Unit 11 Solutions Unit 11 Problem Solutions 11.1 Z responds to X and to Y after 10 ns; Y responds to Z after 5 ns. See FLD p. 713 for answer. 11.3 P and Q will oscillate. See FLD p. 713 for timing chart. 11.4 11.5 11.2 See FLD p. 713 for solution. For part (b), also use the following Karnaugh map. Don’t cares come from the restriction in part (a). H Q See FLD p. 714 for solution. See FLD p. 714 for solution. R 0 1 00 0 X 01 0 X 11 1 1 10 0 1 Q+ = R + H Q 11.6 (a) SRQ Q+ 00 0 0 00 1 1 01 0 0 01 1 0 10 0 1 10 1 1 11 0 0 11 1 0 R Q S 11.6 (b) See FLD p. 714 for solution. 0 1 00 0 1 01 1 1 11 0 0 10 0 0 Q+ = R'Q + S R' 11.11 11.7 See FLD p. 714 for solution. 11.8 See FLD p. 714 for solution. 11.9 See FLD p. 715 for solution. 11.10 See FLD p. 715 for solution. 11.12 For every input/state combination with the condition SR = 0 holding, each circuit obeys the next-state equation Q+ = S + R'Q. When S = R = 1, in (a), both outputs are 1, and in (b), the latch holds its state. S R Q 11.13 (a) Present State Q 0 1 AB 00 0 0 Next State Q+ AB AB 01 11 0 1 1 1 11.13 (b) AB 10 0 1 A B Q P Q+ = AB + QA + QB Q+ = AB + Q(A + B) 11.13 (c) A change between AB = 01 and 10 can cause Q to change depending on the inverter delays. 11.13 (d) P = Q' + A'B' equals Q' in all stable states. 109 © 2010 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Unit 11 Solutions 11.13 (e) A B 11.13 (e) A change between AB = 01 and 10 can cause Q to change depending on the inverter delays. Q P = Q'(A' + B') equals Q' in all stable states. P Q+ = (AB + Q)(A + B) 11.14 (a) Present State Q 0 1 00 0 0 Next State Q+ A B 01 11 0 0 0 1 11.14 (b) & (c) 10 1 1 Q+ = A(B' + Q) This is a reset dominant latch where A´ acts a reset and B´ acts as a set. 11.15 (a) Q+ = (M + N + G)[Q + (M + N+ G)N'G'] = (M +N + G)[Q + N'G'] = (M + N + G)Q + MN'G' 11.15 (b) Q 0 1 000 0 0 001 0 1 011 0 1 GMN 010 100 1 0 1 1 101 0 1 111 0 1 110 0 1 The stable states are in bold. 11.15 (c) When G = 1, the circuit is always stable. When G = 0, M and N determine the state; N = 1 makes the state stable and with N = 0 the state becomes the value of M. There would be a restriction on M and N if they could cause both inputs to the output latch to be 1 when G = 0. This is not possible so there is no restriction. 11.15 (d) P = Q'[N + G + M'N'G'] = Q'[N + G + M'] For every stable state, P = Q' so P is usable as the complement of Q.. 11.16 (a) Q+ = AB + QB 11.16 (b) 11.16 (c) AB = 01 is a hold input combination, AB = 00 and 10 are reset input combinations, and AB = 11 is a set input combination. This is reset dominant latch where S = A and R = B'. P = Q' + B'. In each stable state P = Q' even for the input combination AB = 10 (SR = 11) so P is usable as Q'. Allowing the input combination AB = 10 (SR = 11) would result in unreliable operation if both A and B could change at the same time, i.e., change to AB = 01 (SR = 00), because the latch could end up in either state 0 or 1 depending upon the delays in the circuit. 110 Present State Q 0 1 00 0 0 Next State Q+ A B 01 11 0 1 1 1 10 0 0 The stable states are in bold. © 2010 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Unit 11 Solutions 11.17 (a) Q+ = R'(S + Q) if SR = 0 (b) Q+ = (G + Q)(G' + D) (c) Q+ = D (d) Q+ = (Q + CE)(CE' + D) (e) Q+ = (J + Q)(K' + Q') (f) Q+ = (T + Q)(T' + Q') 11.18 D G Q S R 11.19 11.20 (a) Clock SRQ 00 0 00 1 01 0 01 1 10 0 10 1 11 0 11 1 D Q P 11.20 (b) A set-dominant FF from an S-R FF—The arrangement will ensure that when S = R = 1, S1 = 1, R1 = 0, and Q+ = 1. S S1 Q R1 Q' 11.21 CK R Q+ 0 1 0 0 1 1 1 1 R Q S 0 1 00 0 1 01 1 1 11 0 1 10 0 1 Q+ = S + R'Q Clock S R Q 11.22 (a) Clock & (b) J K (a) Q (b) Q 111 © 2010 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Unit 11 Solutions 11.23 (a) Clock & (b) Q (a) D (b) T 11.24 Clock Q0 Q1 Q2 5 11.25 10 15 20 25 30 35 40 45 50 t (ns) Clock PreN T Q 11.26 11.27 (a) Clock D ClrN Q R Q' CK D Q1 When D = 0, then S = 0, and R = 1, so Q+ = 0. When D = 1, then S = 1, and R = 0, so Q+ = 1. Q2 11.27 (b) R will not be ready until D goes through the inverter, so we must add the delay of the inverter to the setup time: Setup time = 1.5 + 1 = 2.5 ns S 11.28 +V S Propagation delay for the DFF: 2.5 ns (same as for the S-R flip-flop, since the propagation delay is measured with respect to the clock) Q +V R 112 © 2010 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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