using only two-input NOR gates to implement the following function
F(A,B,C,D)=(A⊕B)'(C⊕D)
=(AB'+A'B)'(CD'+C'D)
=(AB')'(A'B)'(CD'+C'D)
=(A'+B)(A+B')(CD'+C'D)
how to simplify further
boolean-algebra
asked Feb 18 '17 at 13:53
samexmm
143
add a comment |
F(A,B,C,D)=(A⊕B)'(C⊕D)
=(AB'+A'B)'(CD'+C'D)
=(AB')'(A'B)'(CD'+C'D)
=(A'+B)(A+B')(CD'+C'D)
how to simplify further
boolean-algebra
asked Feb 18 '17 at 13:53
samexmm
143
add a comment |
F(A,B,C,D)=(A⊕B)'(C⊕D)
=(AB'+A'B)'(CD'+C'D)
=(AB')'(A'B)'(CD'+C'D)
=(A'+B)(A+B')(CD'+C'D)
how to simplify further
boolean-algebra
asked Feb 18 '17 at 13:53
samexmm
143
F(A,B,C,D)=(A⊕B)'(C⊕D)
=(AB'+A'B)'(CD'+C'D)
=(AB')'(A'B)'(CD'+C'D)
=(A'+B)(A+B')(CD'+C'D)
how to simplify further
boolean-algebra
boolean-algebra
asked Feb 18 '17 at 13:53
samexmm
143
asked Feb 18 '17 at 13:53
samexmm
143
asked Feb 18 '17 at 13:53
samexmm
143
asked Feb 18 '17 at 13:53
samexmm
143
asked Feb 18 '17 at 13:53
samexmm
143
143
add a comment |
add a comment |
2 Answers
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We can write: $$(bar A + B)(A + bar B) = A bar A + bar A bar B + AB + Bbar B = AB + bar A bar B $$ Hope it helps.
answered Feb 18 '17 at 13:59
Rohan
27.7k42444
add a comment |
When rewriting a formula for implementation with NOR gates, we aim for a product-of-sums form, because then the translation is immediate. In this case, the trick is to rewrite $(C oplus D)$ as $(C+D)(C'+D')$ to obtain a product of sums (POS).
Once you have your POS, you get a two-level NOR-NOR circuit with one NOR for each clause of the POS and one NOR gate to combine the outputs of the first-level gates. Of course, you also need four inverters, which you may consider one-input NOR gates.
Note that this procedure is the dual of the one that turns any sum of products into a two-level NAND-gate circuit.
In this case the second-level NOR gate would have four inputs, because there are four clauses in the product of sums. If you can only use two-input gates, then you need to decompose the four-input NOR into three two-input NORs and two inverters. If inverters are off limits, you can use two-input NOR gates with their inputs tied together instead.
If you want to go beyond two-level implementations things get significantly more interesting, but I bet you haven't gotten to that subject yet.
answered Feb 18 '17 at 14:06
Fabio Somenzi
6,41221221
add a comment |
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2 Answers
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2 Answers
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We can write: $$(bar A + B)(A + bar B) = A bar A + bar A bar B + AB + Bbar B = AB + bar A bar B $$ Hope it helps.
answered Feb 18 '17 at 13:59
Rohan
27.7k42444
add a comment |
We can write: $$(bar A + B)(A + bar B) = A bar A + bar A bar B + AB + Bbar B = AB + bar A bar B $$ Hope it helps.
answered Feb 18 '17 at 13:59
Rohan
27.7k42444
add a comment |
We can write: $$(bar A + B)(A + bar B) = A bar A + bar A bar B + AB + Bbar B = AB + bar A bar B $$ Hope it helps.
answered Feb 18 '17 at 13:59
Rohan
27.7k42444
We can write: $$(bar A + B)(A + bar B) = A bar A + bar A bar B + AB + Bbar B = AB + bar A bar B $$ Hope it helps.
answered Feb 18 '17 at 13:59
Rohan
27.7k42444
answered Feb 18 '17 at 13:59
Rohan
27.7k42444
answered Feb 18 '17 at 13:59
Rohan
27.7k42444
answered Feb 18 '17 at 13:59
Rohan
27.7k42444
27.7k42444
add a comment |
add a comment |
When rewriting a formula for implementation with NOR gates, we aim for a product-of-sums form, because then the translation is immediate. In this case, the trick is to rewrite $(C oplus D)$ as $(C+D)(C'+D')$ to obtain a product of sums (POS).
Once you have your POS, you get a two-level NOR-NOR circuit with one NOR for each clause of the POS and one NOR gate to combine the outputs of the first-level gates. Of course, you also need four inverters, which you may consider one-input NOR gates.
Note that this procedure is the dual of the one that turns any sum of products into a two-level NAND-gate circuit.
In this case the second-level NOR gate would have four inputs, because there are four clauses in the product of sums. If you can only use two-input gates, then you need to decompose the four-input NOR into three two-input NORs and two inverters. If inverters are off limits, you can use two-input NOR gates with their inputs tied together instead.
If you want to go beyond two-level implementations things get significantly more interesting, but I bet you haven't gotten to that subject yet.
answered Feb 18 '17 at 14:06
Fabio Somenzi
6,41221221
add a comment |
When rewriting a formula for implementation with NOR gates, we aim for a product-of-sums form, because then the translation is immediate. In this case, the trick is to rewrite $(C oplus D)$ as $(C+D)(C'+D')$ to obtain a product of sums (POS).
Once you have your POS, you get a two-level NOR-NOR circuit with one NOR for each clause of the POS and one NOR gate to combine the outputs of the first-level gates. Of course, you also need four inverters, which you may consider one-input NOR gates.
Note that this procedure is the dual of the one that turns any sum of products into a two-level NAND-gate circuit.
In this case the second-level NOR gate would have four inputs, because there are four clauses in the product of sums. If you can only use two-input gates, then you need to decompose the four-input NOR into three two-input NORs and two inverters. If inverters are off limits, you can use two-input NOR gates with their inputs tied together instead.
If you want to go beyond two-level implementations things get significantly more interesting, but I bet you haven't gotten to that subject yet.
answered Feb 18 '17 at 14:06
Fabio Somenzi
6,41221221
add a comment |
When rewriting a formula for implementation with NOR gates, we aim for a product-of-sums form, because then the translation is immediate. In this case, the trick is to rewrite $(C oplus D)$ as $(C+D)(C'+D')$ to obtain a product of sums (POS).
Once you have your POS, you get a two-level NOR-NOR circuit with one NOR for each clause of the POS and one NOR gate to combine the outputs of the first-level gates. Of course, you also need four inverters, which you may consider one-input NOR gates.
Note that this procedure is the dual of the one that turns any sum of products into a two-level NAND-gate circuit.
In this case the second-level NOR gate would have four inputs, because there are four clauses in the product of sums. If you can only use two-input gates, then you need to decompose the four-input NOR into three two-input NORs and two inverters. If inverters are off limits, you can use two-input NOR gates with their inputs tied together instead.
If you want to go beyond two-level implementations things get significantly more interesting, but I bet you haven't gotten to that subject yet.
answered Feb 18 '17 at 14:06
Fabio Somenzi
6,41221221
When rewriting a formula for implementation with NOR gates, we aim for a product-of-sums form, because then the translation is immediate. In this case, the trick is to rewrite $(C oplus D)$ as $(C+D)(C'+D')$ to obtain a product of sums (POS).
Once you have your POS, you get a two-level NOR-NOR circuit with one NOR for each clause of the POS and one NOR gate to combine the outputs of the first-level gates. Of course, you also need four inverters, which you may consider one-input NOR gates.
Note that this procedure is the dual of the one that turns any sum of products into a two-level NAND-gate circuit.
In this case the second-level NOR gate would have four inputs, because there are four clauses in the product of sums. If you can only use two-input gates, then you need to decompose the four-input NOR into three two-input NORs and two inverters. If inverters are off limits, you can use two-input NOR gates with their inputs tied together instead.
If you want to go beyond two-level implementations things get significantly more interesting, but I bet you haven't gotten to that subject yet.
answered Feb 18 '17 at 14:06
Fabio Somenzi
6,41221221
edited Feb 18 '17 at 14:31
answered Feb 18 '17 at 14:06
Fabio Somenzi
6,41221221
answered Feb 18 '17 at 14:06
Fabio Somenzi
6,41221221
answered Feb 18 '17 at 14:06
Fabio Somenzi
6,41221221
6,41221221
add a comment |
add a comment |
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