Create a 5-bit signed comparator

Question

In Cirsim below, create a 5-bit signed comparator. The inputs are two 5-bit bus inputs A and B. The outputs are three single-bit outputs: G, L, and E. The output G is true if A>B. the output E is true if A=B, and the output L is true if A<B. Note that this is a signed comparison, not an unsigned comparison. The rule to determining the result will be different than it would be for an unsigned comparison.

Below you will see three tabs: main, sub, and adder. You are free to use the tabs if you wish but are not required to. However, it is likely going to be much easier to solve this problem using the tabs than not.

There are 3 tabs: adder, sub, and main. It is in cirsim and you can use the following gates to solve the problem:

0 block, 1 block, IN pin, OUT pin, AND gate, NAND gate, OR gate, NOR gate, XOR gate, Inverter, button, LED, CLK, Bus input, Bus output, Fm Bus, To bus. you can also use the adder and subtractor component box once it is manually made in the adder and sub tab.

I am looking for a diagram (3 different diagrams) drawing this out

Answer 1

Hey Lauren! Here are the suggested steps that you can follow:

Main Comparator Circuit

Objective: Determine if A > B, A = B, or A < B.

Components:

1. Two 5-bit bus inputs (A and B)
2. Subtractor component (output: Difference, Sign Bit)
3. Logic gates for outputs G (A > B), E (A = B), L (A < B)

Logic:

1. Subtract B from A:
2. If the result is positive and not zero, G (A > B) is true.
3. If the result is zero, E (A = B) is true.
4. If the result is negative, L (A < B) is true.

Adder Tab

Objective: (Optional) Add two 5-bit signed numbers.

Components:

1. Two 5-bit bus inputs (A and B)
2. Adder component (output: Sum)

Subtractor Tab

Objective: Subtract B from A using two's complement.

Components:

1. Two 5-bit bus inputs (A and B)
2. Subtractor logic (output: Difference, Overflow)

Circuit Diagrams

1. Main Tab

Components:

1. Bus Input A[4:0]
2. Bus Input B[4:0]
3. Subtractor Output to:
4. Logic for G (A > B)
5. Logic for E (A = B)
6. Logic for L (A < B)

Diagram:

Bus Input A[4:0] ---> Subtractor ---> Result

Bus Input B[4:0] ---> Subtractor ---> Result

Subtractor Output ---> Logic Gates ---> G, L, E

2. Adder Tab

Components:

1. Bus Input A[4:0]
2. Bus Input B[4:0]
3. Adder Output (Sum[4:0])

Diagram:

Bus Input A[4:0] ---> Adder ---> Sum[4:0]

Bus Input B[4:0] ---> Adder ---> Sum[4:0]

3. Subtractor Tab

Components:

1. Bus Input A[4:0]
2. Bus Input B[4:0]
3. Subtractor Output (Difference[4:0], Sign Bit)

Diagram:

Bus Input A[4:0] ---> Subtractor ---> Difference[4:0]

Bus Input B[4:0] ---> Subtractor ---> Difference[4:0]

I hope this helps Lauren!