HWK5 - HWK#5 Solution Problem 5.1 For convenience...

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

View Full Document Right Arrow Icon
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 3-input gates: two OR gates and one NOR gate. How do we implement f? We can't use a 3-input 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 3-input OR gate to compute m2+m4+m6. To implement h, we have only one gate left: a 3-input 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 three-variable 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 K-maps 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,
Background image of page 1

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

View Full Document Right Arrow Icon
using 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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 11

HWK5 - HWK#5 Solution Problem 5.1 For convenience...

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