CSE 370 Solution Set #4
Spring 1999

Total Points 20

1) 0 Points:

c. Here's a possible schematic using only NOR gates:

2) 1 Point:
a. Since AND-OR-Invert gates have inverted output we'll want to find the function's complement in SOP form.
Recall that the 3 input XOR function returns true when an odd # of inputs are 1. So the complement returns true when an even # of inputs are 1.

 F' A'B' A'B AB AB' C' 1 0 1 0 C 0 1 0 1
F' = ABC' + AB'C + A'BC + A'B'C'

Here's the corresponding schematic:

3) 1 Point:
e. F = ACE + ACF + ADE + ADF + BCE + BCF + BDE + BDF
A(CE + CF + DE + DF) + B(CE + CF + DE + DF)
(A + B)(CE + CF + DE + DF)
(A + B)( C(E + F) + D(E + F) )
(A + B)( (C + D)(E + F) )
(A + B)(C + D)(E + F)

4) 3 Points:
3.13 with XOR delays 3x NAND delays, the waveform should have similar shape to the following:

5) 3 Points:
Here's the schematics for F, F2, & F3:

The timing waveforms should look like the ones in the Katz text.

6) 4 Points:
Here's the truth table remembering that AB is a 2-bit number multiplied by CD resulting in the 4-bit answer WXYZ:

 A B C D | W X Y Z 0 0 0 0 0 0 0 0 0*0=0 0 0 0 1 0 0 0 0 0*1=0 0 0 1 0 0 0 0 0 0*2=0 0 0 1 1 0 0 0 0 0*3=0 0 1 0 0 0 0 0 0 1*0=0 0 1 0 1 0 0 0 1 1*1=1 0 1 1 0 0 0 1 0 1*2=2 0 1 1 1 0 0 1 1 1*3=3 1 0 0 0 0 0 0 0 2*0=0 1 0 0 1 0 0 1 0 2*1=2 1 0 1 0 0 1 0 0 2*2=4 1 0 1 1 0 1 1 0 2*3=6 1 1 0 0 0 0 0 0 3*0=0 1 1 0 1 0 0 1 1 3*1=3 1 1 1 0 0 1 1 0 3*2=6 1 1 1 1 1 0 0 1 3*3=9

Implementing with 4 16:1 muxes you simply use 1 mux for each output, WXY&Z, from the truth table, the inputs to the mux are the 16 rows of the
table and the select bits are ABCD.
To implement with 4 8:1 muxes you have to use D & D' as additional inputs to the muxes and use ABC as the select bits:

7) 5 Points total:

 A B C D | F 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 1 1 0 1 0 0 0 0 1 0 1 1 0 1 1 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1

a. (1 point) Implement with a 16:1 Mux:

b. (1 point) Implement with an 8:1 Mux:

c. (1 point) Implement with a 4:1 Mux & gates:

d. (1 point) Implement with a 4:16 decoder & a gate:

e. (1 point) Implement with 2 3:8 decoders & a gate:

Note that typical decoders have inverted outputs so that you need to invert the outputs before running them all into an OR gate, which of course is          equivalent to a NAND gate.

8) 3 Points total:
a. Use the 4 bit output from the Bar coder as the select bits to your decoder...

b. If your favorites change then you simply need to re-wire the inputs to your OR gates so instead of 1, 5, 6, 9, 12, 15, they are 0, 2, 7, 8, 10, 13.