Chapter 3 Problem Solutions -y 3.1. (a) Xy + Xy
Transcript of Chapter 3 Problem Solutions -y 3.1. (a) Xy + Xy
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8/14/2019 Chapter 3 Problem Solutions -y 3.1. (a) Xy + Xy
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- 3.1 -
Chapter 3
Problem Solutions
3.1. (a) xy + xy-+ x-y-= x(y+y
-) + x
-y-
P4b
= x.1 + x
-
y
-
P5a= x + x
-y-
P2b
= x + y-
T7a
(b) (x-z-+ x-y + x
-z + xy)
__________________
= (x-z + x
-z-+ yx + yx
-)
__________________
P3a,P3b
= [x-(z+z
-) + y(x+x
-)]
________________
P4b
= (x-.1 + y.1)
__________
P5a
= (x-+ y)
______
P2b
= (x-)
__
y-
T9a
= xy- T5
(c) (x + y)(x-z-+ z)(y
-+ xz)
_______
= (x + y)(z + z-x-)(y-+ xz)
_______
P3a,P3b
= (x + y)(z + x-)(y-+ xz)
_______
T7a
= (x + y)(z + x-)y(xz)
___
T9a,T5
= (x + y)(z + x-)y(x
-+ z-) T9b
= (x + y)(x-
+ z)(x-+ z-)y P3a,P3b
= (x + y)(x-
+ zz-)y P4a
= (x + y)(x-
+ 0)y P5b
= (x + y)x-y P2a
= x-y(x + y) P3b
= x-[y(x + y)] T8b
= x-[y(y + x)] P3a
= x-y T6b
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- 3.2 -
3.1. (continued)
(d) (xy + yz + xz)
_____________
= (xy)
___
(yz)
___
(xz)
___
T9a
= (x-+ y-)(y-+ z-)(x-
+ z-) T9b
= (x-+ y-)(x-+ z-)(y-
+ z-) P3b
= (x- + y-z-)(y- + z-) P4a
= (x-+ y-z-)y-+ (x
-+ y
-z-)z-
P4b
= y-(x-+ y-z-) + z
-(x-+ y
-z-) P3b
= y-x-+ y-y-z-+ z-x-+ z-y-z-
P4b
= x-y-+ y-y-z-+ x-z-+ y-z-z-
P3b
= x-y-+ y-z-+ x-z-+ y-z-
T4b
= x-y-
+ y-z-+ y-z-+ x-z-
P3a
= x-y-
+ y-z-+ x-z-
T4a
(e) xy + yz + x-z = xy + yz.1 + x
-z P2b
= xy + yz(x + x-) + x
-z P5a
= xy + yzx + yzx-+ x-z P4b
= xyz + xy + x-zy + x
-z P3a,P3b
= xy(z + 1) + x-z(y + 1) P4b
= xy.1 + x-z.1 T2a
= xy + x-z P2b
(f) (x + y)(x-+ z) = (x + y)x
-+ (x + y)z P4b
= x-(x + y) + z(x + y) P3b
= x-x + x
-y + zx + zy P4b
= xx-+ x
-y + zx + zy P3b
= 0 + x-y + zx + zy P5b
= x-y + zx + zy P2a
= x-y + zx + zy.1 P2b
= x-y + zx + zy(x + x
-) P5a
= x-y + zx + zyx + zyx
-P4b
= x-yz + x
-y + xzy + xz P3a,P3b
= x-y(z + 1) + xz(y + 1) P4b
= x-y.1 + xz.1 T2a
= x-y + xz P2b
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- 3.3 -
3.1. (continued)
(g) (x + y)(y + z)(x + z) = [(x + y)y + (x + y)z](x + z) P4b
= [y(x + y) + z(x + y)](x + z) P3b
= (yx + yy + zx + zy)(x + z) P4b
= (yx + y + zx + zy)(x + z) T4b
= (y + yx + yz + zx)(x + z) P3a
= (y + zx)(x + z) T6a
= (y + zx)x + (y + zx)z P4b
= x(y + zx) + z(y + zx) P3b
= xy + xzx + zy + zzx P4b
= xy + xxz + yz + xzz P3b
= xy + xz + yz + xz T4b
= xy + (xz + xz) + yz P3a
= xy + xz + yz T4a
(h) xy-+ yz
-+ x-z = xy
-.1 + yz-.1 + x
-z.1 P2b
= xy-(z + z
-) + yz
-(x + x
-) + x
-z(y + y
-) P5a
= xy-z + xy
-z-+ yz
-x + yz
-x-+ x-zy + x
-zy-
P4b
= x-yz + x
-yz-+ y-zx + y
-zx-+ xz
-y + xz
-y-P3a,P3b
= x-y(z + z
-) + y
-z(x + x
-) + xz
-(y + y
-) P4b
= x-y.1 + y
-z.1 + xz
-.1 P5a
= x-y + y
-z + xz
-P2b
3.2. To prove the cancellation law does not hold, use the method
of contradiction. According to T7b, x(x-+y)=xy. Assuming
the cancellation law holds, it follows that x-+y=y for all x
and y in the Boolean algebra. Since x and y denote any
elements in the Boolean algebra, let x=y. It then follows
that y-+y=y. However, from P5a it is known that y must be 1
in this case and not any arbitrary element in the algebra.
Thus, by contradiction, the cancellation law does not hold.
By applying a dual argument starting with T7a, it can be
shown that x+y=x+z does not imply y=z.
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3.3.
(a) (b) (c)
x y z x-z+xy (x
-+y)(x+z) (x+y+z)
______
x-y-z-
xy+yz+x-z xy+x
-z
0 0 0 o 0 1 1 0 0
0 0 1 1 1 0 0 1 1
0 1 0 0 0 0 0 0 0
0 1 1 1 1 0 0 1 1
1 0 0 0 0 0 0 0 0
1 0 1 0 0 0 0 0 0
1 1 0 1 1 0 0 1 1
1 1 1 1 1 0 0 1 1
3.4. Let B={0,1,a} where a0,1. By Postulate P5 of a Boolean
algebra, the complement of the element a, i.e., a-, must
exist and satisfy the relationships a+a-=1 and a.a
-=0. Since
there are just three elements in B, a-
must be 0, 1, or a.
Suppose a-=a, then a+a
-=a+a=a. However, since a1, Postulate
P5a is not satisfied. Thus, a
-
a. Now suppose a-
=1. Then
a.a-=a.1=a. However, since a0, Postulate P5b is not
satisfied. Finally, suppose a-=0. Then, a+a
-=a+0=a. Again
Postulate P5a is not satisfied since a1. Therefore, a-does
not exist and B cannot be a Boolean algebra.
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3.5. (a)
x y z x-
z-
x-+y x
-+z-
(x-+y)(x
-+z-) yz f
0 0 0 1 1 1 1 1 0 1
0 0 1 1 0 1 1 1 0 1
0 1 0 1 1 1 1 1 0 1
0 1 1 1 0 1 1 1 1 1
1 0 0 0 1 0 1 0 0 0
1 0 1 0 0 0 0 0 0 0
1 1 0 0 1 1 1 1 0 1
1 1 1 0 0 1 0 0 1 1
(b)
x y z x-
xy x-z xy+x
-z (xy+x
-z)
______
yz f
0 0 0 1 0 0 0 1 0 1
0 0 1 1 0 1 1 0 0 0
0 1 0 1 0 0 0 1 0 1
0 1 1 1 0 1 1 0 1 1
1 0 0 0 0 0 0 1 0 1
1 0 1 0 0 0 0 1 0 1
1 1 0 0 1 0 1 0 0 0
1 1 1 0 1 0 1 0 1 1
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3.5. (continued)
(c)
x y z x-
y-
x+y-
y+z x-+z (x+y
-)(y+z) f
0 0 0 1 1 1 0 1 0 0
0 0 1 1 1 1 1 1 1 1
0 1 0 1 0 0 1 1 0 0
0 1 1 1 0 0 1 1 0 0
1 0 0 0 1 1 0 0 0 0
1 0 1 0 1 1 1 1 1 1
1 1 0 0 0 1 1 0 1 0
1 1 1 0 0 1 1 1 1 1
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3.5. (continued)
(d)
w x y z w-
y-
wx wxy y-+z w
-(y-+z) f
0 0 0 0 1 1 0 0 1 1 1
0 0 0 1 1 1 0 0 1 1 1
0 0 1 0 1 0 0 0 0 0 0
0 0 1 1 1 0 0 0 1 1 1
0 1 0 0 1 1 0 0 1 1 1
0 1 0 1 1 1 0 0 1 1 1
0 1 1 0 1 0 0 0 0 0 0
0 1 1 1 1 0 0 0 1 1 1
1 0 0 0 0 1 0 0 1 0 0
1 0 0 1 0 1 0 0 1 0 0
1 0 1 0 0 0 0 0 0 0 0
1 0 1 1 0 0 0 0 1 0 0
1 1 0 0 0 1 1 0 1 0 0
1 1 0 1 0 1 1 0 1 0 0
1 1 1 0 0 0 1 1 0 0 1
1 1 1 1 0 0 1 1 1 0 1
3.6. (a) f(x,y,z) = x-y-z-+ x
-y-z + x
-yz + xy
-z + xyz
-
= m0+ m
1+ m
3+ m
5+ m
6
= 3m(0,1,3,5,6)
(b) f(w,x,y,z) = w-x-y-z + w
-x-yz-+ w-xy-z-+ w
-xyz-+ w-xyz + wx
-yz
+ wxy-z- + wxy-z + wxyz
= m1
+ m2+ m
4+ m
6+ m
7+ m
11+ m
12
+ m13
+ m15
= 3m(1,2,4,6,7,11,12,13,15)
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3.7. (a) f(x,y,z) = 3m(0,2,4,5,7)
= x-y-z-+ x
-yz-+ xy
-z-+ xy
-z + xyz
x y z f
0 0 0 1
0 0 1 0
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 1
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3.7. (continued)
(b) f(w,x,y,z) = 3m(1,3,7,8,9,14,15)
= w-x-y-z + w
-x-yz + w
-xyz + wx
-y-z-+ wx
-y-z
+ wxyz-+ wxyz
w x y z f
0 0 0 0 0
0 0 0 1 1
0 0 1 0 0
0 0 1 1 1
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
1 0 0 1 1
1 0 1 0 0
1 0 1 1 0
1 1 0 0 0
1 1 0 1 0
1 1 1 0 1
1 1 1 1 1
3.8. (a) f(x,y,z) = (x+y-+z)(x
-+y+z)(x
-+y-+z-)
= M2M4M7
= JM(2,4,7)
(b) f(w,x,y,z) = (w+x+y+z)(w+x+y-+z-)(w+x
-+y+z
-)(w-+x+y+z)
.(w-+x+y+z
-)(w-+x+y
-+z)(w
-+x-+y-+z)
= M0M3M5M8M9M10M14
= JM(0,3,5,8,9,10,14)
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3.9. (a) f(x,y,z) = JM(0,1,2,5,7)
= (x+y+z)(x+y+z-)(x+y
-+z)(x
-+y+z
-)(x-+y-+z-)
x y z f
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 0
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3.9. (continued)
(b) f(w,x,y,z) = JM(0,3,6,7,9,10,12,13,15)
= (w+x+y+z)(w+x+y-+z-)(w+x
-+y-+z)(w+x
-+y-+z-)
.(w-+x+y+z
-)(w-+x+y
-+z)(w
-+x-+y+z)
.(w-+x-+y+z-)(w-+x-+y-+z-)
w x y z f
0 0 0 0 0
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0
0 1 0 0 1
0 1 0 1 1
0 1 1 0 0
0 1 1 1 0
1 0 0 0 1
1 0 0 1 0
1 0 1 0 0
1 0 1 1 1
1 1 0 0 0
1 1 0 1 0
1 1 1 0 1
1 1 1 1 0
3.10. (a) [(w+x+y)(w-+xz)+y
-z-]
_________________
= [(w+x+y)(w-+xz)]
______________
(y-z-)
___
= [(w+x+y)
______
+(w-+xz)
_____
](y+z)
= [w
-
x
-
y
-
+w(xz)
___
](y+z)= [w
-x-y-+w(x
-+z-)](y+z)
(b) [x-(w-y-+xyz
-)]
__________
= x + (w-y-+xyz
-)
_______
= x + (w-y-)
___
(xyz-)
____
= x + (w+y)(x-+y-+z)
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3.10. (continued)
(c) {wy-[(wy
-)
___
+xz-]}
____________
= w-+ y + [(wy
-)
___
+xz-]
________
= w-+ y + (wy
-)(xz
-)
___
= w
-
+ y + wy
-
(x
-
+z)
(d) {wx+z[(w+x-+y-)
______
+(x+y)
____
]}
____________________
= (wx)
___
{z[(w+x-+y-)
______
+(x+y)
____
]}
_________________
= (w-+x-){z-+[(w+x
-+y-)
______
+(x+y)
____
]
______________
}
= (w-+x-)[z-+(w+x
-+y-)(x+y)]
3.11. (a) f = x[w-.1.yz
-+z(1.y
-+w.1
-)] + x
-[w-.0.yz
-+z(0.y
-+w.0
-)]
= x(w-yz-+y-z) + x
-(wz)
(b) f = [z+w-xy.0-+0.(xy-+wx-)][z-+w-xy.1-+1.(xy-+wx-)]
= (z+w-xy)(z
-+xy-+wx-)
3.12. (a) yz + x-y-z + xyz
-= 1.yz + x
-y-z + xyz
-
= (x+x-)yz + x
-y-z + xyz
-
= xyz + x-yz + x
-y-z + xyz
-
= xy(z+z-) + x
-z(y+y
-)
= xy + x-z
(b) x-y-z-
+ x-y-z + x
-yz-+ xy
-z-+ xy
-z + xyz
-
= y-(x-z-+x-z+xz
-+xz) + yz
-(x-+x)
= y-[x-(z-+z)+x(z
-+z)] + yz
-
= y-(x-+x) + yz
-
= y-+ yz
-
= y-+ z-
(c) xy-+ xz + x
-z-+ xyz
-+ x-y-z = x(y
-+z+yz
-) + x
-(z-+y-z)
= x(y-+z+y) + x
-(z-+y-)
= x + x-(y-+z-)
= x + y-+ z-
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3.12. (continued)
(d) (x+yz)
_____
+ y-z = x
-(y-+z-) + y
-z
= x-y-+ x-z-+ y-z
= x-y-(z+z
-) + x
-z-+ y-z
= x-y-z + x-y-z- + x-z- + y-z
= x-z-(y-+1) + y
-z(x-+1)
= x-z-
+ y-z
(e) w-x-y-z + wx
-y-z + xz + xyz
-= x-y-z(w-+w) + x(z+yz
-)
= x-y-z + x(z+y)
= x-y-z + xz + xy
= (x-y-+x)z + xy
= (y-+x)z + xy
= y-z + xz + xy
= y-z + xz(y+y
-) + xy
= y-z + xyz + xy
-z + xy
= y-z(1+x) + xy(z+1)
= y-z + xy
(f) w-x-yz + wxy + w
-y-+ xy
-+ x-y-= w-x-yz + wxy + y
-(w-+x+x
-)
= w-x-yz + wxy + y
-
= (w-x-z+wx)y + y
-
= w-x-z + wx + y
-
(g) (w+x+y-+z)(w+x+y
-+z-)(w+x
-+y-+z)(w+x
-+y-+z-)(w
-+x-+y-+z)(w
-+x-+y-+z-)
= (w+x+y-+zz
-)(w+x
-+y-+zz-)(w-+x-+y-+zz-)
= (w+x+y-)(w+x
-+y-)(w-+x-+y-)
= (w+x+y-)(ww
-+x-+y-)
= (w+x+y-)(x
-+y-)
= y-+ x-(w+x)
= y- + wx- + xx-
= y-+ wx
-
(h) (x+z)(w+x)(y-+z)(w+y
-) = (x+wz)(y
-+wz)
= wz + xy-
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3.13. xi.f(x
1,x2,...,x
i,...,x
n)
= xi[xi.f(x
1,x2,...,1,...,x
n)+x-i.f(x
1,x2,...,0,...,x
n)]
= xi.f(x
1,x2,...,1,...,x
n)+x
ix-i.f(x
1,x2,...,0,...,x
n)
= xi.f(x
1,x2,...,1,...,x
n)
Equation (b) follows from the duality principle.
3.14. (a) f(x,y,z) = x-(y-+z) + z
-
= x-y-+ x
-z + z
-
= x-y-(z+z
-) + x
-z(y+y
-) + z
-(x+x
-)(y+y
-)
= x-y-z + x
-y-z-+ x-yz + xyz
-+ xy
-z-+ x-yz-
(b) f(x,y,z) = (x+y-)(x+z)
= x + y
-
z
= x(y+y-)(z+z
-) + (x+x
-)y-z
= xyz + xyz-+ xy
-z + xy
-z-+ x
-y-z
3.15. (a) f(x,y,z) = (y+z-)(xy
-+z)
= (y+z-)(x+z)(y
-+z)
= (xx-+y+z
-)(yy
-+x+z)(xx
-+y-+z)
= (x+y+z-)(x
-+y+z
-)(x+y+z)(x+y
-+z)(x
-+y-+z)
(b) f(x,y,z) = x + x-z-(y+z)
= (x+x-)(x+z
-)(x+y+z)
= (x+z-)(x+y+z)
= (x+z-+yy-)(x+y+z)
= (x+y+z-)(x+y
-+z-)(x+y+z)
3.16. (a) f-(x,y,z) = 3m(1,3,4,6,7) = JM(0,2,5)
(b) f-(x,y,z) = 3m(1,2,5,7) = JM(0,3,4,6)
(c) f-(w,x,y,z) = 3m(0,2,3,5,9,10,11,13,15)
= JM(1,4,6,7,8,12,14)
(d) f-(w,x,y,z) = 3m(3,7,8,10,12,13)
= JM(0,1,2,4,5,6,9,11,14,15)
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3.17. (a) f(x,y,z) = 3m(1,3,5)
f-(x,y,z) = JM(1,3,5)
f=(x,y,z) = f(x,y,z) =JM(0,2,4,6,7)
(b) f(x,y,z) = JM(3,4)
f-(x,y,z) = 3m(3,4)
f=(x,y,z) = f(x,y,z) = 3m(0,1,2,5,6,7)
(c) f(w,x,y,z) = 3m(0,1,2,3,7,9,11,12,15)
f-(w,x,y,z) = 3m(4,5,6,8,10,13,14)
f=(w,x,y,z) = f(w,x,y,z) = JM(4,5,6,8,10,13,14)
(d) f(w,x,y,z) = JM(0,2,5,6,7,8,9,11,12)
f-(w,x,y,z) = JM(1,3,4,10,13,14,15)
f=(w,x,y,z) = f(w,x,y,z) = 3m(1,3,4,10,13,14,15)
3.18. (a) f = v(wx-z+y)(w+z) + w
-
(b) f = [(w+x)(v-+y+z)+w
-][(w+x)(v
-+y+z)+v]
(c) f = [(w+x-)(w-+y)]
___________
z + (w-+y)z
-
3.19. (a)
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3.19. (continued)
(b)
(c)
3.20. (a) f = xy-z + (xy-z+w)
______
+ wy-
= A + (A+w)
____
+ wy-
where A=xy-z
= A + (0+w)
____
+ wy-
by T3.12b
= A + w-+ wy
-
= xy-z + w
-+ wy
-
= xy-z + w
-+ y-
= w-+ y-(xz+1)
= w-+ y-
-
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3.20. (continued)
(b) f = [(w-xy-z)
_____
z+x-z-]w-+ wyz
= [(w+x-+y+z
-)z+x
-z-]w-+ wyz
= (wz+x-z+yz+x
-z-)w-+ wyz
= w-x-z + w-yz + w-x-z- + wyz
= w-x-(z+z
-) + yz(w
-+w)
= w-x-+ yz
3.21. (a) f = w[(x+y)z+x-z-]
(b)
3.22. (a) f(w,x,y,z) = 3m(2,3,6,10,15) + dc(0,1,5,9,11,13,14)
= JM(4,7,8,12) + dc(0,1,5,9,11,13,14)
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- 3.18 -
3.22. (continued)
(b)
w x y z f f-
0 0 0 0 - -
0 0 0 1 - -
0 0 1 0 1 0
0 0 1 1 1 0
0 1 0 0 0 1
0 1 0 1 - -
0 1 1 0 1 0
0 1 1 1 0 1
1 0 0 0 0 1
1 0 0 1 - -
1 0 1 0 1 0
1 0 1 1 - -
1 1 0 0 0 1
1 1 0 1 - -
1 1 1 0 - -
1 1 1 1 1 0
(c) f-(w,x,y,z) = 3m(4,7,8,12) + dc(0,1,5,9,11,13,14)
= JM(2,3,6,10,15) + dc(0,1,5,9,11,13,14)
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- 3.19 -
3.23. (a) Using perfect induction:
x y z nand(y,z) nand[x,nand(y,z)] nand(x,y) nand[nand(x,y),z]
0 0 0 1 1 1 1
0 0 1 1 1 1 0
0 1 0 1 1 1 1
0 1 1 0 1 1 0
1 0 0 1 0 1 1
1 0 1 1 0 1 0
1 1 0 1 0 0 1
1 1 1 0 1 0 1
Since the columns for nand[x,nand(y,z)] and
nand[nand(x,y),z] are dissimilar, the nand-operation is
not associative.
(b) Using perfect induction:
x y z nor(y,z) nor[x,nor(y,z)] nor(x,y) nor[nor(x,y),z]
0 0 0 1 0 1 0
0 0 1 0 1 1 0
0 1 0 0 1 0 1
0 1 1 0 1 0 0
1 0 0 1 0 0 1
1 0 1 0 0 0 0
1 1 0 0 0 0 1
1 1 1 0 0 0 0
Since the columns for nor[x,nor(y,z)] and
nor[nor(x,y),z] are dissimilar, the nor-operation is
not associative.
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- 3.20 -
3.24. (a) f = {[(u-v-)
___
(wx)
___
y]
__________
z}
_____________
= (u+v)(w-+x-)y + z
-
(b) f = {[x(xy)
___
]
______
[y(xy)
___
]
______
}
_______________
= x(x-+y-) + y(x-+y-)
= xy-+ x-y
(c) f = {[(x-+y+z
-)
______
+w-+z]
____________
+[(w+x-)
____
+y]
________
}
________________________
= (xy-z+w-+z)(w
-x+y)
= (w-+z)(w
-x+y)
(d) f = {[(v+w)
____
+(x-+y)
____
]
____________
+[(x-+y)
____
+z]
________
}
________________________
= (v-w-+xy-)(xy-+z)
(e) f = {[(u-+v-)
____
+(w+x)
____
+y-]z}
_________________
= (u-+v-)(w+x)y + z
-
3.25. (a)
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- 3.21 -
3.25. (continued)
(b)
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- 3.22 -
3.25. (continued)
(c)
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- 3.23 -
3.25. (continued)
(d)
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- 3.24 -
3.26. (a)
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- 3.25 -
3.26. (continued)
(b)
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- 3.26 -
3.26. (continued)
(c)
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- 3.27 -
3.26. (continued)
(d)
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- 3.28 -
3.27. (a)
(b)
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- 3.29 -
3.27. (continued)
(c)
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- 3.30 -
3.27. (continued)
(d)
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- 3.31 -
3.28. (a)
(b)
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- 3.32 -
3.28. (continued)
(c)
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- 3.33 -
3.28. (continued)
(d)
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- 3.35 -
3.29. (continued)
(b) Nor realization
f = [(w-+x)y+u
-+v](w+z)y
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- 3.36 -
3.30. (x+y)r(x+z) = (x+y)
____
(x+z) + (x+y)(x+z)
____
= x-y-(x+z) + (x+y)x
-z-
= x-y-z + x
-yz-
= x-(y-z+yz
-)
= x-(yrz)
Alternatively, perfect induction could be used.
3.31. Using perfect induction
x y z yz xryz xry xrz (xry)(xrz)
0 0 0 0 0 0 0 0
0 0 1 0 0 0 1 0
0 1 0 0 0 1 0 0
0 1 1 1 1 1 1 1
1 0 0 0 1 1 1 1
1 0 1 0 1 1 0 0
1 1 0 0 1 0 1 0
1 1 1 1 0 0 0 0
Since the columns for xryz and (xry)(xrz) are dissimilar,
the exclusive-or-operation is not distributive over theand-operation.
3.32. (a) xu1 = x-.1-+ x.1
= x-.0 + x
= x
(b) xu0 = x-.0-+ x.0
= x-.1
= x-
(c) xux = x-x-+ xx
= x-+ x
= 1
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- 3.37 -
3.32. (continued)
(d) xux-= x
-x + xx
-
= 0 + 0
= 0
(e) x-uy- = xy + x-y-
= xuy
(f) x-uy = xy
-+ x-y
= xry
= (xuy)
____
(g) (x+y)u(x+z) = (x+y)
____
(x+z)
____
+ (x+y)(x+z)
= x-y-x-z-+ x + yz
= x-y-z- + x + yz
= y-z-+ x + yz
= x + (yuz)
(h) xuyu(x+y) = xu[yu(x+y)]
= xu[y-(x+y)
____
+y(x+y)]
= xu(y-x-+xy+y)
= xu(x-y-+y)
= xu(x-+y)
= x-(x-+y)
____+ x(x
-+y)
= x-xy-
+ xx-+ xy
= xy
3.33. From the definition of the dual of a function, xry should
first be complemented by DeMorgan's law and then each of
the variables in the complemented expression should be
complemented. Applying DeMorgan's law gives
(xry)____
=(x-y+xy
-)
______=(x+y
-)(x-+y). Complementing each variable in
(x+y-)(x-+y) gives (x
-+y)(x+y
-). Therefore,
(xry)dual
= (x-+y)(x+y
-)
= (xry)
____
= xuy
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- 3.38 -
3.34. (a)
According to statement 1, bn-1
should be connected to
gn-1
. According to statement 2, for n-1$k$1,
bk-1
=f(bk,gk-1
). This function is given by the truth
table
bk
gk-1
bk-1
0 0 0
0 1 1
1 0 1
1 1 0
which is described by bk-1
=bkrgk-1
. Thus, each of the
outputs bk-1
for n-1$k$1 is the exclusive-or of the
next higher order binary digit (bk) and the present
order Gray bit (gk-1
). The corresponding logic diagram
is
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- 3.39 -
3.34. (continued)
(b) From part (a) we have
bn-1
= gn-1
bk-1
= bkrgk-1
for n-1$k$1
We now need expressions for gn-1 and gk-1 for n-1$k$1.
Clearly, gn-1
=bn-1
. For the remaining expression,
consider property (x) in Table 3.16. Letting X=gk-1
,
Y=bk, and Z=b
k-1, b
k-1=bkrgk-1
has the form XrY=Z.
Thus, X=YrZ or gk-1
=bkrbk-1
for n-1$k$1. A logic
diagram for an n-bit binary-to-Gray converter is:
3.35. Position: 7 6 5 4 3 2 1
Code group format: b4
b3
b2
p3
b1
p2
p1
c*1
= 0 if even parity over inputs b4, b
2, b
1, and p
1
c*1
= 1 if odd parity over inputs b4, b
2, b
1, and p
1
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- 3.40 -
3.35. (continued)
Truth table for c*1:
b4
b2
b1
p1
c*1
0 0 0 0 0
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0
0 1 0 0 1
0 1 0 1 0
0 1 1 0 0
0 1 1 1 1
1 0 0 0 1
1 0 0 1 0
1 0 1 0 0
1 0 1 1 1
1 1 0 0 0
1 1 0 1 1
1 1 1 0 1
1 1 1 1 0
c*1
= b-4b-2b-1p1+ b-4b-2b1p-1+ b-4b2b-1p-1+ b
-4b2b1p1+ b
4b-2b-1p-1
+ b4b-2b1p1
+ b4b2b-1p1+ b
4b2b1p-1
= b-4b-2(b-1p1+b1p-1) + b
-4b2(b-1p-1+b1p1) + b
4b-2(b-2p-1+b1p1)
+ b4b2(b-1p1+b
1p-1)
= b-4b-2(b1rp1) + b
-4b2(b1rp1)
______
+ b4b-2(b1rp1)
______
+ b4b2(b1rp1)
= (b1rp1)(b-4b-2+b4b2) + (b1rp1)
______
(b-4b2+b4b-2)
= (b1rp1)(b
4rb2)
______
+ (b1rp1)
______
(b4rb2)
Let b1rp1=X and b
4rb2=Y, then c*
1= XY
-+ X-Y = XrY
Therefore, c*1= (b
1rp1) r (b
4rb2)
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- 3.41 -
3.35. (continued)
Using a similar procedure,
c*2
= (b1rp2) r (b
4rb3)
c*3
= (b2rp3) r (b
4rb3)
Each c*ican be realized with a network of the form:
To correct an appropriate bit, it is first necessary to
detect each of the seven combinations of c*3c*2c*1which can
be done with seven and-gates. To invert a bit if in error,
an exclusive-or-gate can be used. For example, to detect
c*3c*2c*1=011 and correct the corresponding bit, b
1, in
position 3, the following network can be used:
B1=b1if the output of the and-gate is 0 and B
1=b-1if the
output of the and-gate is 1.
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3.35. (continued)
Logic Diagram: