Designing Circuits #
We’ve got a hallway with two lights.
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Build a 4 tall by 11 long wall. – //wand – build 4 tall tower – right click top – left click ground 11 blocks away – //set stone_bricks
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Switches at each end.
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Lights 3 apart in the middle of the “ceiling”.
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Throw redstone dots on blocks outside to pass through switch status
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Throw a “circuit breaker switch” in the middle.
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Show //copy, //paste when that’s built.
Three inputs: A, B, C
One output: Light
Rules:
- If C is off, the lights are off.
- If C is on, then:
- If both switches are off, the lights are off.
- Flipping either switch should always change the state of the lights.
Truth table:
| C | A | B | Lights |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |
Reading out sum of products:
- We can easily come up with an expression to output 1 only when we have an exact input: AND together the inputs that are 1 and the not of the inputs that are zero.
- If multiple inputs give a 1 output, OR those together.
- This is a standard form for boolean expressions, called Sum of Products form.
CA’B + CBA'
Let’s build that circuit.
Simplifying with Boolean Algebra #
If you’ve got a regular algebra expression like awx + ayz,
you can transform it into other equivalent expressions using
algebraic laws.
We can apply the distributive law, and undistribute the a
to get a(wx + yz).
Boolean algebra also has laws:
- Associativity of AND and OR:
- abc = a(bc) = (ab)c, etc
- a+b+c = a+(b+c) = (a+b)+c
- Commutativity of AND and OR:
- ab = ba ; a+b = b+a
- Distribution of AND over OR and OR over AND:
- a(b+c) = ab+ac
- a+bc = (a+b)(a+c)
- Identity:
- 1a = a
- 0+b = b
- Annihilation:
- 0a = 0
- 1+a = 1
- Idempotence:
- a+a = a
- aa = a
- Complementation:
- aa’ = 0
- a+a’ = 1
- De’Morgan’s Laws
- x’y’ = (x+y)'
- x’+y’ = (xy)'
So we can simplify our circuit by transforming CA'B + CBA' into
C(A'B + AB') (which law?)
Let’s build that.
Lab Notes #
- Start with a schematic that gives input and output widgets.
- Two two-bit inputs, one 3 bit output.
- Build a 2-bit adder.
- First, write down the formulas for each of the three output
bits as a function of the four input bits.
- Build a truth table.
- Pull out SoP form.
- Simplify with boolean algebra (if possible), to minimize the number of gates - possibly considering shared gates.
- Then build the circuit.
- Show it working for three cases.
- To submit:
- Truth tables.
- SoP form.
- Simplified form.
- Screenshot of the circuit.
- Screenshot of three cases.
