CSE 370 Solution Set #1
Winter 1999
Total Points 20
1) 1 Point: a. 1387
b. 1001101001
c. binary=1100 0100 0110 1011 decimal=50,283
d. octal=1316 hexadecimal=2CE
2) 2 Points: a. 10000110
b. 111
c. 101010010
3) 1 Point: a. 100110
b. 111111
4) 2 Points: a. 19
b. -3
c. -12
d. -13
b. 65,553
5) 1 Point: a. Largest = 2^16 - 1 = 65,535 Smallest = 0
b. Largest = 2^15 - 1 = 32,767 Smallest = -2^15 = -32,768
6) 2 Points: Sign extend the most significant bit
a. 00000101
b. 11111011
7) 1 Point: One possible answer is...
Ignore the numbers, they are just pinouts assigned by the schematic editor.
8) 2 Points: a. X'&Y' + X&Y + X'&Y
X'&Y' + Y&(X + X')
X'&Y' + Y&(1)
X'&Y' + Y
remember you can distribute across "OR" see Katz page 42
(X' + Y)&(Y' + Y)
(X' + Y)&(1)
(X'+Y) 2 literals is minimized
b. X&Z + (X&Z' + Y)' + Y'&Z + X&Y'&Z + X&Z' + X&Y&Z + X&Y'&Z'
since X&Z + X&Y'&Z + X&Y&Z = X&Z
X&Z + (X&Z' + Y)' + Y'&Z + X&Z' + X&Y'&Z'
since X&Z' + X&Y'&Z' = X&Z'
X&Z + (X&Z' + Y)' + Y'&Z + X&Z'
X&Z + ((X&Z')'&Y') + Y'&Z + X&Z'
X&Z + ((X' + Z)&Y') + Y'&Z + X&Z'
X&Z + X'&Y' + Z&Y' + Y'&Z + X&Z'
X&Z + X'&Y' + Y'&Z + X&Z'
X&Z + X&Z' + X'&Y'+ Y'&Z
X&(Z + Z') + X'&Y' + Y'&Z
X + X'&Y' + Y'&Z
(X + X')&(X + Y') + Y'&Z
X + Y' + Y'&Z
(X + Y') 2 literals is minimized
Hint: You can always work out the truth table to verify if your simplified function is the same as the original.
9) 1 Point: a. F' = (X' + Y + Z)&(X' + Y + Z')&(X' + Y' + Z')
simplifies to (X' + Y&Z') 3 literals is minimized
b. G' = (X' + Y + Z')&(X' + Y')
simplifies to (X' + Y'&Z') 3 literals is minimized
10) 2 Points: | A | B | | | XOR |
| -- | -- | -- | -- |
| 0 | 0 | | | 0 |
| 0 | 1 | | | 1 |
| 1 | 0 | | | 1 |
| 1 | 1 | | | 0 |
To understand what 2 input functions you can't make with only 2-input XOR gates you should first realize that there are 16 possible 2 input functions, see Katz page 44.
There were 2 ways to look at the problem. If you assume your only inputs are A and B, then the only functions you can make with just XOR gates are A, B, A ^ B (A xor B), and A ^ A=0. The other 12 functions you can't make.
However if you assumed you could use ground and +5volts as 0 and 1 inputs to the XOR gates then you can also make 1, A ^ 1=A', B ^ 1=B', and (A ^ B) ^ 1=A xnor B. The other 8 functions you still can't make.
11) 1 Point: | A1 | A0 | B1 | B0 | | | S2 | S1 | S0 |
| ---- | ---- | ---- | ---- | -- | ---- | ---- | ---- |
| 0 | 0 | 0 | 0 | | | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | | | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | | | 0 | 1 | 0 |
| 0 | 0 | 1 | 1 | | | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 | | | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | | | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | | | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 | | | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 | | | 0 | 1 | 0 |
| 1 | 0 | 0 | 1 | | | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | | | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 | | | 1 | 0 | 1 |
| 1 | 1 | 0 | 0 | | | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 | | | 1 | 0 | 0 |
| 1 | 1 | 1 | 0 | | | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | | | 1 | 1 | 0 |
12) 1 Point: Arizona State University- neural network speech recognition
13) 1 Point: B.S. in Computer Science with high distinction from University of Colorado
14) 1 Point: University of California Berkeley
15) 1 Point: True