[SOLVED] CENG232-Lab 1

Description

This laboratory aims to get you familiar with basic logic gates and combinational circuit design. You will simplify the circuit that is explained below and draw the circuit using Logisim tool with the given gates.

1           IC Pool

• 74LS04 (Inverter)
• 74LS08 (AND)
• 74LS32 (OR)

2           Lab Work

In this assignment, you are expected to perform the operations described in the following section.

2.1         Specifications

Suppose A and B are 2-bit binary input numbers and X and Y are 1-bit binary output numbers. A and B are represented with A₁, A₀, B₁ and B₀ bits respectively where A₁ and B₁ are the most significant bits and A₀ and B₀ are the least significant bits of the relevant number. Your circuit will take A and B as inputs and give the outputs X and Y with the following instructions:

PS: ⊕ stands for the XOR operation.

|A − B| stands for the absolute value of A minus B.

You have to use ”input pins” and ”output pins” for your inputs and outputs, respectively, from the Toolbar at the top of Logisim. Set their labels correctly using the following names. Please, only set “label” property of the “pin” objects, do not add a “label” object onto the Logisim canvas.

Input pins: A1, A0, B1, B0. Output pins: X, Y.

Each pin corresponds to a digit in a 2-bit binary number. If it is set, then the value of the digit is 1 if reset, then the value of the digit is 0.

2.2         Input Output Examples

1. Suppose A₁A₀ = 00 and B₁B₀ = 10. In this case, A=0 and B=2 in decimal.

|A − B| = 2

Since |A − B| > 1 holds, the output X is 1.

2. Suppose A₁A₀ = 10 and B₁B₀ = 01. In this case, A=2 and B=1 in decimal. |A − B| = 1

Since |A − B| > 1 does not hold, the output X is 0.

3. Suppose A₁A₀ = 01 and B₁B₀ = 01. In this case, A=1 and B=1 in decimal. (A ∗ B) = 1

Since (A ∗ B) = 1 is not prime, the output Y is (A₀ ⊕ B₀) = (1 ⊕ 1) = 0 .

4. Suppose A₁A₀ = 01 and B₁B₀ = 10. In this case, A=1 and B=2 in decimal. (A ∗ B) = 2

Since (A ∗ B) = 2 is prime, the output Y is 0.