Encode the states, inputs, and outputs; create the transition diagram; create the transition table; express the logic equations; implement with flip-flops and gates.
\(\lceil \log_2 n \rceil\). Three or four states need 2 bits — and each bit needs its own flip-flop.
You must define what it does. If the machine powers up or glitches into an unhandled state code with no transition out, it locks up there permanently.
\(2^{\text{state bits} + \text{input bits}}\) — every combination, including the rows for unused state codes. Three states with one input give 8 rows, not 6: Lesson 24's 6 counted states, while 2 bits hold four codes. (A machine with many inputs is listed one row per (state, condition) pair in priority order instead — the Project 2 form.)
Treat each next-state bit column as the output of a truth table whose inputs are the current state bits and the inputs, then write its SOP expression — exactly as in Objective 2.7.
Copies its input \(D\) to its output \(Q\) at each clock edge, and holds it in between. That hold is the memory.
One per state bit. A machine encoded in 2 bits needs 2 flip-flops.
Next-state logic (combinational) reads current state and inputs; flip-flops latch the result on the clock edge; output logic produces the outputs. The feedback path from the flip-flops back to the next-state logic is what makes it a machine.
The clock. The machine advances exactly one transition per clock edge.
The output expression depends only on the state bits, with no input variable in it.
Hold the state in a variable, switch on it with one case per state, and inside each case test the inputs in a defined priority order. The switch replaces the next-state gates; the variable replaces the flip-flops.