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MICHIGAN STATE UNIVERSITY
DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING
ECE 366
Midterm Examination No. 2
Summer Semester, 2007
00
1. Closed book and notes except for one 8 12 ×1100 page (both sides) of notes and selected Fourier
series-Fourier transform tables.
2. You should have 5 problems on 11 pages PLUS this cover.
3. 80 minute time limit.
4. You must clearly show your method to receive (partial) credit. No credit given for correct
answers without clear justification.
5. Point values are shown at the beginning of each problem. The total possible score is 100.
6. Your exam score will be based on the work shown on the exam paper only. Your written
solution must be a clear, self-sufficient response to the problem. In fairness to everyone,
verbal discussions after the exam of “what was meant” by a particular solution will not be
considered.
Name & Student No.
1
Problem 1 [20 points].
(a) [11 points] A continuous-time periodic signal x(t) is real valued and has a fundamental period
T=8. The nonzero Fourier series coefficients for x(t) are specified as
a1 = a∗−1 = 2j, a5 = a−5 = 3.
Express x(t) in the form
x(t) =
∞
X
Ak cos(ωk t + φk ).
k=0
Clearly show your method to receive (partial) credit. Please BOX your answer.
Here, a∗−1 = 2j ⇒ a−1 = −2j.
x(t) = a−5 ej(−5)ω0 t + a−1 ej(−1)ω0 t + a1 ej(1)ω0 t + a5 ej(5)ω0 t
ejω0 t − e−jω0 t
e5jω0 t + e−5jω0 t
+6
2j
2
= −4 sin(ω0 t) + 6 cos(5ω0 t)
= −4
= 4 cos(ω0 t + π/2) + 6 cos(5ω0 t)
(b) [9 points] Consider three continuous-time periodic signals whose Fourier series representations
are as follows:
x1 (t) =
x2 (t) =
100 k
X
1
k=0
100
X
2
2π
ejk 50 t ,
2π
cos(kπ)ejk 50 t ,
k=−100
x3 (t) =
100
X
k=−100
j sin
kπ
2
2π
ejk 50 t .
Use the property of conjugate symmetry to determine which of these three signals is/are real valued?
Briefly explain your reasoning.
Comparing x1 (t) with the Fourier series synthesis equation we get:
( k
1
0 ≤ k ≤ 100
2
ak =
.
0
otherwise
Clearly, a1 6= a∗−1 . Since the conjugate symmetry is being violated, x1 (t) is not real valued.
2
Comparing x2 (t) with the Fourier series synthesis equation we get:
(
cos(kπ)
−100 ≤ k ≤ 100
ak =
.
0
otherwise
Clearly, ak = a∗−k , ∀k. Since conjugate symmetry holds, x2 (t) is real valued.
Comparing x3 (t) with the Fourier series synthesis equation we get:
(
j sin kπ
−100 ≤ k ≤ 100
2
ak =
.
0
otherwise
Now,
kπ
−kπ
= −j sin
2
2
kπ
= j sin
= ak
2
a−k = j sin
⇒ a∗−k
(∵ sin is an odd function)
Since conjugate symmetry holds, x3 (t) is real valued.
Problem 2 [20 points]. Let
(
t, 0 ≤ t ≤ 2
x(t) =
4 − t, 2 ≤ t ≤ 4
be a periodic signal with fundamental period T=4. Determine the Fourier series coefficients of x(t).
Clearly show your method to receive (partial) credit. Please BOX your answers.
R4
For k = 0 (the d.c. value), a0 = 1/4 0 x(t)dt = 1/4(4) = 1.
For k 6= 0, instead of going directly through the Fourier series analysis equation, it is easier to go
through the property of differentiation.
x(t) =
k=∞
X
ak ejkω0 t
k=−∞
k=∞
k=∞
X
X
dx(t)
jkω0 t
⇒
=
ak jkω0 e
=
bk ejkω0 t
dt
k=−∞
k=−∞
Let us find the Fourier series coefficients bk corresponding to
(
1, 0 ≤ t ≤ 2
dx(t)
=
dt
−1, 2 ≤ t ≤ 4
dx(t)
dt .
,
is also periodic with fundamental period T=4. However, for k 6= 0
Now,
3
(
2,
y(t) =
0,
0≤t≤2
2≤t≤4
,
where y(t) is also periodic with the fundamental period T=4, has the same FS coefficients as
In fact, it is easier to manipulate y(t). For k 6= 0
1
bk =
4
=
Z
4
−jkω0 t
y(t)e
0
h
1
1
dt =
4
i2
Z
2
dx(t)
dt .
2e−jkω0 t dt
0
−jkω0 t
e
2(−jkω0 )
0
h
i
−jkω
0
e
=
ejkω0 − e−jkω0
2(−jkω0 )
e−jkω0
=
sin(kω0 ).
kω0
Then,
ak =
bk
e−jkω0
=
sin(kω0 ),
jkω0
jk 2 ω02
where ω0 = 2π/4 = π/2.
Thus,
ak =



1
k=0
k = odd

0
otherwise
−1
2
 k2 ω0
Problem 3 [10 points]. Let
x[n] = sin
2π
π
n+
6
4
be a periodic signal with fundamental period N = 6. Determine the Fourier series coefficients of
x[n]. Clearly show your method to receive (partial) credit. Please BOX your answer.
We would like to write down x[n] as:
X
x[n] =
ak ejω0 kn ,
k=<6>
where the sum is over 6 consecutive k’s. Now we can use Euler’s relation to write x[n] as the linear
combination of exponentials:
2π
π
2π
π
e( 6 n+ 4 ) − e−( 6 n+ 4 )
x[n] =
2j
π
1
1 π ω0 n
= e4 e
− e− 4 e−ω0 n .
2j
2j
π
π
1 4
1 −4
Thus, a1 = 2j
e , a−1 = − 2j
e , and a0 , a2 , a3 , a4 = 0. Further ak ’s will have periodicity of 6, i.e.,
ak+m6 = ak , m ∈ Z.
4
Problem 4 [25 points].
(a) [17 points] Use Tables 4.1 and 4.2 to help determine the Fourier transform of the following
signal:
sin t 2
.
x(t) = t
πt
Clearly show your method to receive (partial) credit. Please BOX your answer.
(b) [8 points] Use Parseval’s relation and the result of part (a) to determine the numerical value of
Z
∞
A=
t
−∞
2
sin t
πt
4
dt.
Clearly show your method to receive (partial) credit. Please BOX your answer.
5
Problem 5 [25 points]. Suppose g(t) = x(t) cos t and the Fourier transform of g(t) is
(
1, |ω| ≤ 2
G(jω) =
.
0, otherwise
Determine x(t). Clearly show your method to receive (partial) credit. Please BOX your answer.
G(jω) is given so we can use the inverse Fourier transform to obtain g(t). Using The FT pairs
table:
sin(2 · t)
g(t) =
πt
Then,
sin(2 · t)
2 sin t cos t
2 sin t
x(t) =
=
=
.
πt · cos t
πt · cos t
πt