hw3_solutions

hw3_solutions - 2.40. The minimum-cost SOP expression for...

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Unformatted text preview: 2.40. The minimum-cost SOP expression for the function f (x1 , x2 , x3 ) = m(3, 4, 6, 7) is f = x1 x3 + x2 x3 The corresponding circuit implemented using NAND gates is x1 Hx3 omework 3 Solutions ECE 152A Summer 2011 H.O. #8 f x2 2.41. A minimum-cost SOP expression for the function f (x1 , x2 , x3 ) = m(1, 3, 4, 6, 7) is f = x1 x2 + x1 x3 + x1 x3 The corresponding circuit implemented using NAND gates is 3.9. Vf V x 2x 3 x1 Vx 4 f x3 Vx 1 Vx 2 2.42. Minimum SOPcexpressionxfor fsiis n for the function f (x1 , x2 , x3 ) = 3.10. The minimum- ost POS e pres o m(3, 4, 6, 7) is = ( 3 + x1 3 3 x + x ) = f x2 xx1 + xx)(+2x2 x43+ x1 x4 f = (x + x )(x3 + x ) The corresponding circuit implemented using1NOR2gates is 4 which leads to the circuit Chapter 4 x1 VDD x3 f x2 4.1. SOP form: f = x1 x2 + x2 x3 POS form: f = (x1 + x2 )(x2 + x3 ) 4.2. SOP form: f = x1 x2 + x1 x3 + x2 x3 POS form: f = (x1 + x3 )(x1 + x2 )(x2 + x3 ) Vf Vx 2-13 1 4.3. SOP form: f = x1 x2 x3 x4 + x1 x2 x3 x4 + x2 x3 x4 POS form: f = (x1 + x4 )(x2 Vx3x3 )(x2 + x3 + x4 )(x2 + x4 )(x1 + x3 ) + Vx 4.4. SOP form: f = x2 x3 + x2 x4 +x2 2 x3 x4 POS form: f = (x2 + x3 )(x2 V x3 + x4 )(x2 + x4 ) + x4 44.5. IfOPcfh rm: uit=s xmp5e+ entx4 + parx4el5 :+ x1 x3 x4 x5 + x1 x2 x4 x5 .12 S ea o circ f i i 3 x l m x3 ed se x2 at x y f = f 1 x4 f = (x3 + x x 4 POS xorm:+ x1 x2 x3+ x41+2x5 )(x3Cosx= +5 5 )(x2 + x3 + x4 )(x1 + x3 + x4 + x5 )(x1 + x2 + x4 + x5 ) + t4 1 x g = x1 x3 SOP expression for f is Cost = 21 3.11. Minimumx4 + x2 x3 x4 + x1 x3 x4 + x1 x2 x4 4.6. SOPcfombinf d cixcui3:+ x1 x5 + x1 x3 + x3 x4 + x2x4 + x1 x2 x3 In a orm: e = r 2 x t f = x5 POS xormx4 + x1 x3+4x2 +1x3 )(3 x4+ x21+2x4 )(x3 + x4 + x5 ) f = f 2 x3 : f = (x1 x + x 2 x x1 + x x x3 g = x2 x3 x4 + x1 x3 x4 + x1 x2 x3 x4 + x1 x2 x4 The fi st 3 p to uc terms whichrleads rodthet circuit are shared, hence the total cost is 31. 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 ) 4.13. If each circuit is implemented separately: f = x1 x2 x4 + x2 x4 x5 + x3 x4 x5 + x1 x2 x4 x5 Cost = 22 4.8. f = x x (0,x )x + x x x + x x x + x x x g = 3 m + 74 5 Cost = 24 5 124 124 245 f = m(1, 6) 3-4 fn = combined ) ircuit: m(2, 5 c Ia f = x x x0,+, x )x x + x x x + x x x x f = 1 m( 4 1 62 4 5 2 345 1245 f = m( 2 5 ) g = x1 x2 x0,+,x2 x4 x5 + x3 x4 x5 + x1 x2 x4 x5 + x3 x5 4 etc. The first 4 product terms are shared, hence the total cost is 31. Note that in this implementation f ⊆ g , thus g = x1 x r x3 z d x x = f + x x , i x2 x ic 4 4.9. f can be 2eali+ex1as2g 4 + x1 x33x45+n wh3 xh case the total cost is lowered to 28. 4.14. f OP (x3m: g ) = ((1 x2 x3 + x4 ) wh4 re g 1 x3 x4 + (x2 2 x32+ x1((3 x4 ↑ xx) x3x2 ) 0 S = for ↑ f ↑ xg ↑ g ) ↑ x1 x2 x e + x = ( 1 ↑ x1 x↑ x )) ↑ x x1 + 1 2 ↑ 4 POS form: f = (x1 + x2 + x3 )(x1 + x2 + x4 )(x1 + x3 + x4 )(x2 + x3 + x4 )(x1 + x2 + x3 + x4 ) The POS form has lower cost. 4.15. f = (((x3 ↓ x3 ) ↓ g ) ↓ ((g ↓ g ) ↓ (x4 ↓ x4 )), where 4.11. The statemex1 ) s↓fal2 ) .↓As 1 co(x2te↓ example en,nfiderff↓x1., x2 , x3 ) = m(0, 5, 7). g = ((x1 ↓ nt i x se (xa ↓ un r x2 )). Th co s = ( f Then, the minimum-cost SOP form f = x1 x3 + x1 x2 x3 is unique. But, there are two minimum-cost POS forms: 4.16. f = (g ↑+ ) ↑)((g ↑ gx ↑ (x ↑ kx),)where g = (x1 ↑ x1 ) ↑ (x2 ↑ x2 ) ↑ (x5 ↑ x5 ) f = (x1 k x3 (x1 + ) 3 ) (k1 + ) 2 and and k = (x3 ↑ (x4 ↑ x4 )) ↑ ((x3 ↑ x3 ) ↑ x4 ) and k = (x3 ↑ (x4 ↑ x4 )) ↑ ((x3 ↑ x3 ) ↑ x4 ) 4.16. f = (g ↑ k ) ↑ ((g ↑ g ) ↑ (k ↑ k )), where g = (x1 ↑ x1 ) ↑ (x2 ↑ x2 ) ↑ (x5 ↑ x5 ) and k = (x3 ↑ (x4 ↑ x4 )) ↑ ((x3 ↑ x3 ) ↑ x4 ) 4.17. f = (g ↓ k ) ↓ ((g ↓ g ) ↓ (k ↓ k )), where g = x1 ↓ x2 ↓ x5 and k = ((x3 ↓ x3 ) ↓ x4 ) ↓ (x3 ↓ (x4 ↓ x4 )). Then, f = f ↓ f . 4.17. f = (g ↓ k ) ↓ ((g ↓ g ) ↓ (k ↓ k )), where g = x1 ↓ x2 ↓ x5 and k = ((x3 ↓ x3 ) ↓ x4 ) ↓ (x3 ↓ (x4 ↓ x4 )). Then, f = f ↓ f . 4.18. f = x1 (x2 + x3 )(x4 + x5 ) + x1 (x2 + x3 )(x4 + x5 ) 4.18. f = x1 (x2 + x3 )(x4 + x5 ) + x1 (x2 + x3 )(x4 + x5 ) 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. 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. 4.20. f = x1 · g + x1 · g , where g = x3 x4 + x3 x4 4.20. f = x1 · g + x1 · g , where g = x3 x4 + x3 x4 4.21 The assoc+tg · th,of theeXOR x1 xraaind ha= be shown as follows: 5.6. f = g · h ia ivi y wher g = ope2 t on c n x3 + x4 Chapter 5 4.21 f = g · h + g · h, where g = x1⊕2(ynd z )= =3 + x4 (y z + y z ) x a⊕h x x⊕ 4.22. Let D(0, 20) be 0 and D(15, 26) be 1. Then=ecomy z sitiy z )yield(y · z + y z ) d x( po + on + x s: g = x5 (x1 + x2 ) 45.1. Le)t Dx8 , 20+ be x4 )gd+ x1x, 26= be x4 g +ex=e4 g mpx3 x4on yieldsy · z + xy z .22. (a= 47 0x4 ) x3 0 an D( 3 5 4 g ) x3 1. Th n 3 x co + yositi g y z + x : d x· z+x f ( (3 g = 43 Costx5 (9 1 + 8 2 ) 27 (b) 7= x + 1 x = f = 02 x (c) 2(x35 4 + x3 x4 )g + x3 x4 gx ⊕ 3 x4⊕ + x= 4 g xyx3 x4y ) ⊕ z ( = x y) g z 3x ( + + x g Cost4=59 7 18 = 27 (d) 1 6 + = (x2 y + xy )z + (xy + xy )z 4- · ( e) 6 1 6 8 0 = x · y z + xy z + xy z + xy · z 4-2 The two SOP expressions are the same. 5.2. (a) 478 (b) −280 (c − c 5.7. In)the 1 ircuit of Figure 5.5b, we have: si 5.3. (a) 478 (b) −281 (c) −2 5 .4 . 5 .8 . 5 .5 . 5 .9 . = = (xi ⊕ yi ) ⊕ ci xi ⊕ yi ⊕ ci ci+1 = (xi ⊕ yi )ci + xi yi = (xi yi + xi y i )ci + xi yi The numbers are represented as follows: = xi yi ci + xi y i ci + xi yi Decimal Sign and Magnitude 1’s Complement+ xi’si Coxiplement 2 c + m yi = y i ci 73 000001001001 000001001001 000001001001 The expressions for1si1and 010 are the 1am11as0t010 e deri1110n11001e 5.4b. 1906 011 0 110 ci+1 01 s 10 e 1 hos 0 ved i1 Figur 0 −95 100001011111 111110100000 111110100001 −163 We will0give a 111cripti1e 110 of for 10011of10000rstand1ng. 10he 00s 0 omplement of a given number can des 0010v1 pro ease 0 unde 1 i 001 T 10 2’ 1 c be found by adding 1 to the 1’s complement of the number. Suppose that the number has k 0s in the leastsignificant bit positions, bk−1 . . . b0 , and it has bk = 1. When this number is converted to its 1’s complement, Tachrosultesof khbitsphaationsvare:e 1. Adding 1 to this string of 1s produces b k bk−1 bk−2 . . . b0 = 100 . . . 0. e he e f ths e t e o er s the alu This result is equivalent to copying the k 0s and the first 1 (in bit position b k ) encountered when the number bu 0 t 11 10 11 fi ( n 1 t 1 111 33) is(ac: nned00rom1r1ght to 54ft. (S):ppose 1hat0the1most-signi7 cantc): − k1bi0s,1b n−1 bn−(−. . bk , have some s)a f 110 i 0 le 2. + t 1 0 1 ’ 1 co − tt +wi1 10 co le −72 e pattern o+010and 01 , but6bk = 1. In 1he 11s 10 mplemen34 his pattern 10ll1be 00 mp+(ment) d in each bit f 0s 001 1s + 9 01ich wil 1 ud position, wh111011l incl23 e bk = 0. 010w0011 ing 1 to 8he entire n-b1t0numbe1 wil(l−1ake b k = 1, but No 1 , add t3 i 01011 r m 05) no further carries will be generated; therefore, the complemented bits in positions b n−1 bn−2 . . . bk+1 will da 0011011 5 4 ( e) : 01110101 (117) (f ): 11010011 (−45) re(m):in unchanged.0 −00101011 −43 −11010110 −(− 42) −11101100 −(−20) Construct 0he 0r10h 1able 11 t 00 t ut 1 t 10011111 (159) 11100111 (−25) xn−1 yn−1 cn−1 cn sn−1 (sign bit) Overflow 0 0 Arithmetic overflow occurs in example e; n0 te that0the pattern0 0011111 repr0 sents −97 rather than +159. o 1 e 0 0 1 0 1 1 0 1 0 0 1 0 0 1 1 1 0 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 0 1 0 1 1 1 1 1 1 0 Note that overflow cannot occur when two numbers with opposite signs are added. From the truth table the overflow expression is Ov erf low = cn cn−1 + cn cn−1 = cn ⊕ cn−1 5-1 5-2 ...
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