Data Representation and Digital Logic: Conversion and Circuit Analysis

This article covers topics such as conversion of numbers from one base to another, circuit analysis, and reduction. It includes examples and explanations of one's complement, two's complement, and signed magnitude. The article also covers circuit reduction using Boolean algebra.

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[WRITE YOUR FULL NAME]
[WRITE YOUR STUDENT ID]
ITC544
ASSIGNMENT 1: DATA REPRESENTATION AND DIGITAL LOGIC)

Question 1
a) (152)x = (6A)16
x = ?
6A is in hexadecimal, the decimal representation will be
6 * 161 + 10 * 160
= 96 + 10
= 106 in decimal
152 is with base x, the decimal representation will be
1 * x2 + 5 * x1 + 2 * x0
x2 + 5x + 2 in decimal
Therefore, x2 + 5x + 2 = 106
x2 + 5x – 104 = 0
x2 + 13x – 8x – 104 = 0
(x + 13)(x – 8) = 0
x = 8, x = -13
As the value of base cannot be negative.
Therefore x = 8.
b)
i) 0xBED into 3-base representation
BED is in hexadecimal, the decimal representation is:
11 * 162 + 14 * 161 + 13 * 160
= 2816 + 224 + 13
= 3053
Now, 3053 is converted to base 3
3 3053 2
3 1017 0
3 339 0
3 113 2
3 37 1
3 12 0
3 4 1
3 1 1
0
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[WRITE YOUR FULL NAME]
[WRITE YOUR STUDENT ID]
ITC544
ASSIGNMENT 1: DATA REPRESENTATION AND DIGITAL LOGIC)

Now, these remainders are written upside down, therefore,
0xBED = (11012002)3
ii) (321)7 into 2-base (binary) representation
321 is in base 7, the decimal representation is:
3 * 72 + 2 * 71 + 1 * 70
= 147 + 14 + 1
= 162
Now, 162 is converted to binary
2 162 0
2 81 1
2 40 0
2 20 0
2 10 0
2 5 1
2 2 0
2 1 1
0
Now, these remainders are written upside, therefore,
(321)7 = (10100010)2
iii) (123)5 into octal representation
123 is in base 5, the decimal representation is:
1 * 52 + 2 * 51 + 3 * 50
= 25 + 10 + 3
= 38
Now, 38 is converted to base 8
8 38 6
8 4 4
0
Now, these remainders are written upside down, therefore,
(123)5 = (46)8
iv) (21.21)8 into decimal representation
2 * 81 + 1 * 80 + 2 * 8-1 + 1 * 8-2
= 16 +1+ 0.25+0.015625
= (17.265625)10
c)
i) One’s complement
Lowest value (negative)  011
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Highest value (positive)  111
ii) Two’s complement
Lowest value (negative)  100
Highest value (positive)  011
iii) Signed Magnitude
Lowest value (negative)  – 3
Highest value (positive)  + 3
Question 2
a)
In circuit 1,
X1 = (A.B)’
In circuit 2,
X2 = A’ + B’
Considering, R.H.S for circuit 1
(A.B)’
= A’ + B’ (using Demorgan’s law (X.Y)’ = X’ + Y’)
= X2 (Boolean function of circuit 2)
Therefore, both the circuits are equal.
Using truth table,
A B A.
B
X1 A’ B’ X2
0 0 0 1 1 1 1
0 1 0 1 1 0 1
1 0 0 1 0 1 1
1 1 1 0 0 0 0
As the columns of X1 and X2 are equal, therefore, both the circuits are equal.
b)
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According to the circuit,
X = A’.B’ + A.B
To reduce the circuit, consider the L.H.S
A’.B’ + A.B
= ((AB)'(A'B')')' (using demorgan’s law)
= ((A' + B')(A + B))' (using demorgan’s law)
= (A'A + B'A + A'B + B'B)' (using distribution law)
= (0 + AB' + A'B + 0)' (using Inverse law)
= (AB' + A'B)' (using null law)
= (A ^ B)' (A XOR B = AB’ + A’B’)
c) To prove:
X’ + Y’ + XYZ’ = X’ + Y’ + Z’
Consider L.H.S
X’ + Y’ + XYZ’
= X’ + (Y’ + Y).(Y’+ XZ’) (Distribution law)
= X’ + 1.(Y’ + XZ’) (Inverse law)
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= X’ + Y’ + XZ’ (Identity law)
= (X’ + X).(X’ + Z’) + Y’ (Distribution law)
= 1.(X’ + Z’) + Y’ (Inverse law)
= X’ + Z’ + Y’ (Identity law)
= X’ + Y’ + Z’ (Commutative law)
= R.H.S
Hence, proved.
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