Solution Code: 1BBD
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Assignment 1: Data representation & Digital logic
Question 1
Question 2.
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(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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