chapter 9 exercise solution according to edition 8
9.1
Represent the following decimal numbers in both binary sign/magnitude and twos
Complement
using 16 bits:
A)
512
B)-29
BINARY
VALUE OF 512= 0000 0010 0000 0000
BINARY
VALUE OF 29= 0 000 0000 0001 1101
SIGN
MAGNITUDE:
512= 0000 0010 0000
0000 (MSB IS KEPT 0 BECAUSE OF THE POSITIVE SIGN )
-29=
1 000 0000 0001 1101 (MSB IS CHANGED TO 1 BECAUSE NO. HAS A
NEGATIVE SIGN)
TWOS
COMPLEMENT:
512=0000 0010 0000 0000 (2’S
COMPILEMENT IS NOT POSSIBLE BECAUSE IT IS NOT A NEGATIVE NUMBER)
-29=
1 111 1111 1110 0011(AFTER PERFORMING 2’S COMPIMENT)
9.2
Represent the following twos complement values in decimal: 1101011; 0101101.

a. 1101011 = -21
As the msb is showing that it is a negative number and
then we calculate the magnitude as:
·
TAKE
TWOS COMPLIMENT
16 8 4 2 1
0 0
1 0 1
0 1 =21 AND SIGN BIT OF ACTUAL NO. MAKES IT -21
b. 0101101 = 45
Because it starts with zero that
indicates it to be a positive number so no 2s compliment is needed.
9.8: calculate (72530-13250) using 10’s compliment
arithmetic. Assume rules similar to those for twos compliment arithmetic.
Sol:
We subtract [M+(-N)],where M=72532 and
N=13250
M=72532
Tens comp of N= 86750
Sum=
72532
+ 86750
______________
159282
Discard carry digit: -10000
Result: 59282
9.9: Consider the twos complement addition of two
n-but numbers:
Zn-1Zn-2……Zo =
Xn-1 Xn-2 …. Xo + Yn-1 Yn-2 ….. Yo
Assume that bitwise addition is
performed with a carry bit ci generated by the addition of xi+yi and ci-1. Let
v be the binary variable indicating overflow when v=1.
Fill the table.
Input
|
Xn-1
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
Yn-1
|
0
|
0
|
1
|
1
|
0
|
0
|
1
|
1
|
|
Cn-2
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
|
output
|
Zn-1
|
0
|
0
|
1
|
0
|
1
|
0
|
1
|
1
|
V
|
0
|
1
|
0
|
0
|
0
|
0
|
1
|
0
|
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