CSM51Asolution_chapter8

# Abd c a0 a c0 c bd bd a b1 a c d1 bcd figure 82

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: by the system assuming that A is the initial state: cb; cbb; cbc; cd; cdc. a,b,d c A/0 a C/0 c b,d b,d a B/1 a c D/1 b,c,d Figure 8.2: State digram for network on Exercise 8.4 122 Exercise 8.5 Solutions Manual - Introduction to Digital Design - February 22, 1999 The state diagram of the pattern recognizer for the sequence 0101011 is shown in Figure 8.3 and has seven states. Each state was labeled with the sequence that it recognizes". We encode these states using three state variables y2 ; y1 ; y0  so that the state assignment of state Si is the radix-2 representation of i. The correspondence between the state assignment and sequence that it detects is shown in the next table. 1/0 0/0 0/0 1/0 0/0 0/0 01 010 1/0 1/0 1/1 0/0 1/0 010101 1/0 0/0 0101 0/0 start 0 01010 Figure 8.3: State diagram for system in Exercise 8.5 State Sequence S0 start S1 0 S2 01 S3 010 S4 0101 S5 01010 S6 010101 The state and transition table is: PS Input x=0 x=1 001,0 000,0 001,0 010,0 011,0 000,0 001,0 100,0 101,0 000,0 001,0 110,0 101,0 000,1 S0 S1 S2 S3 S4 S5 S6 y2 y1y0 000 001 010 011 100 101 110 Y2 Y1Y0 ; z Solutions Manual - Introduction to Digital Design - February 22, 1999 123 The switching expressions for the next state and output are: 0 Y2 = y2y0x0 + y2y0 x + y1 y0 x 0 00 Y1 = y1y0 x + y2y1 y0x0 Y0 = x0 z = y2y1 x These expressions are implemented by AND-OR networks and the state is stored in a 3-bit register. The corresponding sequential network is shown in Figure 8.4. x’ y0’ y2 x y2 y0 x y1 y0 x’ x y1’ y0 x’ y2’ y1 y0’ Y2 Y1 x y2 y1 y0 z Y0 CK Figure 8.4: Network for Exercise 8.5 : The pattern generator for the sequence abcaba is described by the following transition table: PS NS z A Ba B Cb C Dc D Ea E Fb F Aa Let us de ne the following encoding: y2y1 y0 State 000 A z1 z0 001 B 00 a 010 C 01 b 011 D 10 c E 100 101 F Exercise 8.6 124 Solutions Manual - Introduction to Digital Design - February 22, 1999 From the state table and the previous encoding we get the following table and K-maps: PS NS z1 z0 000 001 00 001 010 01 010 011 10 011 100 00 100 101 01 101 000 00 101 - - - - 101 - - - - - y0 y2 Y2: y2 z1 : 0010 0 1 - -  y0 y2 y0 y1 0001 0 0 - -  Y1 : y2 z0 : 0 1 1 0 0 0 - -   y0 y0 The expressions we get from the K-maps are: 0 Y2 = y1y0 + y2y0 00 0 Y1 = y2y1 y0 + y1 y0 0 Y 0 = y0 0 z1 = y1y0 00 0 z0 = y2y1 y0 + y2 y0 The network that implements a cano...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online