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Solutions Manual for Computer System Architecture, 3rd Edition by Morris Mano

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Complete solutions manual for Computer System Architecture, 3rd Edition by Morris Mano. Provides detailed, step-by-step answers to all end-of-chapter problems covering digital logic, processor organization, memory systems, instruction formats, and input/output architecture. Ideal for computer engineering and computer science students needing help with system-level architecture concepts and exam preparation.

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Uploaded on
September 12, 2025
Number of pages
11
Written in
2025/2026
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Exam (elaborations)
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  • morris mano 3rd edition

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ALL 13 CHAPTERS COVERED




SOLUTIONS MANUAL

, TABLE OF CONTENTS




Chapter 1 ……………………………………………………………………………… 4

Chapter 2 ……………………………………………………………………………… 11

Chapter 3 ……………………………………………………………………………… 16

Chapter 4 ……………………………………………………………………………… 20

Chapter 5 ……………………………………………………………………………… 26

Chapter 6 ……………………………………………………………………………… 34

Chapter 7 ……………………………………………………………………………… 45

Chapter 8 ……………………………………………………………………………… 51

Chapter 9 ……………………………………………………………………………… 59

Chapter 10 ……………………………………………………………………………. 63

Chapter 11 ……………………………………………………………………………. 80

Chapter 12 ……………………………………………………………………………. 89

Chapter 13 ……………………………………………………………………………. 95




-3-

,CHAPTER 1
1.1
ABC A•B•C (A•B•C)' A' B' C' A'+B'+C'
000 0 1 1 1 1 1
001 0 1 1 1 0 1
010 0 1 1 0 1 1
011 0 1 1 0 0 1
100 0 1 0 1 1 1
101 0 1 0 1 0 1
110 0 1 0 0 1 1
111 1 0 0 0 0 0

1.2
ABC A B A B C
000 0 0
001 0 1
010 1 1
011 1 0
100 1 1
101 1 0
110 0 0
111 0 1

1.3
(a) A + AB = A(1 + B) = A
(b) AB + AB' = A(B + B') = A
(c) A'BC + AC = C(A'B + A) = C(A' + A) (B + A) = (A + B)C
(d) A'B + ABC' + ABC = A' B + AB(C' + C) = A'B + AB = B(A' + A) = B

1.4
(a) AB + A (CD + CD') = AB + AC (D + D') = A (B + C)
(b) (BC' + A'D) (AB' + CD')

ABB'C' A'AB'D BCC'D' A'CD'D
= + + +
0 0 0 0
0
1.5
(a) (A + B)' (A' + B') = (A'B') (AB) = 0
(b) A + A'B + A'B' = A + A' (B + B') = A + A'= 1

1.6
(a) F = x’y + xyz’
F' = (x + y') (x' + y' + z) = x'y' + xy' + y' + xz + y'z
= y' (1 + x' + x + z) + xz = y'+ xz
(b) F•F' = (x'y + xyz') (y' + xz) = 0 + 0 + 0 + 0 = 0
(c) F + F' = x'y + xyz' + y' + xz (y + y')
= x'y + xy(z' + z) + y' (1 + xz) = x'y + xy + y'
= y(x' + x) + y' = y + y' = 1


-4-

,1.7
(a)
xyz F
000 0
001 1
010 0
011 0
100 0
101 1
110 0
111 1




(c) F = xy'z + x'y'z + xyz (d) Same as (a)
= y'z(x + x') + xz(y + y')
= y'z + xz




1.8
(a) (b)




(c) (d)




-5-

,1.9
(a) (b)




(c) (d)




1.10




(a) (b)

(1) F = xy + z'
F' = x'z + y'z (1) F = AC' + CD + B'D
(2) F = (x + z’) (y + z') (2) F = (A + D) (C' + D) (A + B'+C)

1.11 (a) (b)




-6-

,1.12




1.13
(a) F = x'z' + w'z
(b) = (x' + z) (w' + z')




1.14
S = x'y'z + x'yz' + xy'z' + xyz
= x'(y'z + yz') + x(y'z' + yz) See Fig. 1.2
= x'(y z) + x(y z)' (Exclusive - NDR)
=x y z



1.15
xyz F
000 0
001 0
010 0
011 1
100 0
101 1
110 1
111 1




-7-

,1.16
xyz ABC
000 001
001 010
010 011
011 100
100 011
101 100
110 101
111 111
c = z'
By inspection



1.17


When D = 0; J = 0, K = 1, Q → 0
When D = 1; J = 1, K = 0, Q → 1




1.18
See text, Section 1.6 for derivation.



1.19
(a) DA = x'y + xA; DB = x'B + xA; z = B




-8-

,(b)
Present state Inputs Next state Output
AB xy AB z

00 00 00 0
00 01 10 0
00 10 00 0
00 11 00 0
01 00 01 1
01 01 11 1
01 10 00 1
01 11 00 1
10 00 00 0
10 01 10 0
10 10 11 0
10 11 11 0
11 00 01 1
11 01 11 1
11 10 11 1
11 11 11 1

1.20




1.21
Count up-down binary counter with table E
Present Inputs Next Flip-flop
state state inputs
AB EX AB JA KA JB KB
00 00 00 0X 0X
00 01 00 0X 0X
00 10 11 1X 1X
00 11 01 0X 1X
01 00 01 0X X0
01 01 01 0X X0
01 10 00 0X X1
01 11 10 1X X1
10 00 10 X0 0X
10 01 10 X0 0X
10 10 01 X1 1X
10 11 11 X0 1X
11 00 11 X0 X0
11 01 11 X0 X0
11 10 10 X0 X1
11 11 00 X1 X1
-9-

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