Data Representation and Digital Logic - Assignment 1

This assignment covers topics like hexadecimal to decimal conversion, base conversion, negative and positive numbers in 3-bit computer, circuit equivalence and more.

IT COMPUTER
Name of the Student
Student ID
Subject Code (ITC544)
Assignment 1: Data Representation and Digital Logic
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IT COMPUTER
Question 1
Answer (a): Firstly, let covert hexadecimal Number (6A) to equivalent decimal
number.
⇒ (6A) 16
= (6) (161) + (10) (160)
= 96 + 10
= 106
(152)x = (6A)16
So we can conclude:-
(152) x = 106
(1) (X2) + (5) (x1) + (2) (x0) = 106
X2 + 5x + 2 = 106
X2 + 5x+ 104 = 0
The value of x is 8 (Answer)
Answer (b):
I. 0xBED is a number with radix 16. in hexadecimal number (B), (E) and (D)
represent following (11), (14) and (13).
The decimal representation of the 0xBED is:
⇒ (11) (162) + (14) (161) + (13) (160)
= 2816 + 224 + 13
= 3053
= (3053)10
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IT COMPUTER
Now, let (3053)10 convert into base 3 number:
=3053
3 3053 2
3 1017 0
3 339 0
3 113 2
3 37 1
3 12 0
3 4 1
3 1
So, (3053)10 = (11012002)3
II. convert (321)7 into 2-base (binary)
First,
Let (321)7 convert to (x)10.
(3) (72) + (2) (71) + (1) (70)
= 162
= (162)10
Now let converting (162)10 to (x)2
2/162 ⇔ 81 – remainder 0
2/81 ⇔ 40 – remainder 1
2/40 ⇔ 20 – remainder 0
2/20 ⇔ 10 – remainder 0
2/10 ⇔ 5 – remainder 0
2/5 ⇔ 2 – remainder 1
2/2 ⇔ 1 – remainder 0
The answer is: (321)7 = (10100010)2
III. 1235 into octal representation:
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IT COMPUTER
8 1235
8 154 3
8 19 2
8 2 3
So, (1235)10 = (2323)8
IV. (21.21)8 = (x)10
= (2) (81) + (2) (80) + (2) (8-1) + (1) (8-2)
= 16 + 1+ 0.25 + 0.015625
= (17. 265625)10
So, the conversion of (21.21)8 = (17.265625)10 (Answer)
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IT COMPUTER
Answer (c)
A tiny computer has a word limit of 3 bits, the lowest negative number and highest
positive no will be-
The lowest negative number of 3 bits is: (100)2
The highest positive number of 3 bits is: (011)2
1’s compliment of (100)2 will be: (011)2
1’s compliment of (100)2 will be: (100)2
2’s compliment of (100)2 will be: (100)2
2’s compliment of (100)2 will be: (110)2
Question 2
Answer (a):
From the first figure the output we can extract
is
Let a = 1 and b = 0,
⇔ ( a ) + ( b )
= ( 1 ) + ( 0 )
= ( 1 + 0 )
From the first figure the output we can extract
is
Let a = 1 and b = 0,
⇔ ( a∗b )
= (1∗0 )
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IT COMPUTER
Hence, proved the two circuit is equivalent.
Answer (b):
(A’ AND B’) OR (A AND B)
= A’B’ + AB
= ((AB)’ (A’B’))…… utilizing De Morgan’s theorem
= (( A’+ B’) (A + B))’
= (A’A + B’A + AB + B’B)’…….after Utilizing De Morgan’s Law
= (1 + AB’ + AB + 1)
= (AB’ + AB)
= (A ^ B)
That implies A XNOR B
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IT COMPUTER
Answer (c):
(x’ + y’ + xyz’)
= (x’ + Y’ + x) (x’ + y’ + y) (x’ + y’+ z’)……….utilisions De Morgan
Theorem
= (1 + y’) (1 + x’) (x’ + y’ + z’)
= (x’ + y’ + z’)…..proved
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