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Digital Design -Solution Manual DIGITAL DESIGN FOURTH EDITION
Digital Design -Solution Manual DIGITAL DESIGN FOURTH EDITION
Abo Amar EL HewatyRelated Papers
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© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
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SOLUTIONS MANUAL
DIGITAL DESIGN
FOURTH EDITION
M. MORRIS MANO
California State University, Los Angeles
MICHAEL D. CILETTI
University of Colorado, Colorado Springs
rev 01/21/2007
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
www.Mohandesyar.com
1
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
www.Mohandesyar.com
CHAPTER 1
1.1
Base-10:
Octal:
Hex:
Base-13
16
20
10
A
17
21
11
B
18
22
12
C
19
23
13
10
20
24
14
11
21
25
15
12
1.2
(a) 32,768
1.3
(4310)5 = 4 * 53 + 3 * 52 + 1 * 51 = 58010
(b) 67,108,864
22
26
16
13
23
27
17
14
24
30
18
15
25
31
19
16
26 27 28 29 30
32 33 34 35 36
1A 1B 1C 1D 1E
17 18 19 23 24
31
37
1F
25
32
40
20
26
(c) 6,871,947,674
(198)12 = 1 * 122 + 9 * 121 + 8 * 120 = 26010
(735)8 = 7 * 82 + 3 * 81 + 5 * 80 = 47710
(525)6 = 5 * 62 + 2 * 61 + 5 * 60 = 19710
1.4
14-bit binary: 11_1111_1111_1111
Decimal:
214 -1 = 16,38310
Hexadecimal: 3FFF16
1.5
Let b = base
(a) 14/2 = (b + 4)/2 = 5, so b = 6
(b) 54/4 = (5*b + 4)/4 = b + 3, so 5 * b = 52 – 4, and b = 8
(c) (2 *b + 4) + (b + 7) = 4b, so b = 11
1.6
(x – 3)(x – 6) = x2 –(6 + 3)x + 6*3 = x2 -11x + 22
Therefore: 6 + 3 = b + 1m so b = 8
Also, 6*3 = (18)10 = (22)8
1.7
68BE = 0110_1000_1011_1110 = 110_100_010_111_110 = (64276)8
1.8
(a) Results of repeated division by 2 (quotients are followed by remainders):
43110 = 215(1); 107(1); 53(1); 26(1); 13(0); 6(1)
Answer: 1111_10102 = FA16
3(0)
1(1)
(b) Results of repeated division by 16:
43110 = 26(15); 1(10) (Faster)
Answer: FA = 1111_1010
1.9
(a) 10110.01012 = 16 + 4 + 2 + .25 + .0625 = 22.3125
(b) 16.516 = 16 + 6 + 5*(.0615) = 22.3125
(c) 26.248 = 2 * 8 + 6 + 2/8 + 4/64 = 22.3125
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
www.Mohandesyar.com
2
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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(d) FAFA.B16 = 15*163 + 10*162 + 15*16 + 10 + 11/16 = 64,250.6875
(e) 1010.10102 = 8 + 2 + .5 + .125 = 10.625
1.10
(a) 1.100102 = 0001.10012 = 1.916 = 1 + 9/16 = 1.56310
(b) 110.0102 = 0110.01002 = 6.416 = 6 + 4/16 = 6.2510
Reason: 110.0102 is the same as 1.100102 shifted to the left by two places.
1.11
1011.11
101 | 111011.0000
101
01001
101
1001
101
1000
101
0110
The quotient is carried to two decimal places, giving 1011.11
Checking: 1110112 / 1012 = 5910 / 510 # 1011.112 = 58.7510
1.12
(a) 10000 and 110111
1011
+101
10000 = 1610
1011
x101
1011
1011
110111 = 5510
(b) 62h and 958h
2Eh
+34 h
62h
1.13
0010_1110
0011_0100
0110_0010 = 9810
2Eh
x34h
B38
2
8A
9 5 8h = 239210
(a) Convert 27.315 to binary:
27/2 =
13/2
6/2
3/2
½
Integer
Quotient
13
6
3
1
0
Remainder
+
+
+
+
+
½
½
0
½
½
Coefficient
a0 = 1
a1 = 1
a2 = 0
a3 = 1
a4 = 1
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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3
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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2710 = 110112
.315 x 2
.630 x 2
.26 x 2
.52 x 2
=
=
=
=
Integer
0
1
0
1
+
+
+
+
Fraction
.630
.26
.52
.04
Coefficient
a-1 = 0
a-2 = 1
a-3 = 0
a-4 = 1
.31510 # .01012 = .25 + .0625 = .3125
27.315 # 11011.01012
(b) 2/3 # .6666666667
.6666_6666_67 x 2
.3333333334 x 2
.6666666668 x 2
.3333333336 x 2
.6666666672 x 2
.3333333344 x 2
.6666666688 x 2
.3333333376 x 2
Integer
= 1
= 0
= 1
= 0
= 1
= 0
= 1
= 0
+
+
+
+
+
+
+
+
Fraction
.3333_3333_34
.6666666668
.3333333336
.6666666672
.3333333344
.6666666688
.3333333376
.6666666752
Coefficient
a-1 = 1
a-2 = 0
a-3 = 1
a-4 = 0
a-5 = 1
a-6 = 0
a-7 = 1
a-8 = 0
.666666666710 # .101010102 = .5 + .125 + .0313 + ..0078 = .664110
.101010102 = .1010_10102 = .AA16 = 10/16 + 10/256 = .664110 (Same as (b)).
1.14
1.15
1.16
1.17
(a)
1000_0000
1s comp: 0111_1111
2s comp: 1000_0000
(b)
0000_0000
1s comp: 1111_1111
2s comp: 0000_0000
(c)
1101_1010
1s comp: 0010_0101
2s comp: 0010_0110
(d)
0111_0110
1s comp: 1000_1001
2s comp: 1000_1010
(e)
1000_0101
1s comp: 0111_1010
2s comp: 0111_1011
(f)
1111_1111
1s comp: 0000_0000
2s comp: 0000_0001
(a)
52,784,630
9s comp: 47,215,369
10s comp: 47,215,370
(b)
63,325,600
9s comp: 36,674,399
10s comp: 36,674,400
(c)
25,000,000
9s comp: 74,999,999
10s comp: 75,000,000
(d)
00,000,000
9s comp: 99,999,999
10s comp: 00,000,000
B2FA
B2FA: 1011_0010_1111_1010
15s comp:
4D05
1s comp: 0100_1101_0000_0101
16s comp:
4D06
2s comp: 0100_1101_0000_0110 = 4D06
(a) 3409 o 03409 o96590 (9s comp) o 96591 (10s comp)
06428 – 03409 = 06428 + 96591 = 03019
(b) 1800 o 01800 o 98199 (9s comp) o 98200 (10 comp)
125 – 1800 = 00125 + 98200 = 98325 (negative)
Magnitude: 1675
Result: 125 – 1800 = 1675
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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5
(c) 6152 o 06152 o 93847 (9s comp) o 93848 (10s comp)
2043 – 6152 = 02043 + 93848 = 95891 (Negative)
Magnitude: 4109
Result: 2043 – 6152 = -4109
(d) 745 o 00745 o 99254 (9s comp) o 99255 (10s comp)
1631 -745 = 01631 + 99255 = 0886 (Positive)
Result: 1631 – 745 = 886
1.18
Note: Consider sign extension with 2s complement arithmetic.
(a)
(c)
1.19
10001
1s comp: 01110
2s comp: 01111
10011
Diff:
00010
(b)
100011
1s comp: 1011100
2s comp: 1011101
0100010
1111111
0000001
-000001
101000
(d)
1s comp: 1010111
1s comp:
2s comp: 1011000
2s comp:
001001
Diff:
1100001 (negative)
0011111 (2s comp)
-011111 (diff is -31)
with sign extension
sign bit indicates that the result is negative
2s complement
result
10101
1101010 with sign extension
1101011
110000
0011011 sign bit indicates that the result is positive
Check: 48 -21 = 27
+9286 o 009286; +801 o 000801; -9286 o 990714; -801 o 999199
(a) (+9286) + (_801) = 009286 + 000801 = 010087
(b) (+9286) + (-801) = 009286 + 999199 = 008485
(c) (-9286) + (+801) = 990714 + 000801 = 991515
(d) (-9286) + (-801) = 990714 + 999199 = 989913
1.20
+49 o 0_110001 (Needs leading zero indicate + value); +29 o 0_011101 (Leading 0 indicates + value)
-49 o 1_001111; -29 o 1_100011
(a) (+29) + (-49) = 0_011101 + 1_001111 = 1_101100 (1 indicates negative value.)
Magnitude = 0_010100; Result (+29) + (-49) = -20
(b) (-29) + (+49) = 1_100011 + 0_110001 = 0_010100 (0 indicates positive value)
(-29) + (+49) = +20
(c) Must increase word size by 1 (sign extension) to accomodate overflow of values:
(-29) + (-49) = 11_100011 + 11_001111 = 10_110010 (1 indicates negative result)
Magnitude: 1_001110 = 7810
Result: (-29) + (-49) = -78
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
www.Mohandesyar.com
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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1.21
+9742 o 009742 o 990257 (9's comp) o 990258 (10s) comp
+641 o 000641 o 999358 (9's comp) o 999359 (10s) comp
(a) (+9742) + (+641) o 010383
(b) (+9742) + (-641) o009742 + 999359 = 009102
Result: (+9742) + (-641) = 9102
(c) -9742) + (+641) = 990258 + 000641 = 990899 (negative)
Magnitude: 009101
Result: (-9742) + (641) = -9101
(d) (-9742) + (-641) = 990258 + 999359 = 989617 (Negative)
Magnitude: 10383
Result: (-9742) + (-641) = -10383
1.22
8,723
BCD:
ASCII:
1000_0111_0010_0011
0_011_1000_011_0111_011_0010_011_0001
1.23
1000 0100 0010 ( 842)
0101 0011 0111 (+537)
1101 0111 1001
0110
0001 0011 0111 0101 (1,379)
1.24
(a)
6
0
0
0
0
0
0
1
1
1
1
3
0
0
0
1
1
1
0
0
0
1
(b)
1
0
0
1
0
1
1
0
1
1
0
1
0
1
0
0
0
1
0
0
1
0
Decimal
0
1
2
3
4 (or 0101)
5
6
7 (or 1001)
8
9
1.25
(a) 5,13710
(b)
(c)
(d)
1.26
5,137 9s Comp:
2421 code:
1s comp:
6
0
0
0
0
0
0
1
1
1
1
BCD:
Excess-3:
2421:
6311:
4
0
0
0
0
1
1
0
0
0
0
2
0
0
1
1
0
0
0
0
1
1
1
0
1
0
1
0
1
0
1
0
1
Decimal
0
1
2
3
4
5
6 (or 0110)
7
8
9
0101_0011_0111
1000_0100_0110_1010
1011_0001_0011_0111
0111_0001_0100_1001
4,862
0100_1110_1100_1000
1011_0001_0011_0111 same as (c) in 1.25
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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1.27
For a deck with 52 cards, we need 6 bits (32 < 52 < 64). Let the msb's select the suit (e.g., diamonds,
hearts, clubs, spades are encoded respectively as 00, 01, 10, and 11. The remaining four bits select the
"number" of the card. Example: 0001 (ace) through 1011 (9), plus 101 through 1100 (jack, queen, king).
This a jack of spades might be coded as 11_1010. (Note: only 52 out of 64 patterns are used.)
1.28
G
(dot)
(space)
B
o
o
l
e
01000111_11101111_01101000_01101110_00100000_11000100_11101111_11100101
1.29
Bill Gates
1.30
73 F4 E5 76 E5 4A EF 62 73
73:
F4:
E5:
76:
E5:
4A:
EF:
62:
73:
0_111_0011
1_111_0100
1_110_0101
0_111_0110
1_110_0101
0_100_1010
1_110_1111
0_110_0010
0_111_0011
s
t
e
v
e
j
o
b
s
1.31
62 + 32 = 94 printing characters
1.32
bit 6 from the right
1.33
(a) 897
1.34
ASCII for decimal digits with odd parity:
(0):
(4):
(8):
1.35
(b) 564
10110000
00110100
00111000
(1):
(5):
(9):
(c) 871
00110001
10110101
10111001
(d) 2,199
(2):
(6):
00110010
10110110
(3):
(7):
10110011
00110111
(a)
a b c
a
f
b
c
g
f
g
1.36
a
b
a
f
g
b
f
g
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
www.Mohandesyar.com
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
8
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CHAPTER 2
2.1
(a)
xyz
x+y+z
000
001
010
011
100
101
110
111
0
1
1
1
1
1
1
1
(x + y + z)' x'
1
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
y'
z'
x' y' z'
xyz
(xyz)
(xyz)'
x'
y'
z'
x' + y' + z'
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
1
0
0
0
0
0
0
0
000
001
010
011
100
101
110
111
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
0
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
1
1
1
1
1
1
1
0
(b)
(c)
xyz
x + yz
(x + y)
(x + z)
(x + y)(x + z)
xyz
x(y + z)
xy
xz
xy + xz
000
001
010
011
100
101
110
111
0
0
0
1
1
1
1
1
0
0
1
1
1
1
1
1
0
1
0
1
1
1
1
1
0
0
0
1
1
1
1
1
000
001
010
011
100
101
110
111
0
0
0
0
0
1
1
1
0
0
0
0
0
0
1
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
1
1
(c)
2.2
(d)
xyz
x
y+z
x + (y + z)
(x + y)
(x + y) + z
xyz
yz
x(yz)
xy
(xy)z
000
001
010
011
100
101
110
111
0
0
0
0
1
1
1
1
0
1
1
1
0
1
1
1
0
1
1
1
1
1
1
1
0
0
1
1
1
1
1
1
0
1
1
1
1
1
1
1
000
001
010
011
100
101
110
111
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
1
(a) xy + xy' = x(y + y') = x
(b) (x + y)(x + y') = x + yy' = x(x +y') + y(x + y') = xx + xy' + xy + yy' = x
(c) xyz + x'y + xyz' = xy(z + z') + x'y = xy + x'y = y
(d) (A + B)'(A' + B') = (A'B')(A B) = (A'B')(BA) = A'(B'BA) = 0
(e) xyz' + x'yz + xyz + x'yz' = xy(z + z') + x'y(z + z') = xy + x'y = y
(f) (x + y + z')(x' + y' + z) = xx' + xy' + xz + x'y + yy' + yz + x'z' + y'z' + zz' =
= xy' + xz + x'y + yz + x'z' + y'z' = x y + (x z)' + (y z)'
2.3
(a) ABC + A'B + ABC' = AB + A'B = B
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(b) x'yz + xz = (x'y + x)z = z(x + x')(x + y) = z(x + y)
(c) (x + y)'(x' + y') = x'y'(x' + y') = x'y'
(d) xy + x(wz + wz') = x(y +wz + wz') = x(w + y)
(e) (BC' + A'D)(AB' + CD') = BC'AB' + BC'CD' + A'DAB' + A'DCD' = 0
(f) (x + y' + z')(x' + z') =xx' + xz' + x'y' + y'z' + x'z' + z'z' = z' + y'(x' + z') = z' + x'y'
(a) A'C' + ABC + AC' = C' + ABC = (C + C')(C' + AB) = AB + C'
2.4
(b) (x'y' + z)' + z + xy + wz = (x'y')'z' + z + xy + wz =[ (x + y)z' + z] + xy + wz =
= (z + z')(z + x + y) + xy + wz = z + wz + x + xy + y = z(1 + w) + x(1 + y) + y = x + y + z
(c) A'B(D' + C'D) + B(A + A'CD) = B(A'D' + A'C'D + A + A'CD)
= B(A'D' + A + A'D(C + C') = B(A + A'(D' + D)) = B(A + A') = B
(d) (A' + C)(A' + C')(A + B + C'D) = (A' + CC')(A + B + C'D) = A'(A + B + C'D)
= AA' + A'B + A'C'D = A'(B + C'D)
(e) ABCD + A'BD + ABC'D = ABD + A'BD = BD
2.5
(a)
x
y
Fsimplified
F
(b)
x
y
Fsimplified
F
(c)
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x
y
z
Fsimplified
F
(d)
A
B
0
Fsimplified
F
(e)
x
y
z
Fsimplified
F
(f)
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x
y
z
F
Fsimplified
2.6
(a)
A
B
C
F
Fsimplified
(b)
x
y
z
F
Fsimplified
(c)
x
y
F
Fsimplified
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(d)
w
x
y
z
F
Fsimplified
(e)
A
B
C
D
Fsimplified = 0
F
(f)
w
x
y
z
F
Fsimplified
2.7
(a)
A
B
C
D
F
Fsimplified
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(b)
w
x
y
z
F
Fsimplified
(c)
A
B
C
D
F
Fsimplified
(d)
A
B
C
D
F
Fsimplified
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(e)
A
B
C
D
F
Fsimplified
2.8
F' = (wx + yz)' = (wx)'(yz)' = (w' + x')(y' + z')
FF' = wx(w' + x')(y' + z') + yz(w' + x')(y' + z') = 0
F + F' = wx + yz + (wx + yz)' = A + A' = 1 with A = wx + yz
2.9
(a) F' = (xy' + x'y)' = (xy')'(x'y)' = (x' + y)(x + y') = xy + x'y'
(b) F' = [(A'B + CD)E' + E]' = [(A'B + CD) + E]' = (A'B + CD)'E' = (A'B)'(CD)'E'
F' = (A + B')(C' + D')E' = AC'E' + A D'E' + B'C'E' + B'D'E'
(c) F' = [(x' + y + z')(x + y')(x + z)]' = (x' + y + z')' + (x + y')' + (x + z)' =
F' = xy'z + x'y + x'z'
2.10
(a) F1 + F2 = 6 m1i + 6m2i = 6 (m1i + m2i)
(b) F1 F2 = 6 mi 6mj where mi mj = 0 if i z j and mi mj = 1 if i = j
2.11
(a) F(x, y, z) = 6(1, 4, 5, 6, 7)
(b) F(x, y, z) = 6(0, 2, 3, 7)
F = xy + xy' + y'z
F = x'z' + yz
xyz
F
xyz
F
000
001
010
011
100
101
110
111
0
1
0
0
1
1
1
1
000
001
010
011
100
101
110
111
1
0
1
1
0
0
0
1
A = 1011_0001
B = 1010_1100
2.12
(a)
(b)
(c)
A AND B = 1010_0000
A OR B = 1011_1101
A XOR B = 0001_1101
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(d)
(e)
2.13
NOT A = 0100_1110
NOT B = 0101_0011
(a)
A B
C
Y = A + B + B'(a + C')
(b)
A B C
D
Y = A(B xor D) + C'
(c)
A B CD
Y = A + CD + ABC
(d)
A B C
Y = (A xor C)' + B
(e)
A
B
C D
Y = (A'+ B')C + D')
(f)
A
B
C
D
Y = (A+ B')C' + D)
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2.14
(a)
x
y
z
F =xy + x'y' + y'z
(b)
x
y
z
F = xy + x'y' + y'z
= (x' + y')' + (x + y)' + (y + z')'
(c)
x
y
z
F = xy + x'y' + y'z
= [(xy)' (x'y')' (y'z)']'
(d)
x
y
z
F = xy + x'y' + y'z
= [(xy)' (x'y')' (y'z)']'
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(e)
x
y
z
F = xy + x'y' + y'z
= (x' + y')' + (x + y)' + (y + z')'
2.15
(a) T1 = A'B'C' + A'B'C + A'BC' = A'B'(C' + C) +A'C'(B' + B) = A'B' +A'C' = A'(B' + C')
(b) T2 =T1' = A'BC + AB'C' + AB'C + ABC' + ABC
= BC(A' + A) + AB'(C' + C) + AB(C' + C)
= BC + AB' + AB = BC + A(B' + B) = A + BC
¦ (3, 5, 6, 7) 3 (0, 1, 2, 4)
T1 = A'B'C' + A'B'C + A'BC'
A'B'
T2 = A'BC + AB'C' + AB'C + ABC' + ABC
A'C'
AC'
AC
T1 = A'B' A'C' = A'(B' + C')
BC
T2 =AC' + BC + AC = A+ BC
2.16
(a) F(A, B, C) = A'B'C' + A'B'C + A'BC' + A'BC + AB'C' + AB'C + ABC' + ABC
= A'(B'C' + B'C + BC' + BC) + A((B'C' + B'C + BC' + BC)
= (A' + A)(B'C' + B'C + BC' + BC) = B'C' + B'C + BC' + BC
= B'(C' + C) + B(C' + C) = B' + B = 1
(b) F(x1, x2, x3, ..., xn) = 6mi has 2n/2 minterms with x1 and 2n/2 minterms with x'1, which can be factored
and removed as in (a). The remaining 2n-1 product terms will have 2n-1/2 minterms with x2 and 2n-1/2
minterms with x'2, which and be factored to remove x2 and x'2. continue this process until the last term is
left and xn + x'n = 1. Alternatively, by induction, F can be written as F = xnG + x'nG with G = 1. So F =
(xn + x'n)G = 1.
2.17
(a) (xy + z)(y + xz) = xy + yz + xyz + xz = 6 (3, 5, 6, 7) = 3 (0, 1, 2, 4)
(b) (A' + B)(B' + C) = A'B' + A'C + BC = 6 (0, 1, 3, 7) = 3 (2, 4, 5, 6)
(c) y'z + wxy' + wxz' + w'x'z = 6 (1, 3, 5, 9, 12, 13, 14) = 3 (0, 2, 4, 6, 7, 8, 10, 11, 15)
(d) (xy + yz' + x'z)(x + z) = xy + xyz' + xyz + x'z
= 6 (1, 3, 9, 11, 14, 15) = 3 (0, 2, 4, 5, 6, 7, 8, 10, 12, 13)
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2.18
(a)
wx y z
F
00 0 0
00 0 1
00 1 0
00 1 1
01 0 0
01 0 1
01 1 0
01 1 1
10 0 0
10 0 1
10 1 0
10 1 1
11 0 0
11 0 1
11 1 0
11 1 1
0
1
0
0
0
1
1
1
0
1
1
1
0
1
1
1
F = xy'z + x'y'z + w'xy + wx'y + wxy
F = 6(1, 5, 6, 7, 9, 10 11, 13, 14, 15 )
(b)
x
y'
z
x'
y'
z
w'
x
y
w
x'
y
w
x
y
5 - Three-input AND gates
2 - Three-input OR gates
Alternative: 1 - Five-input OR gate
4 - Inverters
F
(c)
F = xy'z + x'y'z + w'xy + wx'y + wxy = y'z + xy + wy = ycz + y(w + x)
(d)
F = y'z + yw + yx) = 6(1, 5, 9, 13 , 10, 11, 13, 15, 6, 7, 14, 15)
= 6(1, 5, 6, 7, 9, 10, 11, 13, 14, 15)
(e)
y'
z
x
w
y
F
1 – Inverter, 2 – Two-input AND gates, 2 – Two-input OR gates
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2.19
F = B'D + A'D + BD
ABCD
ABCD
ABCD
-B'-D
0001 = 1
0011 = 3
1001 = 9
1011 = 11
A'--D
0001 = 1
0011 = 3
0101 = 5
0111 = 7
-B-D
0101 = 5
0111 = 7
1101 = 13
1111 = 15
F = 6(1, 3, 5, 7, 9, 11,13, 15) = 3(0, 2, 4, 6, 8, 10, 12, 14)
2.20
(a) F(A, B, C, D) = 6(3, 5, 9, 11, 15)
F'(A, B, C, D) = 6(0, 1, 2, 4, 6, 7, 8, 10, 12, 13, 14)
(b) F(x, y, z) = 3(2, 4, 5, 7)
F' = 6(2, 4, 5, 7)
2.21
(a) F(x, y, z) = 6(2, 5, 6) = 3(0, 1, 3, 4, 7)
(b) F(A, B, C, D) = 3(0, 1, 2, 4, 7, 9, 12) = 6(3, 5, 6, 8, 10, 11, 13, 14, 15)
2.22
(a) (AB + C)(B + C'D) = AB + BC + ABC'D + CC'D = AB(1 + C'D) + BC
= AB + BC (SOP form)
= B(A + C) (POS form)
(b) x' + x(x + y')(y + z') = (x' + x)[x' + (x + y')(y + z')] =
= (x' + x + y')(x' + y + z')
= xc + y + zc
2.23
(a) B'C +AB + ACD
A
B
C
D
F
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(b) (A + B)(C + D)(A' + B + D)
A
B
C
D
F
(c) (AB + A'B')(CD' + C'D)
A
B
C
D
F
(d) A + CD + (A + D')(C' + D)
A
B
C
D
F
2.24
x y = x'y + xy'
and (x y)' = (x + y')(x' + y)
Dual of x'y + xy' = (x' + y)(x + y') = (x y)'
2.25
(a) x| y = xy' z y | x = x'y
(x | y) | z = xy'z' z x | (y | z) = x(yz')' = xy' + xz
Not commutative
Not associative
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(b) (x y) = xy' + x'y = y x = yx' + y'x
(x y) z = ¦(1, 2, 4, 7) = x (y z)
Commutative
Associative
2.26
NAND
(Positive logic)
Gate
xy
z
xy
z
xy
z
LL
LH
HL
HH
H
H
H
L
00
01
10
11
1
1
1
0
11
10
01
00
0
0
0
1
NOR
(Positive logic)
Gate
2.27
NOR
(Negative logic)
NAND
(Negative logic)
xy
z
xy
z
xy
z
LL
LH
HL
HH
H
L
L
L
00
01
10
11
1
0
0
0
11
10
01
00
0
1
1
1
f1 = a'b'c + a'bc + abc' + abc
f2 = a'bc' + a'bc + ab'c' + ab'c + abc'
a'
b'
c'
a'
b
c
a
b
c'
a
b
c
a'
b
c'
a
b'
c'
a
b'
c
f1
f2
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2.28
(a) y = a(bcd)'e = a(b' + c' + d')e
y = a(b' + c' + d')e = ab’e + ac’e + ad’e
= 6( 17, 19, 21, 23, 25, 27, 29)
a bcde
y
a bcde
y
0 0000
0 0001
0 0010
0 0011
0 0100
0 0101
0 0110
0 0111
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1 0000
1 0001
1 0010
1 0011
1 0100
1 0101
1 0110
1 0111
0
1
0
1
0
1
0
1
0
0
1
0
1
0
1
0
0
0 1000
0 1001
0 1010
0 1011
0 1100
0 1101
0 1110
0 1111
1 1000
1 1001
1 1010
1 1011
1 1100
1 1101
1 1110
1 1111
(b) y1 = a (c + d + e)= a'(c + d +e) + a(c'd'e') = a'c + a'd + a'e + ac'd'e'
y2 = b'(c + d + e)f = b'cf + b'df + b'ef
y1 = a (c + d + e) = a'(c + d +e) + a(c'd'e') = a'c + a'd + a'e + ac'd'e'
y2 = b'(c + d + e)f = b'cf + b'df + b'ef
a'-c--001000 = 8
001001 = 9
001010 = 10
001011 = 11
a'--d-000100 = 8
000101 = 9
000110 = 10
000111 = 11
a'---e000010 = 2
000011 = 3
000110 = 6
000111 = 7
001100 = 12
001101 = 13
001110 = 14
001111 = 15
001100 = 12
001101 = 13
001110 = 14
001111 = 15
001010 = 10
001011 = 11
001110 = 14
001111 = 15
011000 = 24
011001 = 25
011010 = 26
011011 = 27
010100 = 20
010101 = 21
010110 = 22
010111 = 23
010010 = 18
010011 = 19
010110 = 22
010111 = 23
011100 = 28
011101 = 29
011110 = 30
011111 = 31
011100 = 28
011101 = 29
011110 = 30
011111 = 31
011010 = 26
011001 = 27
011110 = 30
011111 = 31
a-c'd'e'100000 = 32
100001 = 33
110000 = 34
110001 = 35
-b' c--f
-b' -d-f
-b' --ef
001001 = 9
001011 = 11
001101 = 13
001111 = 15
101001 = 41
101011 = 43
101101 = 45
101111 = 47
001001 = 9
001011 = 11
001101 = 13
001111 = 15
101001 = 41
101011 = 43
101101 = 45
101111 = 47
000011 = 3
000111 = 7
001011 = 11
001111 = 15
100011 = 35
100111 = 39
101011 = 51
101111 = 55
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y1 = 6 (2, 3, 6, 7, 8, 9, 10 ,11, 12, 13, 14, 15, 18, 19, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 32, 33, 34, 35 )
y2 = 6 (3, 7, 9, 13, 15, 35, 39, 41, 43, 45, 47, 51, 55)
ab cdef
y1 y 2
ab cdef
y1 y2
ab cdef
y1 y2
ab cdef
y 1 y2
00 0000
00 0001
00 0010
00 0011
00 0100
00 0101
00 0110
00 0111
0
0
1
1
0
0
1
1
0
0
0
1
0
0
0
1
01 0000
01 0001
01 0010
01 0011
01 0100
01 0101
01 0110
01 0111
0
0
1
1
0
0
1
1
0
0
0
0
0
0
0
0
10 0000
10 0001
10 0010
10 0011
10 0100
10 0101
10 0110
10 0111
1
1
1
1
0
0
0
0
0
0
0
1
0
0
0
1
11 0000
11 0001
11 0010
11 0011
11 0100
11 0101
11 0110
11 0111
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
1
00 1000
00 1001
00 1010
00 1011
00 1100
00 1101
00 1110
00 1111
1
1
1
1
1
1
1
1
0
1
0
0
0
1
0
1
01 1000
01 1001
01 1010
01 1011
01 1100
01 1101
01 1110
01 1111
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
10 1000
10 1001
10 1010
10 1011
10 1100
10 1101
10 1110
10 1111
0
0
0
0
0
0
0
0
0
1
0
1
0
1
0
1
11 1000
11 1001
11 1010
11 1011
11 1100
11 1101
11 1110
11 1111
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
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Chapter 3
3.1
y
yz
x
00
m0
0
11
m5
m7
00
01
x
1
0
1
m4
1
1
m5
m7
m6
z
F = z' + x'y
y
00
01
11
m1
m2
1
m5
1
z
F = x' + yz
10
m3
0
x
1
1
m4
m6
1
(c)
m5
m0
1
m7
1
1
yz
m2
10
m2
1
x
10
m3
11
m3
1
(b)
11
m1
m1
m4
1
y
m0
x
0
z
F = xy + x’z'
yz
01
m0
m6
1
(a)
00
m2
1
1
x
10
m3
1
m4
x
01
m1
y
yz
x
m7
m6
1
1
1
z
F = xy + xz + yz
(d)
3.2
y
yz
x
00
m0
0
1
m4
x
01
11
m1
m7
1
m6
0
01
1
m4
11
m1
m3
10
m5
m7
m6
00
x
1
01
11
m1
1
m4
1
1
y
m3
1
10
m2
1
m7
m5
1
m6
1
z
z
(c)
m7
z
F = y + x'z
0
m6
1
1
1
m0
m2
1
1
m5
1
yz
x
10
m2
1
(b)
y
00
11
m3
1
z
F = x'y' + xz
m0
01
m1
m4
x
1
yz
x
m0
0
m5
(a)
00
m2
1
1
x
10
m3
y
yz
x
F = x'y' + xy
(d)
F = y' + x'z
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
www.Mohandesyar.com
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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y
yz
x
00
m0
01
m7
x
1
m5
1
1
10
m2
m7
m6
1
1
1
z
z
F=z
(e)
11
m3
1
m4
m6
1
01
m1
0
1
m5
1
00
m0
m2
1
m4
10
m3
0
x
11
m1
y
yz
x
F = x + y' z
(f)
3.3
y
yz
x
00
m0
0
m1
m5
m7
00
0
11
m3
1
m4
(c)
01
m1
m7
1
10
m6
1
z
F = x'y' + yz + x'yz'
F = x' + yz
y
00
m0
m2
1
z
F = x'y + yz' + y'z'
F = = x' y + z'
x
11
m3
10
m2
1
m4
m6
01
m1
0
1
1
m7
yz
1
m5
1
1
m5
(b)
x
10
m2
1
y
m0
1
x
11
m3
1
m4
1
01
m1
0
z
F =xy + x'y'z' + x'yz'
F = xy + x' z'
yz
x
m0
m6
1
(a)
00
m2
1
1
x
x
10
m3
y
yz
11
1
m4
x
01
1
(d)
m5
m7
m6
1
1
z
F = xyz + x'y'z + xyz'
F = x'y'z + xy
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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3.4
C
CD
AB
00
01
m0
11
m1
m3
m5
m7
10
m2
00
x
00
0
x
1
m4
y
yz
01
11
1
01
m1
m3
1
m2
1
m4
m5
m7
1
m6
1
m12
m13
A
m8
m9
m0
m1
11
m4
m5
m6
m13
m8
m9
B
m14
1
m5
m12
11
1
m11
10
w
m10
m7
m13
1
m8
m15
m9
yz
m0
m1
00
m4
m3
m5
1
m11
m12
m13
m0
m6
1
m15
m14
m9
01
m1
1
m4
1
x
11
10
m3
m2
m7
m6
1
m5
1
01
m12
m13
m15
m14
m8
m9
m11
m10
x
11
1
m8
m10
y
00
m2
m7
m11
10
(e)
yz
wx
10
00
1
11
w
11
1
01
x
1
F = w'x'y +wx
(d)
y
01
m14
1
z
F =CD + ABD + ABC
00
m6
1
D
wx
1
10
1
(c)
m2
01
m15
1
10
m3
1
m4
1
m12
m1
11
00
m7
11
01
m0
1
01
y
00
m2
00
A
yz
wx
10
m3
m10
D
F = BCD + A' BD'
(b)
01
m11
10
C
00
B
m14
1
z
CD
m15
11
F=y
AB
1
10
m0
(a)
m6
1
m10
w
10
z
F = w'x + w'y'z
(f)
1
1
z
F = x'y' + w'y'z
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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3.5
y
yz
wx
00
m0
01
m1
00
10
m4
m5
m0
m12
11
m6
1
m15
1
m8
m9
A
m10
z
F =xz' + w'y'z+ wxy
00
m0
01
m1
00
1
m4
1
m9
m12
10
m7
m13
m15
x
01
m1
m4
m14
m9
10
1
m11
m5
1
m12
A
m10
z
F =w'y' + wx' y' + w'xy
(c)
10
m2
m7
1
m13
m8
m9
m6
1
1
m15
1
10
1
11
m3
1
11
m8
1
1
01
1
m10
1
C
m0
m6
1
B
1
D
F = A'C + A' C'D + B'C'D
00
m2
m14
m11
1
00
1
m15
CD
11
w
m8
AB
1
m5
01
11
m3
m6
1
(b)
y
yz
wx
m13
10
(a)
m7
1
m12
10
10
m2
11
1
m11
m5
01
x
m14
1
11
m3
1
m4
1
m13
01
m1
00
m7
1
00
m2
1
01
w
11
m3
C
CD
AB
m14
B
1
m11
m10
1
1
D
F =BD + A'B + B' D'
or = BD + B'D' + A'D'
(d)
3.6
CD
AB
m0
00
01
m1
m4
m6
m4
B
m10
m12
1
D
F = B' D' +A'BD + ABC'
w
m13
1
m8
10
m2
1
m7
m6
m15
m14
m11
m10
1
x
1
m9
1
10
(b)
m3
m5
1
11
m11
m1
11
1
01
m14
01
00
1
m9
1
m0
1
m15
y
00
m2
m7
m13
1
m8
wx
10
1
m5
m12
(a)
m3
1
11
10
11
1
01
A
yz
C
00
1
1
z
F = xy' +x'z + wx'y
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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CD
00
01
m0
00
11
m1
m4
m5
m12
m15
1
00
1
01
m8
m9
m11
m5
m13
m8
m9
m14
m11
B
m10
1
1
D
F = A'B'D' + BC'D + ACD' + AB'C
(d)
F = B'D' + BCD + A'BD + A'BC
m15
1
10
D
(c)
m6
1
A
1
m7
1
m12
m10
m2
1
11
1
10
10
m3
1
B
m14
11
m1
m4
1
11
01
m0
m6
1
m13
00
m2
m7
1
C
AB
10
m3
1
01
A
CD
C
AB
3.7
y
yz
wx
00
m0
01
m1
00
m4
m5
m12
w
m6
m15
m8
m9
(a)
z
F = z + x'y
00
00
01
m1
m4
m5
m13
11
m9
1
10
1
m15
m11
1
m7
m6
m15
m14
m11
m10
1
1
01
m1
11
m3
00
m5
m7
11
m10
w
10
D
F = B'D' + AC + A'BD + CD (or B'C)
(d)
m13
1
m8
1
1
m6
1
m12
1
10
m2
1
m4
B
1
y
01
m14
B
D
F = C'D + B'C + ABC'
00
m6
1
1
m9
m0
1
1
m8
wx
m2
m7
1
m12
(c)
11
1
yz
1
01
m8
(b)
m3
1
m13
1
10
C
m0
10
1
m2
1
m12
A
m10
1
CD
A
x
10
m3
m5
11
m11
1
AB
m14
11
1
m4
1
10
m1
01
1
1
01
00
1
m7
m13
m0
1
1
11
00
m2
1
01
AB
10
m3
C
CD
11
1
m9
m15
m14
1
m11
x
1
m10
1
1
z
F = wx + x'y + yz
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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3.8
(a) F(x, y, z) = 6(3, 5, 6, 7)
y
yz
x
00
m0
01
11
m1
m3
0
1
m4
x
1
10
m2
m5
m7
1
m6
1
1
z
(b) F = 6(1, 3, 5, 9, 12, 13, 14)
C
CD
AB
00
m0
01
m1
00
1
m4
m5
01
10
m2
1
m7
m6
m15
m14
1
m12
11
A
11
m3
m13
1
m8
1
m9
10
B
1
m11
m10
1
D
(c)
F = 6(0, 1, 2, 3, 11, 12, 14, 15)
y
wx
00
m0
00
01
m1
1
m4
11
m3
1
10
m2
1
m5
m7
m13
m15
1
m6
01
m12
11
w
1
m8
1
m9
m11
10
m14
x
1
m10
1
z
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
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F= 6(3, 4, 5, 7, 11, 12)
(d)
CD
AB
C
00
m0
01
m1
11
m3
00
1
m4
m5
1
01
m7
1
m12
m6
1
m13
m15
m14
m9
m11
m10
B
1
11
A
10
m2
m8
1
10
D
3.9
(a)
(b)
yz
wx
00
01
m0
00
11
m1
m4
m5
1
m12
m15
1
11
m8
1
00
1
01
x
m9
m11
m10
m5
1
m7
1
m6
1
m13
m15
m9
1
m11
1
10
m10
1
z
B
m14
1
m8
1
m2
1
11
A
10
m3
1
m12
1
11
m1
m4
m14
1
10
01
m0
m6
1
m13
00
m2
m7
1
C
AB
10
m3
1
01
w
CD
y
1
D
Essential: xz, x'z'
Non-essential: w'x, w'z'
F = xz + x'z' + (w'x or w'z')
Essential: B'D', AC, A'BD
Non-essential: CD, B'C
F = B'D' + AC + A'BD + (CD OR B'C)
(c)
(d)
CD
AB
00
m0
01
m1
00
11
m3
1
m4
m5
1
01
m12
m8
wx
10
m0
1
m7
m13
01
m1
00
m6
11
m3
1
m4
m5
m15
1
m11
1
10
m14
B
m7
1
m10
1
m13
1
11
w
m6
1
m12
m8
10
m15
1
m9
1
10
m2
1
01
1
m9
y
00
m2
1
1
11
A
yz
C
1
m11
1
m14
1
m10
1
z
D
Essential: BC', AC, A'B'D
F = BC' + AC + A'B'D
Essential: wy', xy, w'x'z
F = wy' + xy + w'x'z
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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(e)
(f)
CD
yz
C
AB
00
01
m0
00
11
m1
1
m4
m5
m12
m8
m6
m9
11
B
m14
m7
m10
1
m13
1
m8
10
1
m6
1
11
m11
m2
1
m5
m12
w
10
m3
1
m4
1
1
01
m1
01
m15
1
10
00
1
m13
11
m0
1
m7
1
00
m2
1
01
A
10
m3
y
wx
m15
1
m9
1
x
m14
1
1
m11
m10
1
1
z
D
Essential: BD, B'C, B'C'D'
F = BD + B'C + B'C'D'
Essential: wy', wx, x'z', xyz
F = wy' + wx + x'z' + xyz
3.10
(a)
(b)
yz
wx
00
m0
00
01
m1
m4
m5
1
m12
m13
10
m15
m8
m9
00
1
00
1
01
m14
x
m11
11
m3
1
m5
m7
1
m12
m10
A
m13
1
m6
1
m15
m14
1
m8
10
10
m2
1
11
1
m9
m11
1
z
Essential: xz, w'x, x'z'
F = xz + w'x + x'z'
01
m1
m4
1
1
C
m0
m6
1
1
AB
m2
m7
1
11
10
11
m3
1
01
w
CD
y
1
m10
1
1
D
Essential: AC, B'D', CD, A'BD
F = AC + B'D' + CD + A'BD
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
www.Mohandesyar.com
B
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
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(c)
(d)
yz
C
CD
AB
00
01
m0
11
m1
00
1
m4
01
11
m13
1
A
m6
m15
m14
m8
1
m9
m5
m7
m12
m13
1
11
w
x
m14
1
m9
1
1
m15
1
m8
10
1
m6
1
B
m10
1
10
m2
1
01
1
m11
10
11
m3
1
m4
m7
1
01
m1
00
1
m12
00
m0
m2
1
m5
1
10
m3
y
wx
1
m11
m10
1
z
D
Essential: wy', xy
Non-essential: wx, x'y'z, w'wz, w'x'z
F = wy' + xy + w'x'z
Essential: BC', AC
Non-essential: AB, AcBcD, BcCD, AcCcD
F = BCc + AC + AcBcD
(e)
(f)
00
m0
01
m1
11
m3
00
m4
m5
m12
m13
10
m8
m9
00
1
m11
m14
m10
1
11
m3
B
1
1
m5
m7
m12
m13
1
m8
10
m6
1
11
w
10
m2
1
01
1
1
01
m1
m4
m6
m15
y
00
m0
1
1
11
10
m7
1
wx
m2
1
01
A
yz
C
CD
AB
m9
1
m15
1
m14
1
m11
1
m10
1
1
z
D
Essential: BD, B'C, AB'C
Non-essential: CD
F = BD + B'C + AB'C
Essential: wy', wx, xyz, x'yz'
F = wy' + wx + xyz + x=yz'
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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3.11
(a) F(A, B, C, D, E) = ¦ (0, 1, 4, 5, 16, 17, 21, 25, 29
F = AcBcDc + ADcE + BcCcDc
A'B'C'D'E'
A'B'C'D'E
A'B'CD'E'
A'B'CD'E
AB'C'D'E'
AB'C'D'E
AB'CD'E
ABC'D'E
ABCD'E
m 0:
m 1:
m 4:
m 5:
m16:
m17:
m21:
m25:
m29:
= 00000
= 00001
= 00100
= 00101
= 10000
= 10001
= 10101
= 11001
= 11101
A=0
D
DE
BC
AB'D'
00
01
1
00
1
01
11
10
1
1
C
11
B
10
B'C'D'
E
A=1
D
DE
BC
AD'E
00
00
1
01
11
1
1
01
1
11
10
C
B
10
1
E
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(b) F(A, B, C, D, E) = A'B'CE' + B'C'D'E' + A'B'D' + B'CD' + A'CD + A'BD
F(A, B, C, D, E) = A'B'D' + B'D'E' + B'CD' + A'CD + A'BD
A'B'CE':
B'C'D'E':
A'B'D':
B'CD':
A'CD:
A'BD:
AB'CDE' + A'B'CD'E'
AB'C'D'E' + A'B'C'D'E'
A'B'CD'E + A'B'CD'E' + A'B'C'D'E + A'B'C'D'E'
AB'CD'E + AB'CD'E' + A'B'CD'E + A'B'CD'E'
A'BCDE + A'BCDE' + A'B'CDE + A'B'CDE'
A'BCDE + A'BCDE' + A'BC'DE + A'BC'DE'
A'B'D'
A=0
D
B'D'E'
DE
BC
B'CD'
00
01
1
00
01
1
1
1
1
1
1
11
10
A'CD
1
C
11
B
1
10
1
A'BD
E
A=1
D
DE
BC
00
01
1
00
01
11
10
1
1
C
11
B
10
E
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
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3.12
(a)
y
yz
wx
00
m0
00
01
m1
1
m4
m5
wx
10
m0
1
m7
m12
m13
m6
m4
m14
x
m8
m9
m11
w
m10
1
m5
10
m2
m7
m8
m6
0
m13
m15
0
11
1
m3
0
m12
1
10
m1
11
0
01
m15
01
00
1
11
00
m2
1
01
w
m3
y
yz
11
0
m9
m11
0
10
z
0
m14
x
0
m10
0
z
F =6(0, 1, 2, 5, 8, 10, 13)
F = x'z' + w'x'y' + w'y'z
F' = yz + xz' + xy + wx'z
F = (y' + z')(x' + z)(x' + y')(w' + x + z')
(b)
CD
AB
00
m0
01
m1
00
11
m4
m5
m7
m13
11
m9
m6
0
m15
0
m8
m2
0
0
m12
10
m3
0
01
A
F = 3(1, 3, 5, 7, 13, 15)
F' = A'D + B'D
F = (A + Dc)(Bc + Dc)
F = C'D' + AB' + CD'
C
m14
B
0
m11
m10
10
D
(c)
CD
AB
C
00
m0
01
m1
00
11
m3
0
m4
m5
0
m7
m6
01
0
m12
11
A
m13
m15
0
m8
10
F = 3(1, 3, 6, 9, 11, 12, 14)
F' = B'D + BCD' + ABD'
F = (B + D')(B' + C' + D)(A' + B' + D)
F = BD + B'D' + A'C'D'
10
m2
m14
B
0
m9
m11
0
m10
0
D
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3.13
(a) F = xy + zc = (x + zc)(y + zc)
(b)
CD
AB
m0
00
01
m1
0
m4
01
m13
1
m8
m9
m15
0
00
0
01
B
m11
1
m8
m2
0
0
m7
1
m13
m9
m6
0
0
m15
1
1
10
m3
m5
1
10
11
1
0
m12
A
m10
1
0
11
0
01
m1
m4
m14
1
1
m0
m6
0
1
1
10
m7
C
00
m2
0
1
AB
10
m3
m5
m12
11
1
0
11
A
CD
C
00
m14
1
m11
1
D
B
0
m10
1
1
D
F = AC' + AD + C'D + AB'C
F' A'D' + A'C + BCD'
F = (A + D)(A + C')(B' + C' + D)
(c)
CD
AB
00
m0
01
m1
11
m3
00
m4
10
m5
m7
m13
m0
00
m6
11
m15
0
m8
m9
m4
m14
B
11
m11
0
m10
0
m5
10
m2
1
m7
1
m13
m6
1
m15
1
m8
10
11
m3
1
1
m12
A
01
m1
1
01
0
0
C
00
m2
0
m12
10
AB
0
01
A
CD
C
m14
B
1
m9
m11
m10
1
D
D
F = (A + C' + D')(A' + B' + D')(A' + B + D')(A' + B + C')
F' = A'CD + ABD + AB'D + AB'C
F = A'C + A'D' + BD' + C'D'
F' = AD + CD +AB'C
F = (A' + D')(C + D')(A' + B + C')
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
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(d)
CD
CD
C
AB
00
m0
01
m1
11
10
m3
00
m2
m0
00
00
m4
m5
m7
01
m6
m4
11
m13
1
m15
1
m8
m14
10
m2
0
m5
0
m7
m6
0
0
m13
m15
m14
11
m11
1
11
m3
0
0
m12
1
m9
10
B
01
m1
0
01
1
m12
A
C
AB
A
m10
1
B
0
m8
m9
m11
m10
0
10
0
D
D
F =ABC' + AB'D + BCD
F = AD + ABC' + BCD
F' = A'C' + A'B' + CD' + B'C'D'
F = (A + C)(A + B)(C' + D)(B + C + D)
3.14
CD
AB
m0
00
01
m1
m4
m3
AB
10
m5
m2
m7
1
m12
m0
m13
m6
01
m1
m4
m15
m14
B
m9
m11
m10
1
m5
m12
A
m3
10
m2
0
m7
0
m6
0
11
1
11
0
01
1
1
m8
C
00
00
11
10
11
1
01
A
CD
C
00
0
m13
m15
0
m8
10
m14
0
m9
m11
0
D
B
0
m10
0
D
SOP form (using 1s):
F = B'C'D' + AB'D' + BC'D + A'BD
F = B'D'(A + C') + BD(A' + C')
POS form (using 0s):
F' = BD' + B'D + A'CD' + ACD
F = [(B' + D)(B + D')][(A + C' + D)(A' + C' + D')]
Alternative POS:
F' = BD' + B'D + A'CD' + A'B'C
F = [(B' + D)(B + D')][(A + C' + D)(A' + B + C)]
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
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3.15
(a)
(b)
C
CD
AB
00
m0
m1
00
x
00
m0
m1
m4
10
m5
1
m7
A
1
x
m7
m13
m14
m9
B
1
m11
m10
1
10
x
z
D
F=1
F = 6(0,1, 2, 3, 4, 5, 6, 7)
F = B'D' + ABC'D
F = 6(0, 2, 6, 8, 10, 13, 14)
(c)
(d)
CD
00
m0
01
m1
11
m3
00
AB
10
m4
m5
m0
m12
11
m6
1
m13
1
m8
00
m7
1
m15
m9
B
m14
x
m11
x
01
m1
x
11
m3
1
10
m2
1
m4
m5
m7
m12
m13
m15
x
m6
01
1
1
10
C
00
m2
x
01
A
CD
C
AB
11
1
A
m10
x
m8
10
m14
B
1
m9
1
D
m11
m10
x
1
D
F = BC' + BD + AB
F = 6(4, 5, 7, 12, 13, 14, 15)
3.16
m15
1
m8
1
m6
1
11
m6
x
m2
x
m12
1
10
m3
m5
01
m2
x
1
1
11
m3
x
0
x
01
11
1
m4
y
yz
01
F = B'D' + A'B' + ABCD
F = F = 6(0, 1, 2, 3, 8, 10, 15)
(a)
CD
AB
C
00
m0
00
01
m1
1
11
m3
1
10
F = A + A'B'
F = (A'(A'B')')'
m2
1
1
m4
m5
m7
m6
m12
m13
m15
m14
01
11
A
1
m8
10
1
m9
1
1
m11
1
1
B
A'
F
1
m10
1
A'
B'
D
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(b)
C
CD
AB
00
01
m0
11
m1
00
F = BC + AB + A'B'C'D
F = ((BC)'(AB)'(A'B'C' D)')'
m2
1
m4
m5
m7
m6
1
01
m12
11
A
10
m3
m13
1
m15
1
m8
B
m14
1
m9
B
C
1
A
B
1
m11
F
m10
A'
10
B'
C'
D
D'
(c)
CD
C
AB
00
01
m0
00
11
m1
10
m3
m2
1
1
m4
01
m5
1
m12
11
m14
1
m9
1
1
m15
1
m8
10
m6
1
m13
1
A
m7
1
F
1
m11
1
B
F' = A'B'D
F = (A'B'D)'
A'
B'
D
m10
1
1
D
(d)
B
BC
A
00
01
m0
m1
m4
m5
11
m3
10
m2
0
A
1
m7
1
m6
1
C
F = AC + AB
F = ((AC)' (AB)' )'
1
A
C
A
B
F
3.17
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CD
AB
00
m0
01
m1
00
m4
11
1
1
m7
11
m6
m8
11
10
m1
m3
m2
m4
m5
m7
m6
m12
m13
m15
m14
m8
m9
m11
m10
01
m13
m15
B
m14
B
11
1
m9
10
01
m0
00
1
m12
C
00
m2
1
m5
AB
10
m3
1
01
A
CD
C
m11
1
A
m10
10
1
D
D
F = A'B' + C'D' + B'C'
F' = BC + AC + BD
F = (BC)'(AC)'(BD)'
A
C
B
D
F'
B
C
3.18
F = (A )B'(C D) = (AB' + A'B)(CD' + C'D) = AB'CD' + AB'C'D + A'BCD' + A'BC'D
C
CD
AB
00
01
m0
m1
m4
m5
11
10
m3
m2
m7
m6
A
B
00
1
01
m12
m13
m15
m14
m8
m9
m11
m10
11
A
10
A'
B'
1
1
1
F
B
C
D
C'
D'
D
F = AB'CD' + AB'C'D + A'BCD' + A'BC'D and F' = A'B' + AB + C'D' + CD
F = (A'B')'(AB)'(C'D')'(CD)' = (A + B)(A' + B') (C' + D')(C + D)
F' = [(A + B)(A' + B')]' + [(C'+ D')(C + D)]'
F = ([(A + B)(A' + B')]' + [(C'+ D')(C + D)]')'
F = ([(A + B)' + (A' + B')'] + [(C'+ D')' + (C + D)'])'
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3.19
(a) F = (w + zc)(xc + zc)(wc + xc + yc)
yz
wx
y
00
m0
11
10
m3
m2
1
00
m4
m5
m12
y
z
1
m7
m6
1
01
1
m13
m15
m14
m9
m11
m10
x
1
11
w
01
m1
m8
1
10
1
1
w
x
F
w
z
1
z
F = y'z' + wx' + w'z'
F =[(y + z)' + (w' + x)' + (w + z)']
F' =[(y + z)' + (w' + x)' + (w + z)']'
(b)
yz
wx
y
00
m0
01
m1
00
11
m3
10
m2
1
w'
x
1
m4
m5
m7
m6
m12
m13
m15
m14
w
x'
01
11
w
1
m8
m9
y'
z'
1
m11
F
x
m10
y
z
10
z
F = 6(1, 2, 13, 14)
F' = w'x + wx' + y'z' + yz = [(w +x')(w' + x)(y + z)(y' + z')]'
F = (w +x')' + (w' + x)' + (y + z)' + (y' + z')
(c) F = [(x + y)(x' + z)]' = (x + y)' + (x' + z)'
F' = [(x + y)' + (x' + z)']'
x
y
F'
x'
z
3.20
Multi-level NOR:
F = (AB' + CD')E + BC(A + B)
F' = [(AB' + CD')E + BC(A + B)]'
F' = [ [(AB' + CD')' + E']' + [ (BC)' + (A + B)']' ]'
F' = [ [((A' + B)' + (C' + D)')' + E']' + [ (B' + C')' + (A + B)']' ]'
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
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A'
B
C'
D
F
E'
B'
C'
A
B
Multi-level NAND:
F = (AB' + CD')E + BC(A + B)
F' = [ (AB' + CD')E]' [BC(A + B)]'
F' = [ ( (AB')'( CD')' )'E]' [BC(A'B')']'
A
B'
C
D'
E
F
B
C
A'
B'
3.21
F = w(x + y + z) + xyz
F' = [w(x + y + z)]'[xyz]' = [w(x'y'z')')]'(xyz)'
x
y
z
x'
y'
F
z'
w
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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3.22
z
D
C
y
B
x
w
A
3.23
CD
AB
C
00
m0
00
m3
10
m2
x
1
m5
m7
m12
11
1
m13
1
m8
m15
m14
m11
m10
x
F
C'
D
B
x
m9
x
A'
B'
m6
1
01
10
11
m1
m4
A
01
1
D
F = AC' + A'D' + B'CD'
F' = D + ABC
F = [D + ABC]' = [D + (A' + B' + C']')]'
3.24
CD
AB
C
00
m0
00
m4
10
m1
m3
m2
m5
m7
m6
m13
m15
m14
1
m12
11
1
m8
10
11
1
01
A
01
m9
1
B
1
m11
1
1
m10
1
D
(a) F = C'D' + AB' + AD'
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
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F' = (C'D')'(AB')'(AD')'
AND-NAND:
C'
D'
F
A
B'
A
D'
(b) F' = [C'D' + AB' + AD']'
AND-NOR:
C'
D'
F’
A
B'
A
D'
(c) F = C'D' + AB' + AD' = (C + D)' + (A' + B)' + (A' + D)'
F' = (C'D')'(AB')'(AD')' = (C + D)(A' + B)(A' + D)
F = [ (C + D)(A' + B)(A' + D) ]'
OR-NAND:
C
D
F
A'
B
A'
D
(d) F = C'D' + AB' + AD' = (C + D)' + (A' + B)' + (A' + D)'
NOR-OR:
C
D
F
A'
B
A'
D
3.25
A
B
ABCD
C
D
A
B
A+B+C+D
C
D
AND-AND
AND
OR-OR
OR
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© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
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A
B
A
B
(AB CD)'
(A + B + C + D)'
C
D
C
D
AND-NAND
NAND
OR-NOR
A
B
A
B
(A'B'C'D')'
C
D
[(AB)' + (C' D')]'
C
D
A+B+C+D
NOR-NAND
NOR
OR
ABCD
NAND-NOR
AND
A'B'
A
B
A
B
A'B'C'D'
C
D
NOR-AND
A' + B' + C' + D'
(A + B + C + D)'
C
D
C'D' (A + B + C + D)'
NAND-OR
NOR
NAND
The degenerate forms use 2-input gates to implement the functionality of 4-input gates.
3.26
g = (a + b +c' + d')(b' + c' + d)(a'+ c + d')
g' = a'b'cd + bcd' + ac'd
c
cd
ab
00
01
11
10
f = abc' + c'd + a'cd'+ b'cd'
c
cd
ab
00
m0
01
m1
00
m4
m5
m7
1
m12
m13
1
11
m8
10
10
m2
1
01
a
11
m3
m15
m0
1
00
1
01
m6
m4
m14
b
m11
1
m5
m10
m8
1
10
m7
m13
m15
m11
d
0
m14
1
0
1
m6
1
0
m9
1
m2
0
1
1
11
a
m3
1
1
m12
1
m9
m1
1
b
0
m10
1
1
d
fg = ac'd + abc'd + b'cd'
3.27
x y = x'y + xy'; Dual = (x' + y)(x + y') = (x y)'
3.28
x
y
x
y
C
P
z
P
z
(a) 3-bit odd parity generator
(b) 4-bit odd parity generator
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3.29
D=A BC
E = A'BC + AB'C = (A B)C
F = ABC' + (A' + B')C = ABC' + (AB)'C = (AB) C
G = ABC
A
A
Half-Adder
B
B
S
C
C
S
D=A
B
C
C
E = (A
B)C
S
F = (AB)
C
G = ABC
Half-Adder
C
Half-Adder
AB
3.30
F = AB'CD' + A'BCD' + AB'C'D + A'BC'D
F = (A B)CD' + (A B) C'D = (A B)(C D)
A
B
F
C
D
3.31
Note: It is assumed that a complemented input is generated by another circuit that
is not part of the circuit that is to be described.
(a)
module Fig_3_22a_gates (F, A, B, C, C_bar, D);
output F;
input
A, B, C, C_bar, D;
wire
w1, w2, w3, w4;
and
(w1, C, D);
or
(w2, w1, B);
and
(w3, w2, A);
and
(w4, B, C_bar);
or
(F, w3, w4);
endmodule
(b)
module Fig_3_22b_gates (F, A, B, C, C_bar, D);
output F;
input
A, B, C, C_bar, D;
wire
w1, w2, w3, w4;
not
(w1_bar, w1);
not
(B_bar, B);
not
(w3_bar, w3);
not
(w4_bar, w4);
nand
(w1, C, D);
or
(w2, w1_bar, B_bar);
nand
(w3, w2, A);
nand
(w4, B, C_bar);
or
(F, w3_bar, w4_bar);
endmodule
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(c)
module Fig_3_23a_gates (F, A, A_bar, B, B_bar, C, D_bar);
output F;
input A, A_bar, B, B_bar, C, D_bar;
wire
w1, w2, w3, w4;
and
(w1, A, B_bar);
and
(w2, A_bar, B);
or
(w3, w1, w2);
or
(w4, C, D_bar);
or
(F, w3, w4);
endmodule
(d)
module Fig_3_23b_gates (F, A, A_bar, B, B_bar, C_bar, D);
output F;
input
A, A_bar, B, B_bar, C_bar, D;
wire
w1, w2, w3, w4;
nand
(w1, A, B_bar);
nand
(w2, A_bar, B);
not
(w1_bar, w1);
not
(w2_bar, w2);
or
(w3, w1_bar, w2_bar);
or
(w4, C, D_bar);
not
(w5, C_bar);
not
(w6, D);
nand
(F_bar, w5, w6);
not
(F, F_bar);
endmodule
(e)
module Fig_3_26_gates (F, A, B, C, D, E_bar);
output F;
input A, B, C, D, E_bar;
wire
w1, w2, w1_bar, w2_bar, w3_bar;
not
(w1_bar, w1);
not
(w2_bar, w2);
not
(w3_bar, E_bar);
nor
(w1, A, B);
nor
(w2, C, D);
nand (F, w1_bar, w2_bar, w3_bar);
endmodule
(f)
module Fig_3_27_gates (F, A, A_bar, B, B_bar, C, D_bar);
output F;
input
A, A_bar, B, B_bar, C, D_bar
wire
w1, w2, w3, w4, w5, w6, w7, w8, w7_bar, w8_bar;
not
(w1, A_bar);
not
(w2, B_bar);
not
(w3, A);
not
(w4, B_bar);
not
(w7_bar, w7);
not
(w8_bar, w8);
and
(w5 w1, w2);
and
(w6, w3, w4);
nor
(w7, w5, w6);
nor
(w8, C, D_bar);
and
(F, w7_bar, w8_bar);
endmodule
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3.32
Note: It is assumed that a complemented input is generated by another circuit that
is not part of the circuit that is to be described.
(a)
module Fig_3_22a_CA (F, A, B, C, C_bar, D);
output F;
input
A, B, C, C_bar, D;
wire
w1, w2, w3, w4;
assign w1 = C & D;
assign w2 = w1| B;
assign w3 = w2 & A);
assign w4 = B & C_bar);
assign F = w3 | w4);
endmodule
(b)
module Fig_3_22b_CA (F, A, B, C, C_bar, D);
output F;
input
A, B, C, C_bar, D;
wire
w1, w2, w3, w4;
assign w1_bar = ~w1;
assign B_bar = ~B;
assign w3_bar = ~w3;
assign w4_bar = ~w4;
assign w1 = ~(C & D);
assign w2 = w1_bar | B_bar;
assign w3 = ~(w2 & A);
assign w4 = ~(B & C_bar);
assign F = w3_bar | w4_bar;
endmodule
(c)
module Fig_3_23a_CA (F, A, A_bar, B, B_bar, C, D_bar);
output F;
input A, A_bar, B, B_bar, C, D_bar;
wire
w1, w2, w3, w4;
assign w1 = A & B_bar;
assign w2 = A_bar & B;
assign w3 = w1 | w2);
assign w4 = C | D_bar;
assign F = w3 | w4;
endmodule
(d)
module Fig_3_23b_CA (F, A, A_bar, B, B_bar, C_bar, D);
output F;
input A, A_bar, B, B_bar, C_bar, D;
wire
w1, w2, w3, w4;
assign w1 = ~(A & B_bar);
assign w2 = ~(A_bar & B);
assign w1_bar = ~w1;
assign w2_bar = ~w2;
assign w3 = w1_bar | w2_bar;
assign w4, C | D_bar;
assign w5 = ~C_bar;
assign w6 = ~D;
assign F_bar = ~(w5 & w6);
assign F = ~F_bar;
endmodule
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3.32
(e)
module Fig_3_26_CA (F, A, B, C, D, E_bar);
output F;
input
A, B, C, D, E_bar;
wire
w1, w2, w1_bar, w2_bar, w3_bar;
not
w1_bar = ~w1;
not
w2_bar = ~w2;
not
w3_bar = ~E_bar;
nor
w1 = (A | B;
nor
w2 = (C | D;
nand
F = ~( w1_bar & w2_bar & w3_bar);
endmodule
(f)
module Fig_3_27_CA (F, A, A_bar, B, B_bar, C, D_bar);
output F;
input
A, A_bar, B, B_bar, C, D_bar
wire
w1, w2, w3, w4, w5, w6, w7, w8, w7_bar, w8_bar;
not
w1 = ~A_bar;
not
w2 = ~B_bar;
not
w3 = ~A;
not
w4 = ~B_bar;
not
w7_bar = ~w7;
not
w8_bar = ~w8;
assign w5 = w1 & w2;
assign w6 = w3 & w4;
assign w7 = ~(w5 | w6);
assign w8 = ~(C | D_bar);
assign F = w7_bar & w8_bar;
endmodule
49
(a)
w3
x
w1
F=x
y
w
y
w4
Initially, with xy = 00, w1 = w2 = 1, w3 = w4 = 0 and F = 0. w1 should change to 0 4ns after xy
changes to 01. w4 should change to 1 8 ns after xy changes to 01. F should change from 0 to 1 10 ns
after w4 changes from 0 to 1, i.e., 18 ns after xy changes from 00 to 01.
(b)
`timescale 1ns/1ps
module Prob_3_33 (output F, input x, y);
wire w1, w2, w3, w4;
and #8 (w3, x, w1);
not #4 (w1, x);
and #8 (w4, y, w1);
not #4 (w2, y);
or #10 (F, w3, w4);
endmodule
module t_Prob_3_33 ();
reg x, y;
wire F;
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Prob_3_33 M0 (F, x, y);
initial #200 $finish;
initial fork
x = 0;
y = 0;
#20 y = 1;
join
endmodule
(c) To simulate the circuit, it is assumed that the inputs xy = 00 have been applied sufficiently long for
the circuit to be stable before xy = 01 is applied. The testbench sets xy = 00 at t = 0 ns, and xy = 1 at t =
10 ns. The simulator assumes that xy = 00 has been applied long enough for the circuit to be in a stable
state at t = 0 ns, and shows F = 0 as the value of the output at t = 0. The waveforms show the response to
xy = 01 applied at t = 10 ns.
Name
0.000ns
39.290ns
78.580ns
117.870ns
x
w1
y
w2
w3
w4
F
t = 28 ns
t = 18 ns
Note: input change occurs at t = 10 ns.
t = 14 ns
t = 10 ns
' = 18 ns
3.34
module Prob_3_34 (Out_1, Out_2, Out_3, A, B, C, D);
output Out_1, Out_2, Out_3;
input
A, B, C, D;
wire
A_bar, B_bar, C_bar, D_bar;
assign A_bar = ~A;
assign B_Bar = ~B;
assign C_bar = ~C;
assign D_bar = ~D;
assign Out_1 = ~( (C | B) & (A_bar | D) & B );
assign Out_2 = ((C * B_bar) | (A & B & C) | (C_bar & B) ) & (A | D_bar);
assign Out_3 = C & ( (A & D) | B ) | (C & A_bar);
endmodule
3.35
module Exmpl-3(A, B, C, D, F)
inputs A, B, C, Output D, F,
output B
and g1(A, B, B);
not (D, B, A),
OR (F, B; C);
// Line 1
// Line 2
// Line 3
// Line 4
// Line 5
// Line 6
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endofmodule;
// Line 7
Line 1: Dash not allowed, use underscore: Exmpl_3. Terminate line with semicolon (;).
Line 2: inputs should be input (no s at the end). Change last comma (,) to semicolon (;). Output is
declared but does not appear in the port list, and should be followed by a comma if it is intended
to be in the list of inputs. If Output is a mispelling of output and is to declare output ports, C
should be followed by a semicolon (;) and F should be followed by a semicolon (;).
Line 3: B cannot be declared as input (Line 2) and output (Line 3). Terminate the line with a semicolon
(;).
Line 4: A cannot be an output of the primitive if it is an input to the module
Line 5: Too many entries for the not gate (only two allowed).
Line 6: OR must be in lowercase: change to “or”.
Line 7: endmodule is mispelled. Remove semicolon (no semicolon after endmodule).
3.36
(a)
B
C
D
x
d
z
a
A
F
w
y
(b)
A1 A0 B1 B0
w1
w6
w2
w7
w3
A_lt_B
A_gt_B
w4
A_eq_B
w5
(c)
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a
b
y1
y2
3.37
UDP_Majority_4 (y, a, b, c, d);
output y;
input
a, b, c, d;
table
// a b c d : y
0 0 0 0 : 0;
0 0 0 1 : 0;
0 0 1 0 : 0;
0 0 1 1 : 0;
0 1 0 0 : 0;
0 1 0 1 : 0;
0 1 1 0 : 0;
0 1 1 1 : 1;
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
endtable
endprimitive
:
:
:
:
:
:
:
:
0;
0;
0;
0;
0;
0;
1;
1;
3.38
module t_Circuit_with_UDP_02467;
wire E, F;
reg A, B, C, D;
Circuit_with_UDP_02467 m0 (E, F, A, B, C, D);
initial #100 $finish;
initial fork
A = 0; B = 0; C = 0; D = 0;
#40 A = 1;
#20 B = 1;
#40 B = 0;
#60 B = 1;
#10 C = 1; #20 C = 0; #30 C = 1; #40 C = 0; #50 C = 1; #60 C = 0; #70 C = 1;
#20 D = 1;
join
endmodule
// Verilog model: User-defined Primitive
primitive UDP_02467 (D, A, B, C);
output D;
input A, B, C;
// Truth table for D = f (A, B, C) = 6(0, 2, 4, 6, 7);
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table
A B C : D // Column header comment
0 0 0 : 1;
0 0 1 : 0;
0 1 0 : 1;
0 1 1 : 0;
1 0 0 : 1;
1 0 1 : 0;
1 1 0 : 1;
1 1 1 : 1;
endtable
endprimitive
// Verilog model: Circuit instantiation of Circuit_UDP_02467
module Circuit_with_UDP_02467 (e, f, a, b, c, d);
output e, f;
input
a, b, c, d;
//
UDP_02467 M0 (e, a, b, c);
and
(f, e, d);
//Option gate instance name omitted
endmodule
Name
0
30
60
A
B
C
D
E
F
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CHAPTER 4
4.1 (a)
T1 = B'C, T2 = A'B, T3 = A + T1 = A + B'C,
T4 = D T2 = D (A'B) = A'BD' + D(A + B') = A'BD' + AD + B'D
F1 = T3 + T4 = A + B'C + A'BD' + AD + B'D
With A + AD = A and A + A'BD' = A + BD':
F1 = A + B'C + BD' + B'D
Alternative cover: F1 = A + CD' + BD' + B'D
F2 = T2 + D = A'B + D
ABCD T1 T2 T3 T4 F1 F2
CD
0000
0001
0010
0011
0100
0101
0110
0111
0
0
1
1
0
0
0
0
0
0
0
0
1
1
1
1
0
0
1
1
0
0
0
0
0
1
0
1
1
0
1
0
0
1
1
1
1
0
1
0
0
1
0
1
1
1
1
1
1000
1001
1010
1011
1100
1101
1110
1111
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
0
1
0
1
0
1
0
1
1
1
1
1
1
1
1
1
0
1
0
1
0
1
0
1
00
m0
01
m1
00
1
m4
01
m5
1
m12
m4
m13
1
m8
10
m15
m9
D
F2 = A'B + D
1
m6
1
m13
m15
1
m8
m9
m14
1
m11
1
1
B
1
m10
1
1
D
F1 = A + B'C+ B'D + BD'
AB
C
00
01
m0
m1
m4
1
m5
m7
m13
1
11
1
m8
10
1
1
m6
1
m12
A
10
m2
1
01
B
11
m3
1
00
m10
1
1
m7
1
10
10
m14
m11
m2
CD
1
1
m5
m12
m6
1
10
m3
1
11
m2
m7
11
1
01
1
1
11
A
11
m3
m1
00
C
CD
01
m0
A
AB
C
00
m9
1
m15
1
m11
1
1
m14
B
1
m10
1
D
F1 = A + CD' + B'D + BD'
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4.2
[(A'D)' A']'= A + D
A'
A
F
B
C
BC + A'
BC
G
D
(A'D)' = A + D’
F = (A + D)(A' + BC) = A'D + ABC + BCD += A'D + ABC
F = (A + D')(A' +BC) = A'D' + ABC + BCD' = A'D' + ABC
CD
AB
m0
01
11
m1
00
m4
m5
m6
m13
m15
m9
B
m3
A
m10
10
m2
1
m5
m7
m6
1
m12
10
1
m13
m15
1
m8
m9
m11
B
m14
1
m10
10
D
D
G = A'D' + ABC + BCD' = A'D' + ABC
F = A'D + ABC + BCD = A'D + ABC
4.3
11
11
1
m11
m1
m4
m14
01
1
01
1
1
m8
m0
00
m7
11
C
00
m2
1
1
m12
AB
10
m3
1
01
A
CD
C
00
(a) Yi = (AiS' + BiS)E' for i = 0, 1, 2, 3
(b) 1024 rows and 14 columns
4.4
(a)
xyz
F
000
001
010
011
100
101
110
111
1
1
1
0
0
0
0
0
y
yz
x
00
m0
1
0
m4
x
01
11
m1
m3
10
m2
1
1
m5
m7
1
m6
x'
y'
F
x'
y'
z
F = x'y' + x'z'
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
www.Mohandesyar.com
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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(b)
xyz
F
000
001
010
011
100
101
110
111
0
1
0
0
0
0
0
0
y
yz
x
00
01
m0
m4
m2
1
m5
z
m7
F
m6
1
1
10
m3
1
0
x
11
m1
1
z
F=z
4.5
xyz
ABC
000
001
010
011
100
101
110
111
010
011
100
101
001
010
011
100
A
yz
x
00
y
01
m0
11
m1
1
m4
m5
x'
y
m2
0
x
10
m3
1
m7
A
m6
y
z
1
1
z
A = x'y + yz
B
yz
x
00
01
m0
10
m3
m2
m7
m6
m5
1
1
y
00
y
01
11
m1
m4
10
m3
1
0
x
z'
C
m0
m2
1
m5
m7
x
z
m6
1
1
B
x
B = x'y' + y'z + xyz'
yz
y'
z
1
z
x
x
y'
1
m4
x
11
m1
1
0
y
1
C
z
C= x'z + xz'
4.6
xyz
F
000
001
010
011
100
101
110
111
0
0
0
1
0
1
1
1
A
yz
x
00
m0
01
m1
y
11
m3
1
0
m4
x
1
10
m2
m5
m7
1
m6
1
1
x
z
y
z
x
y
F
z
F = xz + yz + xy
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
www.Mohandesyar.com
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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module Prob_4_6 (output F, input x, y, z);
assign F = (x & z) | (y & z) | (x & y);
endmodule
4.7
(a)
ABCD
0000
0001
0011
0010
0110
0111
0101
0100
1100
1101
1111
1110
1010
1011
1001
1000
wxyz
CD
0000
0001
0010
0011
0100
0101
0110
0111
AB
C
00
m0
01
m1
11
10
m3
00
m2
m0
00
m4
m5
m7
m6
m4
11
m13
1
m8
10
m15
1
m9
1
m14
1
A
m10
1
m0
m1
11
m3
00
m13
m15
m14
m8
m9
m11
m10
m4
m5
1
m12
m7
m13
10
m9
1
m0
m15
m11
01
m1
m4
B
m14
1
C
y
1
m7
m6
1
m13
m15
m14
m9
1
m11
1
z=A B
=y D
B
10
m2
1
m8
10
1
m3
m5
m12
A
m10
11
1
11
1
w
x
D
1
1
01
D
y = A'B'C A'BC' + ABC + AB'C'
= A'(A B) + A(B C)'
=A B C
= X C
A
1
00
1
m6
1
m8
B
C
00
m2
1
11
1
CD
AB
10
1
01
1
m12
10
1
C
01
1
D
x = AB' + A'B = A B
CD
00
m6
1
11
1
m11
1
B
w=A
A
10
m2
m7
1
D
AB
11
m3
m5
01
m12
1000
1001
1010
1011
1100
1101
1110
1111
01
m1
00
01
A
C
CD
m10
1
D
C
D
z
(b)
module Prob_4_7(output w, x, y, z, input A, B, C, D);
always @ (A, B, C, D)
case ({A, B, C, D})
4'b0000:
{w, x, y, z} = 4'b0000;
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
www.Mohandesyar.com
B
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
58
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4'b0001:
4'b0010:
4'b0011:
4'b0100:
4'b0101:
4'b0110:
4'b0111:
{w, x, y, z} = 4'b1111;
{w, x, y, z} = 4'b1110;
{w, x, y, z} = 4'b1101;
{w, x, y, z} = 4'b1100;
{w, x, y, z} = 4'b1011;
{w, x, y, z} = 4'b1010;
{w, x, y, z} = 4'b1001;
4'b1000:
4'b1001:
4'b1010:
4'b1011:
4'b1100:
4'b1101:
4'b1110:
4'b1111:
endcase
endmodule
{w, x, y, z} = 4'b1000;
{w, x, y, z} = 4'b0111;
{w, x, y, z} = 4'b0110;
{w, x, y, z} = 4'b0101;
{w, x, y, z} = 4'b0100;
{w, x, y, z} = 4'b0011;
{w, x, y, z} = 4'b0010;
{w, x, y, z} = 4'b0001;
Alternative model:
module Prob_4_7(output w, x, y, z, input A, B, C, D);
assign w = A;
assign x = A ^ B);
assign y = x ^ C;
assign z = y ^ D;
endmodule
4.8
ABCD
0000
0001
0011
0010
0110
0111
0101
0100
1100
1101
1111
1110
1010
1011
1001
1000
wxyz
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
CD
C
CD
AB
00
m0
01
m1
00
11
x
m4
10
m3
m5
00
m2
x
m0
x
m7
m6
11
m13
x
m8
10
m15
m4
x
m9
m14
1
B
m11
m10
m8
10
1
D
m0
01
m1
00
11
m3
x
m4
m5
m6
m13
m15
m14
x
m9
x
m11
1
1
m10
1
m12
x
m7
m13
x
11
m8
w = AB+AC'D'
x = B'C + B'D +BC'D'
y = CD' +C'D
z=D
x
m6
1
m15
x
m9
10
m2
1
01
10
m7
C
00
A
x
D
CD
AB
x
m5
x
11
A
10
m2
1
m12
x
11
m3
x
01
m12
01
m1
00
01
A
C
m14
B
x
m11
1
m10
1
D
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
www.Mohandesyar.com
B
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
59
www.Mohandesyar.com
Alternative model:
module Prob_4_8(output w, x, y, z, input A, B, C, D);
assign w = (A&B) | (A & (~C)) & (~D) ;
assign x = ( (~B) & C) | ((~B) & D) | (B & (~C)) & (~D);
assign y = C ^ D;
assign z = D;
endmodule
4.9
ABCD a b c
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1
0
1
1
0
1
1
1
1
1
1
1
1
1
1
0
0
1
1
1
1
1
0
1
1
1
1
1
1
1
d
e f
g
1
0
1
1
0
1
1
0
1
1
1
0
1
0
0
0
1
0
1
0
0
0
1
1
1
1
1
0
1
1
1
0
0
0
1
1
1
0
1
1
CD
CD
C
AB
00
01
m0
11
m1
00
m3
1
m4
m5
m7
01
1
m12
m0
1
00
1
01
m6
1
m13
00
m2
1
m15
C
AB
10
m1
1
m4
B
m14
11
01
m8
m9
m11
1
10
A
m10
1
m5
m7
00
01
m12
m13
m15
m14
m8
m9
m11
m10
1
m3
1
m4
m12
m0
00
m7
1
m6
1
m13
00
m2
1
m5
1
C
AB
10
m15
01
m1
m4
B
m14
m8
m9
1
m11
A
m10
1
m5
00
m3
m7
m4
m5
m7
01
m12
m13
m15
AB
m0
1
00
1
01
m6
B
A
11
m12
m13
m15
m14
m8
m9
m11
m10
1
m9
m11
1
m10
m3
A
C
00
m2
m0
1
01
m1
11
m3
00
m5
1
m12
m7
1
m13
m6
m15
1
m4
1
m14
m5
1
01
B
10
m2
m7
1
1
m6
1
m12
m13
m15
m14
m8
m9
m11
m10
11
m8
10
D
e = A'CD' + B'C'D'
B
1
CD
AB
10
11
m8
10
01
m1
m4
m14
1
d = A'CD' + A'B' C+ B'C'D' + AB'C' + A'BC'D
C
00
m2
11
m6
D
CD
1
1
1
10
c = A'B + A'D + B'C'D' + AB'C'
m1
m2
1
D
m0
10
m3
11
10
10
11
1
01
1
11
C
B
1
CD
11
m1
1
01
11
m6
1
b = A'B' + A'C'D' + A'CD + AB'C'
C
00
m0
01
1
D
CD
AB
00
1
1
10
a = A'C + A'BD + B'C'D' + AB'C'
CD
m2
1
D
AB
10
m3
11
A
A
11
m9
1
m11
m10
A
1
D
f = A'BC' + A'C'D' + A'BD + AB'C'
10
1
1
D
g = A'CD' + A'B'C + A'BC' + AB'C'
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
www.Mohandesyar.com
B
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
60
www.Mohandesyar.com
4.10
ABCD wxyz
0000 0000
0001 1111
0010 1110
0011 1101
0100 1100
0101 1011
0110 1001
0111 1000
1000
1001
1010
1011
1100
1101
1110
1111
CD
C
CD
AB
00
m0
01
m1
00
11
m3
1
m4
01
m5
1
m7
m13
m0
1
00
1
01
m6
1
m15
B
m14
m8
10
m9
m11
00
m0
01
m1
m7
m6
m13
m15
m14
m9
m11
m10
1
m3
m4
m5
m7
1
01
m12
m13
m15
1
11
m8
m9
m11
m0
1
00
1
01
m6
B
m1
m5
m12
m13
1
A
m8
1
m9
1
m6
1
m15
B
m14
1
m11
1
10
10
m2
m7
1
11
m10
m3
m10
1
D
z=D
y = CD' + C'D = C
D
For a 5-bit 2's complementer with input E and output v:
4.11
1
11
1
m4
m14
01
D
v=E
B
C
00
m2
1
1
10
1
CD
AB
10
1
00
A
11
1
D
x = B'(C + D) + CB'D'
= B (C + D)
C
CD
AB
1
m5
10
D
w = A'(B + C + D) + AB'C'D'
= A (B + C + D)
m2
1
m8
1
10
m3
1
m12
A
m10
11
1
11
A
01
m1
m4
11
1000
0111
0110
0101
0100
0011
0010
0001
C
00
m2
1
1
m12
10
(A + B + C + D)
(a)
A2
A3
A1
A0
1
x
y
Half Adder
C
S
x
y
Half Adder
C
S
x
y
Half Adder
C
S
x
y
Half Adder
C
S
Note: 5-bit output
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
www.Mohandesyar.com
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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(b)
A3
1
A2
x
x
y
Full Adder
A1
y
Full Adder
D
B
1
x
y
Full Adder
D
B
1
B
D
A0
1
x
y
Half Adder
B
D
Note: To decrement the 4-bit number, add -1 to the number. In 2's complement format ( add F h ) to
the number. An attempt to decrement 0 will assert the borrow bit. For waveforms, see solution to
Problem 4.52.
4.12
(a)
x
0
0
1
1
y
0
1
0
1
B
0
1
0
0
D
0
1
1
0
D = x'y + xy'
B = x'y
(b)
x y Bin
0
0
0
0
1
1
1
1
Sum
C
V
(a)
1101
0
1
(b)
0001
1
1
(c)
0100
1
0
(d)
1011
0
1
(e)
1111
0
0
xor
AND OR
4.13
4.14
10
+ 5
+ 5
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
B
0
1
1
1
0
0
0
1
D
0
1
1
0
1
0
0
1
Diff = x y z
Bout = x'y + x'z + yz
XOR
+ 10
= 30 ns
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
www.Mohandesyar.com
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
www.Mohandesyar.com
4.15
C4 = G3 + P3C3 = G3 + P3(G2 + P2G1 + P2P1G0 + P2P1P0C0)
= G3 + P3G2 + P3P2G1 + P3P2P1G0 + P3P2P1P0C0
4.16
(a)
(C'G'i + p'i)' = (Ci + Gi)Pi = GiPi + PiCi
= AiBi(Ai + Bi) + PiCi
= AiBi + PiCi = Gi + PiCi
= AiBi + (Ai + Bi)Ci = AiBi + AiCi + BiCi = Ci+1
(PiG'i) Ci = (Ai + Bi)(AiBi)' Ci = (Ai + Bi)(A'i + B'i) Ci
= (A'iBi + AiB'i) Ci = Ai Bi Ci = Si
(b)
Output of NOR gate = (A0 + B0)' = P'0
Output of NAND gate = (A0B0)' = G'0
S1 = (P0G'0) C0
C1 = (C'0G'0 + P'0)' as defined in part (a)
4.17
(a)
(C'iG'i + P'i)' = (Ci + Gi)Pi = GiPi + PiCi = AiBi(Ai + Bi) + PiCi
= AiBi + PiCi = Gi + PiCi
= AiBi + (Ai + Bi)Ci = AiBi + AiCi + BiCi = Ci+1
(PiG'i)Ci = (Ai + Bi)(AiBi)'Ci = (Ai + Bi)(A'i + B'i)Ci
= (A'iBi + AiB'i)Ci = AiBiCi = Si
(b)
Output of NOR gate = (A0 + B0)' = P'0
Output of NAND gate = (A0B0)' = G'0
S0 = (P0G'0)C0
C1 = (C'0G'0 + P'0)'
as defined in part (a)
4.18
Inputs Outputs
ABCD wxyz
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1001
1000
0111
0110
0101
0100
0011
0010
0001
0000
d(A, b, c, d) = 6(10, 11, 12, 13, 14, 15)
Digital Design – Solution Manual. M. Mano. M.D. Ciletti, Copyright 2007, All rights reserved.
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62
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
63
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C
CD
AB
00
m0
00
01
m1
1
m4
11
m3
CD
AB
10
m2
00
m0
m7
m6
m13
x
11
A
m8
m15
x
m9
m4
m11
00
m0
m1
m3
00
x
m4
m5
m12
11
m13
x
m8
m15
x
m9
x
m11
x
10
CD
AB
10
m6
m15
m14
x
x
m9
1
00
1
01
m11
m10
x
B
x
m3
10
m2
1
m5
1
m7
1
m13
m15
x
m14
x
m9
m11
1
D
y=C
m6
1
x
m8
10
11
m1
m12
A
m10
01
1
11
x
C
C
00
m4
m14
B
x
x
m0
m6
1
m7
10
m2
m7
m13
m8
1
D
x = BC' + B'C = B
1
01
10
m2
1
x
11
A
m10
m5
1
m12
x
x
D
w = A'B'C'
C
01
11
B
m14
x
10
CD
11
m3
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01
01
m12
01
m1
00
1
m5
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B
x
m10
x
x
D
z = D'
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