Lecture 14 Finite state machines

Transcription

1 Lecture 14 Finite state machines Finite state machines are the foundation of nearly all digital computation. The state diagram captures the desired system behavior A formulaic process turns this diagram into state table and then into a logic circuit. The example in these notes is taken from All About Circuits at See this site for additional explanations

2 From goal to states Intellectually most difficult step Goal: A circuit which sends a single high pulse when a button is pressed. Slightly tricky bit: The button could be held down for many clock cycles, but the circuit must not send out another pulse until the button is released and pressed again. So there are 3 states the circuit might be in: Waiting for button to be pressed Pressed! Sending out pulse Waiting for button to be released 141

3 State diagram Add transitions Now connect these states with arrows that indicate how the button value would move the system from state to state. These transitions will occur each clock cycle. Also define the starting (entering) state Not pressed You can only stay in this state one cycle since all transitions lead away. Start Waiting for button press. Send Low Pressed Not pressed Pressed! Send High Each state includes what to do for both button conditions. Not pressed Waiting for button release. Send Low Pressed Pressed If button release happens in the clock period immediately following the press event. 142

4 Turn it into numbers Boolean states and transitions Number states Wait, button not pressed, sending low = 00 Active, sending high = 01 Wait, button pressed, sending low = 10 Button pressed/not = 1/0 Send High/Low = 1/0 0 The state of the circuit, encoded as a 2 bit number. Will need flip-flops to remember this value. 1 Start The button value The output of the circuit

5 Create the state table 3 states 2 button values = 6 rows. Name variables State = A,B Input button = I Sent output = Y I AB Construct a state table that captures design Each row is one transition (arrow on diagram) Inputs are states and button condition at one time Outputs are circuit output and next state 3 input variables (A,B,I) 3 output variables (A next, B next,y) Y Don t care 144

6 Add flip-flops The circuit needs to remember what state it is in. This is a job for a flip-flop. J K Func Q next 0 0 Hold Q 0 1 Reset Set Toggle Q We need one for each of A,B. Add to state table as more outputs Four possible conditions of A/B and A next /B next Row A/B A/B next JK function J K 1A 0 0 Reset or Hold 0 X 4A 0 1 Set or Toggle 1 X 3B 1 0 Reset or Toggle X 1 6A 1 1 Set or Hold X 0 145

7 Karnaugh maps! J A = BI K A = I J B = AI K B = 1 Y = AB 146

8 Create the circuit Assuming no bounce I J A = BI Note use of ~Q for logic where ~A needed. K A = I Y = AB J B = AI K B =1 Clock The system will wait in state A,B,Y = 0,0,0 When the button is pressed, it will go to A,B,Y = 0,1,1 for one clock cycle. If the button is immediately released, the system will return to A,B,Y = 0,0,0. If the button is held, the system will go to A,B,Y = 1,0,0 until the button is released, then it will go to A,B,Y = 0,0,0 147

9 Quiz 14.1 Q: For the state diagram shown, what are the initial states and conditions that cause the turnstile to be unlocked A: State = locked, transition = coin B: State = locked, transition = push C: State = unlocked, transition = coin D: A and B E: A and C No matter what state you are in, if you put in a coin, the next state is unlocked. Why would somebody put a coin into an unlocked turnstile??? It doesn t matter. If the designer didn t plan for this condition, the device could freeze up and the London underground could come to a halt. Bad. 148

10 Quiz 14.2 Q: Assume the state is labeled locked=0 and unlocked=1. Assume the input is labeled push = 0 and coin = 1. Which is a row in the state table? A: Current = 0, Input = 0, Next = 1 B: Current = 1, Input = 0, Next = 1 C: Current = 0, Input = 1, Next = 0 D: Current = 1, Input = 1, Next = 0 E: Current = 0, Input = 0, Next = 0 Only locked, push = locked is a valid state transition. 149

11 Quiz 14.3 Q: Consider the state table row: Current = 0, Input = 1, Next = 1. Only one JK flip flop will be required for this device. What are the optimal J and K values to use in the design to implement this transition? A: J = 0, K = 0. B: J = 0, K = 1. C: J = 1, K = 0. D: J = 1, K = 1. E: J = 1, K = X. Reading from the JK flip flop truth table, if we start in 0 and want to end in 1, we can either set (J=1,K=0) or toggle (J=1, K=1), so E is the best choice. Both C and D would work, but are not optimal because X s on the Karnaugh map help us reduce the circuit implementation. 150