hw2_solutions

hw2_solutions - Homework 2 Solutions ECE 152A Summer 2011...

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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 lowest-cost 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; 2-12 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 3-4 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 ) 3-3 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 minimum-cosx SOP form f = x1 x3 + x1 x2 x3 is unique. t But, there are two minimum-cost 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 ) 4-1 4-2 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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