Unformatted text preview: Homework 2 Solutions ECE 152A
Summer 2011
H.O. #6 2.25. The simplest SOP expression for the function is
f = x1 x3 x5 + x1 x3 x4 + x1 x4 x5 + x1 x2 x3 x5
= x1 x3 x5 + x1 x3 x4 + x1 x4 x5 + x1 x3 x5 + x1 x2 x3 x5
= x1 x3 + x1 x3 x4 + x1 x4 x5 + x1 x2 x3 x5
= x1 x3 + x1 x4 x5 + x1 x2 x3 x5
= x1 x3 + x1 x4 x5 + x2 x3 x5 2.26. The simplest POS expression for the function is
f =
= (x1 + x3 + x4 )(x2 + x3 + x4 )(x1 + x2 + x3 )
(x1 + x3 + x4 )(x2 + x3 + x4 )(x1 + x2 + x3 )(x1 + x2 + x3 ) = (x1 + x3 + x4 )(x2 + x3 + x4 )(x2 + x3 )
= (x1 + x3 + x )(x2 + x )
2.38. Using the ciruit in Figure 2.27b as a sta4 ting poin3, the function in Figure 2.26 can be implemented using
r
t
NOR gates as follows:
2.27. The simplest POS exx 3ression for the function is
p
x2 = (xx 1+ x3 + x5 )(x1 + x3 + x5 )(x1 + x2 + x5 )(x1 + x4 + x5 )
2 =
= (x2 + x3 + x5 )(x1 + x3 + x5 )(x1 + x2 + x5 )(x1 + x2 + x5 )(x1 + x4 + x5 )
(x2 + x3 + x5 )(x1 + x3 + x5 )(x1 + x5 )(x1 + x4 + x5 ) =
= (x2 + x3 + x5 )(x1 + x5 )(x1 + x5 (x4 + x5 ))
(x2 + x3 + x5 )(x1 + x5 )(x1 + x5 x4 ) = f (x2 + x3 + x5 )(x1 + x5 )(x1 + x4 ) f 2.28. The lowestcost circuit is deﬁned by
f (x1 , x2 , x3 ) = x1 x2 + x1 x3 + x2 x3 2
f i n u on, f, of th s c 3 ca t b m a o 0 ted n ei g er none of t NOR ga o al t lree i :
2.39. The curcctit in Figure i2.3ircuin iseeiqupletmenwheusinthNAND and he inputstesr as lfohlowsnputs are equal to 0;
otherwise, f is equal to 1. Therefore, using the POS form, the desired circuit can be realized as
x1 f (x1 , x2 , x3 ) = ΠM(0, 3) = x1
2.48. The simplest cirx 3uit is obtained in the POS form(as + x2 + x3 )(x1 + x2 + x3 )
c
g f = (x1 + x2 + x3 )(x1 + x2 + x3 ) x2
x
Verilog code that4implements the circuit is 2.30. The circuit can be implemented as
f =
=
= f mod + x pr 2 b 3 4 ( 1, x2, x3, +
x1 x2 x3 x4 + x1 x2 x3 x4 ule 1 xox2x48+xx1 x2 x3 x4 f); x1 x2 x3 x4
t 1 x2, x3;
x1 x2 x3 (x4 + x4 ) + xinpux3x+, x3 )x4 + x1 (x2 + x2 )x3 x4 + (x1 + x1 )x2 x3 x4
1 x2 (
output f;
x1 x2 x3 + x1 x2 x4 + x1 x3 x4 + x2 x3 x4
or (g, x1, x2, x3);
or (h, ∼x1,8∼x2, ∼x3);
h
2and (f, g, h);
endmodule 2.49. The simplest circuit is obtained in the SOP form as
f = x2 + x1 x3 + x1 x3
Verilog code that implements the circuit is
module prob2 49 (x1, x2, x3, f);
input x1, x2, x3;
output f; 212
assign f = ∼x2  (∼x1 & x3)  (x1 & ∼x3);
endmodule (b) The canonical SOP expression is
V
x2 f = x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3
The number of transistors required using only AND, OR, and NOT gates is
#transistors = NOT gates × 2 + AND gates × 8 + OR gates × 12
3.10. Minimum SOP expression for f is
= 3 × 2 + 5 × 8 + 1 × 12 = 58
f = x2 x3 + x1 x3 + x2 x4 + x1 x4
= (x1 + x2 )(x3 + x4 )
3.7. (a)
f
x1 x2 x3 x4
f
x
which leads to the circuit 1 x 2 x 3 x 4 0
0
0
0
0
0
0
0 0
0
0
0
1
1
1
1 0
0
1
1
0
0
1
1 0
1
0
1
0
1
0
1 1
0
0
0
1
0
0
0 1
1
1
1
1
1
1
1 0
VDD0
0
0
0
1
1
1
1 0
1
1
0
0
1
1 0
1
0
1
0
1
0
1 1
0
0
0
0
0
0
0 Vf (b)
Vx
f 1 Vx 3 = x1 x2 x3 x4 + x1 x2 x3 x4 + x1 x2 x3 x4
= x1 x3 x4 + x2 x3 x4 The number of transistors required using only AND, OR, and NOT gates is
Vx 2 #transistors = NOT gates × 2 + AND gates × 8 + OR gates × 4
Vx 4
= 4 × 2 + 2 × 8 + 1 × 4 = 28 3.8.
3.11. Minimum SOP expression for f is Chapter 4 4.12. If each circuit is implemented separately: f = x4 + x1 VDD3
x2 x
f = x1 x4 + x1 x2 x3 + x1 x2 x4
Cost= 15
g = x1 x3 x4 + x2 x3 x4 + x1 x3 x4 + x1 x2 x4
Cost = 21 which leads to the circuit
Vx
In a combined circuit:
4.1. SOP xormx f + x 1 x2 ++ 2 x3 x x1 + x x x
f = f 2 x3 : 4 = x x3 x4 x x1 x2 3 4
1
123
POSxoxmx f+ x(x1 x x2 )xxx + x3 ) + x x x
f r : = x++( 2 xx
g= 2 3 4
134
1234
124
VDD
Vx
3, hence the total cost is 31.
The ﬁrst 3 product terms are shared
Vx 4.2. SOP form: f = x1 x2 + x1 x3 + x22x3
34
P ea o circ f i im le xente 1 + x rate y+
4.13. IfOScfh rm: uit=s (x1p+ m3 )(xd sepa2 )(xl2 : x3 ) Vf f = x1 x2 x4 + x2 x4 x5 + x3 x4 x5 + x1 x2 x4 x5
Cost = 22
g = f3 r 5 + 4 5 x x x 4 x x x x + x x x
4.3. SOPxoxm: fx=xx1+2x13x24x++1x12x23x44 + x22x43x54 Cost = 24
POS form: f = (x1 + x4 )(x2 + x3 )(x2 + x3 + x4 )(x2 + x4 )(x1 + x3 )
In a combined circuit:
f = x1 x2 x4 + x2 x4 x5 + x3 x4 x5 + x1 x2 x4 x5
g = f1 r 2 : f = 2 x4 x + x x3 x + x + x1 x
4.4. SOPxoxmx4 + xx2 x3 5 + 2 x4 4 x5 2 x3 x4 2 x4 x5 + x3 x5
POS form: f = (x2 + x3 )(x2 + x3 + x4 )(x2 + x4 )
33
The ﬁrst 4 product terms are shared, hence the total cost is 31. Note that in this implementation f ⊆ g , thus
g can be realized as g = f + x3 x5 , in which case the total cost is lowered to 28.
4.5. SOP form: f = x3 x5 + x3 x4 + x2 x4 x5 + x1 x3 x4 x5 + x1 x2 x4 x5
Vf
POS form: f = (x3 + x4 + x5 )(x3 + x4 + x5 )(x2 + x3 + x4 )(x1 + x3 + x4 + x5 )(x1 + x2 + x4 + x5 )
Vx
4
4.14. f = (x3 ↑ g ) ↑ ((g ↑ g ) ↑ x4 ) where g = (x1 ↑ (x2 ↑ x2 )) ↑ ((x1 ↑ x1 ) ↑ x2 ) 4.6. SOP form: f = x2 x3 + x1 x5 + x1 x3 + x3 x4 + x2 x5
Vx
1
POS form: f = (x1 + x2 + x3 )(x1 + x2 + x4 )(x3 + x4 + x5 )
Vx 4.15. f = (((x3 ↓ x3 ) ↓ g ) ↓ ((g ↓ 2g ) ↓ (x4 ↓ x4 )), where
g = orm: x ) x x2 ) ↓ x V3 (x2 x2 1 x x5 n f x x f x
4.7. SOP(f(x1 ↓ f 1= ↓ 3 x4 x5 (+1x↓3x4 x5↓+ x)). 4The+, x1=2f 4↓+ . 3 x4 x5 + x2 x3 x4 + x2 x3 x4 x5
x
POS form: f = (x3 + x4 + x5 )(x3 + x4 + x5 )(x1 + x2 + x3 + x4 + x5 ) Chapter 4 4.16. f = (g ↑ k ) ↑ ((g ↑ g ) ↑ (k ↑ k )), where g = (x1 ↑ x1 ) ↑ (x2 ↑ x2 ) ↑ (x5 ↑ x5 )
ad
(( 3 7 (
4.8. fn= k =mx0, ↑ ) x4 ↑ x4 )) ↑ ((x3 ↑ x3 ) ↑ x4 )
f=
m(1, 6)
m(2, 5)
3.12. f =
4.17. f = (g ↓ k ) ↓ ((g ↓ g ) ↓ (k ↓ k )), where g = x1 ↓ x2 ↓ x5
4.1. SOP form(0, 1, 6)1 x2 + x2 x3
m: f = x
f=
and k = ((x3 ↓ x3 ) ↓ x4 ) ↓ (x3 ↓ (x4 ↓ x4 )). Then, f = f ↓ f .
VDD
VDD
P=
m: f = (x
f OS form(0, 2, 5) 1 + x2 )(x2 + x3 )
etc.
4.18. f = x1 (x2 + x3 )(x4 + x5 ) + x1 (x2 + x3 )(x4 + x5 )
4.2. SOP form: f = x1 x2 + x1 x3 + x2 x3
P = f 1 x2 : f = (x1 x + ( 1 1 + x2 ) x2 3 x4
4.9. f OS xormx3 + x1 x2+4x3 )xxx3 x4 + (x2x+ x3 )
4.19. f = x1 x3 x4 + x2 x3 x4 + x1 x3 x4 + x2 x3 x4 = (x1 + x2 )x3 x4 + (x1 + x2 )x3 x4
This requires 2 OR and 2 AND gates.
44.3. SOP form: f = x1 x2 x3 x4 x1 x1 x2 x3 x41 x3 x4 x3 x41 x2 x3 + x1 x3 x4 + x2 x3 x4
.1 0
++ 2 4+x + 2+x
POS form: f = (x1 + x4 )+ x3+(x1 )+ x2+ xx4+(x1 )+ x3+ xx4(xx2+ xx3 + x4 )(x1 + x2 + x3 + x4 )
2 (x2 ) 3 (x2 + 3 ) 4 (x2 + 4 ) )( 1 + 3 )
T = x1 · f + x h · g , wh r e g .
4.20. f he POS g orm 1 as loweercost= x3 x4 + x3 x4 Vf
4.4. SOP form: f = x2 x3 + x2 x4 + x2 x3 x4
P = t · m:
411. f he g are +fg ·i hfal ee As)a count2 r exah )ple2 3o+sid
wh . e = 1 e n 4 (
4..21 TOSsfothment=s(,x2s+rx3g (x2x+xx3a+dxm=xxc+nxx)er f (x1 , x2 , x3 ) =
m(0, 5, 7).
44
V
Then, the minimumcosx SOP form f = x1 x3 + x1 x2 x3 is unique.
t
But, there are two minimumcost POS forms:
44.5. LOP (x0m:0f bex 3a+d+yx15,4+6xx2ea4.xThen dxcomx5 sitix1 x2ix4ds5
.22 S e=Dor ,+ x = x x5 xD( 3 x 2+) b) x1 d5 + x1 e 3 x4 po + on y el x :
t f( 2 ) )( 0 n V )(x
f
n
1
3
1
3
1
2
POS form: f =2 )x + x4 + x5 )(x3 + x4 + x5 )(x2 + x3 + x4 )(x1 + x3 + x4 + x5 )(x1 + x2 + x4 + x5 )
g = (5 + x ) ( +
f = xx(x1 +3x(x1 3 x3 )(x2 + x3 )
1
f = (x3 x4 + x3 x4 )g + x3 x4 g = x3 x4 g + x3 x4 g + x3 x4 g
Cost = 9 + 18 = 27 V
4.6. SOP form: f = x2 x3 +zx1 x5 + x1 x3 + x3 x4 + x2 x5
POS form: f = (x1 + x2 + x3 )(x1 + x2 + x4 )(x3 + x4 + x5 )
41
42 4.7. SOP form: f = x3 x4 x5 + x3 x4 x5 + x1 x4 x5 + x1 x2 x4 + x3 x4 x5 + x2 x3 x4 + x2 x3 x4 x5
POS form: f = (x3 + x4 + x5 )(x3 + x4 + x5 )(x1 + x2 + x3 + x4 + x5 ) ...
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