Sample Design: A Controller for a Simple Traffic Light | CPSC 5155G, Study notes of Computer Architecture and Organization

Download Sample Design: A Controller for a Simple Traffic Light | CPSC 5155G and more Study notes Computer Architecture and Organization in PDF only on Docsity! CPSC 5155 Chapter 7 Slide 1 of 17 slides Sample Design: A Controller for a Simple Traffic Light CPSC 5155 Chapter 7 Slide 2 of 17 slides Assumption: Two Linked Pairs of Traffic Lights If one light is Green, the “cross light” must be Red. CPSC 5155 Chapter 7 Slide 5 of 17 slides Step 1a: State Diagram for the System Notation: L1L2, so RG  Light 1 is Red and Light 2 is Green The six–state design is more easily implemented. CPSC 5155 Chapter 7 Slide 6 of 17 slides Step 1b: Define the State Table Present State Next State Number Alias Number Alias 0 RR 1 RG 1 RG 2 RY 2 RY 3 RR 3 RR 4 GR 4 GR 5 YR 5 YR 0 RR At the moment, this is just a modulo–6 counter with unusual output. We shall add some additional circuitry to allow for safety constraints. The choice of Red – Red as state 0 is arbitrary, but convenient. CPSC 5155 Chapter 7 Slide 7 of 17 slides Step 2: Count the States and Determine the Flip–Flop Count There are six states, so we have N = 6. Solve 2P–1 < N  2P for P, the number of flip–flops. 2P–1 < 6  2P gives P = 3, because 22 < 6  23. We denote the states by Q2Q1Q0, because the symbol “Y” is taken to indicate the color Yellow. CPSC 5155 Chapter 7 Slide 10 of 17 slides Step 4a: Derive the Output Equations. Alias Q2Q1Q0 R1 G1 Y1 R2 G2 Y2 0 RR 0 0 0 1 0 0 1 0 0 1 RG 0 0 1 1 0 0 0 1 0 2 RY 0 1 0 1 0 0 0 0 1 3 RR 0 1 1 1 0 0 1 0 0 4 GR 1 0 0 0 1 0 1 0 0 5 YR 1 0 1 0 0 1 1 0 0 6 RR 1 1 0 1 0 0 1 0 0 7 RR 1 1 1 1 0 0 1 0 0 Here are the output equations G1 = Q2Q1’Q0’ G2 = Q2’Q1’Q0 Y1 = Q2Q1’Q0Y2 = Q2’Q1Q0’ R1 = (G1 + Y1)’ R2 = (G2 + Y2)’ CPSC 5155 Chapter 7 Slide 11 of 17 slides Step 4a: Derive the Output Equations. (page 2) Here are the equations again. G1 = Q2Q1’Q0’ G2 = Q2’Q1’Q0 Y1 = Q2Q1’Q0 Y2 = Q2’Q1Q0’ R1 = (G1 + Y1)’ R2 = (G2 + Y2)’ We derive the Green and Yellow signals, which are easier. We stipulate that if a light is not Green or Yellow, it must be Red. Now add a safety constraint: If a light is Green or Yellow, the cross light must be Red. R1 = (G1 + Y1)’ + G2 + Y2, and R2 = (G2 + Y2)’ + G1 + Y1 These equations may lead to a light showing two colors. This is obviously an error situation. CPSC 5155 Chapter 7 Slide 12 of 17 slides Step 4b: Derive the State Transition Table. Present State Next State Q2Q1Q0 Q2Q1Q0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 2 0 1 0 0 1 1 3 0 1 1 1 0 0 4 1 0 0 1 0 1 5 1 0 1 0 0 0 6 1 1 0 0 0 0 7 1 1 1 0 0 0 CPSC 5155 Chapter 7 Slide 15 of 17 slides Step 7: Derive the Input Tables Flip-Flop 2 Flip-Flop 1 Flip-Flop 0 PS NS Input PS NS Input PS NS Input Q2Q1Q0 Q2 J2 K2 Q2Q1Q0 Q1 J1 K1 Q2Q1Q0 Q0 J0 K0 0 0 0 0 0 d 0 0 0 0 0 d 0 0 0 1 1 d 0 0 1 0 0 d 0 0 1 1 1 d 0 0 1 0 d 1 0 1 0 0 0 d 0 1 0 1 d 0 0 1 0 1 1 d 0 1 1 1 1 d 0 1 1 0 d 1 0 1 1 0 d 1 1 0 0 1 d 0 1 0 0 0 0 d 1 0 0 1 1 d 1 0 1 0 d 1 1 0 1 0 0 d 1 0 1 0 d 1 1 1 0 0 d 1 1 1 0 0 d 1 1 1 0 0 0 d 1 1 1 0 d 1 1 1 1 0 d 1 1 1 1 0 d 1 CPSC 5155 Chapter 7 Slide 16 of 17 slides Step 8: Derive the Input Equations Here they are J2 = Q1  Q0 J1 = Q2’  Q0 J0 = Q2’ + Q1’ K2 = Q1 + Q0 K1 = Q2 + Q0 K0 = 1 There is no need to summarize the equations. CPSC 5155 Chapter 7 Slide 17 of 17 slides Step 10: Draw the Circuit