This preview shows pages 1–3. Sign up to view the full content.
ee102_midterm2_Sp2005.fm
3/29/06
EE102L Midterm #2  Spring 2005
1 /
10
C Copyright 2005 Gandhi Puvvada
Spring 2005
EE102L
Instructor: Gandhi Puvvada
Midterm Exam II (30%)
Date: April 29, 2005, Friday
OpenBook OpenNotes Exam
Time: 4:00  6:00PM in SGM124
Name:
Total points: 180
Perfect score: 160 / 180
1
( 5 + 6 + 9 = 20
points)
5
min.
Combinational logic design:
1.1
Complete the KMap for the functions
F(A,B,C,D)
produced by the combinational logic given
below. Also produce F using two XOR gates and another 2input gate.
1.2
On the side, we provided
three PROMs. For each,
state the size and the
total
number of bits. Note:
total
,
not just in one location,
total bits in the whole
PROM!
1.3
Build two noninverting 2to1 muxes by completing the following designs. You can use
additional inverters if you need.
Are the two designs functionally identical (meaning one can be used to replace the other)?
YES
/
NO
5
pts
AB
CD
00
01
11
10
00 01 11 10
A
B
C
D
F Map
A
B
C
D
F
A
B
C
D
F
2input gate
XOR
XNOR
A[3:0]
D[3:0]
4
4
Size: __________
Total bits: ______
A[1:0]
D[3:0]
2
4
Size: __________
Total bits: ______
A[3:0]
D[1:0]
4
2
Size: __________
Total bits: ______
6
pts
3+6
pts
Y
I0
I1
S
VDD
/Y
/I0
/I1
S
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document ee102_midterm2_Sp2005.fm
3/29/06
EE102L Midterm #2  Spring 2005
2 /
10
C Copyright 2005 Gandhi Puvvada
The previous two design are again given below with the
VDD/GND
connections
removed
.
Now again build two noninverting 2to1 muxes
with low active enable
by completing the
designs. You can use additional inverters if you need.
Are the two designs functionally identical (meaning one can be used to replace the other)?
YES
/
NO
2
( 4+18+8+18+8+16 = 72 points)
50
min.
Small System Design
: Find the
smallest
number between A and B
which is evenly divisible by
20
10
(decimal 20). The range of your search is A to B, with both A and B
included
.
A and B are two 7 bit binary numbers. A is always
less
than B.
Since
twenty
is
four times five
(20 = 4 x 5), both 4 and 5 should divide the number evenly. A binary number divisible by 4 will
have
its
two
least significant bits
zeros
. If you find the number, you go to the FND (FND = Found)
state, otherwise you go to NSN (NSN = No Such Number) state.
Algorithm: Accept
Ain
into
A
and
Bin
into
B
.
A
= A
6
A
5
A
4
A
3
A
2
A
1
A
0
.
Let us treat this 7bit
register
A
as actually made up of a 5bit register
Au
(
A upper
) holding the upper 5 bits and a 2
bit register
Al
(
A lower
) holding the lowest 2 bits.
Since we are looking for a number divisible by twenty, we will start with
A
6
A
5
A
4
A
3
A
2
0 0
(upper
five bits concatenated with two zeros) which is divisible by 4. Before checking to see if it is also
divisible by 5, we need to see if the
A
1
A
0
are actually two zeros or anything other than two zeros.
If they are two zeros, then
A
6
A
5
A
4
A
3
A
2
0 0
(symbolically written as
Au00
)
is in fact A itself and can
be tested for divisibility by 5. Otherwise, we want to increment this 7bit number
Au00
(
A
6
A
5
A
4
A
3
A
2
0 0
)
by 4 (
=100
), which is same as incrementing the 5bit number
Au
(
A
6
A
5
A
4
A
3
A
2
)
by 1 (
Au <= Au +1
). We then test to see if the resulting 7bit number
Au00
is divisible by 5. If we
fail, we increment the
Au
by 1 again. Of course, every time after incrementation, we need to see
if
Au
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 09/12/2009 for the course EE 201L taught by Professor Puvvada during the Spring '08 term at USC.
 Spring '08
 Puvvada

Click to edit the document details