Data Representation - Desklib

This document contains solutions to questions related to Data Representation. It includes conversion of numbers from one base to another, finding the smallest and largest numbers in one's and two's complement and signed magnitude, and simplification of logic circuits using De-Morgan's Law.

Running head: DATA REPRESENTATION
DATA REPRESENTATION
Name of Student-
Name of University-
Author’s Note-

1DATA REPRESENTATION
Question 1:
a) Given (152) x= (6A) 16
 X2 + (5 * X1) + (2 * X0) = (6 * 161) + (10 * 160)
 X2 + 5X + 2 = 106
 X2 + 5X - 104 = 0
 X2 + 13X - 8X – 104 = 0
 X(X + 13) – 8(X + 13) = 0
 (X - 8) (X + 13) = 0
 X = 8 and X = -13
Therefore, X is 8.
(152)8= (6A) 16
b) i) BED = (B * 162) + (E * 161) + (D * 160)
= 2816 + 224 + 13
= (3053)10
(3053)10 =
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2DATA REPRESENTATION
BED16 = (11012002)3
ii) (321)7 = (3 * 72) + (2 * 71) + (1 * 70)
= (162)10
Again, (162)10 =
(162)10 = (10100010)2
iii) Conversion
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3DATA REPRESENTATION
Therefore, (1235)10 = (2323)8
iv) As per given question 21.218 = (2 * 81) + (1 * 80). (2 * 8-1) + (1 * 7=8-2)
= 17 + 0.25 + 0.015625
= 17.265625
(21.21)8 = 17.265625
c) i) Negative smallest number for one’s compliment = 100
Positive largest number for one’s compliment = 100
ii) Negative smallest number for two’s compliment = 101
Positive largest number for two’s compliment = 011
iii) Negative smallest number for signed magnitude = 111
Positive largest number for signed magnitude = 011
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4DATA REPRESENTATION
Question 2:
a) After generating the output from L.H.S.
A B C D
0 0 0 1
0 1 0 1
1 0 0 1
1 1 1 0
After generating the output from R.H.S.
A B C D E
0 0 1 1 1
0 1 1 0 1
1 0 0 1 1
1 1 0 0 0
So, L.H.S. is same as R.H.S.
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5DATA REPRESENTATION
b) The two not gate and the and can be minimized as
Can be represented by
Therefore, finalized circuit is
c) X’ + Y’ + XYZ’
= X’ + Y’ + (X’ + Y’ + Z)’ [From De-Morgan’s Law]
= (XY (X’ + Y’ + Z))’ [From De-Morgan’s Law]
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6DATA REPRESENTATION
= (XX’Y + XYY’ + XYZ)
= (0 + 0 + XYZ)
= (XYZ)’
= X’ + Y’ + Z’ [From De-Morgan’s Law]
= X’ + Y’ + XYZ’ = X’ + Y’ + Z’ [HENCE, PROVED]
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