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

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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
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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
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// 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
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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
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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
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This note was uploaded on 03/26/2008 for the course ECEN 248 taught by Professor Lu during the Spring '08 term at Texas A&M.

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

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