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Solutions to homework 4 of enee 244 spring 2006, focusing on implementing boolean expressions using exclusive-or and and gates, proving the functional completeness of {and, exclusive-or} gates, and designing combinational circuits such as a 2's complementer and a decoder using {and, not, xor} gates.
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F = AB’CD’ + A’BCD’ + AB’C’D + A’BC’D = AB’(CD’ + C’D) + A’B( CD’ + C’D) = (AB’ + A’B) ( CD’ + C’D) = ( A ⊕ B) ( C ⊕ D)
We know that {AND, OR, NOT} is functionally complete.
To prove that another set is functionally complete, it is enough to show that AND, OR and NOT can each be implemented in terms of the new set.
Here that implies that OR and NOT need to be implemented using AND, XOR.
To implement NOT: a’ = a’.1 + a.0 = a ⊕ 1
To implement OR: A + b = (a + b)’’ = (a’.b’)’ //Using Demorgan’s law. This last term contains only AND and NOT gates, each of which have already been shown to be expressible in the set {AND, XOR}.
Hence both NOT and OR can be implemented in terms of {AND,XOR}. ⇒ {AND,XOR} is functionally complete.
We need:
But we have:
We can add NOT gates at all inputs and the output to convert the second into the first: This is because, at inputs:
This is a NAND gate with an extra NOT gate at output. We can move this extra NOT gate between the two levels of the circuit to the inputs of the second level. This gives:
Since a NOT gate is added at the output as well, this second-level gate becomes a NAND gate as well. Thus, adding NOT gates at all inputs and the output is equivalent to converting all the internal NOR gates to NAND gates, as desired.
To design a 6:64 decoder let the inputs be a,b,c,d,e,f and let the outputs be D0−D63. The design is as shown below. E: enable input o/p: output
We can write F directly by observing the minterms whose numerical values have two or more bits which are 1. F = Σ m (011,101,110,111) = Σ m (3,5,6,7) To implement a function using a MUX, we provide the function inputs as the select inputs to the MUX, and connect the minterms in the canonical form to 1. Thus the circuit is:
A decoder turns its i th^ output high if the input abc=i. A MUX transmits the value of its j th^ data input to the output if the select input def=j.
Thus a 1 will be transmitted to the output by the MUX only if i=j; and 0 otherwise. Hence the output function F is 1 if the number abc = def. More formally:
8 to 1 MUX
0
1
0 0
0
1
1 1
F
a b c