Combi Problems

8/20/2019 Combi Problems

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COMBINATIONAL LOGIC: EXEMPLAR QUESTIONS AND SOLUTIONS

COMBINATIONAL LOGIC

Outcome 1

(a) Convert the following decimal numbers into 8-bit binary:

(i) 10 (ii) 59 (iii) 10 (iv) 18

(b) Convert the following 8-bit binary numbers to decimal:

(i) 10101010 (ii) 11100011 (iii) 00111110 (iv) 10010011

(c) Convert the following he!adecimal numbers to 8-bit binary:

(i) "# (ii) $ (iii) %8 (iv) 0"

(d) Convert the following 8-bit binary numbers to he!adecimal:

(i) 00111011 (ii) 11011000 (iii) 01001110 (iv) 10010111

(e) Convert the following decimal numbers into two-digit he!adecimal

numbers:

(i) % (ii) 5 (iii) 51 (iv) 19

(f) Convert the following he!adecimal numbers into decimal:

(i) &C (ii) "' (iii) 5% (iv) $$

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1##

3 earning and *eaching 2cotland 00'

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Solution

(a) *here a number of different ways in which decimal to binary

conversion can be done4 *he uic6est and sim7lest is to use an

engineering calculator that lets you convert between bases4 ven

the windows-based calculator su77orts base conversion4 iven

below is an illustration of how this is done using the windows

calculator4

$lternatively we can calculate the binary value of a decimal number

using the binary weightings for 8 bits as shown below4

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1#'


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Only 6 binary digits appear as this is all that is required to represent 59 in binary,however to convert to an 8-bit representation just add two eros, i!e! ""111"11

#inary weighting e$pressed as power o% &


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"inally we can calculate the result using the divide by techniue4

$gain using 59 as an e!am7le:

( i) 10 converts to 00001010

(ii) 59 converts to 00111011 (see above)

(iii) 10 converts to

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1#5


'(#

'(#

)(#

)(#


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(iv) 18 converts to

(b) (i)

(ii)

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1#%




)(#

'(#


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(iii)

(iv)

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1#;





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(c)


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(ii) "or $:

> 0010 $ > 1010

so the binary re7resentation for $ is 001010104

( ii i) "or %8

% > 0110 8 > 1000

so the binary re7resentation for %8 is 011010004

(iv) "or 0"

0 > 0000 " > 1111

so the binary re7resentation for 0" is 000011114

(d) *he 7rocess of converting from binary to he!adecimal is sim7ly a

case of reversing the 7revious 7rocess4 /ivide the binary 7attern

into grou7s of four bits and read off the euivalent he!adecimal

digit4

.n e!am7le (i) 00111011is bro6en into 0011 > # and 1011 > &4

*he he!adecimal re7resentation of 00111011 is #&4

(ii) 11011000 > 1101 1000 > /8

(iii) 01001110 > 0100 1110 > '

(iv) 10010111 > 1001 0111 > 9;

Note: if te !in"#$ %"tte#n &oe' not &i(i&e into )#ou%' of fou# !it'

"&& *' to te left+"n& 'i&e, i-e- te MSB 'i&e until te %"tte#n &oe'

&i(i&e-

.o# e/"m%le, 11**1 onl$ cont"in' fi(e !in"#$ &i)it' 0!it'- If 2e "&&

t#ee *' to te left+"n& 'i&e 2e &o not "ffect te ("lue !ut 2e c"n

no2 &i(i&e te %"tte#n into )#ou%' of fou# !it':

***11**1 3 14 in e/"&ecim"l

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1#9



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(e) *here are three methods that we can use to convert decimal to

he!adecimal4

*he sim7lest of all is to use an engineering calculator that has the

facility to convert between bases4 $gain if we loo6 at the

calculator 7rovide in icrosoft ?indows we can see how this is

done4

*he second method is to convert the decimal number into binary

and then convert the binary number into he!adecimal4 $ll of this

has been demonstrated in the 7revious e!am7les (a) @ (d)4

*he third method involves conversion using a table containing the

weighted values of the he!adecimal digits4 *his is similar to the

conversion between binary and decimal4

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1'0


*e$adeci+al digit position

*e$adeci+al digit weighting


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(ii) Convert the decimal number 5 into he!adecimal4

(iii) Convert decimal number 51 into he!adecimal4

(iv) Convert the decimal number 19 to he!adecimal4

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1'1



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(f) (i) Convert the he!adecimal number &C to decimal4

(ii) Convert the he!adecimal number "' to decimal4

(iii) Convert the he!adecimal number 5% to decimal4

(iv) Convert the he!adecimal number $$ to decimal4

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1'



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(a) Aerform the following additions in 8-bit binary4 Convert bac6 to

decimal to chec6 your answer4

(b) Aerform the following subtractions in binary4 Convert bac6 todecimal to chec6 your answer4

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1'#


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Solution

(a) *he decimal numbers in this uestion have already been converted

to binary in 2$B 14

*he rules a77lied to binary addition are identical to those a77lied

to decimal e!ce7t that we must always remind ourselves that we

are wor6ing in binary4

(i)

(ii)

(iii)

(iv)

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1''


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(b) *he decimal numbers in this uestion have already been converted

to binary in 2$B 14

*he rules a77lied to binary subtraction are identical to those

a77lied to decimal e!ce7t that we must always remind ourselves

that we are wor6ing in binary4

(i)

(ii)

(iii)

(iv)

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1'5


1 1 "

. & 1

& cannot besubtracted %ro+ "there%ore a 1 isborrowed %ro+ thene$t colu+n! /hisupper digit is now &

& 1 1


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(a) +e7resent the following numbers as s com7lement 8-bit binary

numbers:

(i) @10 (ii) 10 (iii) 59 (iv) @59

(v) 10 (vi) @10

(b) !7lain why the decimal numbers 18 and @18 cannot be

re7resented using 8-bit s com7lement re7resentation4

(c) Aerform the following calculations using 8-bit s com7lement

binary re7resentation:

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1'%


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Solution

(a) s com7lement is a binary method of re7resenting negative

numbers4 .n s com7lement re7resentation the 2& is used as a

sign bit with 1 indicating a negative number and 0 re7resenting a

7ositive number4

?e will use (i) to illustrate the s com7lement system4

(i) *o re7resent @10 sim7ly write down the binary value for 10:

?e then invert all the bits:

*hen we add 1 to this value:

*he answer is the s com7lement re7resentation of @10:

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1';


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.t is difficult to tell the value of this 8-bit s com7lement

number sim7ly by loo6ing at it4 *o chec6 the value of a

negative number sim7ly re7eat the 7revious o7erations= i4e4invert all bits and add 1:

(ii)

otice that there is no difference in re7resentation= i4e4 this

loo6s li6e a standard 8-bit binary re7resentation of 104

(iii)

(iv)

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1'8



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(v)

(vi)

(b) .f we use 8-bit s com7lement re7resentation then the sign bit

effectively reduces the sie of the number to ; bits4 .n s

com7lement form the range of numbers that can be re7resented is:

$s can be seen= 18 and @18 fall outwith this range and so cannot

be re7resented using 8-bit s com7lement4

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1'9



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(c) (i)

(ii)

(iii)

(iv)

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 150


0% a carry is generated into the ninth bit si+ply ignore the 8-bit answer will be correct


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Outcome 5

(a) .dentify the $2. standard logic symbols shown and com7lete the

truth tables in "ig 2$B 1 (i)4

2ig (34 1i

(b) /raw the &ritish 2tandard symbols for the logic gates shown "ig

2$B 1(i)4

(c) ?rite down the &oolean e!7ression for each of the gates shown in

"ig 2$B 1(i)4

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 151


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Solution

(a)

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 15


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(b)

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 15#



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"or the $2. logic gates given in "ig 2$B :

(a) write down the logic e!7ression

(b) write down the logic out7ut for the in7uts shown4

2ig (34 &

"or the &2 logic gates given in "ig 2$B #:

(a) write down the logic e!7ression

(b) write down the logic out7ut for the in7uts shown4

2ig (34 .

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 15'


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Solution

(a)= (b)

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 155


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Solution

(a)= (b)

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 15%


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Outcome 6

(a) "or the circuit shown in "ig 2$B 1(i) write down the &oolean

e!7ression for D4

(b) sing the integrated circuit diagram sheet identify the .ntegrated

Circuits used in this circuit and 7lace a77ro7riate 7in numbers

beside the gates4

(c) Com7lete the truth table in "ig 2$B 1(ii)4

2ig (34 1i

2ig (34 1ii

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 15;


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Solution

(a)= (b)

2ig (34 1i

(c)

2ig (34 1ii

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 158


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(a) "or the circuit shown in "ig 2$B (i) write down the &oolean

e!7ression for D4

(b) sing the integrated circuit diagram sheet identify the .Cs used in

this circuit and 7lace a77ro7riate 7in numbers beside the gates4

(c) Com7lete the truth table in "ig 2$B 1(ii)4

2ig (34 &i

2ig (34 &ii

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 159


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Solution

(a)= (b)

(c)

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1%0


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Solution

(i) ?or6ing from the out7ut bac6wards:

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1%


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(ii)

(b)

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1%#



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*he truth table for a two-in7ut logic circuit is shown in "ig 2$B '4

(a) "rom the truth table derive the &oolean e!7ression that describes

this table4

(b) sing $2. symbols draw the logic circuit that meets the

reuirements of this table4 Eour schematic should clearly show

.Cs used and 7in numbers4

2ig (34 7

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1%'


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Solution

.n this 7roblem you are trying to generate a circuit from a truth table4

*o do this consider each in7ut combination that 7rovides a logic 1

out7ut and wor6 out how this could be achieved using an $/ gate4

*hese combinations are indicated by arrows in the diagram below4

*o illustrate consider the first combination= $ > 0 and & > 04 .n order

to generate a logic 1 at the out7ut an $/ gate needs all its in7uts to be

logic 1 therefore we need to invert $ and &:

.f we follow this argument for the second combination= $ > 0 and & > 1=

we need to invert $ only to obtain a 1 from the $/ gate out7ut:

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1%5


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*he final ste7 is to combine all the $/ combinations into an ,+ gate4

+ecall that the ,+ gate will generate a logic 1 out7ut when either of its

in7uts are logic= which is e!actly what is needed for this circuit4

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1%%



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*he truth table for a three-in7ut logic circuit is shown in "ig 2$B 54

(a) "rom the truth table derive the &oolean e!7ression that describes

this table4

(b) sing $2. symbols draw the logic circuit that meets the

reuirements of this table4 Eour schematic should clearly show

.Cs used and 7in numbers4

2ig (34 5

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1%;


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Solution

*he only difference between this 7roblem and 2$B ' is the number of

in7uts4 *he way in which we solve the 7roblem is identical4

.dentify the in7ut combinations that give a logic 1 out7ut and wor6 out

how a logic 1 could be achieved using an $/ gate:

*he first combination is $&C > 0014 $s we have three in7uts we will

reuire a three-in7ut $/ gate4 *o obtain a logic 1 out7ut for this

combination we need to invert $ and &:

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1%8


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*he second combination is $&C > 1004 *o obtain a logic 1 out7ut for

this combination we need to invert & and C:

$s before the final ste7 is to combine all the $/ combinations into an

,+ gate4 +ecall that the ,+ gate will generate a logic 1 out7ut when

either of its in7uts are logic 1:

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1%9



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Outcome 7

$ sim7le logic control system for a central heating system has the

following in7uts:

timer cloc6 (C) 0 > outwith time 7eriod for central heating to be on

1 > within time 7eriod for central heating to be on

room thermostat (*) 0 > room tem7erature above the set value

1 > room tem7erature below set value

fros t 7rotection (") 0 > room tem7erature above 'FC

1 > room tem7erature below 'FC

?hen the out7ut from the control system > 1 then the boiler will be

ignited= allowing the system to 7roduce heat4 *he frost 7rotection will

override the timer cloc6 and the room thermostat should the

tem7erature fall below 'FC4

/esign the logic control system to meet the above reuirements4 Eour

design should include the following:

• truth table

• &oolean logic e!7ression

• logic diagram using $2. symbols (this diagram must include .C

identification and 7in numbering)4

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1;0


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Solution

*his outcome advances the combinational design 7rocesses one ste7

further4 .n this outcome we are given a s7ecification for a digital

system and from that a 7ractical circuit must be designed4

*he first ste7 in this 7rocess is to generate a truth table4 *his 7roblem has three

in7uts= C= * and "= therefore we need a three-in7ut truth table4 ach in7ut

condition must be carefully considered before 7lacing a 1 or 0 in the out7ut

column4

/ 2 O/" " " " Output will be " as %rost setting is above 7: and ti+er outwith

O; period!

" " 1 1 Output will be 1 as the %rost setting is indicating that thete+perature is below 7:! /his overrides everything else!

" 1 " " Output will be " as the %rost setting is above 7: and ti+eroutwith O; period! ;ote it doesn


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*he remaining ste7s are identical to those carried out in ,utcome #= i4e4

generate an $/ e!7ression for each ,* > 1 condition and then ,+

all of these $/ e!7ression together:

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1;



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C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1;#



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C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1;'



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*he control logic for a sim7le alarm system in a house has three in7uts

and a single out7ut= which activates an alarm4 *he three in7uts are:

• windows sensor

• doors sensor

• master 6ey4

*he sensors o7erate as follows:

window sensor (?) 0 > all windows closed

1 > a window is o7en

door sensor (/) 0 > all doors closed

1 > a door is o7ened

master 6ey (G) 0 > alarm system is disarmed

1 > alarm system is armed

.f the alarm system is disarmed then the logic signals from the sensors

are ignored and the alarm will not sound4 *he alarm is activated by a

logic 1 from the out7ut of the control logic4

/esign the circuit to meet the reuirements of this control logic4 Eour

design should include the following:

• truth table

• &oolean logic e!7ression

• logic diagram using $2. symbols (this diagram must include .C

identification and 7in numbering)4

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1;5


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Solution

$s with the 7revious e!am7le a truth table must be generated and all

in7ut 7ermutations carefully considered4 *he reuired out7ut and

associated reasons are shown below4

= > O/" " " " 3lar+ is disar+ed so it doesn 1 condition and then ,+

all of these $/ e!7ressions together to create the &oolean e!7ression

that fully describes the reuirements of the truth table4

C*+,.C $/ C*+.C$ "/$*$2 (.* ) 1;%


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