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

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HWK #2 Solutions Problem 2.1 (a) By inspection, we can write: f(a,b,c 1 ,c 0 ) = ac 1 'c 0 ' + (b)c 1 'c 0 + (a+b)c 1 c 0 ' + (a'b + ab')c 1 c 0 This can be simplified as follows: f(a,b,c 1 ,c 0 ) = ac 1 'c 0 ' + bc 1 'c 0 + (a+b)c 1 c 0 ' + (a'b + ab')c 1 c 0 = ac 0 '(c 1 ' + c 1 ) + bc 1 'c 0 + bc 1 c 0 ' + a'c 1 c 0 ' + (a'bc 1 c 0 + ab'c 1 c 0 ) = a(b'c 0 (c 1 + c 0 ') + bc 1 (c 0 ' + a'c 0 ) + bc 1 'c 0 = ac 0 ' + ab'c 1 + bc 1 c 0 ' + a'bc 1 + bc 1 'c 0 = ac 0 ' + ab'c 1 + a'bc 1 + bc 1 'c 0 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.

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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') consensus = (abe) (ae + b') + ((ae)' cd) (ae + b') distributive = (abe) + ((ae)' cd) (ae + b') absorption = (abe) + (ae)' cd((ae)'' + b') involution = (abe) + (ae)' cdb' no-name Problem 2.3
Problem 2.4 a. A key observation here is that multiplying a binary number by 2 amounts to shifting the bits to the left and inserting a zero. Therefore, since Z=z

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

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