{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sol8 - Chapter 8 8.1 The expressions for the inputs of the...

This preview shows pages 1–6. Sign up to view the full content.

Chapter 8 8.1. The expressions for the inputs of the flip-flops are D 2 = Y 2 = wy 2 + y 1 y 2 D 1 = Y 1 = w y 1 y 2 The output equation is z = y 1 y 2 8.2. The excitation table for JK flip-flops is Present Flip-flop inputs state w = 0 w = 1 Output y 2 y 1 J 2 K 2 J 1 K 1 J 2 K 2 J 1 K 1 z 00 1 d 0 d 1 d 1 d 0 01 0 d d 0 0 d d 1 0 10 d 0 1 d d 1 0 d 0 11 d 0 d 1 d 1 d 0 1 The expressions for the inputs of the flip-flops are J 2 = y 1 K 2 = w J 1 = wy 2 + w y 2 K 1 = J 1 The output equation is z = y 1 y 2 8.3. A possible state table is Present Next state Output z state w = 0 w = 1 w = 0 w = 1 A A B 0 0 B E C 0 0 C E D 0 0 D E D 0 1 E F B 0 0 F A B 0 1 8-1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
8.4. Verilog code for the solution given in problem 8.3 is module prob8 4 (Clock, Resetn, w, z); input Clock, Resetn, w; output z; reg z; reg [3:1] y, Y; parameter [3:1] A = 3’b000, B = 3’b001, C = 3’b010, D = 3’b011, E = 3’b100, F = 3’b101; // Define the next state and output combinational circuits always @(w or y) case (y) A: if (w) begin Y = B; z = 0; end else begin Y = A; z = 0; end B: if (w) begin Y = C; z = 0; end else begin Y = E; z = 0; end C: if (w) begin Y = D; z = 0; end else begin Y = E; z = 0; end D: if (w) begin Y = D; z = 1; end else begin Y = E; z = 0; end E: if (w) begin Y = B; z = 0; end else begin Y = F; z = 0; end F: if (w) begin Y = B; z = 1; end else begin Y = A; z = 0; end default: begin Y = 3’bxxx; z = 0; end endcase 8-2
// Define the sequential block always @( negedge Resetn or posedge Clock) if (Resetn == 0) y < = A; else y < = Y; endmodule 8.5. A minimal state table is Present Next State Output state w = 0 w = 1 z A A B 0 B E C 0 C D C 0 D A F 1 E A F 0 F E C 1 8.6. An initial attempt at deriving a state table may be Present Next state Output z state w = 0 w = 1 w = 0 w = 1 A A B 0 0 B D C 0 0 C D C 1 0 D A E 0 1 E D C 0 0 States B and E are equivalent; hence the minimal state table is Present Next state Output z state w = 0 w = 1 w = 0 w = 1 A A B 0 0 B D C 0 0 C D C 1 0 D A B 0 1 8-3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
8.7. For Figure 8.51 have (using the straightforward state assignment): Present Next state state w = 0 w = 1 Output y 3 y 2 y 1 Y 3 Y 2 Y 1 Y 3 Y 2 Y 1 z A 0 0 0 0 0 1 0 1 0 1 B 0 0 1 0 1 1 1 0 1 1 C 0 1 0 1 0 1 1 0 0 0 D 0 1 1 0 0 1 1 1 0 1 E 1 0 0 1 0 1 0 1 0 0 F 1 0 1 1 0 0 0 1 1 0 G 1 1 0 1 0 1 1 1 0 0 This leads to Y 3 = wy 3 + y 1 y 2 + wy 1 y 3 Y 2 = wy 3 + w y 1 y 2 + wy 1 y 2 + wy 1 y 2 y 3 Y 1 = y 3 w + y 1 w + wy 1 y 2 z = y 1 y 3 + y 2 y 3 For Figure 8.52 have Present Next state state w = 0 w = 1 Output y 2 y 1 Y 2 Y 1 Y 2 Y 1 z A 0 0 0 1 1 0 1 B 0 1 0 0 1 1 1 C 1 0 1 1 1 0 0 F 1 1 1 0 0 0 0 This leads to Y 2 = wy 2 + y 1 y 2 + w y 2 Y 1 = y 1 w + wy 1 y 2 z = y 2 Clearly, minimizing the number of states leads to a much simpler circuit. 8-4
8.8. For Figure 8.55 have (using straightforward state assignment): Present Next state state DN=00 01 10 11 Output y 4 y 3 y 2 y 1 Y 4 Y 3 Y 2 Y 1 z S1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 - 0 S2 0 0 0 1 0 0 0 1 0 0 1 1 0 1 0 0 - 0 S3 0 0 1 0 0 0 1 0 0 1 0 1 0 1 1 0 - 0 S4 0 0 1 1 0 0 0 0 - - - 1 S5 0 1 0 0 0 0 1 0 - - - 1 S6 0 1 0 1 0 1 0 1 0 1 1 1 1 0 0 0 - 0 S7 0 1 1 0 0 0 0 0 - - - 1 S8 0 1 1 1 0 0 0 0 - - - 1 S9 1 0 0 0 0 0 1 0 - - - 1 The next-state and output expressions are Y 4 = Dy 3 Y 3 = Dy 1 + Dy 2 + Ny 2 + Dy 3 y 2 y 1 Y 2 = N y 2 + y 3 y 1 + N y 3 y 2 y 1 Y 1 = Ny 2 + D y 2 y 1 + D y 2 y 1 z = y 4 + y 1 y 2 + y 1 y 3 Using the same approach for Figure 8.56 gives Present Next state state DN=00

This preview has intentionally blurred sections. Sign up to view the full version.

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

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern