Master of Computer Application (MCA) – Semester 1

MC0062 – Digital Systems, Computer Organization and Architecture

(Book ID: B0680 & B0684)

Assignment Set – 1

Q.-1. Describe the concept of Binary arithmetic with suitable numerical examples.

Answer.

Binary arithmeticArithmetic in binary is much like arithmetic in other numeral systems. Addition, subtraction, multiplication, and division can be performed on binary numerals.AdditionThe circuit diagram for a binary half adder, which adds two bits together, producing sum and carry bits.The simplest arithmetic operation in binary is addition. Adding two single-digit binary numbers is relatively simple, using a form of carrying: 0 + 0 → 0 0 + 1 → 1 1 + 0 → 1 1 + 1 → 10, carry 1 (since 1 + 1 = 0 + 1 × binary 10)

Adding two "1" digits produces a digit "0", while 1 will have to be added to the next column. This is similar to what happens in decimal when certain single-digit numbers are added together; if the result equals or exceeds the value of the radix (10), the digit to the left is incremented: 5 + 5 → 0, carry 1 (since 5 + 5 = 0 + 1 × 10) 7 + 9 → 6, carry 1 (since 7 + 9 = 6 + 1 × 10)This is known as carrying. When the result of an addition exceeds the value of a digit, the procedure is to "carry" the excess amount divided by the radix (that is, 10/10) to the left, adding it to the next positional value. This is correct since the next position has a weight that is higher by a factor equal to the radix. Carrying works the same way in binary: 1 1 1 1 1 (carried digits) 0 1 1 0 1+ 1 0 1 1 1-------------= 1 0 0 1 0 0In this example, two numerals are being added together: 011012 (1310) and 101112 (2310). The top row shows the carry bits used. Starting in the rightmost column, 1 + 1 = 102. The 1 is carried to the left, and the 0 is written at the bottom of the rightmost column. The second column from the right is added: 1 + 0 + 1 = 102 again; the 1 is carried, and 0 is written at the bottom. The third column: 1 + 1 + 1 = 112. This time, a 1 is carried, and a 1 is written in the bottom row. Proceeding like this gives the final answer 1001002 (36 decimal).When computers must add two numbers, the rule that: x xor y = (x + y) mod 2 for any two bits x and y allows for very fast calculation, as well.

A simplification for many binary addition problems is the Long Carry Method or Brookhouse Method of Binary Addition. This method is generally useful in any binary addition where one of the numbers has a long string of “1” digits. For example the following large binary numbers can be added in two simple steps without multiple carries from one place to the next.

(carried digits) Versus 1 1 1 1 1 1 1 1 (Long Carry Method) 1 1 1 0 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 0+ 1 0 1 0 1 1 0 0 1 1 + 1 0 1 0 1 1 0 0 1 1 (add

crossed out digits first)

----------------------- + 1 0 0 0 1 0 0 0 0 0 0 = (sum of crossed out digits)

(now add remaining digits)= 1 1 0 0 1 1 1 0 0 0 1 ---------------------------

1 1 0 0 1 1 1 0 0 0 1

In this example, two numerals are being added together: 1 1 1 0 1 1 1 1 1 02 (95810) and 1 0 1 0 1 1 0 0 1 12 (69110). The top row shows the carry bits used. Instead of the standard carry from one column to the next, the lowest place-valued "1" with a "1" in the corresponding place value beneath it may be added and a "1" may be carried to one digit past the end of the series. These numbers must be crossed off since they are already added. Then simply add that result to the uncanceled digits in the second row. Proceeding like this gives the final answer 1 1 0 0 1 1 1 0 0 0 12 (164910).

Addition table0 1

0 0 11 1 10

SubtractionSubtraction works in much the same way: 0 − 0 → 0 0 − 1 → 1, borrow 1 1 − 0 → 1 1 − 1 → 0

Subtracting a "1" digit from a "0" digit produces the digit "1", while 1 will have to be subtracted from the next column. This is known as borrowing. The principle is the same as for carrying. When the result of a subtraction is less than 0, the least possible value of a digit, the procedure is to "borrow" the deficit divided by the radix (that is, 10/10) from the left, subtracting it from the next positional value.

* * * * (starred columns are borrowed from) 1 1 0 1 1 1 0− 1 0 1 1 1----------------= 1 0 1 0 1 1 1

Subtracting a positive number is equivalent to adding a negative number of equal absolute value; computers typically use two's complement notation to represent negative values. This notation eliminates the need for a separate "subtract" operation. Using two's complement notation subtraction can be summarized by the following formula:A − B = A + not B + 1For further details, see two's complement.

MultiplicationMultiplication in binary is similar to its decimal counterpart. Two numbers A and B can be multiplied by partial products: for each digit in B, the product of that digit in A is calculated and written on a new line, shifted leftward so that its rightmost digit lines up with the digit in B that was used. The sum of all these partial products gives the final result.

Since there are only two digits in binary, there are only two possible outcomes of each partial multiplication:

* If the digit in B is 0, the partial product is also 0 * If the digit in B is 1, the partial product is equal to A

For example, the binary numbers 1011 and 1010 are multiplied as follows:

1 0 1 1 (A) × 1 0 1 0 (B) --------- 0 0 0 0 ← Corresponds to a zero in B + 1 0 1 1 ← Corresponds to a one in B + 0 0 0 0 + 1 0 1 1 --------------- = 1 1 0 1 1 1 0

Binary numbers can also be multiplied with bits after a binary point:

1 0 1.1 0 1 (A) (5.625 in decimal) × 1 1 0.0 1 (B) (6.25 in decimal) ------------- 1 0 1 1 0 1 ← Corresponds to a one in B + 0 0 0 0 0 0 ← Corresponds to a zero in B + 0 0 0 0 0 0 + 1 0 1 1 0 1 + 1 0 1 1 0 1 ----------------------- = 1 0 0 0 1 1.0 0 1 0 1 (35.15625 in decimal)

See also Booth's multiplication algorithm.Multiplication table

0 10 0 01 0 1DivisionSee also: Division (digital)Binary division is again similar to its decimal counterpart:

Here, the divisor is 1012, or 5 decimal, while the dividend is 110112, or 27 decimal. The procedure is the same as that of decimal long division; here, the divisor 1012 goes into the first three digits 1102 of the dividend one time, so a "1" is written on the top line. This result is multiplied by the divisor, and subtracted from the first three digits of the dividend; the next digit (a "1") is included to obtain a new three-digit sequence:

1 ___________1 0 1 ) 1 1 0 1 1 − 1 0 1 ----- 0 1 1

The procedure is then repeated with the new sequence, continuing until the digits in the dividend have been exhausted: 1 0 1 ___________1 0 1 ) 1 1 0 1 1 − 1 0 1 ----- 0 1 1 − 0 0 0 ----- 1 1 1 − 1 0 1 ----- 1 0

Q-2. Explain the Boolean rules with relevant examples.

Ans-

Boolean algebra finds its most practical use in the simplification of logic circuits. If we translate a logic circuit's function into symbolic (Boolean) form, and apply certain algebraic rules to the resulting equation to reduce the number of terms and/or arithmetic operations, the simplified equation may be translated back into circuit form for a logic circuit performing the same function with fewer components. If equivalent function may be achieved with fewer components, the result will be increased reliability and decreased cost of manufacture.

To this end, there are several rules of Boolean algebra presented in this section for use in reducing expressions to their simplest forms. The identities and properties already reviewed in this chapter are very useful in Boolean simplification, and for the most part bear similarity to many identities and properties of "normal" algebra. However, the rules shown in this section are all unique to Boolean mathematics.

Example:

This rule may be proven symbolically by factoring an "A" out of the two terms, then applying the rules of A + 1 = 1 and 1A = A to achieve the final result:

Please note how the rule A + 1 = 1 was used to reduce the (B + 1) term to 1. When a rule like "A + 1 = 1" is expressed using the letter "A", it doesn't mean it only applies to expressions containing "A". What the "A" stands for in a rule like A + 1 = 1 is any Boolean variable or collection of variables. This is perhaps the most difficult concept for new students to master in Boolean simplification: applying standardized identities, properties, and rules to expressions not in standard form.

For instance, the Boolean expression ABC + 1 also reduces to 1 by means of the "A + 1 = 1" identity. In this case, we recognize that the "A" term in the identity's standard form can represent the entire "ABC" term in the original expression.

The next rule looks similar to the first one shown in this section, but is actually quite different and requires a more clever proof:

Note how the last rule (A + AB = A) is used to "un-simplify" the first "A" term in the expression, changing the "A" into an "A + AB". While this may seem like a backward step, it certainly helped to reduce the expression to something simpler! Sometimes in mathematics we must take "backward" steps to achieve the most elegant solution. Knowing when to take such a step and when not to is part of the art-form of algebra, just as a victory in a game of chess almost always requires calculated sacrifices.

Another rule involves the simplification of a product-of-sums expression:

And, the corresponding proof:

To summarize, here are the three new rules of Boolean simplification expounded in this section:

Definition of Boolean Expressions

A Boolean expression is defined as the expression in which constituents have one of two rates and the algebraic processes defined on the set are logical OR, a type of addition, and logical AND, a type of multiplication.

Boolean expression varies from the ordinary expression in three ways: the values that variables may presume, which are a logical instead of a numeric quality, perfectly 0 and 1; in the processes related to these values; and in the properties of those processes, that is, the laws that they follow. Expressions contain mathematical logical, digital logics, computer programming, set theory, and statistics.

Laws of Boolean Algebra:

      Laws of Boolean algebra:   

                 (1a)       x · y = y · x

                 (1b)       x + y = y + x

                 (1c)       1 + x = 1

                 (2a)       x · (y · z) = (x · y) · z

                 (2b)       x + (y + z) = (x + y) + z

                 (3a)       x · (y + z) = (x · y) + (x · z)

                 (3b)       x + (y · z) = (x + y) · (x + z)

                 (4a)       x · x =  x

                 (4b)       x + x = x

                 (5a)       x · (x + y) = x

                 (5b)       x + (x · y) = x

                 (6a)       x · x1 = 0

                 (6b)       x + x1 = 1

                 (7)          (x1)1 = x

                 (8a)        (x · y)1 = x1 + y1

                 (8b)        (x + y) 1 = x1 · y1

          Thus the above rules are used to translate the logic gates directly. Although some of the rules can be derived from simpler identities derived in the packages.                                   

Example for Boolean Expressions:

Example 1: Determine the OR operations using Boolean expressions.

Solution: Construct a truth table for OR Operation,

               x         y          x V y

               F         F             F

               F         T             T

               T         F             T

               T         T             T

Example 2: Determine the AND operations using Boolean expressions.

Solution: Construct a truth table for AND Operation,

               x         y          x Λ y

               F         F             F

               F         T             F

               T         F             F

               T         T             T

Example3 : Determine the NOT operations using Boolean expressions.

Solution: Construct a truth table for NOT Operation,

               x         ¬x

               F          T

               T          F

Q-3 Explain Karnaugh maps with your own examples illustrating the formation.Ans-

Karnaugh map

K-map showing minterms and boxes covering the desired minterms. The brown region is an overlapping of the red (square) and green regions.

The input variables can be combined in 16 different ways, so the Karnaugh map has 16 positions, and therefore is arranged in a 4 × 4 grid.

The binary digits in the map represent the function's output for any given combination of inputs. So 0 is written in the upper leftmost corner of the map because ƒ = 0 when A = 0, B = 0, C = 0, D = 0. Similarly we mark the bottom right corner as 1 because A = 1, B = 0, C = 1, D = 0 gives ƒ = 1. Note that the values are ordered in a Gray code, so that precisely one variable changes between any pair of adjacent cells.

After the Karnaugh map has been constructed the next task is to find the minimal terms to use in the final expression. These terms are found by encircling groups of 1s in the map. The groups must be rectangular and must have an area that is a power of two (i.e. 1, 2, 4, 8…). The rectangles should be as large as possible without containing any 0s. The optimal groupings in this map are marked by the green, red and blue lines. Note that groups may overlap. In this example, the red and green groups overlap. The red group is a 2 × 2 square, the green group is a 4 × 1 rectangle, and the overlap area is indicated in brown.

The grid is toroidally connected, which means that the rectangular groups can wrap around edges, so is a valid term, although not part of the minimal set—this covers Minterms 8, 10, 12, and 14.

Perhaps the hardest-to-visualize wrap-around term is which covers the four corners—this covers minterms 0, 2, 8, 10.

Examples:

Karnaugh maps are used to facilitate the simplification of Boolean algebra functions. The following is an unsimplified Boolean Algebra function with Boolean variables A, B, C, D, and their inverses. They can be represented in two different functions:

f(A,B,C,D) = ∑(6,8,9,10,11,12,13,14) Note: The values inside ∑ are the minterms to map (i.e. which rows have output 1 in the truth table).

Truth table

Using the defined minterms, the truth table can be created:

# A B C D f(A,B,C,D)0 0 0 0 0 01 0 0 0 1 02 0 0 1 0 03 0 0 1 1 04 0 1 0 0 05 0 1 0 1 0

6 0 1 1 0 17 0 1 1 1 08 1 0 0 0 19 1 0 0 1 110 1 0 1 0 111 1 0 1 1 112 1 1 0 0 113 1 1 0 1 114 1 1 1 0 115 1 1 1 1 0

Parity generator

This picture show the function : parity generator.The input is 1(0) if is input with cans (or odd) number of 1(0). This function is MAX lengt of 4 input karnaugh map - as you can see there is no way how make simple this function - no 1(0) has not 1(0) as neighbor

Karnaugh map for 5 (five) inputsMore people do not know how create map for 5 (five) inputs. Next picture showing how you can do it. And show you different rules on 5 inputs karnaugh map:

The adderThe basic circuit is a adder. There is a example of basic component of adder (also you can use half-adders).

Q-4. With a neat labeled diagram explain the working of binary half and full adders.Ans.

A half adder is a logical circuit that performs an addition operation on two binary digits. The half adder produces a sum and a carry value which are both binary digits.

A full adder is a logical circuit that performs an addition operation on three binary digits. The full adder produces a sum and carry value, which are both binary digits. It can be combined with other full adders or work on its own.

Q.-5. Describe the concept of timing diagrams and synchronous logic.

Ans-

Timing Diagrams:

Timing diagrams are the main key in understanding digital systems. Timing diagrams explain digital circuitry functioning during time flow. Timing diagrams help to understand how digital circuits or sub circuits should work or fit in to larger circuit system. So learning how to read Timing diagrams may increase your work with digital systems and integrate them.Bellow is a list o most commonly used timing diagram fragments:

Low level to supply voltage:

Transition to low or high level:

Bus signals – parallel signals transitioning from one level to other:

High Impedance state:

Bus signal with floating impedance:

Conditional change of on signal depending on another signal transition:

Transition on a signal causes state changes in a BUS:

More than one transition causes changes in a BUS:

Sequential transition – one signal transition causes another signal transition and second signal transition causes third signal transition.

As you see timing diagrams together with digital circuit can completely describe the circuits working. To understand the timing diagrams, you should follow all symbols and transitions in timing diagrams. You can find plenty of symbols in timing diagrams. It depends actually on designer or circuit manufacturer. But once you understand the whole picture, you can easy read any timing diagram like in this example:

Synchronous sequential logic

Nearly all sequential logic today is 'clocked' or 'synchronous' logic: there is a 'clock' signal, and all internal memory (the 'internal state') changes only on a clock edge. The basic storage element in sequential logic is the flip-flop.

The main advantage of synchronous logic is its simplicity. Every operation in the circuit must be completed inside a fixed interval of time between two clock pulses, called a 'clock cycle'. As long as this condition is met (ignoring certain other details), the circuit is guaranteed to be reliable. Synchronous logic also has two main disadvantages, as follows.

1. The clock signal must be distributed to every flip-flop in the circuit. As the clock is usually a high-frequency signal, this distribution consumes a relatively large amount of power and dissipates much heat. Even the flip-flops that are doing nothing consume a small amount of power, thereby generating waste heat in the chip.

2. The maximum possible clock rate is determined by the slowest logic path in the circuit, otherwise known as the critical path. This means that every logical calculation, from the simplest to the most complex, must complete in one clock cycle. One way around this limitation is to split complex operations into several simple operations, a technique known as 'pipelining'. This technique is prominent within microprocessor design, and helps to improve the performance of modern processors.

Q.6- Discuss the Quine McCluskey method.

Answer-

Quine–McCluskey algorithmThe Quine–McCluskey algorithm (or the method of prime implicants) is a method used for minimization of boolean functions which was developed by W.V. Quine and Edward J. McCluskey. It is functionally identical to Karnaugh mapping, but the tabular form makes it more efficient for use in computer algorithms, and it also gives a deterministic way to check that the minimal form of a Boolean function has been reached. It is sometimes referred to as the tabulation method.

The method involves two steps:

1. Finding all prime implicants of the function.2. Use those prime implicants in a prime implicant chart to find the essential prime

implicants of the function, as well as other prime implicants that are necessary to cover the function.

Complexity:Although more practical than Karnaugh mapping when dealing with more than four variables, the Quine–McCluskey algorithm also has a limited range of use since the problem it solves is NP-hard: the runtime of the Quine–McCluskey algorithm grows exponentially with the number of variables. It can be shown that for a function of n variables the upper bound on the number of prime implicants is 3n/n. If n = 32 there may be over 6.5 * 1015 prime implicants. Functions with a large number of variables have to be minimized with potentially non-optimal heuristic methods, of which the Espresso heuristic logic minimizer is the de-facto standard

Step 1: finding prime implicants

Minimizing an arbitrary function:

A B C D f

m0 0 0 0 0 0

m1 0 0 0 1 0

m2 0 0 1 0 0

m3 0 0 1 1 0

m4 0 1 0 0 1

m5 0 1 0 1 0

m6 0 1 1 0 0

m7 0 1 1 1 0

m8 1 0 0 0 1

m9 1 0 0 1 x

m10 1 0 1 0 1

m11 1 0 1 1 1

m12 1 1 0 0 1

m13 1 1 0 1 0

m14 1 1 1 0 x

m15 1 1 1 1 1

One can easily form the canonical sum of products expression from this table, simply by summing the minterms (leaving out don't-care terms) where the function evaluates to one:

fA,B,C,D = A'BC'D' + AB'C'D' + AB'CD' + AB'CD + ABC'D' + ABCD.

Of course, that's certainly not minimal. So to optimize, all minterms that evaluate to one are first placed in a minterm table. Don't-care terms are also added into this table, so they can be combined with minterms:

Number of 1s Minterm Binary Representation

1m4 0100

m8 1000

2

m9 1001

m10 1010

m12 1100

3m11 1011

m14 1110

4 m15 1111

At this point, one can start combining minterms with other minterms. If two terms vary by only a single digit changing, that digit can be replaced with a dash indicating that the digit doesn't matter. Terms that can't be combined any more are marked with a "*". When going from Size 2 to Size 4, treat '-' as a third bit value. Ex: -110 and -100 or -11- can be combined, but not -110 and 011-. (Trick: Match up the '-' first.)

Number of 1s Minterm 0-Cube Size 2 Implicants Size 4 Implicants

1m4 0100 m(4,12) -100* m(8,9,10,11) 10--*

m8 1000 m(8,9) 100- m(8,10,12,14) 1--0*

-- -- -- m(8,10) 10-0 --

2

m9 1001 m(8,12) 1-00 m(10,11,14,15) 1-1-*

m10 1010 -- --

m12 1100 m(9,11) 10-1 --

-- -- -- m(10,11) 101- --

3m11 1011 m(10,14) 1-10 --

m14 1110 m(12,14) 11-0 --

4m15 1111 m(11,15) 1-11 --

m(14,15) 111-

Note: In this example, none of the terms in the size 4 implicants table can be combined any further. Be aware that this processing should be continued otherwise (size 8 etc).

Step 2: prime implicant chartNone of the terms can be combined any further than this, so at this point we construct an essential prime implicant table. Along the side goes the prime implicants that have just been generated, and along the top go the minterms specified earlier. The don't care terms are not placed on top - they are omitted from this section because they are not necessary inputs.

4 8 10 11 12 15 => A B C Dm(4,12)* X X -100 => - 1 0 0m(8,9,10,11) X X X 10-- => 1 0 - -m(8,10,12,14) X X X 1--0 => 1 - - 0m(10,11,14,15)* X X X 1-1- => 1 - 1 -

Here, each of the essential prime implicants has been starred - the second prime implicant can be 'covered' by the third and fourth, and the third prime implicant can be 'covered' by the second and first, and neither is thus essential. If a prime implicant is essential then, as would be expected, it is necessary to include it in the minimized boolean equation. In some cases, the essential prime implicants do not cover all minterms, in which case additional procedures for chart reduction can be employed. The simplest "additional procedure" is trial and error, but a more systematic way is Petrick's Method. In the current example, the essential prime implicants do not handle all of the minterms, so, in this case, one can combine the essential implicants with one of the two non-essential ones to yield one of these two equations: Both of those final equations are functionally

Q-7. With appropriate diagrams explain the bus structure of a computer system.

Ans-

BUS structure : A group of lines that serves as a connecting path for several devices is called bus.In addition to the lines that carry the data, the bus must have lines for address and control purposes.In computer architecture, a bus is a subsystem that transfers data between computer components inside a computer or between computers

A computer bus structure is provided which permits replacement of removable modules during operation of a computer wherein means are provided to precharge signal output lines to within a predetermined range prior to the usage of the signal output lines to carry signals, and further, wherein means are provided to minimize arcing to pins designed to carry the power and signals of a connector. In a specific embodiment, pin length, i.e., separation between male and female components of the connector, are subdivided into long pin length and short pin length. Ground connections and power connections for each voltage level are assigned to the long pin lengths. Signal connections and a second power connection for each voltage level is assigned to the short pin lengths. The precharge/prebias circuit comprises a resistor divider coupled between a power source and ground with a high impedance tap coupled to a designated signal pin, across which is coupled a charging capacitor or equivalent representing the capacitance of the signal line. Bias is applied to the precharge/prebias circuit for a sufficient length of time to precharge the signal line to a desired neutral signal level between expected high and low signal values prior to connection of the short pin to its mate..

Early computer buses were literally parallel electrical buses with multiple connections, but the term is now used for any physical arrangement that provides the same logical functionality as a parallel electrical bus. Modern computer buses can use both parallel and bit-serial connections, and can be wired in either a multidrop (electrical parallel) or daisy chain topology, or connected by switched hubs, as in the case of USB.

Description of BUS :At one time, "bus" meant an electrically parallel system, with electrical conductors similar or identical to the pins on the CPU. This is no longer the case, and modern systems are blurring the lines between buses and networks.Buses can be parallel buses, which carry data words in parallel on multiple wires, or serial buses, which carry data in bit-serial form. The addition of extra power and control connections, differential drivers, and data connections in each direction usually means that most serial buses have more conductors than the minimum of one used in the 1-Wire and UNI/O serial buses. As data rates increase, the problems of timing skew, power consumption, electromagnetic interference and crosstalk across parallel buses become more and more difficult to circumvent. One partial solution to this problem has been to double pump the bus. Often, a serial bus can actually be operated at higher overall data rates than a parallel bus, despite having fewer electrical connections, because a serial bus inherently has no timing skew or crosstalk. USB, FireWire, and Serial ATA are examples of this. Multidrop connections do not work well for fast serial buses, so most modern serial buses use daisy-chain or hub designs.Most computers have both internal and external buses. An internal bus connects all the internal components of a computer to the motherboard (and thus, the CPU and internal memory). These types of buses are also referred to as a local bus, because they are intended to connect to local devices, not to those in other machines or external to the computer. An external bus connects external peripherals to the motherboard.Network connections such as Ethernet are not generally regarded as buses, although the difference is largely conceptual rather than practical. The arrival of technologies such as InfiniBand and HyperTransport is further blurring the boundaries between networks and buses. Even the lines between internal and external are sometimes fuzzy, I²C can be used as both an internal bus, or an external bus (where it is known as ACCESS.bus), and InfiniBand is intended to replace both internal buses like PCI as well as external ones like Fibre Channel. In the typical desktop application, USB serves as a peripheral bus, but it also sees some use as a networking utility and for connectivity between different computers, again blurring the conceptual distinction.

Q.8- Explain the CPU organization with relevant diagrams.

Answer.

CPU which is the heart of a computer consists of Registers, Control Unit and Arithmetic Logic Unit. The following tasks are to be performed by the CPU:1. Fetch instructions: The CPU must read instructions from the memory.2. Interpret instructions: The instructions must be decoded to determine what action is required.3. Fetch data: The execution of an instruction may require reading data from memory or an I/O module.4. Process data: The execution of an instruction may require performing some arithmetic or logical operations on data.5. Write data: The results of an execution may require writing data to the memory or an I/O module.

Function of computer organization:1.(a) A CPU can be defined as a general purpose instruction set processor responsible for program execution.(b) A CPU consists of address bus, data bus and control bus.(c) A computer with one CPU is called microprocessor and with more than one CPU in called multiprocessor.(d) The address bus in used to transfer addresses from the CPU to main memory or to I/O devices(e) Data bus is the main path by which information is transferred to and from the CPU.(f) Control bus is used by CPU to control various devices connected and to synchronise their operations with those of the CPU.2. (a) A control unit take the instructions one by one to execute. It takes data from input devices and store it in memory. And also sends data from memory to the output device.(b) All arithmetic and logical operations are carried out by Arithmetic Logical Unit(c) A control unit and the arithmetic logical unit together is known as CPU3. (a) The accumulator is the main register of the ALU.(b) In execution of the most of the instructions the accumulator is used to store a input data or output result.(c) Instructions register holds the opcode of the current instruction(d) Memory address register holds the address of the current instructions.4. A accumulator based CPU consists of (a) data processing unit (b) program control unit and (c) memory and I/O interface unit.(a) (i) In the data processing unit, data is processed to get some results.(ii) the accumulator in the main operand register of the ALU.(b) (i) Program control unit controls various parts of CPU.(ii) Program counter holds the address of the next instructions to be read from memory after the current instruction is executed.(iii) Instruction register holds the opcode of the current instruction.(iv) Control circuits hold the responsibility of every operation of the CPU.(c) (i) Data registers of the memory and I/O interface unit acts as a buffer between the CPU and main memory.

(ii) Address register contains the address of the present instructions obtained from the program control unit.5. (a) Stack pointer and flag register are special registers of CPU.(b) Stack pointers holds the address of the most recently entered item into the stack.(c) Flag registers indicates the status which depends on the results of the operation.6. (a) Micro operations are the operations executed on data stored in registers.(b) A set of micro operations specified by an instruction is known as macro operation.7. (a) A sequence of operations involved in processing an instruction constitutes an instruction cycle(b) Fetch cycle in defined as the time required for getting the instruction code from main memory to CPU(c) Executed cycle is the time required to decode and execute an instruction.(d) Fetch cycle requires a fixed time slot and execute cycle requires variable time slot.8. (a) a word length indicates the number of bits the CPU can process at a time.(b) A memory size indicates the total storage capacity of the CPU.(c) Word length is the indication of bit length of each register.9. A computer is said to be operated based on stored program concept if it stores the instructions as well as data of a program in main memory when they are waiting to execute.10. A stored program in main memory is executed instruction after instruction in a successive method by using program counter.

Q.9: Explain the organization of 8085 processor with the relevant diagrams.

Ans- The Intel 8085 is an 8-bit microprocessor introduced by Intel in 1977. It was binary-compatible with the more-famous Intel 8080 but required less supporting hardware, thus allowing simpler and less expensive microcomputer systems to be built.The "5" in the model number came from the fact that the 8085 required only a +5-volt (V) power supply rather than the +5V, -5V and +12V supplies the 8080 needed. Both processors were sometimes used in computers running the CP/M operating system, and the 8085 later saw use as a microcontroller, by virtue of its low component count. Both designs were eclipsed for desktop computers by the compatible Zilog Z80, which took over most of the CP/M computer market as well as taking a share of the booming home computer market in the early-to-mid-1980s.The 8085 had a long life as a controller. Once designed into such products as the DECtape controller and the VT100 video terminal in the late 1970s, it continued to serve for new production throughout the life span of those products (generally longer than the product life of desktop computers).The 8085 Architecture follows the "von Neumann architecture", with a 16-bit address bus, and a 8-bit data bus.The 8085 used a multiplexed Data Bus i.e.the address was split between the 8-bit address bus and 8-bit data bus. (For saving Number of Pins)

Registers:

The 8085 can access 216 (= 65,536) individual 8-bit memory locations, or in other words, its address space is 64 KB. Unlike some other microprocessors of its era, it has a separate address space for up to 28 (=256) I/O ports. It also has a built in register array which are usually labeled A (Accumulator), B, C, D, E, H, and L. Further special-purpose registers are the 16-bit Program Counter (PC), Stack Pointer (SP), and 8-bit flag register F. The microprocessor has three maskable interrupts (RST 7.5, RST 6.5 and RST 5.5), one Non-Maskable interrupt (TRAP), and one externally serviced interrupt (INTR). The RST n.5 interrupts refer to actual pins on the processor-a feature which permitted simple systems to avoid the cost of a separate interrupt controller chip.

Buses:* Address bus - 16 line bus accessing 216 memory locations (64 KB) of memory.* Data bus - 8 line bus accessing one (8-bit) byte of data in one operation. Data bus width is the traditional measure of processor bit designations, as opposed to address bus width, resulting in the 8-bit microprocessor designation.* Control buses - Carries the essential signals for various operations.

Q-10. Describe the theory of addressing modes.

Answer.

Addressing modes are an aspect of the instruction set architecture in most central processing unit (CPU) designs. The various addressing modes that are defined in a given instruction set architecture define how machine language instructions in that architecture identify the operand (or operands) of each instruction. An addressing mode specifies how to calculate the effective memory address of an operand by using information held in registers and/or constants contained within a machine instruction or elsewhere.In computer programming, addressing modes are primarily of interest to compiler writers and to those who write code directly in assembly language.How many Addressing modes :Different computer architectures vary greatly as to the number of addressing modes they provide in hardware. There are some benefits to eliminating complex addressing modes and using only one or a few simpler addressing modes, even though it requires a few extra instructions, and perhaps an extra register.[1] It has proven[citation needed] much easier to design pipelined CPUs if the only addressing modes available are simple ones.Most RISC machines have only about five simple addressing modes, while CISC machines such as the DEC VAX supermini have over a dozen addressing modes, some of which are quite complicated. The IBM System/360 mainframe had only three addressing modes; a few more have been added for the System/390.When there are only a few addressing modes, the particular addressing mode required is usually encoded within the instruction code (e.g. IBM System/390, most RISC). But when there are lots of addressing modes, a specific field is often set aside in the instruction to specify the addressing mode. The DEC VAX allowed multiple memory operands for almost all instructions, and so reserved the first few bits of each operand specifier to indicate the addressing mode for

that particular operand. Keeping the addressing mode specifier bits separate from the opcode operation bits produces a orthogonal instruction set.

Even on a computer with many addressing modes, measurements of actual programs [indicate that the simple addressing modes listed below account for some 90% or more of all addressing modes used. Since most such measurements are based on code generated from high-level languages by compilers, this reflects to some extent the limitations of the compilers being used.

Simple addressing modes for codeAbsolute +----+------------------------------+ |jump| address | +----+------------------------------+ (Effective PC address = address)The effective address for an absolute instruction address is the address parameter itself with no modifications.PC-relative +----+------------------------------+ |jump| offset | jump relative +----+------------------------------+

(Effective PC address = next instruction address + offset, offset may be negative)The effective address for a PC-relative instruction address is the offset parameter added to the address of the next instruction. This offset is usually signed to allow reference to code both before and after the instruction.This is particularly useful in connection with jumps, because typical jumps are to nearby instructions (in a high-level language most if or while statements are reasonably short). Measurements of actual programs suggest that an 8 or 10 bit offset is large enough for some 90% of conditional jumps[citation needed].Another advantage of program-relative addressing is that the code may be position-independent, i.e. it can be loaded anywhere in memory without the need to adjust any addresses.Some versions of this addressing mode may be conditional referring to two registers ("jump if reg1==reg2"), one register ("jump unless reg1==0") or no registers, implicitly referring to some previously-set bit in the status register. See also conditional execution below.Register indirect +-------+-----+ |jumpVia| reg | +-------+-----+ (Effective PC address = contents of register 'reg')

The effective address for a Register indirect instruction is the address in the specified register. For example, (A7) to access the content of address register A7.The effect is to transfer control to the instruction whose address is in the specified register.Many RISC machines have a subroutine call instruction that places the return address in an address register—the register indirect addressing mode is used to return from that subroutine call.

Sequential addressing modessequential execution +------+ | nop | execute the following instruction +------+(Effective PC address = next instruction address)The CPU, after executing a sequential instruction, immediately executes the following instruction.Sequential execution is not considered to be an addressing mode on some computers.Most instructions on most CPU architectures are sequential instructions. Because most instructions are sequential instructions, CPU designers often add features that deliberately sacrifice performance on the other instructions—branch instructions—in order to make these sequential instructions run faster.Conditional branches load the PC with one of 2 possible results, depending on the condition—most CPU architectures use some other addressing mode for the "taken" branch, and sequential execution for the "not taken" branch.Many features in modern CPUs -- instruction prefetch and more complex pipelineing, Out-of-order execution, etc. -- maintain the illusion that each instruction finishes before the next one begins, giving the same final results, even though that's not exactly what happens internally.Each "basic block" of such sequential instructions exhibits both temporal and spatial locality of reference.CPUs that do not use sequential executionCPUs that do not use sequential execution with a program counter are extremely rare. In some CPUs, each instruction always specifies the address of next instruction. Such CPUs have a instruction pointer that holds that specified address, but they do not have a complete program counter. Such CPUs include some drum memory computers, the SECD machine, and the RTX 32P.Other computing architectures go much further, attempting to bypass the von Neumann bottleneck using a variety of alternatives to the program counter.conditional executionSome computer architectures (e.g. ARM) have conditional instructions which can in some cases obviate the need for conditional branches and avoid flushing the instruction pipeline. An instruction such as a 'compare' is used to set a condition code, and subsequent instructions include a test on that condition code to see whether they are obeyed or ignored. +------+-----+-----+ |skipEQ| reg1| reg2| skip the following instruction if reg1=reg2 +------+-----+-----+ (Effective PC address = next instruction address + 1)Skip addressing may be considered a special kind of PC-relative addressing mode with a fixed "+1" offset. Like PC-relative addressing, some CPUs have versions of this addressing mode that only refer to one register ("skip if reg1==0") or no registers, implicitly referring to some previously-set bit in the status register. Other CPUs have a version that selects a specific bit in a specific byte to test ("skip if bit 7 of reg12 is 0").Unlike all other conditional branches, a "skip" instruction never needs to flush the instruction pipeline.

Q-11. Describe the following elements of Bus Design: A) Bus Types B) Arbitration Methods C) Bus TimingAns.-

BUS TYPES

Bus channels can be separated into two general types, namely a dedicated and multiplexed. A bus line dedicated permanently assigned a function or a subset of physical computer components.

As an example of dedication to the function is the use of separate dedicated address and data lines, which is a common thing for the bus.

However, this is not important. For example, the address and data information can be transmitted through the same number of channels using control channel address is invalid. In the early transfer of data, address bus and address placed on a valid control activated. At this time, each module has a specific time period to copy the address and determine whether the address is a module located. Then address removed from the bus and the bus connection is used for reading or writing data transfer next. Methods use the same line for various purposes is known as time multiplexing.

The advantage is time multiplexing requires fewer channels, which saves space and cost. The disadvantage is the need for a more complex circuit within each module. There's also a fairly large decrease in performance due to certain events that use the channels together cannot function in parallel.

Physical dedication associated with the use of multiple buses, each bus it is connected with only a subset of the modules. A common example is the use of bus I / O to interconnect all I / O module, then this bus is connected to the main bus through a kind of adapter module I / O. The main advantage of physical dedication is a high throughput; because of the traffic congestion is only small data. The disadvantage is the increased size and cost of the system.

b) ARBITRATION METHOD

In all systems except the simplest system, more than one module is required to control the bus. For example, an I / O module may be required to read or write directly to memory, without sending data to the CPU. Because at one point is just a unit that will successfully transmit data through the bus, then take a few method of arbitration. The various methods by and large can be classified as a method centralized and distributed methods. In the centralized method, a hardware device, known as bus controllers or arbitrary, is responsible for the allocation of time on the bus. Perhaps it is CPU device module shaped or a separate section.

In a distributed method, there is no central controller. Rather, each module consists of access control logic and modules work together to put on a bus together. In the second method of arbitration, the goal is to assign a device, the CPU or I / O module, acting as master. Then the master can initiate data transfer (e.g., read or write) by using other devices, which work as slave for this particular data exchange.

C) Bus timing:

The timing diagram example on the right describes the Serial Peripheral Interface (SPI) Bus. Most SPI master nodes have the ability to set the clock polarity (CPOL) and clock phase (CPHA) with respect to the data. This timing diagram shows the clock for both values of CPOL as well as the values for the two data lines (MISO & MOSI) for each value of CPHA. Note that when CPHA=1 then the data is delayed by one-half clock cycle.

SPI operates in the following way:

The master determines an appropriate CPOL & CPHA value The master pulls down the slave select (SS) line for a specific slave chip

The master clocks SCK at a specific frequency

During each of the 8 clock cycles the transfer is full duplex:

The master writes on the MOSI line and reads the MISO line

o The slave writes on the MISO line and reads the MOSI line

When finished the master can continue with another byte transfer or pull SS high to end the transfer

When a slave's SS line is high then both of its MISO and MOSI line should be high impedance so to avoid disrupting a transfer to a different slave. Prior to SS being pulled low, the MISO & MOSI lines are indicated with a "z" for high impedance. Also prior to the SS being pulled low the "cycle #" row is meaningless and is shown greyed-out.

Note that for CPHA=1 the MISO & MOSI lines are undefined until after the first clock edge and are also shown greyed-out before that.

A more typical timing diagram has just a single clock and numerous data lines

Q-12. Explain the following types of operations:

Types of Operations Data transfer

ArithmeticAns.-

Types of OperationsA computer has a set of instructions that allows the user to formulate any data-processing task. To carry out tasks, regardless of whether a computer has 100 instructions or 30U instructions, its instructions must be capable of performing following basic operations :

Data movement Data processing

Program sequencing and controlThe details of above operations are discussed here. The standard notations used in the discussion* are as listed below :

Processor registers are represented by notations R^ K-, R2/ ... and so on, The addresses of the memory locations are represented by names such as LOC, PLACE,

MEM, etc I/O registers are represented by names such as DATA IN, DATA OUT and so on. Tlie contents of register or memory location are denoted by placing square brackets

around the name of the register or memory location-

Data Movement OperationsThese operations include the following data transfer operations :

Data transfer between memory and CPU register Data transfer between CPU registers Data transfer between processor and input/output devices.

Data Transfer between Memory and CPU RegistersArithmetic and logical operations are performed primarily on data in CPU registers. Therefore it is necessary to transfer data from memory to CPU registers before operation and transfer data from CPU registers to memory after operation.We know that both program instructions and data operands are stored in the memory. To execute an instruction processor has to fetch instruction and read the operand or operands if necessary from the memory. After execution of instruction processor may store the result or modified operand back in the memory. Thus, there are two basic operations required in the memory access : 1. Load (Read or fetch) 2. Store (write).Load Operation : In the load operation the contents of the specified memory location are read by the processor. For load operation processor sends the address of the memory location whose contents are to be read and generates the appropriate read signal to indicate that it is a read operation. The memory identifies the addressed memory location and sends the contents of that memory location to the processor. This operation is illustrated in Fig. 3.1

Store Operation : In the store operation die data from the processor is stored in the

specified memory location. For store operation processor sends the address of the memorylocation where the data is to be written on the address lines, the data to be written on the data lines and generates the appropriate write signal to indicate the store operation.The memory identifies the addressed memory location and writes the data sent by the processor at that location destroying the former contents of that location. This operation is illustrated in Fig. 3.2*

Fig. 3.2 Store operation

The data items or instructions which are transferred between memory and processor may be either byte or word- They can be transferred using a single operation* Mainly this transfer takes place between processor registers and the memory locations. We know that processor has small number of built-in registers. Each register is capable of holding a word data- These registers are either the source or the destination of a transfer to or from the memory.

Example : R2 <- |LOG]

This expression states that the contents of memory location LOC are transferred into the processor register R2.

Data Transfer between CPU RegistersAs per the requirement of registers, the CPU registers can be made free by data transfer operations between CPU registers*Example : R3 <- [R2]This expression states that the contents of processor register R2 are transferred into processor register R3.

Data Transfer between Processor and Input/Output DevicesMany applications of a processor based system requires the transfer of data between external circuitry to the processor and processor to the external circuitry, e.g. - user can give information to the processor using keyboard and user can see the result or output information from the processor with the help of display device. The transfer of data between keyboard and processor and display device is called input/output (I/O) data transfer.Example : R1 <- [DATA IN]This expression states that the contents of I/O register, DATA IN are transferred into processor register R1.The Fig. 3.3 shows the typical bus connection for processor, keyboard and display. The DATA IN and DATA OUT are the registers by which the processor reads the contents from keyboard and sends the data for display, respectively. We know that the rate of data transfer from the keyboard is limited by typing speed of the user and the rate of data transfer to the display device is determined by the rate at which characters can be transmitted over the link between the computer and the display device. The rate of output data transfer to display is much higher than the input data rate from the keyboard; however both of these rates are much slower than the speed of a processor that can execute many millions of instructions per second. Due to the speed difference between these devices we have to use synchronisation mechanism for proper transfer of data between them.As shown in the Fig. 3.3, SIN and SOUT status bits are used to synchronize data transfer between keyboard and processor, and data transfer between display and processor, respectively.

When key is pressed, the corresponding character code is stored in the DATA IN register and S IN

status bit is set to indicate that the valid character code is available in the DATA IN register. Under program control processor checks the S IN bit, and when it finds SIN = 1, it reads the contents of the DATA IN register. After completion of read operation SIN is automatically cleared to 0. If another key is pressed, the corresponding character code is entered into the DATA IN register, SIN is again set to 1 and the process repeats.

When the character is to be transferred from the processor to the display, DATA OUT register and SOUT status bit are used. Under program control, processor checks SOUT bit SOUT = 1, indicates that the display is ready to receive character. Therefore, when processor wants to transfer data to the display device and SOUT = 1, processor transfers data to be displayed to the DATA OUT

register and dears SOUT status bit to 0. The display device then reads the character from DATA OUT register. After acceptance of the data by the display device, the SOUT bit is automatically set to 1 and display device is ready to accept the next data byte.

Data Processing OperationsThese operations include the following types of operations.

Arithmetic operations Logical operations Shift and Rotate operations

Arithmetic OperationsThese operations include basically the following operations. Add Subtract Multiply Divide §

Increment Decrement Negate (Change sign of operand)

Example : ADO R1, R2, R3This expression states that the contents of processor registers Rl and R2 are added and the result is stored in the register R3.The multiplication instruction does the multiplication of two operands and stores the result in the destination operand. On the other hand the division instruction does the division of two operands and stores the result in the destination operand. For example

MUL R1, RO : R1 <-R1 x RO

DIV R1, RO : R1<-R1 + R0

Unfortunately, all processors do not provide these instructions. On those processors MUL and DIV instruction are implemented by performing basic operations such as add, subtract shift and rotate, repeatedly.

Logical Operations

These operations include basically the following operations.

AND OR NOT EXOR Compare Shift Rotate

Example : AND R1f R2, R3

This expression states that the contents of processor registers Rl and R2 are logically ANDed and the result is stored in the register R3.

Shift and Rotate Operations

These operations shift or rotate the bits of the operand right or left by some specified number of bit positions.

Logical Shift : There are two logical shift instructions logical shift left (LShiftL) and logical shift right (LShiftR). These two instructions shift an operand by a number of the positions specified in a count operand contain in the instruction. The general syntax for these instructions are:

LshiftL dst, countLshiftR dst count

The count operand may be an immediate number, or it may be contains of the processor register. The Fig. 3.4 (see on next page) shows the operation of LShiftL and LShiftR instructions.

Arithmetic Shift : In logical shift operation we have seen that the vacant positions created within the register due to shift operation arc filled with zeroes. In arithmetic right shift it is necessary to repeat the sign bit as the fill-in bit for the vacated position. This requirement on right shifting distinguishes arithmetic shifts from logical shifts. Otherwise two shift operations are very similar. The Fig. 3.5 shows the example of an arithmetic right shift (AshiftR). The arithmetic left shift (AshiftL) is same as the logical left shift.

Master of Computer Application (MCA) – Semester 1

MC0062 – Digital Systems, Computer Organization and Architecture

(Book ID: B0680 & B0684)

Assignment Set – 1

Q. 1. With a neat labeled diagram explain the working of Gated Latches.

Answer.

Gated SR Latch

Two possible circuits for gated SR latch are shown in Figure 1. The graphical symbol for gated SR latch is shown in Figure 2.

R SQ Q

Clk Clk

Q Q S R

(a) Gated SR latch with NOR and AND gates. (b) Gated SR latch with NAND gates.

Figure 1. Circuits for gated SR latch.

S Q

Clk

R Q

Figure 2. The graphical symbol for gated SR latch

The characteristic table for a gated SR latch which describes its behavior is as follows.

Clk S R Q+Comments

0 × × Q No change, typically stable states Q = 0, Q = 1 or Q = 1, Q = 01 0 0 Q No change, typically stable states Q = 0, Q = 1 or Q = 1, Q = 01 0 1 0 Reset1 1 0 1 Set

1 1 1 × Avoid this setting

Figure 3 shows an example timing diagram for gated SR latch (assuming negligible propagation delays through the logic gates). Notice that during the last clock cycle when Clk = 1, both R = 1 and S = 1. So as Clk returns to 0, the next state will be uncertain. This explains why we need to avoid the setting in the last row of the above characteristic table in normal operation of a gated SR latch.

1Clk

0

1R

0

1S

0

Q 1

0

Q 10

?

?

Time

Figure 3. An example timing diagram for gated SR latch.

Gated D Latch

A possible circuit for gated D latch is shown in Figure 4. The graphical symbol for gated D latch is shown in Figure 5.

DS

(Data) Q

Clk

R Q

Figure 4. A circuit for gated D latch.

D Q

Clk Q

Figure 5. The graphical symbol for gated D latch

The characteristic table for a gated D latch which describes its behavior is as follows.

Clk D Q+Comments

0 × Q No change1 0 01 1 1

Figure 6 shows an example timing diagram for gated D latch (assuming negligible propagation delays through the logic gates).

t 1

t 2

t 3

t 4

1Clk

0

1D

0

1Q

0

Time

Figure 6. An example timing diagram for gated D latch.

A Negative-edge-triggered Master-Slave D Flip-Flop

A possible circuit for a negative-edge-triggered master-slave D flip-flop is shown in Figure 7. The graphical symbol for a negative-edge-triggered D flip-flop is shown in Figure 8.

Master Slave

Q m Q sD D Q D Q Q

Clock Clk Q Clk Q Q

. A negative−edge−triggered master−slave D flip−flop.

D Q

Q

The graphical symbol for negative−edge−triggered D flip−flop.

Note that unlike latches which are level sensitive (i.e., the output of a latch is controlled by the level of the clock input), flip-flops are edge triggered (i.e., the output changes only at the point in time when the clock changes from one value to the other). The master-slave D flip-flop shown in Figure 7 responds on the negative edge (i.e., the edge where the clock signal changes from 1 to 0) of the clock signal. Hence it is negative-edge-triggered. The circuit can be changed to respond to the positive clock edge by connecting the slave stage directly to the clock and the master stage to the complement of the clock.

Figure 9 shows an example timing diagram for negative-edge-triggered D flip-flop (assuming negligible propagation delays through the logic gates).

t 1 t 2 t 3

1Clock

0

1D

0

1

Q m 0

1Q = Q s

0

Time

An example timing diagram for negative−edge−triggered master−slave D flip−flop.

A Positive-edge-triggered D Flip-Flop

Besides building a positive-edge-triggered master-slave D flip-flop as mentioned in our preceding discussion, we can accomplish the same task by a circuit presented in Figure 10. It requires only six NAND gates and, hence, fewer logic gates.

1 P3

2P1

5 Q

Clock

3P2 6 Q

D4 P4

The operation of the circuit in Figure 10 is as follows. When Clock = 0, the outputs of gates 2 and 3 are high. Thus P 1 = P 2 = 1, which maintains the output latch, comprising gates 5 and 6, in its present state. At the same time, the signal P 3 is equal to D, and P 4 is equal to its complement D. When Clock changes to 1, the following changes take place. The values of P 3 and P 4 are transmitted through gates 2 and 3 to cause P 1 = D and P 2 = D, which sets Q = D and Q = D. To operate reliably, P 3 and P 4 must be stable when Clock changes from 0 to 1. Hence the setup time of the flip-flop is equal to the delay from the D input through gates 4 and 1 to P 3. The hold time is given by the delay through gate 3 because once P 2 is stable, the changes in D no longer matter.

For proper operation it is necessary to show that, after Clock changes to 1, any further changes in D will not a ect the output latch as long as Clock = 1. We have to consider two cases. Suppose first that D = 0 at the positive edge of the clock. Then P 2 = 0, which will keep the output of gate 4 equal to 1 as long as Clock = 1, regardless of the value of the D input. The second case is if D = 1 at the positive edge

of the clock. Then P 1 = 0, which forces the outputs of gates 1 and 3 to be equal to 1, regardless of the D input. Therefore, the flip-flop ignores changes in the D input while Clock = 1.

Figure 11 shows the graphical symbol for a positive-edge-triggered D flip-flop.

D Q

Q

Figure 11. The graphical symbol for positive−edge−triggered D flip−flop.

Figure 12 shows an example timing diagram for positive-edge-triggered D flip-flop (assuming negligible propagation delays through the logic gates).

t 1 t 2 t 3

1Clock

0

1D

0

1Q

0

Time

Figure 12. An example timing diagram for positive−edge−triggered master−slave

D flip−flop.

D Flip-Flops with Clear and Preset

In using flip-flops, it is often necessary to force the flip-flops into a known initial state. A simple way of providing the clear and present capability is to add extra input to the flip-flops. Figure 12 shows a master-slave D flip-flop with Clear and Preset. Placing a 0 on the Clear input will force the flip-flop into the state Q = 0. If Clear = 1, then this input will have no e ect on the NAND gates. Similarly, Preset = 0 forces the flip-flop into the state Q = 1, while Preset = 1 has no e ect. To denote that the Clear and Preset inputs are active when their value is 0, we placed an overbar on the names in Figure 13. Note that the circuit that uses this flip-flop should not try to force both Clear and Preset to 0 at the same time. A graphical symbol for this flip-flop is shown in Figure 14.

Preset

DQ

Clock

Q

Clear

A circuit for master−slave D flip−flop with Clear and Preset.

Preset

D Q

Q

Clear

Figure 14. The graphical symbol for master−slave D flip−flop with Clear and Preset.

A similar modification can be done on the positive-edge-triggered D flip-flop of Figure 10, as indicated in Figure 15. A graphical symbol for this flip-flop is shown in Figure 16. Again, both Clear and Preset inputs are active low. They do not disturb the flip-flop when they are equal to 1.

Preset

Q

Clock

Q

D

Clear

Figure 15. A positive−edge−triggered D flip−flop with Clear and Preset.

Preset

D Q

Q

Clear

Figure 16. The graphical symbol for positive−edge−triggered D flip−flop with Clear and Preset.

In the circuits in Figures 13 and 15, the e ect of a low signal on either the Clear or Preset input is immediate. For example, if Clear = 0 then the flip-flop goes into the state Q = 0 immediately, regardless of the value of the clock signal. In such a circuit, where the Clear signal is used to clear a flip-flop without regard to the clock signal, we say that the flip-flop has an asynchronous clear. In practice, it is often preferable to clear the flip-flops on the active edge of the clock. Such synchronous clear can be accomplished as shown in Figure 17. The flip-flop operates normally when the Clear input is equal to 1. But if Clear goes to 0, then on the next positive edge of the clock the flip-flop will be cleared to 0.

Clear D Q QD

Q QClock

Figure 17. Synchronous reset for a D flip−flop.

Q. 2. With a neat labeled diagram explain the working of Master Slave JK Flip Flop.Answer.The Master-Slave JK Flip-flopThe Master-Slave Flip-Flop is basically two gated SR flip-flops connected together in a series configuration with the slave having an inverted clock pulse. The outputs from Q and Q from the "Slave" flip-flop are fed back to the inputs of the "Master" with the outputs of the "Master" flip-flop being connected to the two inputs of the "Slave" flip-flop. This feedback configuration from the slave's output to the master's input gives the characteristic toggle of the JK flip-flop as shown below.The Master-Slave JK Flip-Flop

The input signals J and K are connected to the gated "master" SR flip-flop which "locks" the input condition while the clock (Clk) input is "HIGH" at logic level "1". As the clock input of the "slave" flip-flop is the inverse (complement) of the "master" clock input, the "slave" SR flip-flop does not toggle. The outputs from the "master" flip-flop are only "seen" by the gated "slave" flip-flop when the clock input goes "LOW" to logic level "0". When the clock is "LOW", the outputs from the "master" flip-flop are latched and any additional changes to its inputs are ignored. The gated "slave" flip-flop now responds to the state of its inputs passed over by the "master" section. Then on the "Low-to-High" transition of the clock pulse the inputs of the "master" flip-flop are fed through to the gated inputs of the "slave" flip-flop and on the "High-

to-Low" transition the same inputs are reflected on the output of the "slave" making this type of flip-flop edge or pulse-triggered.Then, the circuit accepts input data when the clock signal is "HIGH", and passés the data to the output on the falling-edge of the clock signal. In other words, the Master-Slave JK Flip-flop is a "Synchronous" device as it only passes data with the timing of the clock signal.

Q.3. Describe the functioning of Negative edge triggered 2 – Bit Ripple up counters with relevant diagram(s).

Answer.Two bit ripple counter used two flip-flops. There are four possible states from 2 – bit up-counting I.e. 00, 01, 10 and 11.

The counter is initially assumed to be at a state 00 where the outputs of the tow flip-flops are noted as Q1Q0. Where Q1 forms the MSB and Q0 forms the LSB.

For the negative edge of the first clock pulse, output of the first flip-flop FF1 toggles its state. Thus Q1 remains at 0 and Q0 toggles to 1 and the counter state are now read as 01.

During the next negative edge of the input clock pulse FF1 toggles and Q0 = 0. The output Q0 being a clock signal for the second flip-flop FF2 and the present transition acts as a negative edge for FF2 thus toggles its state Q1 = 1. The counter state is now read as 10.

For the next negative edge of the input clock to FF1 output Q0 toggles to 1. But this transition from 0 to 1 being a positive edge for FF2 output Q1 remains at 1. The counter state is now read as 11.

For the next negative edge of the input clock, Q0 toggles to 0. This transition from 1 to 0 acts as a negative edge clock for FF2 and its output Q1 toggles to 0. Thus the starting state 00 is attained.

Q.4 Describe the functioning of Three and Four bit Synchronous Binary Up-counter.Answer. 3-bit Synchronous Binary Up-counter.Q0 changes on each clock pulse. Observation on F1 indicates that it changes or toggles its output Q1 state only when Q0 is 1 or HIGH. This occurs at the negative edges of second, fourth, sixth, eighth clock pulses. Therefore Q0 is connected to the inputs J and K of the F1.Similarly Q2 changes state only when Q0 = 1 and Q1 = 1 . This is been realized with an AND gate logic. Inputs J and K are connected through an AND gate which has the function Q0 Q1.

Clock pulse Q2 Q1 Q0

0 0 0 0

1 0 0 1

2 0 1 0

3 0 1 1

4 1 0 0

5 1 0 1

6 1 1 0

7 1 1 1

8 0 0 0

Three bit synchronous up counter

4- bit Synchronous Binary Up-counter.The J and K input control for the first three flip-flops is the same as presented in three stage counter. For the fourth stage. Q3, changes only twice in the sequence. And notice that during both of these transitions occur following the times, QA, QB and QC are all HIGH. This condition is denoted by a three input AND gate. For all other times inputs J and K of Q3 and LOW, and it is in a no-change condition.

Time diagram for 4-bit synchronous up-counter

Q.5 Explain the design of modulus counters.

Answer.

A single flip-flop used gives two states output and is referred to as mod-2 counter. With two flip-flops four output states can be counted in ascending or in descending way and is referred to as mod-4 or mod-22 counter. With ‘n’ flip-flops a mod-2n counting is possible either of ascending or of descending type.To design an asynchronous counter to count till M or mod-M counter where M is not a power of 2, following procedure is used.

Find the number of flip-flops required n = log2 M. calculated value is not an integer value if the M # 2n then select n by rounding to the next integer value.

First writer the sequence of counting till M either in ascending or in descending way. Tabulate the value to reset the flip-flops in a mod-M count. Find the flip-flop outputs which are to reset from the tabulated value. Tap the output from these flip-flops and feed it to an NAND gate whose output is

connected to the clear pin.

Q.6 Describe Serial in and Serial out shift registers.

Ans.

Serial-in, serial-out shift registers delay data by one clock time for each stage. They will store a bit of data for each register. A serial-in, serial-out shift register may be one to 64 bits in length, longer if registers or packages are cascaded. Below is a single stage shift register receiving data which is not synchronized to the register clock. The "data in" at the D pin of the type D FF (Flip-Flop) does not change levels when the clock changes for low to high. We may want to synchronize the data to a system wide clock in a circuit board to improve the reliability of a digital logic circuit.

The obvious point (as compared to the figure below) illustrated above is that whatever "data in" is present at the D pin of a type D FF is transferred from D to output Q at clock time. Since our example shift register uses positive edge sensitive storage elements, the output Q follows the D input when the clock transitions from low to high as shown by the up arrows on the diagram above. There is no doubt what logic level is present at clock time because the data is stable well before and after the clock edge. This is seldom the case in multi-stage shift registers. But, this was an easy example to start with. We are only concerned with the positive, low to high, clock edge. The falling edge can be ignored. It is very easy to see Q follow D at clock time above. Compare this to the diagram below where the "data in" appears to change with the positive clock edge.

Since "data in" appears to changes at clock time t1 above, what does the type D FF see at clock time? The short over simplified answer is that it sees the data that was present at D prior to the clock. That is what is transfered to Q at clock time t1. The correct waveform is QC. At t1 Q goes to a zero if it is not already zero. The D register does not see a one until time t2, at which time Q goes high.

Since data, above, present at D is clocked to Q at clock time, and Q cannot change until the next clock time, the D FF delays data by one clock period, provided that the data is already synchronized to the clock. The QA waveform is the same as "data in" with a one clock period delay.

.

Three type D Flip-Flops are cascaded Q to D and the clocks paralleled to form a three stage shift register above.

Type JK FFs cascaded Q to J, Q' to K with clocks in parallel to yield an alternate form of the shift register above.

A serial-in/serial-out shift register has a clock input, a data input, and a data output from the last stage. In general, the other stage outputs are not available Otherwise, it would be a serial-in, parallel-out shift register..

Q. 7. Explain the main memory system.

Answer.

Main Memory

The main memory stores data and instructions. Main memories are usually built from dynamic IC’s known as dynamic RAMs. These semiconductor ICs can also implement static memories referred to as static RAMs (SRAMs). SRAMs are faster but cost per bit is higher. These are oten used to build caches.

Types of Random-Access Semiconductor Memory

Dynamic RAM: for example: Charge in capacitor. It requires periodic refreshing. Static RAM: for example: Flip-flop logic-gate. Applying power is enough no need for refreshing). Dynamic RAM is simpler and hence smaller than the static RAM. Therefore more dense and less expensive. But it requires supporting refresh circuitry. Static RAM/s are faster than dynamic RAM’s.

ROM: The data is actually wired in the factory. Can never be altered.

PROM: Programmable ROM. It can only be programmed once after its fabrication. It requires special device to program.

EPROM: Erasable Programmable ROM. It can be programmed multiple times. Whole capacity need to be erased by ultraviolet radiation before a new programming activity. It can be partially programmed.

EEPROM: Electrically Erasable Programmable ROM. Erased and programmed electrically. It can be partially programmed. Write operation takes considerably longer time compared to read operation.

Each more functional Rom is more expensive to build, and has smaller capacity then less functional ROM’s.

Q. 8. Describe various Cache replacement algorithms.

Ans.

For and set associative mapping a replacement algorithm is needed. The most common algorithms are discussed here. When a new block is to be brought into the cache and all the positions that it may occupy are full, then the cache controller must decide which of the old blocks to overwrite. Because the programs usually stay in localized areas for a reasonable period of time, there is high probability that the blocks that have been referenced recently will be referenced again soon. Therefore when a block is to be overwritten, the bock that has not referenced for the longest time is overwritten. This block is called least-recently-used block

(LRU), and the technique is called the LRU replacement algorithm. In order to use the LRU algorithm, the cache controller must track the LRU block as computation proceeds.

There are several replacement algorithms that require less overhead than LRU method. One method is to remove the oldest block from a full set when a new block must be brought in. this method is referred to as FIFO. In this technique no updating is needed when hit occurs. However, because the algorithm does not consider the recent patterns of access to blocks in the cache, it is not effective as LRU approach in choosing the best block to remove. There is another method called least frequently used (LFU) that replaces that block in the set which has experienced the fewer references. It is implemented by associating a counter with each slot. Yet another simplest algorithm called random, is to choose a block to be overwritten in random.

Q. 9. Explain I/O module and its usage.

Ans.

Input-output interface provides a method for transferring information between internal storage and external I/O devices. Peripherals connected to a computer need special communication links for interfacing them with the central processing unit. The purpose of the communication link is to resolve the differences that exist between the central computer and each peripheral. The major differences are:

1. Peripherals are electromechanical and electromagnetic devices and their manner of operation is different from the operation of the CPU and memory, which are electronic devices. Therefore, a conversion of signal values may be required.

2. The data transfer rate of peripherals is usually slower than the transfer rate of the CPU, and consequently, a synchronization mechanism may be needed.

3. Data codes and formats in peripherals differ from the word format in the CPU and memory.

4. The operating modes of peripherals are different from each other and each must be controlled so as not to disturb the operation of other peripherals connected to the CPU.

Input/Output Module• Interface to CPU and Memory• Interface to one or more peripherals• GENERIC MODEL OF I/O DIAGRAM 6.1

I/O Module Function• Control & Timing• CPU Communication• Device Communication• Data Buffering• Error Detection

I/O Steps• CPU checks I/O module device status• I/O module returns status• If ready, CPU requests data transfer• I/O module gets data from device• I/O module transfers data to CPU• Variations for output, DMA, etc.

I/O Module Decisions• Hide or reveal device properties to CPU• Support multiple or single device• Control device functions or leave for CPU• Also O/S decisions

e.g. Unix treats everything it can as a fileInput Output Techniques

• Programmed

• Interrupt driven• Direct Memory Access (DMA)

Q. 10. Write about the following with respect to CPU Control:

Functional Requirements Control Signals

Answer.

Functional RequirementsThe functional requirements of control unit are those functions that the control unit must perform. And these are the basis for the design and implementation of the control unit.

A three step process that lead to the characterization of the Control Unit: Define the basis elements of the processor Describe the micro-operations that the processor performs Determine the functions that the control unit must perform to cause the micro-

operations to be performed.

1. Basic Elements of ProcessorThe following are the basic functional elements of a CPU:

ALU: is the functional essence of the computer. Registers: are used to store data internal to the CPU.

2. Types of Micro-operationThese operations consist of a sequence of micro operations. All micro instructions fall into one of the following categories:

Transfer data between registers Transfer data from register to external Transfer data from external to register Perform arithmetic or logical ops

3. Functions of Control UnitNow we define more explicitly the function of control unit. The control unit perform two tasks:

Sequencing: The control unit causes the CPU to step through a series of micro-operations in proper sequence based on the program being executed.

Execution: The control unit causes each micro-operation to be performed.

For the control unit to perform its function, it must have inputs that allow it to determine the state of the system and outputs that allow it to control the behavior of the system. These are the external specifications of the control unit. Internally, the control unit must have the logic required to perform sequencing and execution functions.

Control unit InputsThe possible inputs for the control units are:Clock: The control unit uses clock to maintain the timings.Instruction register: Op-code of the current instruction is used to determine which micro-instructions to be performed during the execution cycle.Flags: These are needed by the control unit to determine the status of the CPU and outcome of previous ALU operations.Example: As seen earlier the instruction ISZ, which is increment and skip if zero, the control unit will increment the PC if the zero flag is set.Control signals from control bus: the control bus portion of the system bus provides signals to the control unit, such as interrupt signals and acknowledgements.Control Signals – outputThe following are the control signals which are output of the control unit:Control signals within CPU: There are two types

1. Signals that cause data to be moved from one register to another.2. Signals that activate specific ALU functions

Control signals to control bus: There are two types1. Signals to memory2. Signals to I/O modules

11. Discuss and Differentiate Hardware and Micro-programmed control unit.Ans. Hardware Control Unit

There are two major types of control organization: hardwired control and microprogrammed control. In the hardwired organization, the control logic is implemented with gates, flip-flops, decoders, and other digital circuits. It has the advantage that it can be optimized to produce a fast mode of operation. In the microprogrammed organization, the control information is stored in a control memory. The control memory is programmed to initiate the required sequence of microoperations. A hardwired control, as the name implies, requires changes in the wiring among the various components if the design has to be modified or changed. In the microprogrammed control, any required changes or modifications can be done by updating the microprogram in control memory.

The block diagram of the control unit is shown in Fig. 5-6. It consists of two decoders, a sequence counter, and a number of control logic gates. An instruction read from memory is placed in the instruction register (IR). The position of this register in the common bus system is indicated in Fig. 5-4. The instruction register is shown again in Fig. 5-6, where it is divided into three parts: the I bit, the operation code, and bits 0 through 11. The operation code in bits 12 through 14 are decoded with a 3 x 8 decoder. The eight outputs of the decoder are designated by the symbols D0 through D7. The subscripted decimal number is equivalent to the binary value of the corresponding operation code. Bit 15 of the instruction is transferred to a flip-flop designated by the symbol I. Bits 0 through 11 are applied to the control logic gates. The 4-bit sequence counter can count in binary from 0 through 15. The outputs of the counter are decoded into 16 timing signals T0 through T15.

Micro Programed Control Unit

The control memory is assumed to be a ROM, within which all control information is permanently stored. The control memory address register specifies the address of the microinstruction, and d the control data register holds the microinstruction read from memory.

The microinstruction contains a control word that specifies one or more micro-operations for the data processor. Once these operations are executed, the control must determine the next address. The location of the next microinstruction may be the one next in sequence, or it may be located somewhere else in the control memory. For this reason it is necessary to use some bits of the present microinstruction to control the generation of the address of the next microinstruction. The next address may also be a function of external input conditions. While the microoperations are being executed, the next address is computed in the next address generator circuit and then transferred into the control address register to read the next microinstruction. Thus a microinstruction contains bits for initiating microoperations in the data processor part and bits that determine the address sequence for the control memory.

Next address

generator (sequencer)

Control address register

Control memory (ROM)

Control

data

register

Control word

External input

Next address information

The next address generator is sometimes called a microprogram sequencer, as it determines the address sequence that is read from control memory. The address of the next microinstruction can be specified in several ways, depending on the sequencer inputs. Typical functions of a microprogram sequencer are incrementing the control address register by one, loading into the control address register an address from control memory, transferring an external address, or loading an initial address to start the control operations.

The control data register holds the present microinstruction while the next address is computed and read from memory. The data register is some-times called a pipeline register. It allows the execution of the microoperations specified by the control word simultaneously with the generation of the next microinstruction. This configuration requires a two-phase clock, with one clock applied to the address register and the other to the data register.

The system can operate without the control data register by applying a single-phase clock to the address register. The control word and next-address information are taken directly from the control memory. It must be realized that a ROM operates as a combinational circuit, with the address value as the input and the corresponding word as the output. The content of the specified word in ROM remains in the output wires as long as its address value remains in the address register. No read signal is needed as in a random-access memory. Each clock pulse will execute the microoperations specified by the control word and also transfer a new address to the control address register. In the example that follows we assume a single-phase clock and therefore we do not use a control data register. In this way the address register is the only component in the control system that receives clock pulses. The other two components: the sequencer and the control memory are combinational circuits and do not need a clock.

The main advantage of the microprogrammed control is the fact that once the hardware configuration is established, there should be no need for further hardware or wiring changes. If we want to establish a different control sequence for the system, all we need to do is specify a different set of microinstructions for control memory. The hardware configuration should not be changed for different operations; the only thing that must be changed is the microprogram residing in control memory.

Q. 12. Explain:

RISC and CISC

An important aspect of computer architecture is the design of the instruction set for the processor. The instruction set chosen for a particular computer determines the way that machine language programs are constructed. Early computers had small and simple instruction sets, forced mainly by the need to minimize the hardware used to implement them. As digital hardware became cheaper with the advent of integrated circuits, computer instructions tended to increase both in number and complexity. Many computers have instruction sets that include more than 100 and sometimes even more than 200 instructions. These computers also employ a variety of data types and a large number of addressing modes. The trend into computer hardware complexity was influenced by various factors, such as upgrading existing models to provide more customer applications, adding instructions that facilitate the translation from high-level language into machine language programs, and striving to develop machines that move functions from software implementation into hardware implementation. A computer with a large number of instructions is classified as a complex instruction set computer, abbreviated CISC. In the early 1980s, a number of computer designers recommended that computers use fewer instructions with simple constructs so they can be executed much faster within the CPU without having to use memory as often. This type of computer is classified as a reduced instruction set computer or RISC. In this section we introduce the major characteristics of CISC and RISC architectures and then present the instruction set and instruction format of a RISC processor.

CISC Characteristics

The design of an instruction set for a computer must take into consideration not only machine language constructs, but also the requirements imposed on the use of high-level programming languages. The translation from high-level to machine language programs is done by means of a compiler program. One reason for the trend to provide a complex instruction set is the desire to simplify the compilation and improve the overall computer performance. The task of a compiler is to generate a sequence of machine instructions for each high-level language statement. The task is simplified if there are machine instructions that implement the statements directly. The essential goal of a CISC architecture is to attempt to provide a single machine instruction for each statement that is written in a high-level language. Examples of CISC architectures are the Digital Equipment Corporation VAX computer and the IBM 370 computer.

Another characteristic of CISC architecture is the incorporation of variable-length instruction formats. Instructions that require register operands may be only two bytes in length, but instructions that need two memory addresses may need five bytes to include the entire instruction code. If the computer has 32-bit words (four bytes), the first instruction occupies half a word, while the second instruction needs one word in addition to one byte in the next word. Packing variable instruction formats in a fixed-length memory word requires

special decoding circuits that count bytes within words and frame the instructions according to their byte length.

The instructions in a typical CISC processor provide direct manipulation of operands residing in memory. For example, an ADD instruction may specify one operand in memory through index addressing and a second operand in memory through a direct addressing. Another memory location may be included in the instruction to store the sum. This requires three memory references during execution of the instruction. Although CISC processors have instructions that use only processor registers, the availability of other modes of operations tend to simplify high-level language compilation. However, as more instructions and addressing modes are incorporated into a computer, the more hardware logic is needed to implement and support them, and this may cause the computations to slow down. In summary, the major characteristics of CISC architecture are:

1. A large number of instructions—typically from 100 to 250 instructions2. Some instructions that perform specialized tasks and are used infrequently3. A large variety of addressing modes—typically from 5 to 20 different modes4. Variable-length instruction formats5. Instructions that manipulate operands in memory

RISC Characteristics

The concept of RISC architecture involves an attempt to reduce execution time by simplifying the instruction set of the computer. The major characteristics of a RISC processor are:

1. Relatively few instructions2. Relatively few addressing modes3. Memory access limited to load and store instructions4. All operations done within the registers of the CPU5. Fixed-length, easily decoded instruction format6. Single-cycle instruction execution7. Hardwired rather than microprogrammed control

The small set of instructions of a typical RISC processor consists mostly of register-to-register operations, with only simple load and store operations for memory access. Thus each operand is brought into a processor register with a load instruction. All computations are done among the data stored in processor registers. Results are transferred to memory by means of store instructions. This architectural feature simplifies the instruction set and encourages the optimization of register manipulation. The use of only a few addressing modes results from the fact that almost all instructions have simple register addressing. Other addressing modes may be included, such as immediate operands and relative mode.

By using a relatively simple instruction format, the instruction length can be fixed and aligned on word boundaries. An important aspect of RISC instruction format is that it is easy to decode. Thus the operation code and register fields of the instruction code can be accessed

simultaneously by the control. By simplifying the instructions and their format, it is possible to simplify the control logic. For faster operations, a hardwired control is preferable over a microprogrammed control.

A characteristic of RISC processors is their ability to execute one instruction per clock cycle. This is done by overlapping the fetch, decode, and execute phases of two or three instructions by using a procedure referred to as pipelining. A load or store instruction may require two clock cycles because access to memory takes more time than register operations. Efficient pipelining, as well as a few other characteristics, are sometimes attributed to RISC, although they may exist in non-RISC architectures as well. Other characteristics attributed to RISC architecture are:

A relatively large number of registers in the processor unit Use of overlapped register windows to speed-up procedure call and return Efficient instruction pipeline

Compiler support for efficient translation of high-level language programs into machine language programs. A large number of registers is useful for storing intermediate results and for optimizing operand references. The advantage of register storage as opposed to memory storage is that registers can transfer information to other registers much faster than the transfer of information to and from memory. Thus register-to-memory operations can be minimized by keeping the most frequent accessed operands in registers. Studies that show improved performance for RISC architecture do not differentiate between the effects of the reduced instruction set and the effects of a large register file. For this reason a large number of registers in the processing unit are sometimes associated with RISC processors.