.
In modern personal computers, most secondary and tertiary
storage media are also used for off-line storage. Optical discs and
flash memory devices are most popular, and to much lesser extent
removable hard disk drives. In enterprise uses, magnetic tape is
predominant. Older examples are floppy disks, Zip disks, or
punched cards.
Characteristics of storage
A 1GB DDR RAM module (detail)
Storage technologies at all levels of the storage hierarchy can be
differentiated by evaluating certain core characteristics as well as
measuring characteristics specific to a particular implementation.
These core characteristics are volatility, mutability, accessibility,
and addressibility. For any particular implementation of any
storage technology, the characteristics worth measuring are
capacity and performance.
Volatility
Non-volatile memory
Will retain the stored information even if it is not constantly
supplied with electric power. It is suitable for long-term storage of
information.
Volatile memory
Requires constant power to maintain the stored information. The
fastest memory technologies of today are volatile ones (not a
universal rule). Since primary storage is required to be very fast, it
predominantly uses volatile memory.
Dynamic random-access memory
A form of volatile memory which also requires the stored
information to be periodically re-read and re-written, or refreshed,
otherwise it would vanish.
Static random-access memory
A form of volatile memory similar to DRAM with the exception that
it never needs to be refreshed as long as power is applied. (It
loses its content if power is removed).
Mutability
Read/write storage or mutable storage
Allows information to be overwritten at any time. A computer
without some amount of read/write storage for primary storage
purposes would be useless for many tasks. Modern computers
typically use read/write storage also for secondary storage.
Read only storage
Retains the information stored at the time of manufacture,
and write once storage (Write Once Read Many) allows the
information to be written only once at some point after
manufacture. These are called immutable storage. Immutable
storage is used for tertiary and off-line storage. Examples
include CD-ROM and CD-R.
Slow write, fast read storage
Read/write storage which allows information to be overwritten
multiple times, but with the write operation being much slower than
the read operation. Examples include CD-RW and flash memory.
Accessibility
Random access
Any location in storage can be accessed at any moment in
approximately the same amount of time. Such characteristic is well
suited for primary and secondary storage.
Sequential access
The accessing of pieces of information will be in a serial order, one
after the other; therefore the time to access a particular piece of
information depends upon which piece of information was last
accessed. Such characteristic is typical of off-line storage.
Addressability
Location-addressable
Each individually accessible unit of information in storage is
selected with its numerical memory address. In modern
computers, location-addressable storage usually limits to primary
storage, accessed internally by computer programs, since
location-addressability is very efficient, but burdensome for
humans.
File addressable
Information is divided into files of variable length, and a particular
file is selected with human-readable directory and file names. The
underlying device is still location-addressable, but the operating
system of a computer provides the file system abstraction to make
the operation more understandable. In modern computers,
secondary, tertiary and off-line storage use file systems.
Content-addressable
Each individually accessible unit of information is selected based
on the basis of (part of) the contents stored there. Content-
addressable storage can be implemented
using software (computer program) or hardware (computer
device), with hardware being faster but more expensive option.
Hardware content addressable memory is often used in a
computer's CPU cache.
Capacity
Raw capacity
The total amount of stored information that a storage device or
medium can hold. It is expressed as a quantity
of bits or bytes (e.g. 10.4 megabytes).
Memory storage density
The compactness of stored information. It is the storage capacity
of a medium divided with a unit of length, area or volume (e.g. 1.2
megabytes per square inch).
Performance
Latency
The time it takes to access a particular location in storage. The
relevant unit of measurement is typically nanosecond for primary
storage, millisecond for secondary storage, and second for tertiary
storage. It may make sense to separate read latency and write
latency, and in case of sequential access storage, minimum,
maximum and average latency.
Throughput
The rate at which information can be read from or written to the
storage. In computer data storage, throughput is usually
expressed in terms of megabytes per second or MB/s, though bit
rate may also be used. As with latency, read rate and write rate
may need to be differentiated. Also accessing media sequentially,
as opposed to randomly, typically yields maximum throughput.
Energy use
Storage devices that reduce fan usage, automatically shut-down
during inactivity, and low power hard drives can reduce energy
consumption 90 percent.
2.5 inch hard disk drives often consume less power than larger
ones. Low capacity solid-state drives have no moving parts and
consume less power than hard disks. Also, memory may use more
power than hard disks. computer utilizingain to the primary
storage, which is shared between multiple processors in a much
lesser degreeDirect-attached storage (DAS) is a traditional mass
storage, that does not use any network. This is still a most popular
approach. This retronym was coined recently, together with NAS
and SAN.Network-attached storage (NAS) is mass storage
attached to a computer which another computer can access at file
level over a local area network, a private wide area network, or in
the case ofonline file storage, over the Internet. NAS is commonly
associated with the NFS and CIFS/SMB protocolsSAN) is a
specialized network, that provides other computers with storage
capacity. The crucial difference between NAS and SAN is the
former presents and manages file systems to client computers,
whilst the latter provides access at block-addressing (raw) level,
leaving it to attaching systems to manage data or file systems
within the provided capacity. SAN is commonly associated
with Fibre Channel networks.
Basic hardware components
Apart from the physical communications media themselves as
described above, networks comprise additional basic hardware
building blocks interconnecting their terminals, such as network
interface cards (NICs), hubs, bridges, switches, and routers.
Network interface cards
A network card, network adapter, or NIC (network interface card)
is a piece of computer hardware designed to allow computers to
physically access a networking medium. It provides a low-level
addressing system through the use of MAC addresses.
Each Ethernet network interface has a unique MAC address which
is usually stored in a small memory device on the card, allowing
any device to connect to the network without creating an address
conflict. Ethernet MAC addresses are composed of six octets.
Uniqueness is maintained by the IEEE, which manages the
Ethernet address space by assigning 3-octet prefixes to equipment
manufacturers. The list of prefixes is publicly available.
Repeaters and hubs
A repeater is an electronic device that receives a signal, cleans it
of unnecessary noise, regenerates it, and retransmits it at a higher
power level, or to the other side of an obstruction, so that the
signal can cover longer distances without degradation. In most
twisted pair Ethernet configurations, repeaters are required for
cable that runs longer than 100 meters. A repeater with multiple
ports is known as a hub. Repeaters work on the Physical Layer of
the OSI model. Repeaters require a small amount of time to
regenerate the signal. This can cause a propagation delay which
can affect network communication when there are several
repeaters in a row. Many network architectures limit the number of
repeaters that can be used in a row (e.g. Ethernet's 5-4-3 rule).
Today, repeaters and hubs have been made mostly obsolete by
switches (see below).
Repeaters Hub
Bridges
A network bridge connects multiple network segments at the data
link layer (layer 2) of the OSI model. Bridges broadcast to all ports
except the port on which the broadcast was received. However,
bridges do not promiscuously copy traffic to all ports, as hubs do,
but learn which MAC addresses are reachable through specific
ports. Once the bridge associates a port and an address, it will
send traffic for that address to that port only.
Bridges learn the association of ports and addresses by examining
the source address of frames that it sees on various ports. Once a
frame arrives through a port, its source address is stored and the
bridge assumes that MAC address is associated with that port.
Bridges come in three basic types:
Local bridges: Directly connect LANs
Remote bridges: Can be used to create a wide area network
(WAN) link between LANs. Remote bridges, where the
connecting link is slower than the end networks, largely have
been replaced with routers.
Wireless bridges: Can be used to join LANs or connect remote
stations to LANs.
Switches
A network switch is a device that forwards and filters OSI layer
2 datagrams (chunks of data communication) between ports
(connected cables) based on the MAC addresses in the
packets.[15]
A switch is distinct from a hub in that it only forwards
the frames to the ports involved in the communication rather than
all ports connected. A switch breaks the collision domain but
represents itself as a broadcast domain. Switches make
forwarding decisions of frames on the basis of MAC addresses. A
switch normally has numerous ports, facilitating a star topology for
devices, and cascading additional switches.[16]
Some switches are
capable of routing based on Layer 3 addressing or additional
logical levels; these are called multi-layer switches. The
term switch is used loosely in marketing to encompass devices
including routers and bridges, as well as devices that may
distribute traffic on load or by application content (e.g., a
Web URL identifier).
Routers
A router is an internetworking device that
forwards packets between networks by processing information
found in the datagram or packet (Internet protocol information
from Layer 3 of the OSI Model). In many situations, this
information is processed in conjunction with the routing table (also
known as forwarding table). Routers use routing tables to
determine what interface to forward packets (this can include the
"null" also known as the "black hole" interface because data can
go into it, however, no further processing is done for said data).
Firewalls
A firewall is an important aspect of a network with respect to
security. It typically rejects access requests from unsafe sources
while allowing actions from recognized ones. The vital role
firewalls play in network security grows in parallel with the constant
increase in 'cyber' attacks for the purpose of stealing/corrupting
data, planting viruses, etc.
A computer network, often simply referred to as a network, is a
collection of hardware components and computers interconnected
by communication channels that allow sharing of resources and
information.
Networks may be classified according to a wide variety of
characteristics such as the medium used to transport the
data, communications protocol used, scale, topology, and
organizational scope.
The rules and data formats for exchanging information in a
computer network are defined by communications protocols. Well-
known communications protocols are Ethernet, a hardware
and Link Layer standard that is ubiquitous in local area networks,
and the Internet Protocol Suite, which defines a set of protocols for
internetworking, i.e. for data communication between multiple
networks, as well as host-to-host data transfer, and application-
specific data transmission formats.
NETWORKS
A local area network (LAN) is a computer network that
interconnects computers in a limited area such as home, school,
computer laboratory or office building. The defining characteristics
of LANs, in contrast to wide area networks (WANs), include their
usually higher data-transfer rates, smaller geographic area, and
lack of a need for leased telecommunication lines.
ARCNET, Token Ring and other technology standards have been
used in the past, but Ethernet over twisted pair cabling, and Wi-
Fi are the two most common technologies currently used to build
LANs.
A wide area network (WAN) is a telecommunication network that
covers a broad area (i.e., any network that links across
metropolitan, regional, or national boundaries). Business and
government entities utilize WANs to relay data among employees,
clients, buyers, and suppliers from various geographical locations.
In essence this mode of telecommunication allows a business to
effectively carry out its daily function regardless of location.
This is in contrast with personal area networks (PANs), local area
networks (LANs), campus area networks (CANs), or metropolitan
area networks(MANs) which are usually limited to a room, building,
campus or specific metropolitan area (e.g., a city) respectively.
A metropolitan area network (MAN) is a computer network that
usually spans a city or a large campus. A MAN usually
interconnects a number of local area networks (LANs) using a
high-capacity backbone technology, such as fiber-optical links, and
provides up-link services to wide area networks (or WAN) and the
Internet.
Description of OSI layers
According to recommendation X.200, there are seven layers, each
generically known as an N layer. An N+1 entity requests services
from the layer-N entity.
At each level, two entities (N-entity peers) interact by means of the
N protocol by transmitting protocol data units (PDU).
A Service Data Unit (SDU) is a specific unit of data that has been
passed down from an OSI layer to a lower layer, and which the
lower layer has not yet encapsulated into a protocol data unit
(PDU). An SDU is a set of data that is sent by a user of the
services of a given layer, and is transmitted semantically
unchanged to a peer service user.
The PDU at any given layer, layer N, is the SDU of the layer
below, layer N-1. In effect the SDU is the 'payload' of a given PDU.
That is, the process of changing a SDU to a PDU, consists of an
encapsulation process, performed by the lower layer. All the data
contained in the SDU becomes encapsulated within the PDU. The
layer N-1 adds headers or footers, or both, to the SDU,
transforming it into a PDU of layer N-1. The added headers or
footers are part of the process used to make it possible to get data
from a source to a destination.
OSI Model
Data unit Layer Function
Host
layers
Data
7. Application Network process to application
6. Presentation
Data representation, encryption
and decryption, convert
machine dependent data to
machine independent data
5. Session Interhost communication
Segments 4. Transport End-to-end connections,
reliability and flow control
Media
layers
Packet/Datagram 3. Network Path determination andlogical
addressing
Frame 2. Data link Physical addressing
Bit 1. Physical Media, signal and binary
transmission
Some orthogonal aspects, such as management and security,
involve every layer.
Security services are not related to a specific layer: they can be
related by a number of layers, as defined by ITU-T X.800
Recommendation.
These services are aimed to improve the CIA
triad(confidentiality, integrity, and availability) of transmitted data.
Actually the availability of communication service is determined
by network design and/or network management protocols.
Appropriate choices for these are needed to protect againstdenial
of service.
Layer 1: physical layer
The physical layer defines electrical and physical specifications for
devices. In particular, it defines the relationship between a device
and a transmission medium, such as a copper or optical cable.
This includes the layout
ofpins, voltages, cable specifications, hubs, repeaters, network
adapters, host bus adapters (HBA used in storage area networks)
and more.
The major functions and services performed by the physical layer
are:
Establishment and termination of a connection to
a communications medium.
Participation in the process whereby the communication
resources are effectively shared among multiple users. For
example, contention resolution and flow control.
Modulation, or conversion between the representation of digital
data in user equipment and the corresponding signals
transmitted over a communications channel. These are signals
operating over the physical cabling (such as copper and optical
fiber) or over a radio link.
Parallel SCSI buses operate in this layer, although it must be
remembered that the logical SCSI protocol is a transport layer
protocol that runs over this bus. Various physical-layer Ethernet
standards are also in this layer; Ethernet incorporates both this
layer and the data link layer. The same applies to other local-area
networks, such as token ring, FDDI, ITU-T G.hn and IEEE 802.11,
as well as personal area networks such as Bluetooth and IEEE
802.15.4.
Layer 2: data link layer
The data link layer provides the functional and procedural means
to transfer data between network entities and to detect and
possibly correct errors that may occur in the physical layer.
Originally, this layer was intended for point-to-point and point-to-
multipoint media, characteristic of wide area media in the
telephone system. Local area network architecture, which included
broadcast-capable multiaccess media, was developed
independently of the ISO work in IEEE Project 802. IEEE work
assumed sublayering and management functions not required for
WAN use. In modern practice, only error detection, not flow control
using sliding window, is present in data link protocols such
as Point-to-Point Protocol (PPP), and, on local area networks, the
IEEE 802.2 LLC layer is not used for most protocols on the
Ethernet, and on other local area networks, its flow control and
acknowledgment mechanisms are rarely used. Sliding window flow
control and acknowledgment is used at the transport layer by
protocols such as TCP, but is still used in niches where X.25 offers
performance advantages.
The ITU-T G.hn standard, which provides high-speed local area
networking over existing wires (power lines, phone lines and
coaxial cables), includes a complete data link layer which provides
both error correction and flow control by means of a selective
repeat Sliding Window Protocol.
Both WAN and LAN service arrange bits, from the physical layer,
into logical sequences called frames. Not all physical layer bits
necessarily go into frames, as some of these bits are purely
intended for physical layer functions. For example, every fifth bit of
the FDDI bit stream is not used by the layer.
Layer 3: network layer
The network layer provides the functional and procedural means of
transferring variable length data sequences from a source host on
one network to a destination host on a different network, while
maintaining the quality of service requested by the transport layer
(in contrast to the data link layer which connects hosts within the
same network). The network layer performs
network routingfunctions, and might also perform fragmentation
and reassembly, and report delivery errors. Routers operate at this
layer, sending data throughout the extended network and making
the Internet possible. This is a logical addressing scheme – values
are chosen by the network engineer. The addressing scheme is
not hierarchical.
The network layer may be divided into three sublayers:
Subnetwork access – that considers protocols that deal with the
interface to networks, such as X.25;
Subnetwork-dependent convergence – when it is necessary to
bring the level of a transit network up to the level of networks on
either side
Subnetwork-independent convergence – handles transfer across
multiple networks.
Feature Name TP0 TP1 TP2 TP3 TP4
Connection oriented network Yes Yes Yes Yes Yes
Connectionless network No No No No Yes
Concatenation and separation No Yes Yes Yes Yes
Layer 4:
transport
layer
The transport
layer provides
transparent
transfer of
data between
end users,
providing
reliable data
transfer
services to the
upper layers.
The transport
layer controls the reliability of a given link through flow control,
segmentation/desegmentation, and error control. Some protocols
are state- and connection-oriented. This means that the transport
layer can keep track of the segments and retransmit those that fail.
The transport layer also provides the acknowledgement of the
successful data transmission and sends the next data if no errors
occurred.
OSI defines five classes of connection-mode transport protocols
ranging from class 0 (which is also known as TP0 and provides the
least features) to class 4 (TP4, designed for less reliable networks,
similar to the Internet). Class 0 contains no error recovery, and
was designed for use on network layers that provide error-free
connections. Class 4 is closest to TCP, although TCP contains
Segmentation and reassembly Yes Yes Yes Yes Yes
Error Recovery No Yes Yes Yes Yes
Reinitiate connection (if an
excessive number of PDUs are
unacknowledged)
No Yes No Yes No
Multiplexing and
demultiplexing over a
single virtual circuit
No No Yes Yes Yes
Explicit flow control No No Yes Yes Yes
Retransmission on timeout No No No No Yes
Reliable Transport Service No Yes No Yes Yes
functions, such as the graceful close, which OSI assigns to the
session layer. Also, all OSI TP connection-mode protocol classes
provide expedited data and preservation of record boundaries.
Detailed characteristics of TP0-4 classes are shown in the
following table:
Perhaps an easy way to visualize the transport layer is to compare
it with a Post Office, which deals with the dispatch and
classification of mail and parcels sent. Do remember, however,
that a post office manages the outer envelope of mail. Higher
layers may have the equivalent of double envelopes, such as
cryptographic presentation services that can be read by the
addressee only. Roughly speaking, tunneling protocols operate at
the transport layer, such as carrying non-IP protocols such
as IBM's SNA or Novell's IPX over an IP network, or end-to-end
encryption with IPsec. While Generic Routing
Encapsulation (GRE) might seem to be a network-layer protocol, if
the encapsulation of the payload takes place only at endpoint,
GRE becomes closer to a transport protocol that uses IP headers
but contains complete frames or packets to deliver to an
endpoint. L2TP carries PPP frames inside transport packet.
Although not developed under the OSI Reference Model and not
strictly conforming to the OSI definition of the transport layer,
the Transmission Control Protocol (TCP) and the User Datagram
Protocol(UDP) of the Internet Protocol Suite are commonly
categorized as layer-4 protocols within OSI.
Layer 5: session layer
The session layer controls the dialogues (connections) between
computers. It establishes, manages and terminates the
connections between the local and remote application. It provides
for full-duplex,half-duplex, or simplex operation, and establishes
checkpointing, adjournment, termination, and restart procedures.
The OSI model made this layer responsible for graceful close of
sessions, which is a property of the Transmission Control Protocol,
and also for session checkpointing and recovery, which is not
usually used in the Internet Protocol Suite. The session layer is
commonly implemented explicitly in application environments that
use remote procedure calls.
Layer 6: presentation layer
The presentation layer establishes context between application-
layer entities, in which the higher-layer entities may use different
syntax and semantics if the presentation service provides a
mapping between them. If a mapping is available, presentation
service data units are encapsulated into session protocol data
units, and passed down the stack.
This layer provides independence from data representation
(e.g., encryption) by translating between application and network
formats. The presentation layer transforms data into the form that
the application accepts. This layer formats and encrypts data to be
sent across a network. It is sometimes called the syntax layer.
The original presentation structure used the basic encoding rules
of Abstract Syntax Notation One (ASN.1), with capabilities such as
converting an EBCDIC-coded text file to an ASCII-coded file,
orserialization of objects and other data structures from and
to XML.
Layer 7: application layer
The application layer is the OSI layer closest to the end user,
which means that both the OSI application layer and the user
interact directly with the software application. This layer interacts
with software applications that implement a communicating
component. Such application programs fall outside the scope of
the OSI model. Application-layer functions typically include
identifying communication partners, determining resource
availability, and synchronizing communication. When identifying
communication partners, the application layer determines the
identity and availability of communication partners for an
application with data to transmit. When determining resource
availability, the application layer must decide whether sufficient
network or the requested communication exist. In synchronizing
communication, all communication between applications requires
cooperation that is managed by the application layer.
Cross-layer functions
There are some functions or services that are not tied to a given
layer, but they can affect more than one layer. Examples are
security service (telecommunication)[3]
as defined by ITU-
T X.800 Recommendation.
management functions, i.e. functions that permit to configure,
instantiate, monitor, terminate the communications of two or
more entities: there is a specific application layer
protocol, common management information protocol (CMIP) and
its corresponding service, common management information
service (CMIS), they need to interact with every layer in order to
deal with their instances.
MPLS operates at an OSI-model layer that is generally
considered to lie between traditional definitions of layer 2 (data
link layer) and layer 3 (network layer), and thus is often referred
to as a "layer-2.5" protocol. It was designed to provide a unified
data-carrying service for both circuit-based clients and packet-
switching clients which provide a datagram service model. It can
be used to carry many different kinds of traffic, including IP
packets, as well as native ATM, SONET, and Ethernet frames.
ARP is used to translate IPv4 addresses (OSI layer 3) into
Ethernet MAC addresses (OSI layer 2)
Interfaces
Neither the OSI Reference Model nor OSI protocols specify any
programming interfaces, other than as deliberately abstract service
specifications. Protocol specifications precisely define the
interfaces between different computers, but the software interfaces
inside computers are implementation-specific.
For example Microsoft Windows' Winsock, and Unix's Berkeley
sockets and System V Transport Layer Interface, are interfaces
between applications (layer 5 and above) and the transport (layer
4).NDIS and ODI are interfaces between the media (layer 2) and
the network protocol (layer 3).
A set of techniques where by a sequence of information-carrying
quantities occurring at discrete instances of time is encoded into
a corresponding regular sequence of electromagnetic carrier
pulses. Varying the amplitude, polarity, presence or absence,
duration, or occurrence in time of the pulses gives rise to the
four basic forms of pulse modulation: pulse-amplitude
modulation (PAM), pulse-code modulation (PCM), pulse-width
modulation (PWM, also known as pulse-duration
modulation, PDM), and pulse-position modulation (PPM).
Analog-to-digital conversion
An important concept in pulse modulation is analog-to-digital
(A/D) conversion, in which an original analog (time- and
amplitude-continuous) information signal s(t) is changed at the
transmitter into a series of regularly occurring discrete pulses
whose amplitudes are restricted to a fixed and finite number of
values. An inverse digital-to-analog (D/A) process is used at the
receiver to reconstruct an approximation of the original form
of s(t). Conceptually, analog-to-digital conversion involves two
steps. First, the range of amplitudes of s(t) is divided or
quantized into a finite number of predetermined levels, and each
such level is represented by a pulse of fixed amplitude. Second,
the amplitude of s(t) is periodically measured or sampled and
replaced by the pulse representing the level that corresponds to
the measurement. See also Analog-to-digital converter; Digital-
to-analog converter.
According to the Nyquist sampling theorem, if sampling occurs at
a rate at least twice that of the bandwidth of s(t), the latter can
be unambiguously reconstructed from its amplitude values at the
sampling instants by applying them to an ideal low-pass filter
whose bandwidth matches that of s(t).
Quantization, however, introduces an irreversible error, the so-
called quantization error, since the pulse representing a sample
measurement determines only the quantization level in which
the measurement falls and not its exact value. Consequently, the
process of reconstructing s(t) from the sequence of pulses yields
only an approximate version of s(t).
Pulse-amplitude modulation
In PAM the successive sample values of the analog signal s(t) are
used to effect the amplitudes of a corresponding sequence of
pulses of constant duration occurring at the sampling rate. No
quantization of the samples normally occurs (Fig. 1a, b). In
principle the pulses may occupy the entire time between
samples, but in most practical systems the pulse duration, known
as the duty cycle, is limited to a fraction of the sampling
interval. Such a restriction creates the possibility of interleaving
during one sample interval one or more pulses derived from
other PAM systems in a process known as time-
division multiplexing (TDM). See also Multiplexing and multiple
access. sine wave. (a) Analog signal, s(t). (b) Pulse-
amplitude modulation. (c) Pulse-width modulation. (d) Pulse-
position modulation.">
Forms of pulse modulation for the case where the analog
signal, s(t), is a sine wave. (a) Analog signal, s(t). (b) Pulse-
amplitude modulation. (c) Pulse-width modulation. (d) Pulse-
position modulation.
Pulse-width modulation
In PWM the pulses representing successive sample values of s(t)
have constant amplitudes but vary in time duration in direct
proportion to the sample value. The pulse duration can be
changed relative to fixed leading or trailing time edges or a fixed
pulse center. To allow for time-division multiplexing, the
maximum pulse duration may be limited to a fraction of the time
between samples (Fig. 1c).
Pulse-position modulation
PPM encodes the sample values of s(t) by varying the position of
a pulse of constant duration relative to its nominal time of
occurrence. As in PAM and PWM, the duration of the pulses is
typically a fraction of the sampling interval. In addition, the
maximum time excursion of the pulses may be limited (Fig. 1d).
Pulse-code modulation
Many modern communication systems are designed to transmit
and receive only pulses of two distinct amplitudes. In these so-
called binary digital systems, the analog-to-digital conversion
process is extended by the additional step of coding, in which
the amplitude of each pulse representing a quantized sample
of s(t) is converted into a unique sequence of one or more pulses
with just two possible amplitudes. The complete conversion
process is known as pulse-code modulation.
Figure 2a shows the example of three successive quantized
samples of an analog signal s(t), in which sampling occurs
every T seconds and the pulse representing the sample is limited
to T/2 seconds. Assuming that the number of quantization levels
is limited to 8, each level can be represented by a unique
sequence of three two-valued pulses. In Fig. 2b these pulses are
of amplitude V or 0, whereas in Fig. 2c the amplitudes are V and
−V.
Pulse-code modulation. (a) Three successive quantized samples
of an analog signal. (b) With pulses of amplitude V or 0. (c) With
pulses of amplitude V or −V.
PCM enjoys many important advantages over other forms of pulse
modulation due to the fact that information is represented by a
two-state variable. First, the design parameters of a PCM
transmission system depend critically on the bandwidth of the
original signal s(t) and the degree of fidelity required at the
point of reconstruction, but are otherwise largely independent of
the information content of s(t). This fact creates the possibility
of deploying generic transmission systems suitable for many
types of information. Second, the detection of the state of a
two-state variable in a noisy environment is inherently simpler
than the precise measurement of the amplitude, duration, or
position of a pulse in which these quantities are not constrained.
Third, the binary pulses propagating along a medium can be
intercepted and decoded at a point where the
accumulated distortion and attenuation are sufficiently low to
assure high detection accuracy. New pulses can then be
generated and transmitted to the next such decoding point. This
so-called process of repeatering significantly reduces the
propagation of distortion and leads to a quality of transmission
that is largely independent of distance.
Time-division multiplexing
An advantage inherent in all pulse modulation systems is their
ability to transmit signals from multiple sources over a common
transmission system through the process of time-division
multiplexing. By restricting the time duration of a pulse
representing a sample value from a particular analog signal to a
fraction of the time between successive samples, pulses derived
from other sampled analog signals can be accommodated on the
transmission system.
One important application of this principle occurs in the
transmission of PCM telephone voice signals over a digital
transmission system known as a T1 carrier. In standard T1 coding,
an original analog voice signal is band-limited to 4000 hertz by
passing it through a low-pass filter, and is then sampled at the
Nyquist rate of 8000 samples per second, so that the time
between successive samples is 125 microseconds. The samples
are quantized to 256 levels, with each of them being represented
by a sequence of 8 binary pulses. By limiting the duration of a
single pulse to 0.65 microsecond, a total of 193 pulses can be
accommodated in the time span of 125 microseconds between
samples. One of these serves as asynchronization marker that
indicates the beginning of such a sequence of 193 pulses, while
the other 192 pulses are the composite of 8 pulses from each of
24 voice signals, with each 8-pulse sequence occupying a
specified position. T1 carriers and similar types of digital carrier
systems are in widespread use in the world's telephone networks.
Bandwidth requirements
Pulse modulation systems may incur a significant bandwidth
penalty compared to the transmission of a signal in its analog
form. An example is the standard PCM transmission of an analog
voice signal band-limited to 4000 hertz over a T1 carrier. Since
the sampling, quantizing, and coding process produces 8 binary
pulses 8000 times per second for a total of 64,000 binary pulses
per second, the pulses occur every 15.625 microseconds.
Depending on the shape of the pulses and the amount
of intersymbol interference, the required transmission bandwidth
will fall in the range of 32,000 to 64,000 hertz. This compares to
a bandwidth of only 4000 hertz for the transmission of the signal
in analog mode. See alsoBandwidth requirements
(communications).
Applications
PAM, PWM, and PPM found significant application early in the
development of digital communications, largely in the domain of
radio telemetry for remote monitoring and sensing. They have
since fallen into disuse in favor of PCM.
Since the early 1960s, many of the world's telephone network
providers have gradually, and by now almost completely,
converted their transmission facilities to PCM technology. The
bulk of these transmission systems use some form of time-
division multiplexing, as exemplified by the 24-voice channel T1
carrier structure. These carrier systems are implemented over
many types of transmission media, including twisted pairs of
telephone wiring, coaxial cables, fiber-optic cables, and
microwave. See also Coaxial cable; Communications
cable; Microwave; Optical communications; Optical
fibers; Switching systems (communications).
The deployment of high-speed networks such as the Integrated
Service Digital Network (ISDN) in many parts of the world has also
relied heavily on PCM technology. PCM and various modified
forms such as delta modulation (DM) and adaptive differential
pulse-code modulation (ADPCM) have also found significant
application in satellite transmission systems. See
also Communications satellite; Data communications; Electrical
communications; Integrated services digital network
(ISDN); Modulation.
Sampled signal (discrete signal): discrete time, continuous values.
Quantized signal: continuous time, discrete values.
Digital signal (sampled, quantized): discrete time, discrete values.
Quantization, in mathematics and digital signal processing, is the
process of mapping a large set of input values to a smaller set –
such as rounding values to some unit of precision. A device
oralgorithmic function that performs quantization is called
a quantizer. The error introduced by quantization is referred to
as quantization error or round-off error. Quantization is involved
to some degree in nearly all digital signal processing, as the
process of representing a signal in digital form ordinarily involves
rounding. Quantization also forms the core of essentially all lossy
compression algorithms.
Because quantization is a many-to-few mapping, it is an
inherently non-linear and irreversible process (i.e., because the
same output value is shared by multiple input values, it is
impossible in general to recover the exact input value when given
only the output value).
The set of possible input values may be infinitely large, and may
possibly be continuous and therefore uncountable (such as the set
of all real numbers, or all real numbers within some limited range).
The set of possible output values may be finite or countably
infinite. The input and output sets involved in quantization can be
defined in rather general way. For example, vector quantization is
the application of quantization to multi-dimensional (vector-valued)
input data.[1]
There are two substantially different classes of applications where
quantization is used:
The first type, which may simply be called rounding quantization,
is the one employed for many applications, to enable the use of
a simple approximate representation for some quantity that is to
be measured and used in other calculations. This category
includes the simple rounding approximations used in everyday
arithmetic. This category also includes analog-to-digital
conversion of a signal for a digital signal processing system
(e.g., using a sound card of a personal computer to capture an
audio signal) and the calculations performed within most digital
filtering processes. Here the purpose is primarily to retain as
much signal fidelity as possible while eliminating unnecessary
precision and keeping the dynamic range of the signal within
practical limits (to avoid signal clipping or arithmetic overflow). In
such uses, substantial loss of signal fidelity is often
unacceptable, and the design often centers around managing
the approximation error to ensure that very little distortion is
introduced.
The second type, which can be called rate–distortion
optimized quantization, is encountered insource coding for
"lossy" data compression algorithms, where the purpose is to
manage distortion within the limits of the bit rate supported by a
communication channel or storage medium. In this second
setting, the amount of introduced distortion may be managed
carefully by sophisticated techniques, and introducing some
significant amount of distortion may be unavoidable. A quantizer
designed for this purpose may be quite different and more
elaborate in design than an ordinary rounding operation. It is in
this domain that substantial rate–distortion theory analysis is
likely to be applied. However, the same concepts actually apply
in both use cases.
The analysis of quantization involves studying the amount of data
(typically measured in digits or bits or bit rate) that is used to
represent the output of the quantizer, and studying the loss of
precision that is introduced by the quantization process (which is
referred to as thedistortion). The general field of such study of rate
and distortion is known as rate–distortion theory.
Module
4
Signal Representation
and Baseband
Processing
Version 2 ECE IIT,
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Lesson
21
Nyquist Filtering and
Inter Symbol
Interference
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After reading this lesson, you will learn about:
Power spectrum of a random binary
sequence; Inter symbol interference
(ISI); Nyquist filter for avoiding
ISI; Practical improvisation of ideal
Nyquist filter; Raised Cosine (RC) filter and Root–Raised Cosine
(RRC) filtering; Nyquist’s sampling theorem plays a significant role in the
design of pulse- shaping filters, which enable us to restrict the bandwidth of
information-bearing pulses. In this lesson, we start with a short discussion on the
spectrum of a random sequence and then focus on the concepts of Nyquist Filtering.
Specifically, we develop an idea about the narrowest (possible) frequency band that will be needed
for transmission of information at a given symbol
rate. Consider a random binary sequence shown in Fig.4.21.1
(a) following a common style (NRZ: Non-Return-to-Zero Pulses) for its
representation. Note that a binary random sequence may be represented in several
ways. For example, consider Fig.4.21.1 (b). The impulse sequence of Fig.4.21.1 (b) is
an instantaneously sampled version of the NRZ sequence in Fig.4.21.1 (a) with a
sampling rate of one sample/pulse. The information embedded in the random binary sequence
of Fig.4.21.1 (a) is fully preserved in the impulse sequence of Fig.4.21.1 (b). Verify
that you can easily read out the logical sequence only by looking at the impulses. So,
the two waveforms are equivalent so far as their information content is concerned.
The obvious difference, from practical standpoint, is the energy carried
by a pulse. 1 0 0 1 0 1
1 +a
0 Time
-a T = t Fig.4.21.1(a) Sketch of a NRZ ( Non-Return-to-Zero)
waveform for a random binary sequence 10 0 1011 +1 0 -1 Fig.4.21.1 (b): Instantaneously sampled version of the
NRZ sequence in Fig.4.21.1 (a) Version 2 ECE IIT,
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Power Spectrum of Random Binary Sequence
Let us consider the NRZ representation of Fig.4.21.2(a)
where in a pulse is of height ‘a’ and duration Tb. We wish to get an idea about the spectrum
of such sequences. The sequence may be viewed as a sample function of a random
process, say X(t). So, our approach is to find the ACF of X(t) first and then take its
Fourier Transform. Now, the starting instant of observation need not synchronize with
the start time of a pulse. So, we assume that the starting time of the first pulse (i.e. the initial
delay)’td’ is equally likely to lie anywhere between 0 and
Tb, i.e., 1 ,0 tT = = () T pt = 4.21.1 d b b n 0, elsewhere
Next, the ’0’-s and ‘1’-s are equally likely. So, we can
readily note that, E[X(t)] = 0. +a t -a td Tb Fig. 4.21.2(a): A random NRZ sequence representing
an information sequence ACF of X(t) Let Rx (tk,ti) denote the ACF
of X(t). Rx (tk,ti) = E[X(tk) · X(ti)]
4.21.2 Two distinct cases are to be considered: a) when the shift,
i.e. ¦tk - ti¦ is greater than the bit duration Tb and b) when ¦tk -
ti¦= Tb. Case-I: Let, ¦tk - ti¦>
Tb. In this case, X (tk) and X (ti) occur in different bit
intervals and hence they are independent of each other. This
implies, E [X (tk) · X (ti)] = E [X (ti)] · E [X
(tk)] = 0; 4.21.3 Case-II: ¦tk - ti¦<Tb . For simplicity, let us set tk
= 0 and ti <tk = 0. In this case, the random variables X(tk) and X(ti) occur
in the same pulse interval iff td < Tb - ¦tk - ti¦. Further, both X(tk) and X(ti) are of
same magnitude ‘a’ and same polarity.
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Thus, we get a conditional
expectation: at T t t < -- i 4.21.4 ·= dbk EXt Xt elsewhere () () 2 , k it 0, d Averaging this result over all possible values of
td, we get, - - Ttt ·= () () () EXt Xt apt dt b ki 2 d k i d 0 - - - tt a dt a Ttt == - - 1, t t = b ki
2
2 k i k T TT b i d b b 0 4.21.5 By similar argument, it can be shown that for other values
of tk , the ACF of a binary waveform is a function of the time shift t = tk - ti. So, the
autocorrelation function Rx(t) can be expressed as
[Fig.4.21.2(b)]: t t -< aT T t = R 4.21.6 b () 2 1, b x = 0, T t b Rx(t) a2
t Tb -Tb 0 Fig. 4.21.2(b): Auto Correlation Function Rx(t) for a
random binary waveform Now the power spectral density of the random process X(t)
can be obtained by taking the Fourier Transform of
Rx(t ): () ()
t p t T 2 1 exp
2
=- - Sf a j f dt b T x b T - b ( = 4.21.7 22 sin aT c fT) b b A rough sketch of Sx(f) is shown in Fig. 4.21.2(c). Note
that the spectrum has a peak 1 . A value of ‘a2.Tb’. The spectrum stretches towards ±8 and it
has nulls at ± T n. b normalized version of the spectrum is shown in Fig.
4.21.2(d) where the amplitude is normalized with respect to the peak amplitude and the
frequency axis is expressed in terms of ‘fTb’.
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Sx(f) a2 Tb f -3/Tb -2/Tb -1/Tb
1/Tb 2/Tb 3/Tb Fig. 4.21.2(c) : A sketch of the power spectral density,
Sx(f) for a random binary NRZ pulse sequence
Fig. 4.21.2(d) : Normalized power spectral density, Sx(f)
for a random binary NRZ pulse sequence The wide stretch of the spectrum is understandable as the
time pulses are sharply limited within a bit duration. But then, if such a
random NRZ sequence is used to modulate a carrier sinusoid, one can easily imagine that the
modulated signal spectrum will also have an infinite width and hence, such a
modulated signal cannot be transmitted without distortion though a wireless channel, which is
band-limited. If the modulated signal is forced through a band limited channel without
appropriate spectral shaping, the spectrum of the modulated signal at the receiver will be
truncated due to the band pass characteristic of the channel. From the principle of time-
frequency duality, one can now guess that the energy of a transmitted pulse will no more be
limited within its typical duration ‘Tb’.
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Alternatively, the energy of one pulse will spill over the
time slot of one or more subsequent pulses, causing Inter Symbol Interference
(ISI). So, over a specific pulse duration ‘Tb’, the receiver will collect energy due to one
desired and multiple undesired pulses. A typically vulnerable situation is when a negative
pulse appears in a string of positive pulses or vice versa. In general, the received signal
becomes more vulnerable to noise and upon demodulation; the information sequence
may be erroneous. The extent of degradation in the quality of received information depends
on the time spread of energy of a transmitted pulse and how this effect of ISI is addresses
in the receiver. Another reason why sharp rectangular pulses, even though
designed following Gramm-Scmidt orthogonalization procedure, are
not good for band-limited channels is that, one is simply not allowed to use the full
bandwidth that may be presented by a physical channel. Specifically, most
wireless transmissions must have a priori approval from concerned regulatory authority of a
country. It is mandatory that signal transmission is done precisely over the narrow
portion of the allocated bandwidth so that the adjacent bands can be allocated for other
transmission schemes. Further, to conserve bandwidth for various transmission
applications, the narrowest feasible frequency slot only may be allocated. So,
it is necessary to address the issue of ISI in general and the issue of small transmission
bandwidth by shaping pulses. There are several equalization techniques available for
addressing the issue of ISI. Many of these techniques probe the physical channel and use the
channel state information in devising powerful adaptive equalizers. In this course, we
skip further discussion on equalization and focus only on the issue of pulse shaping
for reduction in transmission bandwidth. Nyquist Filter for avoiding ISI Let us recollect the second part of Nyquist’s sampling
theorem for low pass signals which says that a signal band limited to B Hz can be
recovered from a sequence of uniformly spaced and instantaneous samples of the signal
taken at least at the rate of 2B samples per second. Following this theorem, we may
now observe that the impulse sequence of Fig.4.21.1 (b), which contains information
‘1001011’, can be equivalently described by an analog signal, band limited to ½ x
1sample/pulse. That is, if the bit rate is 1 bit/sec and we sample it at 1 sample/sec, the minimum
bandwidth necessary is ½ Hz! An ideal low-pass filter with brick-wall type frequency
response and having a cutoff of 0.5 Hz will generate an equivalent analog waveform (pulse
sequence) when fed with the random impulse
sequence. Recollect that the impulse response of an ideal low-pass
filter is a sinc function [y = sinc(x), Fig. 4. 21.3]. We note that a) sinc(0)
= 1.0 and the impulse response has nulls at x = ±1, ±2, …. sec, b) null-to-null
time-width of the prominent pulse is 2 sec, c) the pulse is symmetric around t = 0 and d)
the peak amplitude of the pulse is of the same polarity as that of the input impulse.
As the filter is a linear network, the output analog waveform is simply superposition of sinc
pulses, where the peaks of adjacent sinc pulses are separated in time by 1 sec, which is
the sampling interval. Version 2 ECE IIT,
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Fig. 4. 21.3(a) Plot of y =
sinc(x) t R t R s 2 s 2 Fig. 4. 21.3(b) Sketch of output of Nyquist filter for
positive and negative impulses In general, if the information symbol rate is Rs
symbols/sec, i.e. the symbol 1 second, the single-sided bandwidth of the low-
pass filter, known interval is Ts = R
s R Hz. popularly as the equivalent Nyquist Bandwidth,
is BN = 2 s
A simple extension of the above observations implies that
for instantaneous samples of random information-bearing pulse (or symbol)
sequence (@ 1 sample per symbol) will exhibit nulls at ±nTs seconds at the filter
output. Now, assuming an ideal noise less baseband channel of bandwidth BN and zero (or
fixed but known) delay, we can comment that the same filter output waveform will
appear at the input of the receiver. Though the waveform will appear much differ ent from the
initial symbol sequence, the information (embedded in the polarity and magnitude) can
be retrieved without any effect of the other pulses by sampling the received baseband
signal exactly at peak position of each shaped pulse as at those time instants all other pulses
have nulls. This ideal brick- wall type filter is known as the Nyquist Filter for zero-ISI
condition. Fig. 4.21.4 highlights some important features of
Nyquist Filter.
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Fig. 4.21.4 Typical impulse response of a Nyquist filter
[5 samples /symbol duration have been used to obtain the responses using a digital computer.
±32 sample duration around the normalized peak value of 1.0 is
shown] Mathematical Explanation Let us consider an ideal lowpass filter with single-sided
bandwidth W. It is usually convenient to deal with the normalized impulse response
h(t) of the filter where in, Wt p ) 2 sin 4.21.8 h(t)=sinc (2Wt) = ( ) Wt p 2
Now if an impulse of strength ‘A’ is applied at the input of
the filter at t = t1, the filter o/p may be expressed as, y(t) = Ah(t- t1)=A.sinc 2W (t- t1) 4.21.9 An information-carrying symbol sequence may be
represented as ( - . t A d ), where Ai = ± A and ti = i.Ts 4.21.10 8 i t i = 0 i The response of the low-pass filter to this
sequence is, 4.21.11 - = )} ( 2 { sin . ) ( t W c A t y 8 i t i = 0 i 1 = Now, if we set /2 R = s 2.T W and sample the output of the filter at t =
m.Ts, it is easy s to see that,
m m () 4.21.12 = = = = A i) - .sinc(m A i - m .sinc2W A ) m.T ( T t y m i i s s = = 0 i 0 i So, we obtain the peak of the m-th pulse clean and devoid
of any interference from previous pulses.
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Practical improvisations: The above discussion on Nyquist bandwidth is ideal
because neither a low-pass filter (LPF) with brick-wall type response can be realized
physically nor can an ideal impulse sequence be generated to represent a discrete information
sequence. Further, note that if there is any error in the sampling instant (which, for a
practical system is very likely to occur from time to time due to the effects of thermal noise
and other disturbances), contribution from the other adjacent pulses will creep into
the sample value and will cause Inter Symbol Interference. For example, the second
lobe peak is only about 13 dB lower compared to the main lobe peak and the decay of the
side-peaks of a sin x/x function is not very rapid with increase in x. So, the
contribution of these peaks from adjacent symbols may be significant. Fortunately, the
desired features of pulse epoch and zero crossings are not unique to a ‘sinc’ pulse. Other pulse
shapes are possible to design with similar features, though the bandwidth requirement for
transmitting a discrete information sequence will be more compared to the
corresponding Nyquist bandwidth (BN). Two such relevant constructs are known as a) Raised
Cosine (RC) filter and b) Root-Raised Cosine (RRC)
filter. Raised Cosine Filter
Let us denote a normalized pulse shape which avoids ISI
as x(t). then, = 1 x(0) and 0 n 0, ) t x 4.21.13 ( = ± nT t = s
Some of the practical requirements on x(t) are
the following: (a) Energy in the main pulse is as much as possible
compared to the total energy distributed beyond the first nulls around the main peak.
This ensures better immunity against noise at the receiver for a given signal
transmission power. So, it is desired that the magnitde of the local maxima of the i-th pulse of
x(t) between iT= t < (i+1)T decreases monotonically and rapidly
with time. (b) The pulse shape x(t) should be so chosen that some
error in instants of sampling at the receiver does not result in appreciable ISI. These two
requirements are usually depicted in the form of a mask in technical
standards. Out of several mathematical possibilities, the following
amplitude mask is very useful for application on the ideal Nyquist filter
impulse response h(t): t cos ßp T ) s t m ß = 4.21.14 ( 2 ) / (4 - 1 2 2 T t s The resulting pulse shape is known as a Raised Cosine
pulse with a roll-off of ‘ß’( 0 ß =1). he Raised Cosine pulse is
described as: t ßp cos T pRC (t) =
sinc(t/Ts). )
2 s ß / (4 - 1 T t 2 2 s 4.21.15 The normalized spectrum of Raised Cosine
pulse is:
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1 ß - H(f) = 1, for () = , f 2 T s ß ß ß p T 1 1 1 + - - = () ( ) ( ) s = = - f f for , cos 2 ß T T T 2 2 2 2 s s s + 1 = 0, for () f 2 ß > 4.21.16 T s
Figs. 4.21.5 (a)-(c) highlight features of a Raised Cosine
(RC) filter. The roll off factor ‘ß’ is used to find a trade off between the absolute
bandwidth that is to be used and the difficulty (in terms of the order of the filter and the
associated delay and inaccuracy) in implementing the filter. The minimum usable bandwidth
is obviously the Nyquist bandwidth, BN (for ß =0) for which the filter is unrealizable
in practice while a maximum absolute bandwidth of 2BN (for ß = 1.0) makes it much
easier to design the filter. ß lies between 0.2 and 0.5 for most of the practical systems where
transmission bandwidth is not a luxury. Use of analog components was dominant
earlier though the modern trend is to use digital filters. While digital FIR structure usually
ensures better accuracy and performance, IIR structure is also used to reduce the
necessary hardware.
Fig. 4.21.5 (a) Typical impulse response of a Raised
Cosine filter with a roll off factor ß = 0.5[5 samples /symbol duration have been used to obtain
the responses using a digital computer. ±32 sample duration around the normalized
peak value of 1.0 is shown]
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Fig. 4.21.5 (b) Typical amplitude spectrum of a Raised
Cosine filter with a roll off factor ß = 0.5. The magnitude is not normalized. Multiply the
normalized frequency values shown above by a factor of 5 to read the frequency
normalized to symbol rate. For example, i) 0.1(from the above figure)x5 = 0.5 = (1 -ß).Rs,
where ß =0.5 and Rs = 1. The transition band of the filter starts here and ii) 0.3(from the
above figure)x5 = 1.5 = (1 +ß).Rs. The stop band of the filter
starts here.
Fig. 4.21.5 (c) Typical phase spectrum of a Raised
Cosine filter with a roll off factor ß = 0.5. Multiply the normalized frequency values shown above
by a factor of 5 to read the frequency normalized to symbol
rate. The side lobe peaks in the impulse response of a Raised
Cosine filter decreases faster with time and hence results in less ISI compared to
the ideal Nyquist filter in case of sampling error in the
receiver. There is another interesting issue in the design of pulse
shaping filters when it comes to applying the concepts in a practical
communication transceiver. From our discussion so far, it may be apparent that the pulse-shaping
filter is for use in the transmitter only, before the signal is modulated by a carrier
or launched in the physical channel. However, there is always a need for an equivalent
lowpass filter in the receiver to eliminate out-of-band noise before demodulation and
decision operations are carried Version 2 ECE IIT,
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out. This purpose is accomplished in a practical receiver by
splitting a Raised Cosine filter in two parts. Each part is known as a Root Raised
Cosine (RRC) filter. One RRC filter is placed in the transmitter while the other part is
placed in the receiver. The transmit RRC filter does the job of pulse shaping and
bandwidth-restriction fully while not ensuring the zero-ISI condition completely. In case of a
linear time invariant physical channel, the receiver RRC filter, in tandem with the
transmit RRC filter, fully ensures zero-ISI condition. Additionally, it filters out undesired out-
of-band thermal noise. On the whole, this approach ensures zero-ISI condition in the
demodulator where it is necessary and it also effectively ensures that the equivalent noise-
bandwidth of the received signal is equal to the Nyquist bandwidth BN. The overall
complexity of the transceiver is reduced without any degradation in performance compared
to a system employing a RC filter in the transmitter and a different out-of-band noise
eliminating filter in the receiver. Figs. 4.21.6 (a)-(c) summarize some features of a Root-
Raised Cosine filter.
Fig. 4.21.6 (a) Typical impulse response of a Root Raised
Cosine filter with a roll off factor ß = 0.5[5 samples /symbol duration have been used
to obtain the responses using a digital computer. ±32 sample duration around the
normalized peak value of 0.5 is shown]
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Fig. 4.21.6 (b) Typical amplitude spectrum of a Raised
Cosine filter with a roll off factor ß = 0.5. Multiply the normalized frequency values shown
above by a factor of 5 to read the frequency normalized to
symbol rate.
Fig. 4.21.6 (c) Typical phase spectrum of a Raised
Cosine filter with a roll off factor ß = 0.5. Multiply the normalized frequency values shown above
by a factor of 5 to read the frequency normalized to symbol
rate. We will mention another practical issue related to the
design and implementation of pulse-shaping filters. Usually, an information sequence is
presented at the input of a pulse shaping filter in the form of pulses (.e.g. bipolar
NRZ) and the width of such a pulse is not negligibly small (as in that case the energy per
pulse would be near zero and hence some pulses may even go unrecognized). This finite
non-zero width of the pulses causes a distortion in the RC or RRC pulse shape. The
situation is somewhat analogous to what happens when we go for flat-top sampling of a band-
limited analog signal instead of instantaneous sampling. The finite pulse-width results in
amplitude fluctuation and again introduces ISI. It needs a relatively simple pulse magnitude
correction which is x commonly referred as
‘ x sin amplitude compensation’ to restore the
quality of shaped Version 2 ECE IIT,
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% Sample the signal 100 times per
second, for 2 seconds.
Fs = 100;
t = [0:2*Fs+1]'/Fs;
Fc = 10; % Carrier frequency
pulses. The transfer function of the RRC shaping filter in
the transmitter is scaled appropriately to incorporate the amplitude
correction. Problems Q4.21.1) Mention possible causes of Inter Symbol
Interference (ISI) in a digital communication receiver. Q4.21.2) Sketch the power spectrum of a long random
binary sequence when the two logic levels are represented by +5V
and 0V. Q4.21.3) Sketch the frequency response characteristics of
an ideal Nyquist low pass filter. Q4.21.4) What are the practical difficulties in
implementing an ideal Nyquist low pass filter?
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x = sin(2*pi*t); % Sinusoidal
signal
% Modulate x using single- and
double-sideband AM.
ydouble = ammod(x,Fc,Fs);
ysingle = ssbmod(x,Fc,Fs);
% Compute spectra of both
modulated signals.
zdouble = fft(ydouble);
zdouble =
abs(zdouble(1:length(zdouble)/2+1)
);
frqdouble = [0:length(zdouble)-
1]*Fs/length(zdouble)/2;
zsingle = fft(ysingle);
zsingle =
abs(zsingle(1:length(zsingle)/2+1)
);
frqsingle = [0:length(zsingle)-
1]*Fs/length(zsingle)/2;
% Plot spectra of both modulated
signals.
figure;
subplot(2,1,1);
plot(frqdouble,zdouble);
title('Spectrum of double-sideband
signal');
subplot(2,1,2);
plot(frqsingle,zsingle);
title('Spectrum of single-sideband
signal');
FM MODULATION
[r, c] = size(x);
if r*c == 0 y = []; return; end; if (r == 1) x = x(:); len = c; else len = r; end; Df = 2*pi*beta*100; x = interp(x,20); x = x*Df; Fs = 20*8000; Fc = 40000; t = (0:1/Fs:((size(x,1)-1)/Fs))'; t = t(:, ones(1, size(x, 2))); x = 2 / Fs * pi * x; x = [zeros(1, size(x, 2)); cumsum(x(1:size(x,1)-1, :))]; y = cos(2 * pi * Fc * t + x ); plot(y)
FM DEMODULATION
function x = FMdemod(y,beta) [r, c] = size(y); if r*c == 0 y = []; return; end; if (r == 1) y = y(:); len = c; else len = r; end; Fc = 40000; Fs = 8000*20; pi2 = 2*pi; [num, den] = butter(5, Fc * 2 / Fs); sen = 2*pi*beta*100; %pre-process the filter. if abs(den(1)) < eps
error('First denominator filter coefficient must be non-zero.'); else num = num/den(1); if (length(den) > 1) den = - den(2:length(den)) / den(1); else den = 0; end; num = num(:)'; den = den(:)'; end; len_den = length(den); len_num = length(num); x = y; y = 2 * y; ini_phase = pi/2; for ii = 1 : size(y, 2) z1 = zeros(length(den), 1); s1 = zeros(len_num, 1); intgl = 0; memo = 0; for i = 1:size(y, 1)
%start with the zero-initial condition integer. vco_out = cos(pi2 * intgl+ini_phase); if len_num > 1 s1 = [y(i, ii) * vco_out; s1(1:len_num-1)]; else s1 = y(i, ii); end tmp = num * s1 + den * z1; if len_den > 1 z1 = [tmp; z1(1:len_den-1)]; else z1 = tmp; end; intgl = rem(((tmp*sen + Fc)/ Fs + intgl), 1); x(i, ii) = tmp; end; end; x = x; x = decimate(x,20);
KARPAGA VINAGA COLLEGE OF ENGINERING AND TECHNOLOGY
DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
EC1351- DIGITAL COMMUNICATION
QUESTION BANK
SEM/YEAR/SEC: VI/III/B STAFF I/C: A.JOTHIMANI
UNIT-I PULSE MODULATION
Part A
1. Give the mathematical form of sampling process
2. List out uses of sampling theorem
3. Define instantaneous sampling
4. What is anti aliasing effect
5. What is PWM
6. List out the types of Quantization
7. What is
8. Transfer 01101001 in to Manchester code
9. What is noise consideration in PCM
10. Define processing gain.
11. What are the two limitations of Delta Modulation? (Apr/May-04)
12. What you understand by the term ‘aliasing’ ? (Nov/Dec05)
13. A band pass signal has the spectral range that extends from 20 KHz to82
KHz. Find the acceptable range of sampling frequency fs . (Nov/Dec05)
14. State Sampling Theroem ( May/June -07)
15. Define Quantization error ( May/June -07)
16. What is SNR of PCM system if the number of quantization levels? (Apr/May-06)
17. State band pass sampling theorem. (Apr/May-06)
18. Plot the magnitude spectrum of the ideally sampled version of the signal
M(t)=2cos(200t)+40sin(290t).Assuming that the sampling rate is 1khz.
(Apr/May - 08) 19.Define position modulation scheme with a suitable diagram (Apr/May -08)
20.State band pass sampling theorem. (Apr/May-04)
Part B
1. Drive the expression for the sampling process in time domain. (16)
2. What are all the types of sampling technique and explain about any two. (16)
3. a. Explain the generation of PPM and PWM with neat circuit diagram. (8)
b. Explain the quantization process with PCM block diagram. (8)
4. Write brief notes on 1) TDMA (8 )
2) FDMA (8)
5. Compare DM with ADM and explain linear prediction filter. (16)
What is meant by a compander?What are the two types of compression?(8)
(ii) Explain the frame format and signaling scheme used in T1 carrier system?(8)
(Apr/May-04)
6..(i) Derive expressions quantization noise and signal-to-noise in a PCM system using a
Uniform quantiser.(10).
(ii) A sinusoidal signal is transmitted using PCM.An output SNR of 55.8dB is required
Find the number of representation levels required to achieve this performance.
( Nov/Dec-05)
7. (i) Explain with a neat diagram, the direct sequence spread binary PSK system?(8)
(ii) Obtain the expression for the processing gain(PG) in this system.
(8) (Apr/May-06)
8.. (i) Derive the expression for SNR in PCM system and compare it with
Delta modulation. Explain how can the SNR be improved in a PCM system.(10)
(ii) Show that prediction error variance is less than the variance of the predictor
input Of the predictor of order one. (Apr/May-06) 9.
(i) With supporting derivation, prove that if a signal contains no frequencies higher than W
Hertz,it may be reconstructed from its samples at a sequence of points spaced ½ W seconds
apart. (8)
(ii) Explain the principle of Delta modulation and derive an expression for average output
noise power in delta modulation.(8) (Apr/May-06)
10. Explain the process of quantization, encoding and decoding in pulse code
Modulation?In what way differential PCM is better than PCM?(16) ( May/June 07)
11. Describe the following systems by presenting appropriate diagrams.
(i) Time Division Multiplexing(8)
(ii) Delta Modulation.(8) ( May/June -07)
12.(a) (i) Give the block diagram of differential pulse code modulation scheme and
explain the principles in detail(9)
(ii) Obtain an expression for the processing gain of a DPCM system.(4)
(iii)Suppose an existing standard PCM system for the voice signal is replaced by
a DPCM processing gain 6dB,while maintaining the (SNR)Q. What will be the
reduction in the bit rate achieved by DPCM?(3) (Apr/May -08)
13.(i) For a Uniform quantizer,discuss the way in which the number of quantization
Levels (L) influence the bandwidth and the Quantization noise.(3)
(ii) Discuss the need for Non-uniform Quantization of speech signal (2).
(iii) Outline the principles of compander used for speech signal.(4)
(iv) Give the bit rate of a Delta modulator fed with samples at a rate of
40ksamples/sec.(2)
(v)How does adaptive delta modulation help in alleviating the problems associated with
Delta modulation scheme.(5) (Apr/May -08)
14. Explain the principle,operation ,signals ,constellation diagram transmitter and
receiver of a MSK system.(16) (May/June-07)
UNIT -2
BASEBAND PULSE TRANSMISSION
Part A
1. Define matched filter
2. What is decision device
3. Draw the block diagram of base band binary data transmission system
4. What is Nyquist channel
5. Draw the frequency and time response of raised cosine spectrum
6. What you mean by correlative level coding
7. Define LMSA
8. What is DFE
9. List out the use of eye patterns
10. Define zero crossing effect
11.. How is the eye pattern obtained on the CRO? (Apr/May-04)
12. What is the condition for zeros Inter symbol Interference? (Apr/May-04)
13.What is the purpose of using an eye pattern? (Nov/Dec05)
14. Why do you need adaptive equalization in switched telephone network?
(Nov/Dec05)
15.What is an ideal Nyquist channel? (Apr/May-06)
16. What is meant by intersymbol interference? (Apr/May-06)
17. Why do we need equalization in base band pulse transmission? ( May/June -07)
18. A message has zero mean value and a peak value of 10V. It is to be quantized using a
Step size of 0.1V with one level coinciding to 0 V. Find number of bits required
for encoding the quantized signal. ( May/June -07)
19. Give the Nyquist criterion for zero ISI.Plot the impulse response of an ideal Nyquist
channel. (Apr/May -08)
20.Draw an eye pattern and respresent the ways in which it could be used to evaluate the
performance a baseband pulse transmission system. (Apr/May -08)
Part B
1. Derive the expression for the matched filter and prove any one property. (16)
2. Prove the effect ISI using necessary block diagram and design. (16)
3. Write brief notes on Nyquist’s criterion for distortion less base
band binary transmission. (16)
4. Explain the six array PAL transmission. (16)
5. a) (i) Explain the cross talk produced due to high frequency cutoff of the channel in
PAM. What is the effect of pulse width on the cross talk factor?(8)
(ii) Compare the power spectra of different binary formats.(8) (Apr/May-06)
6. (i) With neat sketches, explain the duo binary signaling scheme.(8)
(ii) Write briefly about eye pattern and adaptive equalization for data transmission.(8)
(Apr/May-06)
7. a) (i) Generate the code words for (7,4) Hamming code.(8)
(ii) State and prove the properties of syndrome decoding. (8) (Apr/May-06)
8. (i) Explain briefly about linear block codes.(4)
(ii) Evaluate the syndrome S for all five probable single error patterns in (5,1)
Repetition code.(6)
(iii) Briefly explain the Viterbi decoding algorithm .(6) (Apr/May-06)
9. Why do we adoptive equalizer and design any one type of adoptive equalizer. (16)
10..Find the optimum filter frequency response How that maximizes the output SNR
When the input noise is not a white noise.(16) ( May/June -07)
11. Discuss the following:
(i) Adaptive Equalization(8)
(ii) Base band M-ary PAM transmission. (8) ( May/June -07)
12. (a) (i) What are the special features of correlative coding?(2)
(ii) Draw the block diagram of a Duobinary encoder and obtain an expression for
the frequency response of the same.(6)
(iii) What is the need for precoder.(1)
(iv) For the binary sequence 011010100110 give the output of a Duobinary
encoder (in the absence of a precoder).(3)
(v) For the binary sequence 011010100110 give the output of a Duobinary
Encoder with a precoder. (4) (Apr/May-08)
(vi) (i) Obtain the expression for the average probability of symbol error assuming
NRZ signaling,if the noise is modeled as AWGN.(10)
(ii) Compare a baseband binary PAM system with that of M-ary PAM system.(6)
(Apr/May -08)
13. (i) Sketch the time response and frequency response of signal with raised cosine pulse
spectrum.(8)
(ii) Why is the precoding used with duobinary signaling scheme?Draw the block diagram
of precoderand explain its operation (8) (Apr/May-04)
14. (i) Draw the block diagram of an adaptive filter and explain the LMS algorithm.(10)
(ii) Explain how eye pattern is used to study the performance of a data transmission
system.
(6) (Apr/May-04).
15. What do you understand by the term Inter –Symbol Interference (ISI)?Discuss in detail
the Nyquist criterion for minimizing ISI.Explain the difficulties in implementing it in a
practical system.(16). (Nov/Dec05)
16.(i)Discuss the merits and demerits of Duobinary Signalling.(6)
(ii)The binary data(0 1 1 1 0 0 1 0 1 ) are applied to the input of a duobinary
encoder.Construct the duobinary encoder output and corresponding receiver output,without a
precoder .Suppose due to error during transmission ,the level produced by the third digit is
reduced to zero,construct the new receiver output(10). (Nov/Dec05)
UNIT - 3
PASSBAND DATA TRANSMISSION
Part - A
1. Define FSK
2. Draw the Basic block diagram of FSK
3. Define MFSK
4. Draw the wave form of the MPSK
5. What is MSK
6. Define Non-coherent FSK
7. Differentiate coherent FSK from Non-coherent FSK
8. What do you mean by Probability error.
9. Define Passband transmission.
10 Draw the baseband signal.
11. How is the transfer function of the Matched Filter related to the spectrum of the signal?
(Apr/May-04)
12. In minimum shift keying what is the relation between the signal frequencies and bit rate?
(Apr/May-04)
13. Draw an illustrative figure to show the operation of a correlation receiver?
(Nov/Dec05)
14.Highlight the major difference between a QPSK signal and aMSK signal. (Nov/Dec05)
15. Compare the probability of error of PSK with that of FSK. (Apr/May-06)
16. State the difference between coherent and non coherent binary modulation schemes.
(Apr/May-06)
17. Differentiate coherent and non coherent receivers. ( May/June -07)
18. What do you understand by continuous phase frequency shift keying?
( May/June -07)
19.Plot the power spectrum of a BPSK signal operated with a carrier frequency of
140MHz,modulated by data bits at a rate of 2400 bits/sec.What is the bandwidth
requirement? (Apr/May -08)
20.Give the signal space representations of QPSK.How is the performance of the system
related to the distances between the symbols in the signal space ? (Apr/May -08)
Part B
1. Derive the expression for the BPSK technique. (16)
2. Explain the MFSK and derive the pe. (16)
3. Draw the space diagram of MSK and explain about the same. (16)
4. Derive the error probability expression for the MPSK. (16)
5. Compare the all types of Passband transmission. (16)
6. (i) Discuss briefly about Minimum Shift Keying for a Continuous Phase Frequency Shift
Keying(CPFSK) signal.(8)
(ii)With necessary equations and signal space diagram, explain briefly about FSK
system.(8) (Apr/May-06)
7 (i) Explain the Quadriphase Shift keying (QPSK) modulation(8)
(ii) Obtain probability of error interms of Eb and No (Apr/May-06)
8.(i) In a PSK system, the received waveforms S1(t)=A Coswt,S2(t)=-A Coswt are
Coherently detected with a matched filter. The value of A is 20 mV, and the bit
Rate is 1 Mbps. Assume that the noise power spectral density h/2=10-11 W/Hz.
Find the proability of error Pe.(6)
(ii) Enumerate on carrier and symbol synchronization(5+5) ( May/June -07)
9. (i) Give the block diagram for the generation and detection of BFSK signal and
Give a brief explanation of the same.(8)
(ii)What are the special properties of MSK scheme?(4)
(iii) Compare the performance of BPSK with that of BFSK.(4) (Apr/May -08)
10. (i) Discuss the generation and detection of QPSK with suitable block
Diagrams (10) .
Write a note on DPSK.(6) (Apr/May -08)
11. Explain the direct sequence spread spectrum modulation with coherent binary
Phase shift keying.(16) (May/June-07)
12. Describe the following:
(i) Signal space dimensionality and processing gain. (8)
(ii). Maximum length and Gold codes. (8) (May/June-07)
13. (i) Explain the technique used for generating N orthonormal basis functions.(8)
(ii) What do you understand by coherent detection?(2)
(iii)Discuss a method for bit synchronization.(6) (Apr/May-04)
14.(i) Draw the block diagram of QPISIC transmitter and coherent QPSK receriver and
explain their operation.(10)
(ii)Compare the BER of coherent PSK coherent QPSK and coherent FSK. (6) (Apr/May-
04)
Unit- 4
ERROR CONTROL CODING
Part A
1. Define the Channel coding
2. List out the uses of the Channel Coding
3. Differentiate Sourced coding from the Channel coding.
4. Prove any two properties of Block Codes.
5. What is syndrome?
6. Define Code rate.
7. What is BSC?
8. Draw the channel diagram of the BPSK system
9. What is turbo codes
10. Define CRC.
11. What is meant by syndrome of a linear block code? (Apr/May-04)
12. The minimum transmitter power requirement in the absence of channel coding of a
communication System is 2W. If a channel coding scheme of 3 dB is incorporated in the
system give the minimum required transmitter power. (Apr/May-04)
13.Explain the fundamental difference between block codes and convolutional codes.
(Nov/Dec05)
14.How will you define coding gain with reference to error control codes ? (Nov/Dec05)
15. List t7. What is meant by BCH code? (Apr/May-06)
16. What is a convolution code? (Apr/May-06) he advantages of Turbo codes
( May/June -07)
17. Define Minimum distance. ( May/June -07)
Part B
1. Explain the channel coding technique used in the Block codes. (16)
2. What are all the message vectors can be extracted from the
code vector that was generated by 1+X+X3. (16)
3. Explain the Viterbi algorithm for the detection of any two
message vectors (One Word) from the code vectors that
were generated by (111), (101) and (011). (16)
4. Explain the construction of Block Code and
Explain how error syndrome is calculated. (16)
5. Explain in detail about Orthogonal codes, Biorthogonal codes and
Transorthogonal codes. (16)
6. Consider a (7,4) linear block code with the parity-check matrix give by
(i)Construct code words for this (7,4) code
(ii)Show that this code is Hamming code.
(iii)Illustrate the relation between the minimum distance and the structure of the
parity-check matrix H by considering the word 0101100. (May/June-07)
7.Let g(x) be the generator polynomial of a cyclic code C.Find a scheme in
Encoding the data sequence (d0,d1…………dk-1) into an (n,k) system code C.
(16) (May/June-07)
8.(a) (i) Give the properties of cyclic codes.(3)
(ii) Give the details of implementation of cyclic encoder and decoder based on
Linear feedback shift registers.(9)
(iii) A communication system operates at a bit rate of 10kbps with a transmitter
power of 50Watts with a BER of 10^-7 using BPSK.If a channel coder of
coding gain 3 dB is incorporated into the system,calculate the transmitter power required to
maintain the same BER.(assume the modulation scheme remains the same).State the
disadvantage of including the channel coder,if there is any.(4) (Apr/May -08)
9. (i) Draw the block diagram of a convolutional encoder of constraint length 3
And code rate ½.(3)
(ii) Draw the state diagram for the encoder you have given in.(5)
(iii) Discuss the decoder algorithm in detail for the encoder given by you.(8)
(iv) Define Hamming distance between linear block codes.Give the error
correction and error detection capability in terms of Hamming distance.
(Apr/May -08) (v) Give the special features of Trellis codes. (Apr/May -08)
10.(i)Write the generator matrix and parity check matrix of a (7,4) Hamming code.(6)
(ii)Describe a decoding procedure for linear block code.(6)
(iii)Explain the features of RS code (4) (Apr/May-04)
11.(i)Draw the diagram of a rate -1/2 convolutional encoder with constraint length 12.What is
the generator polynomial of the encoder? Find the encoded sequence you have drawn
,corresponding to the message sequence(10011).(12)
(ii)obtain the Trellis diagram of the encoder that you have drawn.(4). (Apr/May-04)
UNIT- 5
SPREAD SPECTRUM MODULATION
Part A
1 What are all the types of SSM
2 What is the use of PN sequence
3 List out the prosperities of PN sequence
4 What is the processing gain of SSM.
5 Define jamming margin
6 Define FHSSM
7 Define SFHSS.
8 Define FFHSM.
9 Draw the PN sequence generator for [6,4,2,1]
10 What you mean by runs in PN sequence
11 State the Gold Theorem
12. When is the PN sequence called as maximal length sequence?
13. What is meant by processing gain of DS spread spectrum system? (Apr/May-04)
14. Write down the properties of PN sequences. (Nov/Dec05)
15How do you define processing gain for aDS-CDMA system? (Nov/Dec05)
16.fine frequency hopping? April/may-06)
17 What are the applications of Spread spectrum modulation? April/may-06
18 What do you mean by Jamming? ( May/June -07)
19List any two applications of spread spectrum modulation. ( May/June -07)
Part B 1. Generate the PN sequence for [5, 4, 3, 1] and prove the all properties of the same. (16)
2. List out and prove the Properties of the PN sequence. (16)
3. Explain the concept of DS/BPSK. (16)
4. List out the uses of the FH. Explain the FH techniques with the frequency spectrum. (16)
5. Write shorts notes on
1. Probability Error. (4)
2. Jamming Margin (4)
3. (S/N) of the DS System. (8)
6.(i). Discuss the configuration of a Direct Sequence spread spectrum system with
BPSK in detail.(13)
(ii) In a direct sequence spread spectrum system with BPSK,the information bit
Duration is 4.095 ms and the PN chip duration is 1us.Calculate the processing
Gain of the system and the feedback shift length used for the PN sequence.
(3) (Apr/May -08) 7.
With suitable block diagrams ,discuss the structure of the transmitter and
Receiver of frequency hopping spread spectrum systems.(10).
(ii)What is the difference between slow FH and fast FH systems?(2)
Assume that a slow FH system employs M-FSK data modulation with N hop bands and
operates with a binary input data rate of R bits/sec.What is the minimum bandwidth
requirement of the system?(4) (Apr/May -08)
8 give the properties of maximal length sequence(PN sequence). (Apr/May -08)
9List any two applications of spread spectrum. (Apr/May -08)
10State and explain the properties of maximal length sequences.(6)
(ii)Draw the block diagram of DS- spread spectrum system transmitter and receiver and
explain the function performed by each block in brief.(10) (Apr/May-04)
11Explain the principle of operation of frequency hopped M-ary FSK spread spectrum
system(10)
(ii)Discuss the ways in which fast hopping scheme and slow frequency hopping (spread
spectrum) schemes could be used to mitigate multipath effect.(6) (Apr/May-04)
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