CSE 370 Solution Set #4
Spring 1999
Total Points 20


1) 0 Points:

    c. Here's a possible schematic using only NOR gates:
        hw4_1.gif (5444 bytes)
  
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:
       hw4_2.gif (3888 bytes)

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:
    hw4_4.gif (1920 bytes)

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:
hw4_6.gif (8958 bytes)

7) 5 Points total:
Start with the truth table for the function:

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:
       hw4_7a.gif (2858 bytes)
    b. (1 point) Implement with an 8:1 Mux:
        hw4_7b.gif (3724 bytes)
    c. (1 point) Implement with a 4:1 Mux & gates:
        hw4_7c.gif (3555 bytes)
    d. (1 point) Implement with a 4:16 decoder & a gate:
        hw4_7d.gif (2914 bytes)
    e. (1 point) Implement with 2 3:8 decoders & a gate:
        hw4_7e.gif (3788 bytes)
        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...
        hw4_8.gif (2906 bytes)
    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.