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
using only two-input NOR gates to implement the following function
0
$\begingroup$
boolean-algebra
3 Answers
1
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.
0
We can write: $$(\bar A + B)(A + \bar B) = A \bar A + \bar A \bar B + AB + B\bar B = AB + \bar A \bar B $$ Hope it helps.
0
18 Draw a logic diagram using only two-input NOR gates to implement the following function: F1A, B, C, D2 = (A { B)'(C { D)