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Systems and Signals
Lee, Fall 201011
EE102
Final Exam
NAME:
You have 3 hours for 6 questions.
•
Show enough (neat) work in the clear spaces on this exam to convince us that you
derived, not guessed, your answers.
•
Put your ﬁnal answers in the boxes at the bottom of the page.
Closed notes, closed book, 1 letter sized handwritten sheets allowed.
Problem
Score
1
2
3
4
5
6
Total
1
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View Full Document Problem 1. Fourier Transforms
(15 Points)
Find the Fourier transform of the following signal
f
(
t
)
t
0
1
2
3
4
5

5

4

3

2

1
sinc
2
(
t
/
3
)
rect
(
t
2
)
This is an inﬁnite sequence of rectangles, weighted by a sinc
2
(). Eliminate convolutions in
your answer.
Solution:
We can write this signal as
f
(
t
) =
rect
(2
t
)
*
h
sinc
2
(
t/
3)
δ
1
(
t
)
i
.
The Fourier transform is then
F
(
jω
) =
1
2
π
"
1
2
sinc
ω/
2
2
π
!
[3Δ(3
ω/
2
π
)
*
2
πδ
2
π
(
ω
)]
#
=
3
2
sinc
±
ω
4
π
²
"
∞
X
n
=
∞
Δ(3
ω/
2
π
)
*
δ
(
ω

2
πn
)
#
=
3
2
sinc
±
ω
4
π
²
"
∞
X
n
=
∞
Δ(3(
ω

2
πn
)
/
2
π
)
#
=
3
2
∞
X
n
=
∞
sinc
±
ω
4
π
²
Δ
3(
ω

2
πn
)
2
π
!
F
(
jω
) =
3
2
∑
∞
n
=
∞
sinc
³
ω
4
π
´
Δ
³
3(
ω

2
πn
)
2
π
´
2
Problem 2. Laplace Transforms
(15 Points)
Find the Laplace transform of the following signal
f
(
t
)
t
p
(
t
)
p
(
t

2)
p
(
t

4)
2
1
0
4
f
(
t
)
This is the sum of an inﬁnite sequence of delayed causal subpulses
f
(
t
) =
∞
X
n
=0
p
(
t

2
n
)
Assume
p
(
t
) =
e

7
t
u
(
t
) and indicate region of convergence.
Hint:
L{
p
(
t

n
)
}
=
e

sn
P
(
s
)
.
Also, recall that
∞
X
n
=0
a
n
=
1
1

a
when simplyﬁng your answer.
Solution:
f
(
t
) =
∞
X
n
=0
p
(
t

2
n
)
so
F
(
s
) =
∞
X
n
=0
L{
p
(
t

2
n
)
}
=
∞
X
n
=0
e

2
sn
P
(
s
) =
P
(
s
)
∞
X
n
=0
(
e

2
s
)
n
=
P
(
s
)
1
1

e

2
s
Then
P
(
s
) is
L
n
e

7
t
u
(
t
)
o
=
1
s
+ 7
Substituting this into the expression for
F
(
s
),
F
(
s
) =
1
(
s
+ 7)(1

e

2
s
)
.
ROC:
σ >
0
F
(
s
) =
1
(
s
+7)(1

e

2
s
)
, ROC:
σ >
0
3
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View Full Document Problem 3. Discrete Time Fourier Transform
(30 Points)
An engineer purchased an expensive, ﬁnelytuned discretetime lowpass ﬁlter with an im
pulse response
h
[
n
] and frequency response
H
(
e
jω
). Sadly, the ﬁlter did not exactly suit the
needs of the application and need to be modiﬁed.
Note: In this problem we adopt an intuitive notion of nonideal lowpass ﬁlters. To be
”lowpass”
H
(
e
jω
)
merely should have larger magnitude at
ω
≈
0
,
±
2
π,
±
4
π
,..., than at
ω
≈ ±
π,
±
3
π,
±
5
π
, etc
Parts a) and b) are independent: you may solve them in any order.
a) Consider a ﬁlter
f
[
n
] with timedomain response:
f
[
n
] =
(
1

h
[0]
if
n
= 0

h
[
n
]
if
n
6
= 0
.
•
Provide an expression for
F
(
e
jω
) in terms of
H
(
e
jω
).
Solution.
Rewrite the expression for
f
[
n
] in closed form:
f
[
n
] =
δ
[
n
]

h
[
n
].
F
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This note was uploaded on 12/13/2011 for the course EE 102 taught by Professor Levan during the Fall '08 term at UCLA.
 Fall '08
 Levan

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