EE202B.

Signals and Systems

Homework 3
Assigned: April 06, 2023
Due: April 14, 2023

Problem 1. (20pts) Complex exponentials as LTI Eigenfunctions:

Consider a causal continuous-time linear time-invariant (LTI) system, whose input x(t) and output
y(t) are related by the following differential equation:
d
y(t) + 4y(t) = x(t). (1)
dt
(a) Suppose x(t) is a periodic signal with period T . Show that the output signal y(t) is also periodic
with period T .
Answer: The system is LTI, so given input x(t + T ) the output would be y(t + T ),
due to time invariance. The periodicity of x(t) tells us that x(t) = x(t + T ) so the
corresponding output y(t) = y(t + T ); therefore, y(t) is periodic with period T .

(b) Let the Fourier series coefficients of x(t) be ak . Express the Fourier series coefficients of y(t),
denoted by bk , as a function of ak and the period T .
Answer:

∞
X
y(t) = bk ejkω0 t
k=−∞
∞
d X
y(t) = jkω0 bk ejkω0 t
dt
k=−∞
∞ ∞ ∞
d X
jkω0 t
X
jkω0 t
X
y(t) + 4y(t) = (4 + jkω0 )bk e = x(t) = jkω0 bk e = ak ejkω0 t
dt
k=−∞ k=−∞ k=−∞

Matching the coefficient for each exponential gives:

(4 + jkω0 )bk = ak
ak
bk =
4 + jkω0
(c) Given that x(t) = cos (2πt), find T, ak , and bk .
Answer:

ej2πt + e−j2πt
x(t) = cos(2πt) =
2
Therefore, T = 1, a1 = 21 , and a−1 1
= 2 . Applying previous results, we find
a1 1
b1 = =
4 + 2πj 8 + 4πj

1
 a−1 1
b−1 = =
4 − 2πj 8 − 4πj
The rest of the ak ’s and bk ’s are zero.

(d) For the input x(t) above, find out explicitly the output y(t) as a function of t.
Answer:

1 8 − 4πj
b1 = =
8 + 4πj 64 + 16π 2
1 8 + 4πj
b−1 = =
8 − 4πj 64 + 16π 2
+ e−j2πt − e−j2πt
   j2πt     j2πt 
j2πt −j2πt 16 e 8π e
y(t) = b1 e + b−1 e = +
64 + 16π 2 2 64 + 16π 2 2j
   
16 8π
= cos(2πt) + sin(2πt)
64 + 16π 2 64 + 16π 2

(e) Repeat steps (c) and (d) above for the input x(t) = cos2 (2πt) + sin (6πt + π/4).
(Hint: cos2 θ = 21 + 12 cos(2θ))
Answer:

1 + cos(4πt)
cos2 (2πt) =
2
π π π
sin(6πt + ) = cos( )sin(6πt) + sin( )cos(6πt)
4 4 4
ejθ +e−jθ ejθ −e−jθ
Applying cos(θ) = 2 and sin(θ) = 2j , we get

1 + j −j6πt 1 −j4πt 1 1 j4πt 1 − j j6πt

x(t) = √ e + e + + e + √ e
2 2 4 2 4 2 2
1+j 1 1 1 1−j
Therefore, T = 1, a−3 = √ ,a
2 2 −2
= 4 , a0 = 2 , a2 = 4, and a3 = √
2 2
Applying results
from Part (b)
a−3 4 − 6π + (6π + 4)j
b−3 = =
4 − 6πj 16 + 36π 2
a−2 1 + πj
b−2 = =
4 − 4πj 16 + 16π 2
a0 1
b0 = =
4 8
a2 1 − πj
b2 = =
4 + 4πj 16 + 16π 2
a3 4 − 6π − (6π + 4)j
b3 = =
4 + 6πj 16 + 36π 2
The rest of the ak ’s and bk ’s are zero.

2
 y(t) = b−3 e−j6πt + b−2 e−j4πt + b0 + b2 ej4πt + b3 ej6πt

       
2 − 3π 3π + 2 1 π 1
= sin(6πt) + sin(6πt) + cos(4πt) + sin(4πt) +
4 + 9π 2 4 + 9π 2 8 + 8π 2 8 + 8π 2 8

Problem 2. (20pts) Constructing a CT Signal from its FS Coefficients:

We are given the following set of information about a signal x(t):

1. x(t) is a real signal,

2. x(t) is periodic with fundamental period T = 6, and has FS coefficients ak ,

3. ak = 0 for k > 2,

4. x(t) = 1 − x(t − 3)
R3
5. 16 −3 |x(t)|2 dt = 38 , and

6. a1 is positive and real.

Show that x(t) = A cos2 (Bt + C), and determine the values of the constants A, B, and C.
Answer:
Information 1 and 3 implies the following:
(i) a0 is real,
(ii) ak = a∗−k , for k = 0, 1, 2.
Therefore, from the synthesis equation, we have the following:

x(t) = a0 + 2Re{a1 ejπt/3 + a2 ej2πt/3 } = 1 − x(t − 3).

Let’s say that the FS coefficient of x(t − 3) are given by bk . So, Equation above implies:
(i) a0 = 1 − b0
(ii) a1 = -b1
(iii) a2 = -b2
(iv) a−1 = -b−1
(v) a−2 = -b−2
Also, by time-shift property of FS coefficient, we have bk = (−1)k ak , for k = 0, 1, 2.
Hence:
(i) a0 = b0
(ii) a1 = -b1
(iii) a2 = b2
(iv) a−1 = -b−1
(v) a−2 = b−2

3
 Given that a1 is real and positive, we know that a1 = a−1 . The above equations
imply a2 = a−2 = 0, and a0 = 1/2. Therefore,
 
1 πt
x(t) = + 2a1 cos
2 3

Now, using Parseval’s relation, we have:

Z 3  2
1 2 1 3
|x(t)| dt = + 2a21 = ,
6 −3 2 8

which gives a1 = 1/4. Therefore,

1 1 πt πt
x(t) = + cos = cos2
2 2 3 6
So, x(t) = Acos2 (Bt + C), with A = 1, B = π/6, and C = 0 (or any integer multiple of π).

4
Problem 3. (20pts)
(a) Determine the Fourier series representations for the following signal:

Answer:
The period of the signal is T = 6.
Z −1
1 3
Z Z 2 
1
a0 = x(t)dt = dt + (−1)dt = 0
T −3 6 −2 1

For k ̸= 0,
Z −1 Z 2    
1 −jk 2π t −jk 2π t j π  2π
ak = e T dt − e T dt = cos k − cos k
T −2 1 kπ 3 3

(b) Use result of part (a) to find the FS representation for the following signal:

Answer:
The period of the signal is T = 6.
Z 2
1 1
a0 = x(t)dt =
6 −2 2

Note that the signal in (a) is the derivative of the signal in (b). We use the result
from part (a) along with the integration property.
    
1 j π  2π
ak = cos k − cos k
jk(2π/T ) kπ 3 3

Problem 4. (20pts) In the following, we specify the Fourier series coefficients of a continuous-time
signal that is periodic with period 4. Determine the signal x(t):
(
0, k=0
ak = k sin kπ/4
(j) kπ , otherwise

5
 Answer:
Here are some of the facts we know about x(t):
a0 = 0 → no DC component in x(t)
T = 4 → ω0 = 2π/4 = π/2
 k
−k sin(−kπ/4) 1 −sin(kπ/4)
a−k = (j) =
−kπ j −kπ

sin(kπ/4)
= (−j)k = a∗k
kπ
Thus x(t) is a real signal.
Noting that j = ejπ/2 → (j)k = (ejπ/2 )k = ejkπ/2 = ejkω0 = e−jkω0 (−1) , we can consider
x(t) to be a time-shifted version of another signal y(t) such that:

sinkπ/4
x(t) = y(t + 1), where, b0 = 0, bk̸=0 = , and, ak = bk ejkω0 (1)
kπ
If we let T = 4 and T1 = 12 , we have ŷ(t) is periodic with period 4 and

|t| < 21

1,
ŷ(t) = 1
0, 2 < |t| < 2

2T1 1
c0 = =
T 4
For k ̸= 0,

sin(kω0 T1 ) sin(kπ/4)
ck = = = bk
kπ kπ
ŷ(t) has a non-zero DC level, but we can shift the signal to eliminate c0 . Define ȳ(t)
periodic with period 4 and
 3
1 4, |t| < 21
ȳ(t) = ŷ(t) − = 1 1
4 − 4 , 2 < |t + 1| < 2

Then ȳ(t) has FS coefficient bk and hence y(t) = ȳ(t). We know x(t) = y(t + 1) = ȳ(t + 1),
and therefore x(t) is periodic with period 4 and
 3
4, |t + 1| < 12
x(t) = 1 1
− 4 , 2 < |t + 1| < 2

Problem 5. (20pts) Suppose we are given the following information about a periodic signal x[n]
with period 8 and Fourier coefficients ak :

1. ak = −ak−4

2. x[2n + 1] = (−1)n .

6
Sketch one period of x[n].
Answer:
According to the frequency shifting property, ak−4 are the Fourier series coefficients
for (−1)n x[n]. Thus, Information 1 is equivalent to

x[n] = −(−1)n x[n] for all n.

For n even, this implies that x[n] = 0. For n odd, the statement is a tautology and,
therefore, gives no information. Information 2 however, determines x[n] for n odd
uniquely. Hence,

0 n even
x[n] =
(−1)(n−1)/2 n odd
which is sketched below for 0 ≤ n ≤ 7