CSM51Asolution_chapter8

CSM51Asolution_chapter8

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Unformatted text preview: d to activate the set reset inputs of an SR latch. C represents the clock input. Exercise 8.14 The table for the combinational network is shown next: Solutions Manual - Introduction to Digital Design - February 22, 1999 135 SR cell J Comb. Circ. K C R S SR Cell A Q’ Q Figure 8.17: Gate implementation of an edge-triggered JK ip- op C 0 1 1 1 1 1 1 1 1 It is easy to see that J 0 0 0 0 1 1 1 1 K 0 0 1 1 0 0 1 1 Q 0 1 0 1 0 1 0 1 S 0 0 0 0 1 1 0 R 0 0 1 0 0 0 1 S = Q0 CJ R = QCK The combinational network is composed of two AND gates only. The timing diagram for the network is shown in Figure 8.18. The timing diagram of the internal signal A, the output of the master module, is also shown. J K C Q A Figure 8.18: Timing diagram for Exercise 8.14 Exercise 8.15 The solution of this exercise is similar to Exercise 8.14, making J = K = T . 136 Exercise 8.16 Solutions Manual - Introduction to Digital Design - February 22, 1999 From the network we obtain the following table, based on the expressions below: JA = KA = xQB JB = KB = x QAQB x = 0 x = 1 x = 0 x = 1 00 01 10 11 0000 0000 0000 0000 0011 1111 0011 1111 00 01 10 11 01 10 11 00 PS Input Input JAKA JB KB NS The outputs are expressed as: z3 = QAQB z2 = QAQ0B z1 = Q0AQB z0 = Q0AQ0B State Name Code S0 00 S1 01 S2 10 11 S3 Giving the following names to the states: we get the transition table: Input Output QA QB x = 0 x = 1 S0 S0 S1 0 S1 S1 S2 1 S2 S2 S3 2 S3 S3 S0 3 NS The state diagram for the given network is presented in Figure 8.19, and corresponds to a modulo-4 counter with decoded output. Exercise 8.17 PS The expression for the ip- op inputs are JA = xQC KA = xQ0B JB = Q0A KB = QA JC = xQB KC = x0 Q0B Solutions Manual - Introduction to Digital Design - February 22, 1999 137 0 1 0 S0/0 S1/1 1 1 S3/3 1 S2/2 0 0 Figure 8.19: State diagram for Exercise 8.16 From the characteristic expressions of the JK ip- op we get the following expressions for the transition functions: QAt + 1 QB t + 1 QC t + 1 z = = = = QAtx0 + QB t + xQ0AtQC t QB tQ0A t + Q0B tQ0At QC tQB t + x + Q0C txQB t Q0C t The corresponding transition table is PS Input Input Output QAQB QC 000 001 010 011 100 101 110 111 00,10,01 00,10,01 00,10,00 00,10,00 00,01,01 00,01,01 00,01,00 00,01,00 x=0 JA KA; JB KB ; JC KC 01,10,00 11,10,00 00,10,10 10,10,10 01,01,00 11,01,00 00,01,10 10,01,10 x=1 x=0 x=1 010 010 010 011 100 100 100 101 010 111 011 111 000 001 101 101 z NS 1 0 1 0 1 0 1 0 To get a high-level description w...
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This note was uploaded on 03/26/2010 for the course CS 187154200 taught by Professor Ercegovac,m.d. during the Winter '09 term at UCLA.

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