computer ion and architecture first home work

8/8/2019 Computer ion and Architecture First Home Work

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ASSIGNMENT: - 1BY: - NARINDER PAL

CLASS: - MSC (cs)

SUBJECT NAME: COMPUTER ORGANIZATION AND ARCHITECTURE

Part A

Q1: Identify at least two application areas (discuss their roles also)

for

(a) Encoder/Decoder(b) Multiplexers/De-multiplexer(c)Flip Flops

ANS: - ENCODER:-

An encoder is combinational circuit that performs the

inverse operation of a decoder. The simplest encoder is a 2n *n binary input lines

and n is the number of output lines.

DECODER:-


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A Decoder Is A Multiple Logic Circuit That Converts Coded Input

Into Code Outputs, Where The Input And Output Codes Are Different.

(a) Application Areas OfEncoder/Decoder:-1) A device or program that uses predefined algorithms to encode, or

compress audio or video data for storage or transmission use. See also

decoder.

(2) A circuit that is used to convert between digital video and analog video.

(3) The radio frequency spectrum is filled with noise and other signals,

especially those frequencies where unlicensed transmitter operation under

FCC part 15 rules is allowed. When using a wireless remote control system

it is desirable to have a way of filtering out or ignoring those unwanted

signals to prevent false data from being received.

(4) A simple way to accomplish this is to use an encoder IC at the

transmitter and a decoder IC at the receiver. The encoder generates serial

codes that are automatically sent three times and must be received at least

twice before data is accepted as valid by the decoder circuit.

(5)In the early days of radio control", before these coding ICs were

available, radio controlled garage doors sometimes opened themselves when

they received transmissions from a plane passing overhead or a two-way

radio operating in the area. Encoding and decoding is now used in most


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wireless control systems to prevent this type of interference.

(d) Multiplexers/De-Multiplexer:-Multiplexers:-

A multiplexer is a combinational circuit that receives

binary information from 1 of 2n

input data lines and drax it to a single output

datelines for the output is determine by a set of selection input. 2n

to 1

multiplexer have 2n

input datelines and input selection lines, whose bit

combination determine which input data are selected for the output.

De-multiplexer:-

A digital de multiplex has 1 input and many output

lines. It performs the reverse operation of a digital multiplexer. It accept a

single input and send it to 1 out of many output lines, which is selected by

select lines.

Application Areas of Multiplexers/De-Multiplexer:-

Telegraphy:-

The earliest communication technology using electrical

wires, and therefore sharing an interest in the economies afforded by

multiplexing, was the electric telegraph. Early experiments allowed two

separate messages to travel in opposite directions simultaneously, first using

an electric battery at both ends, then at only one end.

Telephony:-


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In telephony, a customer's telephone line now typically

ends at the remote concentrator box down the street, where it is multiplexed

along with other telephone lines for that neighborhood or other similar area.

The multiplexed signal is then carried to the central switching office on

significantly fewer wires and for much further distances than a customer's

line can practically go. This is likewise also true for digital subscriber lines

(dsl).

(e)Flip Flops:-In digital circuits, a flip-flop is a term referring to an electronic

circuit (a bitable multi vibrator) that has two stable states and thereby is

capable of serving as one bit of memory. A flip-flop is usually controlled by

one or two control signals and/or a gate or clock signal. The output often

includes the complement as well as the normal output.

Q2: Discuss the basic logic behind counters i.e. how will you obtain

1000(8) from 0111(7)? How will you implement the same?

ANS: - A binary 0 is zero or off and a binary 1 is one or on. Binary

numbers, as the name suggests, have a value in decimal of two because

the binary number can either be on or off. Binary is calculated from the

place value that the 1 holds. An example of this is the binary number:

00000001. Starting from the left we record the position of the 1 as the

decimal number one. The binary number 00000011 is equal to 1 plus 2


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which is three in decimal. The binary number 00000111 is equal to 1

plus 2 plus 4 which seven in decimal.

0 1 1 1 1 1 1 1 = 127

0 1 1 1 1 1 1 0 = 126

0 0 0 0 0 0 1 0 = 2

0 0 0 0 0 0 0 1 = 1

0 0 0 0 0 0 0 0 = 0

1 1 1 1 1 1 1 1 = 1

1 1 1 1 1 1 1 0 = 2

1 0 0 0 0 0 0 1 = 127

1 0 0 0 0 0 0 0 = 128

Q3: State and prove De-Morgan's Theorem by taking a suitable

example.ANS: - De-Morgan's Theorem:-

The de-morgan's theorem says that the

complement of the product of a given set of variables is equal to the sum of the

complements of the individual variables; and the complement of the sum of a given

set of variables is equal to the product of the complements of the individual

variables. De-morgan's theorem applies to any arbitrary number of variables.

Example:-

1)


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=

2)

=

Part - B

Q4: Multiply any 2 binary numbers of 4-bits each, using Booch's

Method and show step by step values.

ANS: -Booths algorithm:-

Booths algorithm is used to multiply to

signed or unsigned numbers. It is invented by Andrew D. booth.

STEPS OF BOOTHS MULTIFICTION:-

1)Change the number into binary coded form. If number isnegative then find out its 2s compliment.

2)Initialized the values to the QR and AC registers.3)AC initialized the zero. Sets the value Q1 = zero.


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4)Initialized the SC registers equal to the number of bits in QR.5)If Q0Q1 are 00 or 11 then apply the arithmetic right shift

operation (ARSH). If Q0Q1 is the values 01then add the contents

BR into register AC and if the contents of Q0Q1 are 10 then

subtract the contents of BR from AC.

6)Detriment the SC by 1.7)This step is continue until the value of sc become 00.8)Then at the end combine the result of BR and AC. Then the

result is the binary number of the multiplicands mortificationnumber.

Example: - we take the example of 2*3.

The binary number of 2 is 0010 and the binary number of 3 is

0011. The 2s complement of 2 is 0010

1101 (1s complement)

+1

-------------

1110 (2s complement)

--------------Q0 Q1 BR= 1110 AC QR Q1 SC

0010

Initial 0000 0011 0 100

1 0 sub BR 1110


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---------

1110

ARSH 1111 0001 1 011

1 1 ARSH 1111 1000 1 0100 1 add BR 0010

---------

0001

ARSH 0000 1100 0 001

0 0 ARSH 0000 0110 0 000

Now multiply the AC and QR. The result is 00000110 which is the binary number

of 6 (2*3).

Q5: Perform BCD addition and subtraction of 1254 and 456.

ANS: - BCD Addition:-

BCD, or binary-coded decimal, represents the 10

decimal digits in terms of binary numbers. It is possible to build digital hardware

that man-ip-ulates BCD directly, and such hardware could be found in early com-

puters and many hand-held calculators. The BCD system was chosen for the

internal number system in these machines because it is easy to convert it to

alphanumeric representations for printouts and displays. The compelling

advantages of BCD have waned over time, and these digits are supported by more

modern hardware simply to provide backward compatibility with earlier

generations of machines. In this section, we briefly examine the approaches for


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constructing BCD arithmetic -elements.

BCD ADDITION

CARRYBIT

1 1 1

1254 0001 0010 0101 0100

+456 0000 0100 0101 0110

1710 0001 0110 1010>9 1010>9

0110 0110

0001 0111 0001 0000

Sum 1 7 1 0

BCD SUBTRACTION

1254 0001 0010 0101 0100

-456 0000 0100 0101 0110

798 1101>9 1111>9 1110>9

0110 0110 0110

0111 1001 1000

sub 0 7 9 8

Q6: Show the contents of E, A, Q and SC during the process of

division of two binary numbers, 10100011 by 1011.

ANS: - First, let me briefly explain BCD. In BCD, each digit of a


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decimal

number is represented by a 4-bit value (similar to the way hexadecimal

digits are represented by 4-bit groups in "pure" binary). Since

decimal only has the digits 0 through 9, only the following values are

valid "4-bit groups" in BCD:

DEC BCD DEC BCD

0 0000 5 0101

1 0001 6 01102 0010 7 0111

3 0011 8 1000

4 0100 9 1001

The combinations 1010, 1011, 1100, 1101, 1110, and 1111 are invalid

because they don't add up to a decimal digit. Since each decimal digit

requires 4 BCD bits, we need 16 bits to represent a 4-digit decimal

value. Your CPU, however, can only work with 8 bits at a time. So we'll

need to break up the numbers into two parts (bytes) and work with one

byte at a time. Let's call the bytes (or the registers where the bytes are

stored) A1 and A0 for the first number - with A1 being the moresignificant byte (MSB) and A0 being the less significant (LSB.) Let's

call the bytes B1 and B0 for the second number - again letting B1 be the

more significant byte and B0 the less significant. For example, let's say


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we want to add the numbers:

2483

+ 1795

------

The 2483 would be represented as 0010 0100 1000 0011 (I'll put spaces

in between each group of 4 bits for readability) and so:

A1 = 0010 0100

A0 = 1000 0011

The 1795 would be represented as 0001 0111 1001 0101 and so:

B1 = 0001 0111

B0 = 1001 0101

When adding large numbers with pencil and paper, we start with the

least significant digits (the units place) and work our way from right to

left. Similarly, our CPU will begin with the LSBs and work over to theMSBs.

First we'll add the LSBs; A0 + B0:


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1 111


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+ 0110 0000

------------

1 0111 1000

That's better. Let's store that as S0. Now the MSBs. We'll add

A1 + B1:

_____ Carry-over from the LSBs (in the C-flag)

/v

1111

0010 0100

+ 0001 0111

-----------

0011 1100

This time, we got the answer 0011 1100, but that doesn't make sense in

BCD (1100 is an invalid 4-bit group in BCD.) Why? Because again the

CPU did the addition in pure binary. When it added 4 + 7 + 1 it got 12

and represented it in binary as 1100. But for BCD it should have carried10 of them over to the tens digit and only kept 0010 (2 in decimal). To

"force" the half-carry (i.e. the carry from the 4th bit to the 5th), we'll add

6 to the units digit (0000 0110 in BCD) to make the group of 10 that


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should have been carried a group of 16 that the CPU will carry:

1111

0010 0100

+ 0001 0111

-----------

0011 1100

+ 0000 0110

-----------0100 0010

Q7: Why is it necessary to consider the case of divide overflow?

ANS: - Divide Overflow:-

In computer science, divide and conquer

(D&C) is an important algorithm design paradigm based on multi-

branched recursion.

The division operation may result in a quotient overflow when we have a fixed

length of register and it cannot hold a number exceeds the standard length. The

divide overflow condition must be avoided in normal computer operation because

the entire quotient will be too long for transfer into a memory until that has words

of standard length that is the same as the length of registers. When the divider is

twice as long as divisor, the condition for overflow is a divide overflow condition

occur if the high order half bits of the dividend constitute a number greater than or

equal to the divisor. Another problem is that division by zero must be avoided.

This occurs because any dividend will be greater than or equal; to a divisor which


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is equal to zero.

The divide-and-conquer paradigm often helps in the discovery of efficient

algorithms. It was the key, for example, to Karatsuba's fast multiplication method,

the quick sort and merge sort algorithms, the Stassen algorithm for matrix

multiplication, and fast Fourier transforms.

computer ion and architecture first home work

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