Question

Create a combinatorial circuit that performs a 3 bit multiplication. The circuit will have two 3-bit inputs A2 A1 A0 and B2 B1 B0 and one 6 bit output.

Create a combinatorial circuit that performs a 3-bit multiplication. The circuit will accept two 3-bit inputs A2 A1 A0 and B2 B1 B0 and generate one 6 bit output.

First, create 3 Integrated circuits (iC). The first IC takes as input, B0, A2, A1, and A0 as input, and it generates a 6-bit output. If B0 is 0, then the output of all the 6 bits is 0. If B0 is 1, the output is 0-bit arithmetic left shift of the binary number (0 0 0 A2 A1 A0).

The second IC takes as input, B1, A2, A1, and A0 as input, and it generates a 6-bit output. If B1 is 0, then the output of all the 6 bits is 0. If B1 is 1, the output is one-bit arithmetic left shift of the binary number (0 0 0 A2 A1 A0).

The third IC takes as input, B2, A2, A1, and A0 as input, and it generates a 6-bit output. If B2 is 0, then the output of all the 6 bits is 0. If B2 is 1, the output is two-bit arithmetic left shift of the binary number (0 0 0 A2 A1 A0).

In an arithmetic left shifter, the existing bits are moved to the left by the specified number, and the new spaces to the left are filled by 0. For example, if the input is 100111 the 1 bit output of the shifter is 001110.

Lastly, create a 6-bit ripple carry added that can add two 6-bit numbers. You can re-use full-adder circuits from assignment 3 for this question. Use the 6-bit ripple carry adders to add the output of IC 1, IC 2, and IC 3, to generate the final result.

An example of the multiplication operation and the outputs of different circuits is shown below.

010 <---- A2 A1 A0

*. 101 <----- B2 B1 B0

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000010 <--- output of circuit 1

000000 <--- output of circuit 2

+ 001000 <--- output of circuit 3

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001010 <--- output after addition / final output

Answer 1