Data Representation and Digital Logic

This assignment delves into data representation techniques, specifically focusing on IEEE-754 floating-point numbers. It explores the binary format of these numbers, including sign, exponent, and mantissa. The assignment then examines different representations of signed numbers using 5 bits: signed magnitude, one's complement, and two's complement. Finally, it presents a problem involving a digital logic circuit design for a main gate system, requiring students to simplify a Boolean expression using Karnaugh maps and prove its equivalence to 1.

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Data Representation and Digital Logic
Part 1
(a) Number conversions
It is important to note that there are two kinds of zeros: +0 and -0
The value of an IEEE-754 number is computed as shown,
sign∗2exponent∗mantissa
The sign is stored in bit 32. The exponent can be computed from bits 24 to 31 by subtracting
the exponent obtained from 127. The mantissa is also referred to as the significand or
fraction. The significand is stored in bits of 1-23.
There is an invisible leading bit that is not stored in the memory but is placed as the first bit,
Therefore, in the given number
0 01111110 10100000000000000000000
0− positive signed number
01111110=12610
e xponent=12610−12710=−110
¿ (−1 )0∗( 1+ 0.625 )∗2−1
¿ 1.625∗1
2
¿ 0.812510
(b) Equivalents of “word” using 5 bits
(i) Signed number
(ii) One’s complement
(iii) Two’s complement
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For a signed number using 5 bits the range is such that,
power 6 5 4 3 2 1 0
Base 2 1000000 100000 10000 1000 100 01 1
Base 10 64 32 16 8 4 2 1
Five bits are represented by the highlighted section,
The signed magnitude, one’s complement and two’s complement are represented as shown
below,
Signed Magnitude ¿+1510 ¿−1510
¿ 011112 ¿111112
One’s complement the range is [−15,15 ]10
¿ 111112 ¿ 100002
Two’s complement the range is ¿ [−16,15 ]10
¿ 011112 ¿100002
Part 2
(a) Part a
Main entrance CSU door activation
The lock system has a 24 hour clock that times the locking schedule.
9:00 am to 12:00 noon main entrance door active
1:00 pm to 4:00 pm main entrance door active
Four inputs namely W, X, Y, Z are to be used
The circuit diagram is as shown below,
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From the output, we can develop a Boolean expression using Karnaugh maps as shown
below,
Grouping the terms in the powers of 2 we obtain,
A' B' D'+ A' BD+C' D=Y
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Inputs Output
A B C D Q
0 0 0 0 1
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0
0 1 0 0 0
0 1 0 1 1
0 1 1 0 0
0 1 1 1 1
1 0 0 0 0
1 0 0 1 1
1 0 1 0 0
1 0 1 1 0
1 1 0 0 0
1 1 0 1 1
1 1 1 0 0
1 1 1 1 0

(b) Using basic Boolean algebra identical ties for Boolean Variables A, B and C, prove that
X’Y + XYZ’ + Y’ + XZ (Y+Y’) = 1. Please show all steps and mention the identities
used. [6 marks]
There are several basic laws that govern the Boolean Algebra computations. These laws
enable a complex system to be simplified to the least number.
Assuming that the Boolean Variables A, B, C are X, Y, Z respectively.
X’Y + XYZ’ + Y’ + XZ (Y+Y’) = 1
Solution
A' B+ AB C' + B'+ AC
Simplifying further, making B a common factor in both terms,
B ( A' + A C' ) + B' + AC
Simplifying by introducing C in the brackets,
B ( A' C + A' C'+ A C' ) + B'+ AC
The first and last term in brackets demonstrate an Exclusive-Or condition of the equation,
A' C+ A C'=A ⊕C … .. exclusive∨¿
B ( A' C' ( A ⊕C )+ AC )+ ABC + B'
A' C ' ( A ⊕C ) =0
B' + AC
From the identities,
Assuming the A, C represents high and B’ represents low,
B'+ AC=1
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