UC Davis
1
Hussain AlAsaad
U
NIVERSITY
OF
C
ALIFORNIA
—D
AVIS
D
EPARTMENT
OF
E
LECTRICAL
OMPUTER
E
NGINEERING
EEC180A—D
IGITAL
S
YSTEMS
I
W
INTER
2011
E
XAM
II
S
TUDENT
I
NFORMATION
I
NSTRUCTIONS
The exam is closed book and notes. A single doublesided cheat sheet is allowed.
Print your name and your ID number.
There are four problems in the exam. Solve all of them and show your work.
If you need more space for your solution, use the back of the sheets.
E
XAM
G
RADE
Name
Hussain AlAsaad
ID Number
xxxxxxxxx
Problem
Maximum
Points
Student
Score
1
25
25
2
25
25
3
25
25
4
25
25
Total
100
100
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2
Hussain AlAsaad
1. A
RITHMETIC
C
IRCUITS
(15 + 10 = 25
POINTS
)
Let
X
be a 3bit unsigned number represented by
X
2
X
1
X
0
,
Y
be a 4bit unsigned number repre
sented by
Y
3
Y
2
Y
1
Y
0
,
Z
be a 5bit unsigned number represented by
Z
4
Z
3
Z
2
Z
1
Z
0
, and let
W
be a 9bit
unsigned number represented by
W
8
W
7
W
6
W
5
W
4
W
3
W
2
W
1
W
0
.
1.1 Design a gatelevel circuit
C
that computes the function
. Assume that the input
combinations (
X
2
X
1
X
0
= 110) and (
X
2
X
1
X
0
= 111) will never appear at the inputs of
C
. Minimize
the number of gates in your design. Assume that the literals and their complements are available.
1.2 Suppose that we have an 8bit adder (shown below), show how we can implement the
function
.
W
25
X
–
2
=
X
W
X
2
X
1
X
0
W
8
W
7
W
6
W
5
W
4
W
3
W
2
W
1
W
0
0
0
0
0
0
0
0
1
1
0
0
1
0
0
1
0
0
0
0
1
1
0
0
0
0
1
0
0
0
0
0
1
0
1
0
1
0
1
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
1
1
0
1
0
0
0
0
0
0
0
0
0
1
1
0
x
x
x
x
x
x
x
x
x
1
1
1
x
x
x
x
x
x
x
x
x
From a Kmap we get:
We choose
, since it yields less number of gates. So, we have
Based on the above, we need three gates to implement
C
.
W
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 Spring '08
 REDINBO
 Logic gate, Flipflop, University of California, Davis

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