Hws05 - COCO 2610 Solution to Homework 5(2*3 = 6 A 1 4.19(c(d and(e(GJ Total score 2(d AB A 00 ~ co A W 0 0111 01 d 11 10,B 00 01 It 1 11 C 11c d

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Unformatted text preview: COCO 2610 . , Solution to Homework 5 (2'*3 = 6') A 1) 4.19 (c), (d), and (e) (GJ Total score: 2} (d) AB A 00 ~ co A. W. 0 0111:! 01 d 11 10 ;\,B 00 01 It 1 11 C 11c d [ 10 A. B c: B F=/t.' C'.D+ABC + A . B' 0 or B' C' . 0 [ 10 B D B F=A'.B+C'.D w (e, W'.Z WX y Z'\. 00 00 r 01 11 10 1 x Y' X.Z W'.y W'.X x F =W'Y +XZ 2) 4.20 for (c), (d), and (e) in 4.19 only (2'*3 = 6') (e) A , C+D rB'O 01 0 (d) , A A'+C A'+ C' N+ B'+ C A+C' - I-----11 0 d 0 d Io B 0 0 1110 I' B+D c [ 1lnOIll o1 11110 II 011 B 1110 lid B + C' C 101 a III B;C'F-IO F =(A + C') eN + B' + C) . (B + OJ F =(B + C') (C + 0) or (A + OJ (C + 0). eN + C'J or (B + C'). (C +0) (A'+ 0) CorB + C') . (B + D) (N + 0) ( Y+Z w (e) W' +Z W+X F = (X + YJ - (v',!' + Y') 3) (d) If:b1 (t\) 'tN'.y' \NX YZ (f-.) w (2'~2 ~If ) (e) o ~111110 01 0 III 0 I 0 1011 II 0 I 0 X X' l 0 AB co"- 00 01 11 10 00 A A' C D Y ["_11 I,'oi'.x,F 0 } Z 01 11111 B 0 B [ 10 ='yV'. Y' + IN'. x: +X'. FB+A'C-D 4) (7')Design a combinational circuit that has three inputs A, B, and C (one bit for each) and two outputs DOand D 1. The output should be the product of A and B if ( false, else the output is the sum of A and B a) produce the truth table for both outputs (2') b) give the canonical sum for each output (2') c) minimize each output expression independently if necessary using K-map(l') d) draw the logic diagram for the combinational circuit.(2') a) A 0 0 1 1 0 0 1 1 B 0 1 0 1 0 1 0 1 C 0 0 0 0 1 1 1 1 D1 0 0 0 0 0 0 0 1 DO 0 0 0 1 0 1 1 0 b) Dl =ABC DO=ABC'+A'BC+AB'C c) D1 = ABC DO= ABC'+A'BC+AB'C (same as in part b) d) A ='=' DC B c D ...
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This homework help was uploaded on 04/22/2008 for the course ECSE 2610 taught by Professor Ji during the Spring '08 term at Rensselaer Polytechnic Institute.

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Hws05 - COCO 2610 Solution to Homework 5(2*3 = 6 A 1 4.19(c(d and(e(GJ Total score 2(d AB A 00 ~ co A W 0 0111 01 d 11 10,B 00 01 It 1 11 C 11c d

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