1. Prelab: Implement the following mathematical equation as follows: - Sign: using Logic gates (AND, OR, and NOT) gates. Xo: using Active Low Decoder 3x8. X1: using Logic gates NAND gates only (using SOP equation). X2: using Logic gates NOR gates only (using POS equation). Hints: - A consists of 2-bit input, B is a single-bit input, and the objective is to substitute all possible combinations of A and B to compute the output X. The bit-width of X depends on the maximum value resulting from the input substitutions. Additionally, please note that during substitution, both positive and negative numbers may be obtained, with the sign bit set to 0 for positive numbers and 1 for negative numbers. X=2A+B – 2 Α1 Α B Sign X2 X1 X0 0 Decimal Output 000 001 010 0 11 100 10 1 11 10 11 1

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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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1. Prelab: Implement the following mathematical equation as follows:
-
Sign: using Logic gates (AND, OR, and NOT) gates.
Xo: using Active Low Decoder 3x8.
X1: using Logic gates NAND gates only (using SOP equation).
X2: using Logic gates NOR gates only (using POS equation).
Hints:
- A consists of 2-bit input, B is a single-bit input, and the objective is to substitute all possible
combinations of A and B to compute the output X. The bit-width of X depends on the
maximum value resulting from the input substitutions.
Additionally, please note that during substitution, both positive and negative numbers may be
obtained, with the sign bit set to 0 for positive numbers and 1 for negative numbers.
X=2A+B – 2
Α1 Α
B Sign X2 X1 X0
0
Decimal
Output
000
001
010
0 11
100
10 1
11 10
11 1
Transcribed Image Text:1. Prelab: Implement the following mathematical equation as follows: - Sign: using Logic gates (AND, OR, and NOT) gates. Xo: using Active Low Decoder 3x8. X1: using Logic gates NAND gates only (using SOP equation). X2: using Logic gates NOR gates only (using POS equation). Hints: - A consists of 2-bit input, B is a single-bit input, and the objective is to substitute all possible combinations of A and B to compute the output X. The bit-width of X depends on the maximum value resulting from the input substitutions. Additionally, please note that during substitution, both positive and negative numbers may be obtained, with the sign bit set to 0 for positive numbers and 1 for negative numbers. X=2A+B – 2 Α1 Α B Sign X2 X1 X0 0 Decimal Output 000 001 010 0 11 100 10 1 11 10 11 1
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