# Switching Circuits - Boolean Algebra | Digital logic & Data representation Assessment Answers

August 04, 2017
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Solution Code: 1BBD

## Question:

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Assignment 1: Data representation & Digital logic

Question 1

1. Determine the value of basexif (211)x= (152)8
2. A computer stores all integers in 8 bits. The computer also uses 2's complement method for representing negative numbers and IEEE 754 single precision representation for storing floating point values. Please show how the following values (in decimal) would be stored in the computer:
3. -35
4. -22.625

Question 2.

1. Express the switching circuit shown in the figure below in binary logic notation:
2. Write Boolean expressions and construct the truth table describing the outputs of the circuit depicted by the following logic diagram:
3. Using basic Boolean algebra identities for Boolean variablesA,BandC, prove thatABC+ABC'+AB'C+A'BC=AB+AC+BC. Please show all steps and mention the identities used.

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## Solution:Data representation & Digital logic

(211) base x= (152) base8..... eq. (1)

Convertingboth sides into decimal, we get:

2*x2+1*x1+1*x0= 1*82+5*81+2*80

2x2+x+1 = 64+40+2

2x2+x+1 = 106

2x2+x+1-106 = 0

2x2+x-105 = 0

This is quadratic equation,and we get x=7, x= -7.5

Considering x=7 we get both sides equal.

Thus the base is 7.

(i)

Float value = -35

Single Precision:

Bits 31 (Sign Bit) = 1; i.e. 1 for negative value

Bits 30 to 23 (Exponent Field) = 10000100; Decimal value of exponent field and exponent = 132 -127 =5

Bits 22 to 0 (Significand) =1 .00011000000000000000000; Decimal value of Significand = 1.0937500

Thus

-35 Decimal = 11000010 00001100 00000000 00000000 IEEE754

(ii)

Float value = -22.625

Single Precision:

Bits 31 (Sign Bit) = 1; i.e. 1 for negative value

Bits 30 to 23 (Exponent Field) = 10000011; Decimal value of exponent field and exponent = 131 -127 =4

Bits 22 to 0 (Significand) = 1 .01101010000000000000000; Decimal value of Significand = 1.4140625

Thus

-22.625 Decimal = 11000001 10110101 00000000 00000000 IEEE754

0 = Off

1 = On

 A B C 0 0 0 ?0 0 0 1 ?0 0 1 0 ?1 0 1 1 ?1 1 0 0 ?0 1 0 1 ?1 1 1 0 ?0 1 1 1 ?1

 A\B C 00 01 11 10 0 0 1 5 2 1 4 5 7 6

= BC + AC

= (A + B) C

Boolean Expression y1 = a ?(c +d +e)

 Truth Table a c d e y1 T T T T F T T T F F T T F T F T T F F F T F T T F T F T F F T F F T F T F F F T F T T T T F T T F T F T F T T F T F F T F F T T T F F T F T F F F T T F F F F F

Boolean Expression y2 = b.f (c+d+e)

 b f c d e y2 T T T T T F T T T T F F T T T F T F T T T F F F T T F T T F T T F T F F T T F F T F T T F F F F T F T T T F T F T T F F T F T F T F T F T F F F T F F T T F T F F T F F T F F F T F T F F F F F F T T T T T F T T T F T F T T F T T F T T F F T F T F T T T F T F T F T F T F F T T F T F F F F F F T T T F F F T T F F F F T F T F F F T F F F F F F T T F F F F T F F F F F F T F F F F F F F

ABC+ABC'+AB'C+A'BC=AB+AC+BC

Solving RHS

=AB+AC+BC

Adding the term which is not present:

= AB(C'+C) + A (B+B') C + (A' + A) BC

We know that(A' + A) = (B+B') = (C'+C) = 1

= ABC’ + ABC + ABC + AB’C + A’BC + ABC

= (ABC + ABC + ABC) + ABC’ + AB’C + A’BC

We know that ABC + ABC + ABC = ABC

= ABC+ABC'+AB'C+A'BC

Hence proved.

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