Computer Architecture Cleous Mccalla

Published Date: 02 Nov 2017

February 4, 2014

Assessment # 4

Derive write both sum of products (SOP) and product of sums (POS) Boolean expressions describing the Output:

Sum of products (SOP):

A

B

C

OUTPUT

EXPRESSION

0

0

0

1

A’ B’ C’

0

0

1

0

0

1

0

1

A’ B C’

0

1

1

0

1

0

0

0

1

0

1

1

A B’ C

1

1

0

1

A B C’

1

1

1

0

SOP = (A’ B’ C’) + (A’ B C’) + (A B’ C) + (A B C’)

Products of Sum (POS):

A

B

C

OUTPUT

EXPRESSION

0

0

0

1

0

0

1

0

A + B + C’

0

1

0

1

0

1

1

0

A + B’ + C’

1

0

0

0

A’ + B + C

1

0

1

1

1

1

0

1

1

1

1

0

A’ + B’ + C’

POS = (A + B + C’) ∙ (A + B’ + C’) ∙ (A’ + B + C) ∙ (A’ + B’ + C’)

Sum of products (SOP):

A

B

C

OUTPUT

EXPRESSION

0

0

0

1

A’ B’ C’

0

0

1

1

A’ B’ C

0

1

0

0

0

1

1

1

A’ B C

1

0

0

0

1

0

1

1

A B’ C

1

1

0

1

A B C’

1

1

1

1

A B C

SOP = (A’ B’ C’) + (A’ B’ C) + (A’ B C) + (A B’ C) + (A B C’) + (A B C)

Products of Sum (POS):

A

B

C

OUTPUT

EXPRESSION

0

0

0

1

0

0

1

1

0

1

0

0

A + B’ + C

0

1

1

1

1

0

0

0

A’ + B + C

1

0

1

1

1

1

0

1

1

1

1

1

POS = (A + B’ + C) ∙ (A’ + B + C)

Determine the sum of products expression for the following function: f(A, B, C) = (0,1,2,6)

If

22 = 4

21 = 2

20 = 1

The binary values for the following are:

0 = 0 0 0

1 = 0 0 1

2 = 0 1 0

6 = 1 1 0

Sum of products (SOP):

A

B

C

OUTPUT

EXPRESSION

0

0

0

1

A’ B’ C’

0

0

1

1

A’ B’ C

0

1

0

1

A’ B C’

0

1

1

0

1

0

0

0

1

0

1

0

1

1

0

1

A B C’

1

1

1

0

SOP = (A’ B’ C’) + (A’ B’ C) + (A’ B C’) + (A B C’)

Assume that X consists of 3 bits, X2,X1,X0. Write a logic function that is true if and only if X contains only one 1.

X2 X1 X0.

Out

0 0 0

0

0 0 1

1

0 1 0

1

0 1 1

0

1 0 0

1

1 0 1

0

1 1 0

0

1 1 1

0

F = X2’ X1’ X0 + X2’ X1 X0’ + X2 X1’ X0’

Give the ASCII code for the letters U and k.

ASCii Code for Upper case U = 85

ASCii Code for lower case k = 107

Why is a flip-flop also called a bistable?

A flip-flop circuit is called bistable because they are digital logic circuits that can be in one of two stable states. They will maintain their state indefinitely until an input pulse called a trigger is received.

How does a SR latch differ from a gated SR latch?

The difference between a SR latch and a Gated SR Latch is that a SR latch will change its state whenever a change is made to the S or R input, but a Gated SR latch will only allow a change in state when the gate (denoted by E below) is high.

SR Latch

http://sub.allaboutcircuits.com/images/04173.png

Gated SR latch

http://sub.allaboutcircuits.com/images/04178.png

List any eight rules of Boolean algebra simplification.

Eight rules of Boolean algebra simplification are:

The Cumulative rule Example: A+B = B+A

The Associative rule Example: A + (B + C) = (A + B) + C

The Null rule Example: A + 1 = 1 & A ∙ 0 = 0

The Absorption rule Example: A + AB = A

The Idenpotency rule Example: A + A = A & A ∙ A =a

The Distributive rule Example: A(B + C) = AB + AC

The Adjacency rule Example: AB + A’B = A

Use either the rules of Boolean algebra or Karnaugh maps to simplify the following:

F = A + ABC +A’C

The karnaugh map table was populated based on the following factors

Highs (1s) for all the outputs where A is 1, no matter what B & C are

Highs (1s) for all the outputs where A, B & C are all 1s

Highs (1s) for all the outputs where A is 0 and C is 1 no matter what B is

Lows (0s) for all other outputs

A B

C

0

1

0 0

0

1

0 1

0

1

1 1

1

1

1 0

1

1

Once A is High (1) the output is high no matter what B & C are

Once C is High (1) the output is high no matter what A & B are

Hence the simplified form of this expression is: F = A + C

F = (A’B + C)’ + C

For this simplification we use Boolean algebra

(A’B + C)’ + C = [(A’B)’ ∙ C’] + C = [(A’’ + B’) ∙ C’] + C =

[(A + B’) ∙ C’] + C = AC’ + B’C’ + C

NOT the entire expression:

(AC’ + B’C’ + C)’ = (AC’)’ ∙ (B’C’)’ ∙ C’ = (A’+C’’) ∙ (B’’+C’’) ∙ C’ =

(A’+C) ∙ (B+C) ∙ C’ = (A’+C) ∙ (BC’+ C’C) = (A’+C) ∙ (BC’) =

(A’B C’) + (BC’C) = (A’B C’) + (0) = (A’B C’)

NOT the entire expression a second time to get back to the original state:

(A’B C’)’ = A’’+ B’ + C’’ = A + B’ + C

Hence the simplified form of this expression is: F = A + B’ + C

F = (A + B)(A’ + B’)

For the first step of this simplification we use Boolean algebra

(A + B) ∙ (A’ + B’) = AA’ + AB’ + A’B + BB’ = AB’ + A’B

We construct a truth table to further simplify this expression

A B

Out

0 0

0

0 1

1

1 0

1

1 1

0

Based on the operation of the above truth table we conclude that this is an Exclusive OR gate and hence the simplified expression is F= A \oplus B

Draw the logic gate equivalent to the following formulas:

AB+C(A’ +B)’

A+B+C(A’+C’)

(A+B’)(BC +A)+D