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 Decimal A₁ A B Sign X X₁ Xo Output oooo 0 0 100 0 0 1 10 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

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Author:James Kurose, Keith Ross
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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
Decimal
A₁ A B Sign X X₁ Xo
Output
oooo
0
0
100
0
0
1
10
0 1 1
1 0 0
1
0 1
1
1 0
1
1 1
Transcribed Image Text: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 Decimal A₁ A B Sign X X₁ Xo Output oooo 0 0 100 0 0 1 10 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
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