HWK2 - HWK #2 Solutions Problem 2.1 (a) By inspection, we...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HWK #2 Solutions Problem 2.1 (a) By inspection, we can write: f(a,b,c 1 ,c ) = ac 1 'c ' + (b)c 1 'c + (a+b)c 1 c ' + (a'b + ab')c 1 c This can be simplified as follows: f(a,b,c 1 ,c ) = ac 1 'c ' + bc 1 'c + (a+b)c 1 c ' + (a'b + ab')c 1 c = ac '(c 1 ' + c 1 ) + bc 1 'c + bc 1 c ' + a'c 1 c ' + (a'bc 1 c + ab'c 1 c ) = a(b'c (c 1 + c ') + bc 1 (c ' + a'c ) + bc 1 'c = ac ' + ab'c 1 + bc 1 c ' + a'bc 1 + bc 1 'c = ac ' + ab'c 1 + a'bc 1 + bc 1 'c Shown below is an implementation with three AND gates, one OR gate, and two inverters. (b) To compute a logical function (such as XOR or AND) on two 4-bit numbers, we simply compute the output on a bit-by-bit basis. So we just need to use four copies of the network N from part (a); no additional gates are needed. Problem 2.2 (a) and (b) (x + y') (w(y + z)yw'z' + x)' = x'y' is proved below. The dual is proved to the right: xy' + ( w + yz + y + w' + z') x )' = x' + y' . Notice that the proofs are dual: each step of the right proof is the dual of that same step in the left proof. (x + y') (w(y + z)yw'z' + x)' xy' + ((w + yz +y + w' + z') x)' = (x + y') ( 0 (y + z)yz' + x)' complementarity = xy' + ((1 + yz +y + z') x)' = (x + y') (0 + x)' null = xy' + (1x)' = (x + y') x' identity = xy' + x' = x'y' no-name = x' + y' (abe + (a' + e') cd + bcd) (ab' + ae + b') = (abe + (a' + e') cd + bcd) (ae + b') absorption = (abe + (ae)' cd + bcd) (ae + b') deMorgan = (abe + (ae)' cd) (ae + b')...
View Full Document

Page1 / 9

HWK2 - HWK #2 Solutions Problem 2.1 (a) By inspection, we...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online