ece202 e2 sp2008 sol.pdf-EE-202 Exam II ...
ece202_e2_sp2008_sol.pdf-EE-202 Exam II March 3, 2008
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ece202 e2 sp2008 sol.pdf-EE-202 Exa...
ece202_e2_sp2008_sol.pdf-EE-202 Exam II March 3, 2008
##### Page 8
Exam 2, ECE 202, Sp 08
(7)
2
!
2
e
0.5
t
(
)
u
(
t
)
(8)
none of above
S
OLUTION
11
.
(5)
I
in
=
(2
s
+
1)
V
out
!
4
sV
out
=
(1
!
2
s
)
V
out
Hence,
V
out
=
1
(1
!
2
s
)
I
in
=
!
0.5
s
(
s
!
0.5)
=
1
s
!
1
s
!
0.5
.
Thus
v
out
(
t
)
=
1
!
e
0.5
t
(
)
u
(
t
)
.
12.
For
t
!
0
the convolution of
f
(
t
)
=
2
u
(
!
t
)
with
h
(
t
)
=
2
e
!
2
t
u
(
t
)
is:
(1)
e
!
2
t
u
(
t
)
(2)
2
(3)
4
e
!
2
t
u
(
t
)
(4)
!
4
e
!
2
t
u
(
t
)
(5)
!
2
e
!
2
t
u
(
t
)
(6) 0
(7)
2
e
!
2
t
u
(
t
)
(8) none of above
Solution 12.
(7)
y
(
t
)
=
f
*
h
=
4
e
!
2
"
u
(
"
!
t
)
d
"
=
2
t
<
0
2
e
!
2
t
t
#
0
\$
%
&
&
0
(
)

##### Page 9
Exam 2, ECE 202, Sp 08
WORKOUT PROBLEM.
(40 points)
There are TWO PARTS to this problem.
Part 2 is on a later
page.
Part 1 (20 points).
Compute the convolution
y
(
t
)
=
f
(
t
)*
h
(
t
)
for
f
(
t
)
and
h
(
t
)
sketched below.
Plot
y
(
t
)
on the graph immediately below this problem statement.
Properly label the graph.
You must
show all work to receive credit for the (correct) plot.
dh
1
dt
=
2
!
(
t
+
1)
"
2
!
(
t
)
and
g
(
t
)
=
f
(
!
)
d
!
"#
t
\$
=
r
(
t
+
1)
"
r
(
t
"
1)
.
Therefore
y
(
t
)
=
2
r
(
t
+
2)
!
2
r
(
t
+
1)
!
2
r
(
t
)
+
2
r
(
t
!
1)
.

##### Page 10
Exam 2, ECE 202, Sp 08
Part 2 (20 points).
This part is a complication of part 1 and requires some thinking to obtain the answer
easily.
Compute the convolution
y
(
t
)
=
f
(
t
)*
h
(
t
)
for
f
(
t
)
and
h
(
t
)
sketched below.
Plot
y
(
t
)
on
the graph immediately below this problem statement.
Properly label the graph.
You must show all
work to receive credit for the (correct) plot.
Hint:
how can you utilize part 1 and the distributive
law of convolution.
Then
h
(
t
)
=
h
1
(
t
)
+
h
2
(
t
)
and
h
*
f
=
h
1
*
f
+
h
2
*
f
.
From part 1,
h
1
*
f
=
2
r
(
t
+
2)
!
2
r
(
t
+
1)
!
2
r
(
t
)
+
2
r
(
t
!
1)
.
It remains to compute
h
2
*
f
.
Graphical convolution seems to be the best approach here.
We will
graphically evaluate
y
2
(
t
)
=
h
2
(
t
!
"
)
f
(
"
)
d
"
!#
t
\$
.
Obviously,
y
(
t
)
=
0
for
t
!"
1
.
Consider
!
1
"
t
"
1
, as per the picture immediately below.
Here
y
(
t
)
=
AreaTriangle
!
Area
!
Sm
!
Triangle
!
on
!
left
=
2
!
(1
!
t
)
2
2
=
1.5
+
t
!
0.5
t
2
.

##### Page 11
Exam 2, ECE 202, Sp 08
For
1
!
t
!
3
,
y
(
t
)
=
(1
!
(
t
!
2))
2
2
=
(3
!
t
)
2
2
=
4.5
!
3
t
+
0.5
t
2
.
Finally for
t
!
3
,
y
(
t
)
=
0
.
Alternate Solution Method:
f
(
t
)
=
u
(
t
+
1)
!
u
(
t
!
1)
"
f
'(
t
)
=
#
(
t
+
1)
!
#
(
t
!
1)
.
h
(
t
)
=
2
u
(
t
+
1)
!
r
(
t
)
+
r
(
t
!
2)
.
Also note that
h
(
!
)
d
!
"#
t
\$
=
2(
t
+
1)
u
(
t
+
1)
"
0.5
tr
(
t
)
+
0.5(
t
"
2)
r
(
t
"
2)
=
2(
t
+
1)
u
(
t
+
1)
"
0.5
t
2
u
(
t
)
+
0.5(
t
"
2)
2
u
(
t
"
2)
Thus
y
(
t
)
=
2(
t
+
2)
u
(
t
+
2)
!
0.5(
t
+
1)
r
(
t
+
1)
+
0.5(
t
!
1)
r
(
t
!
1)
!
2
tu
(
t
)
+
0.5(
t
!
1)
r
(
t
!
1)
!
0.5(
t
!
3)
r
(
t
!
3)
=
2(
t
+
2)
u
(
t
+
2)
!
0.5(
t
+
1)
2
u
(
t
+
1)
+
0.5(
t
!
1)
2
u
(
t
!
1)
!
2
tu
(
t
)
+
0.5(
t
!
1)
2
u
(
t
!
1)
!
0.5(
t
!
3)
2
u
(
t
!
3)
t = -3:.01:4;
y1 = 2*ramp(t+2)-2*ramp(t+1)-2*ramp(t)+2*ramp(t-1);
y2 = (1.5 + t -0.5*t.^2).*(u(t+1)-u(t-1));
y3 = (4.5 – 3*t + 0.5*t.^2).*(u(t-1) - u(t-3));
y = y1 + y2 + y3;
plot(t,y)
grid
xlabel('time')

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