Post-class 11/13

7 files changed, 88 insertions, 0 deletions

diff --git a/20.md b/20.md
index a3af7a4..0d30c32 100644
--- a/20.md
+++ b/20.md
@@ -156,3 +156,6 @@ reg [2:0] y, Y;
 parameter [2:0] A=3'b001, B=3'b010, C=3'b100;
 ```
 
+---
+
+[Multiple control inputs ->](21.md)
diff --git a/21.1.png b/21.1.png
new file mode 100644
index 0000000..eca6441
--- /dev/null
+++ b/21.1.png
diff --git a/21.2.png b/21.2.png
new file mode 100644
index 0000000..3025cee
--- /dev/null
+++ b/21.2.png
diff --git a/21.3.png b/21.3.png
new file mode 100644
index 0000000..13aa2b8
--- /dev/null
+++ b/21.3.png
diff --git a/21.4.png b/21.4.png
new file mode 100644
index 0000000..ebc0050
--- /dev/null
+++ b/21.4.png
diff --git a/21.5.png b/21.5.png
new file mode 100644
index 0000000..dd19944
--- /dev/null
+++ b/21.5.png
diff --git a/21.md b/21.md
new file mode 100644
index 0000000..7449cbe
--- /dev/null
+++ b/21.md
@@ -0,0 +1,85 @@
+[\<- 1-hot encoding and State machines in Verilog](20.md)
+
+---
+
+# Multiple control inputs
+
+## Defining and starting the vending machine example
+
+### Adding control inputs
+
+- So far, we've seen just one control input
+- Next example is a vending machine
+	- Items cost 15 cents and the machine accepts both nickles (N) and dimes (D)
+	- Need to track (via states) how much money has been put in, and whether there is change
+- Every state needs to account for all feasible input combinations
+	- Doesn't necessarily mean a full enumeration of all possible input encodings, but solution space should be complete
+
+### An arc for every input combo
+
+- Start with state A, representing no money
+- Stay in A as long as no money inserted
+- N goes to a state (B) representing 5 cents
+- D goes to a state (C), for 10 cents
+- Note that N\*D can't happen, not in diagram
+	- N really means N\*!D, D really means D\*!N
+
+![diagram](21.1.png)
+
+---
+
+## Thinking thru the rest of the vending machine state diagram
+
+### Now evaluate state B
+
+- Adding possibilities from B results in adding state D, which enables output
+- Note that state C represents 10 cents, whether we got there from a dime or two nickels
+
+![diagram](21.2.png)
+
+### Evaluating state C
+
+- A dime inserted in state C introduces an interesting question: what to do?
+	- Decision here: give them 5 cents credit
+		- Needs a new state (E) to keep track of this
+
+![diagram](21.3.png)
+
+### Finishing Up
+
+- Once a selection is enabled (i.e., the output is asserted), we can unconditionally transition to the next state
+	- State A from state D
+	- State B from state E, to give 5 cents credit
+
+---
+
+## State table and next state equations for 1-hot encoding
+
+### Alternate notation
+
+- Different state names, but same function
+- Different view of the uncoditional transfers
+
+![diagram](21.4.png)
+
+### The state table
+
+- Without assignments, for now
+	- As a reminder: S1=0 cents, S2=10 cents, S3=5 cents, S4=15 cents, S5=20 cents
+- Fully encoded states would need 3 flops
+	- 5 states -> need 3 bits to get 5 unique states
+- 1-hot encoding would need 5 flops
+
+![diagram](21.5.png)
+
+### Next state equations
+
+- State table starts to get hard to manage
+- Next state is the sum of the arcs leading to the state
+- With 1-hot encoding, not too bad:
+	- `D1 = Q1*!D*!N + Q4`
+	- `D2 = Q1*D + Q2*!D*!N + Q3*N`
+	- `D3 = Q1*N + Q3*!D*!N + Q5`
+	- `D4 = Q2*N + Q3*D`
+	- `D5 = Q2*D`
+- Getting more complex than this is where Verilog really helps