{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

CSM51Asolution_chapter8

# 29 network for exercise 826 a to design a cyclic

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: NS 01 0 0- 10 1 01 -0 SR 150 Solutions Manual - Introduction to Digital Design - February 22, 1999 we get the following switching expressions S2 = xQ1 R2 = xQ2 S1 = xQ02Q01 R1 = xQ1 The output is obtained directly from the state register. The sequential network is shown in Figure 8.29. Q2 S CK CK R x Q1 S CK CK R Q’ Q1’ Q Q’ Q2’ Q Figure 8.29: Network for Exercise 8.26 a To design a cyclic counter with the output sequence 0; 1; 3; 7; 6; 4; 0; 1; : : : we need six states. In this rst part, we select the coding so that the output corresponds to the state. The state table is Input PS x=0 x=1 000 000 001 001 001 011 011 011 111 100 100 000 110 110 100 111 111 110 NS = z Since the excitation function of a JK ip- op is PS NS 01 0 0- 11 -1 -0 JK Exercise 8.27 Solutions Manual - Introduction to Digital Design - February 22, 1999 151 we get the following ip- op inputs Q2 Q1 Q0 000 001 011 100 110 111 PS 000-0 -0 -0 x=0 00-0 0-0 -0 Input 0-0 -0 00-0 001-1 -0 -0 x=1 01-0 0-1 -0 J2 K2 J1 K1 J0 K0 1-0 -0 00-1 From K-maps we obtain J2 = xQ1 J1 = xQ0 J0 = xQ02 K2 = xQ01 K1 = xQ00 K0 = xQ2 The sequential network is shown in Figure 8.30. Recall that the output z = z2 ; z1 ; z0  corresponds to the state vector Q2 ; Q1 ; Q0 . z2 z1 J CK 0 K CK x Q’ Q J CK 1 K Q’ Q J CK 2 K Q’ Q z0 Figure 8.30: Network for Exercise 8.27 b In this second case, the state table is PS Input x=0 x=1 S0 S1 S2 S3 S4 S5 S0 S1 S2 S3 S4 S5 NS S1 S2 S3 S4 S5 S0 0 1 3 7 6 4 Output with the following encoding for the state: 152 Solutions Manual - Introduction to Digital Design - February 22, 1999 000 001 010 011 100 101 Using the previously given excitation function for a JK ip- op, the ip- ops inputs are PS Input Output Q2Q1 Q0 x=0 x=1 zt 000 0- 0- 0- 0- 0- 1- 000 001 0- 0- -0 0- 1- -1 001 0- -0 0- 0- -0 1- 011 010 011 0- -0 -0 1- -1 -1 111 -0 0- 0- -0 0- 1- 110 100 101 -0 0- -0 -1 0- -1 100 From K-maps we obtain S0 S1 S2 S3 S4 S5 Q2 Q1 Q0 J2 K2 J1 K1 J0 K0 For the output, J2 = xQ1 Q0 J1 = xQ02 Q0 J0 = x K2 = xQ0...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online