1 / 16

CSE 140 Lecture 10 Sequential Networks: Implementation

Learn how to implement Mealy and Moore machines in sequential networks. Understand the format, tools, and procedures involved in designing these systems. Explore the advantages they offer.

jtrantham
Download Presentation

CSE 140 Lecture 10 Sequential Networks: Implementation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CSE 140 Lecture 10Sequential Networks: Implementation Professor CK Cheng CSE Dept. UC San Diego

  2. Implementation • Format and Tool • Procedure • Excitation Tables • Example

  3. Canonical Form: Mealy and Moore Machines x(t) y(t) Combinational Logic CLK x(t) C2 y(t) x(t) C1 C2 y(t) C1 CLK CLK

  4. Canonical Form: Mealy and Moore Machines Mealy Machine: yi(t) = fi(X(t), S(t)) Moore Machine: yi(t) = fi(S(t)) si(t+1) = gi(X(t), S(t)) x(t) x(t) C1 C2 y(t) C1 C2 y(t) CLK CLK s(t) s(t) Moore Machine Mealy Machine

  5. iClicker • The advantage of Moore machine over Mealy machine is that for Moore machine, • the circuit is smaller • the circuit is faster • the input is synchronized with clock • the output is synchronized with clock • None of the above

  6. Sequential Network Implementation:Format and Tool Canonical Form: Mealy & Moore machines State Table  Netlist Tool: Excitation Table x(t) C1 C2 y(t) CLK s(t) Q(t+1) = h(x(t), Q(t)) y(t) = f(x(t), Q(t))

  7. Implementation: Procedure Given a state table, we have NS: Q(t+1) = h(X(t),Q(t)) We want to derive D(t), T(t), (S(t) R(t)), (J(t) K(t)) as functions of (X,Q(t)). We implement D, T, (S R), (J K) as combinational logic. State Table => Excitation Table

  8. W NS PS PS NS Implementation: Procedure F-F State Table <=> F-F Excitation Table W • W: • D F-F • D(t)= eD(Q(t+1), Q(t)) • T F-F • T(t)= eT(Q(t+1), Q(t)) • SR F-F • S(t)= eS(Q(t+1), Q(t)) • R(t)= eR(Q(t+1), Q(t)) • JK F-F • J(t)= eJ(Q(t+1), Q(t)) • K(t)= eK(Q(t+1), Q(t))

  9. Implementation: Procedure • State table: y(t)= f(Q(t), x(t)), Q(t+1)= h(x(t),Q(t)) • Excitation table of F-Fs: • D(t)= eD(Q(t+1), Q(t)); • T(t)= eT(Q(t+1), Q(t)); • (S, R), or (J, K) • From 1 & 2, we derive excitation table of the system • D(t)= gD(Q(t), x(t))= eD(h(x(t),Q(t)),Q(t)); • T(t)= gT(Q(t), x(t))= eT(h(x(t),Q(t)),Q(t)); • (S, R) or (J, K). • Use K-map to derive optional combinational logic implementation. • T(t)= gT(Q(t), x(t)) • y(t)= f(Q(t), x(t))

  10. JK 00 0 1 11 1 0 10 1 1 01 0 0 0 1 Q(t+1) Q(t) Q(t+1) NS PS 0 0- -1 1 1- -0 0 1 Q(t) JK Excitation Table State table of JK F-F: Excitation table of JK F-F: If Q(t) is 1, and Q(t+1) is 0, then JK needs to be 0-.

  11. Excitation Tables and State Tables State Tables: Excitation Tables: SR SR Q(t+1) NS SR PS PS 0 0- 01 1 10 -0 00 0 1 01 0 0 10 1 1 11 - - 0 1 0 1 Q(t) Q(t) Q(t+1) T T Q(t+1) NS T PS PS 0 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 Q(t) Q(t) Q(t+1)

  12. Excitation Tables and State Tables Excitation Tables: State Tables: JK JK Q(t+1) NS JK PS PS 0 0- -1 1 1- -0 00 0 1 01 0 0 10 1 1 11 1 0 0 1 0 1 Q(t) Q(t) Q(t+1) D D Q(t+1) NS D PS PS 0 0 0 1 1 1 0 0 0 1 1 1 0 1 0 1 Q(t) Q(t) Q(t+1)

  13. iClicker • Given a flip-flop, the relation of its state table and excitation table is • One to one • One to many • Many to one • Many to many • None of the above

  14. J Q Q’ K C1 T Implementation: ExampleImplement a JK F-F with a T F-F Q(t+1) = h(J(t),K(t),Q(t)) = J(t)Q’(t)+K’(t)Q(t) State Table JK JK PS 00 0 1 01 0 0 10 1 1 11 1 0 0 1 Q(t)

  15. Example: Implement a JK flip-flop using a T flip-flop Excitation Table of T Flip-Flop T(t) = Q(t) XOR Q(t+1) Q(t+1) NS PS 0 0 1 1 1 0 0 1 Q(t) T Excitation Table of the Design id 0 1 2 3 4 5 6 7 J(t) 0 0 0 0 1 1 1 1 K(t) 0 0 1 1 0 0 1 1 Q(t) 0 1 0 1 0 1 0 1 Q(t+1) 0 1 0 0 1 1 1 0 T(t) 0 0 0 1 1 0 1 1 T(t) = Q(t) XOR ( J(t)Q’(t) + K’(t)Q(t))

  16. Example: Implement a JK flip-flop using a T flip-flop T(J,K,Q): K 0 2 6 4 0 0 1 1 T = K(t)Q(t) + J(t)Q’(t) 1 3 7 5 Q(t) 0 1 1 0 J J Q Q’ T K

More Related