HWK #5 Solution
Problem 5.1
For convenience throughout this problem, let's start by writing out the K
maps:
yz
00 01 11 10
x
0
1
0
0
0
1
1
1
1
1
f
yz
00 01 11 10
x
0
0
0
0
1
1
0
0
1
1
g
yz
00 01 11 10
x
0
0
1
1
1
1
0
0
1
1
h
a.
By putting x, y, and z at the inputs of our 3:8 decoder, we have available at the
output of the decoder all possible minterms. We then simply OR the minterms
together to realize each function. The tricky part is that we have only 3input
gates: two OR gates and one NOR gate.
How do we implement f? We can't use a 3input OR gates, because f has 5
minterms. So we instead implement f', and invert it using the NOR gate. In other
words, f' = OR(m1,m2,m3). Feeding m1, m2, and m3 into a NOR gate means we
have (m1+m2+m3)' = m1'm2'm3' = M1M2M3 = f.
To implement g, we just use a 3input OR gate to compute m2+m4+m6. To
implement h, we have only one gate left: a 3input OR gate. But h(x,y,z) =
m1+m2+m3+m4+m6 = (m2+m4+m6) + m1 + m3. We can just use an OR gate
with inputs g, m1, and m3.
b.
Any threevariable function can be implemented with a 4:1 MUX plus an inverter.
Since we do not have any inverters, we need to worry about which variables will
serve as the select inputs to the muxes.
Looking at the Kmaps for f and g, we can see that there is no combination of y
and z that gives 1 when x = 0 and 0 when x = 1. Thus, we can use y and z for the
select lines without needing an inverter  all the inputs will be 0, 1, or x. For h,
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View Full Documentusing y and z as select inputs won't work (if z is 1, we need to output x');
however, we can see that for no combination of x and z will we need to output y'.
c.
The ROM size is 8x3 (8 words with 3 bits per word).
ROM programming is analogous to the decoder solution. Each output of the ROM
represents a minterm of the function. We then OR together the minterms to come
up with our function.
ROM Truth Table
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 Fall '08
 BROWN
 Logic gate, Input/output, exclusive or

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