Winter 2024

ECE 102
Prof. Christina Fragouli
TA: Semira Galijasevic

ECE 102: Signals and Systems

Final Solutions
Monday March 18, 2024

NAME: UID:

This exam has 4 problems for a total of 200 points

(and one bonus problem for an additional 50 points).

The exam is for 2.5 hours.

Closed book. No calculators. No electronic devices.
One page, letter-size, two-sided cheat-sheet allowed.
Please answer the questions in the spaces provided on the question sheets. If you
run out of room for an answer, append pages at the end of the exam.
Please, write your name and UID on the top of each loose sheet

Good luck!

1
Problem 1 (50 points): The following three questions are not related to each other.

(a) (20 points) Find the signal y(t) = x1 (t) ∗ x2 (t) by calculating the convolution through inte-
gration, where

1 0 ≤ t < 1
 

t 0 ≤ t < 1 
x1 (t) = and x2 (t) = 3 1 ≤ t < 2 (1)
0 otherwise 

0 otherwise


Solution:

x1 (t) x2 (t)
3

1 1

t t
1 2

R∞
Let y(t) = x1 (t) ∗ x2 (t) = −∞ x1 (τ )x2 (t − τ )dτ .

Case 1: t < 0 =⇒ y(t) = 0.

Case 2: 0 ≤ t < 1
t
t2
Z
y(t) = τ dτ = .
0 2

Case 3: 1 ≤ t < 2
Z t−1 Z 1
1 3
y(t) = 3τ dτ + τ dτ = (t − 1)2 + = t2 − 2t + .
0 t−1 2 2

Case 4: 2 ≤ t < 3
1
−3 2
Z
3 9
y(t) = 3τ dτ = (1 − (t − 2)2 ) = t + 6t −
t−2 2 2 2

Case 5: t > 3 =⇒ y(t) = 0.

(b) (10 points) Assume that z(t) = x(t) ∗ y(t), where the signal z(t) is depicted in the figure
below. Plot the signal w(t) = x(t − 1) ∗ y(t + 3).

2
 z(t)

t
−1 1

Solution:

To solve this problem we can use properties of convolution with delta function. Thus, we can
write

x(t − 1) = x(t) ∗ δ(t − 1) (2)

y(t + 3) = y(t) ∗ δ(t + 3) (3)

Plugging (2) and (3) into w(t) = x(t − 1) ∗ y(t + 3) we get following:

w(t) = (x(t) ∗ δ(t − 1)) ∗ (y(t) ∗ δ(t + 3)) (4)

= δ(t − 1) ∗ x(t) ∗ y(t) ∗ δ(t + 3) (5)
= δ(t − 1) ∗ z(t) ∗ δ(t + 3) (6)
= z(t − 1) ∗ δ(t + 3) (7)
= z(t − 1 + 3) (8)
= z(t + 2) (9)

where (5) follows from commutative property of convolution and (6) follows from z(t) =
x(t) ∗ y(t).

w(t)

t
−3 −1

(c) (20 points) Prove that if h(t) = f (t) ∗ g(t) then

Z +∞ Z +∞  Z +∞ 
h(x) dx = f (x) dx g(x) dx . (10)
−∞ −∞ −∞

3
Informally, areas multiply under convolution.

Solution:

Take the Fourier transform of h = f ∗ g to get F{h} = F{f }F{g} and evaluate at 0:
Z ∞ Z ∞  Z ∞ 
h(x) dx = F{h(0)} = F{f (0)}F{g(0)} = f (x) dx g(x) dx .
−∞ −∞ −∞

4
Problem 2 (50 points): The following questions are not related to each other.

1. (25 points) Find the Fourier Transform for the following three signals:
1
(a) x1 (t) = cos2 (t) − 4

Solution:

1 1 cos(2t) 1 1 cos(2t)
x1 (t) = cos2 (t) − = + − = +
4 2 2 4 4 2
1 11
X1 (ω) = 2πδ(ω) + [2πδ(ω − 2) + 2πδ(ω + 2)]
4 22
π π π
= δ(ω) + δ(ω − 2) + δ(ω + 2)
2 2 2
(b) x2 (t) = x1 (−2t)

Solution:

1 ω
x1 (−2t) → X1 ( )
| − 2| −2
1 ω
X2 (ω) = X1 ( )
| − 2| −2
1 π −ω π −ω π −ω
= [ δ( ) + δ( − 2) + δ( + 2)]
2 2 2 2 2 2 2
π π π
= δ(ω) + δ(ω + 4) + δ(ω − 4)
2 2 2

(c) x3 (t) = x2 (t) − x1 (t)e2jt

Solution:

x3 (t) = x2 (t) − x1 (t)e2jt

X3 (ω) = X2 (ω) − X1 (ω − 2)

Then,
π π π π π π
X3 (ω) = δ(ω) + δ(ω + 4) + δ(ω − 4) − [ δ(ω − 2) + δ(ω − 4) + δ(ω)]
2 2 2 2 2 2
π π
= − δ(ω − 2) + δ(ω + 4)
2 2
1
You may use (without proving) the fact that δ(aω) = |a| δ(ω).

2. (10 points) Consider a periodic signal x(t) with period T0 = 5 sec and FS coefficients

3 cos(3πk) −ik 2π ik 2π
ck = 2π (e
3 + e 3 )].
−ik 3

Is the signal x(t) even, odd, or neither?

5
 We can write ck as follows:
9 1 2π 2π 9
ck = − cos(3πk) (e−ik 3 + eik 3 ) = − cos(3πk) cos(2πk/3)
iπk 2 iπk
9
=i cos(3πk) cos(2πk/3)
πk
Since ck is purely imaginary we know that the signal can’t be even. Now we check if the
signal is odd. The signal is odd if the FS coefficients are purely imaginary and ck = −c−k .
Lets find c−k :
9
c−k = i cos(3π(−k)) cos(2π(−k)/3)
π(−k)
9
= −i cos(3πk) cos(2πk/3)
πk
= −ck

Therefore, signal x(t) is odd.

3. (15 points) Describe how you would design the frequency response (i.e., how |H(ω)| vs. ω
should behave) of an ideal filter, that can take as input a periodic signal of fundamental
period T0 and output a (nonzero if possible) signal that has fundamental period T0 /2.

Solution: The original signal has nonzero coefficients at frequencies who are multiples of the fun-
damental frequency f0 = T10 . We want the output signal to have nonzero coefficients at frequencies
who are multiples of T20 = 2f0 . Thus we need a filter that removes (makes zero) all coefficients ck
for k odd and allows to pass all frequencies corresponding to k even (2f0 , −2f0 , 4f0 , .. etc. Note
that the resulting signal still has T0 as period; but its fundamental period is T0 /2.

6
Problem 3 (40 points): The following two questions are not related to each other.

1
1. (15 points) The transfer function of a causal LTI system is H(s) = s2 +16
.

(a) (10 points) Find the differential equation that relates the input x(t) to the output y(t)
of this system.

Solution:

1
Y (s) = H(s)X(s) = X(s)
+ 16 s2
s2 Y (s) + 16Y (s) = X(s)
Thus,
d2 y(t)
+ 16y(t) = x(t)
dt2
(b) (5 points) Is this system stable?

Solution:
To determine if the system is stable we look into ROC of transfer function H(s) which is
σ > 0 since H(s) has two poles at ±4i. Since ROC does not include iω axis the system
is not stable.

2. (25 points) Consider a system that has the following input-output equation (this system is
called analog averager):
Z t
y(t) = x(τ )dτ (11)
t−T
where T is a given constant.

(a) (7 points) Is this system LTI?

Solution:
Yes, this system is LTI. Proof:
Linearity:
Rt Rt
Let y1 (t) = t−T x1 (τ )dτ , y2 (t) = t−T x2 (τ )dτ and let x3 (t) = αx1 (t) + βx2 (t) . Then,
Z t
y3 (t) = x3 (τ )dτ
t−T
Z t
= (αx1 (τ ) + βx2 (τ ))dτ
t−T
Z t Z t
=α x1 (τ )dτ + β x2 (τ )dτ
t−T t−T

= αy1 (t) + βy2 (t)

7
 Time Invariance:
Let x1 = x(t − t0 ). Then
Z t
S{x1 (t)} = x1 (τ )dτ
t−T
Z t
= x(τ − t0 )dτ
t−T
Z t−t0
= x(u)du - change of variables u = τ − t0
t−T −t0
Z t−t0
= x(τ )dτ - change of variables τ = u
t−T −t0

= y(t − t0 )

(b) (10 points) Is this system BIBO stable?

Solution:
The system is BIBO stable. Proof:
For a LTI system to be BIBO stable, the impulse response must be absolute integrable,
R∞
i.e. −∞ |h(τ )|dτ < ∞. Lets calculate impulse response of this system.
Z t  
t − T /2
h(t) = Sys{δ(t)} = δ(τ )dτ = u(t) − u(t − T ) = Π .
t−T T
The impulse response is rectangular function which is absolute integrable, so the system
is BIBO stable.
(c) (8 points) What is the output of this system when the input is r(t), where r(t) = tu(t)
is the ramp function?

Solution:
Z ∞
y(t) = h(t) ∗ r(t) = h(t − τ )r(τ )dτ
−∞

Case 1: t < 0 :

y(t) = 0

Case 2: 0 < t < T :

t
t2
Z
y(t) = τ dτ = .
0 2
Case 3: t ≥ T :
t
t2 (t − T )2 t2 − t2 + 2tT − T 2
Z  
T
y(t) = τ dτ = − = =T t− .
t−T 2 2 2 2

Alternative Solution: Instead of convolution, this problem can be solved using

input-output equations.

8
Problem 4 (60 points) The following questions are not related to each other.

1. (20 points) Find the signal y(t) that satisfies the equation:
Z t
y(τ )dτ = 3u(t) − 2y(t) (12)
0

where u(t) is the unit step function.

Solution:

Applying the Laplace Tranform on Eqn (12) and using the integration property of Laplace
Transform we have:
Z t 
Y (s) 3
L y(τ )dτ = = − 2Y (s)
0 s s
3
=⇒ Y (s) =
2(s + 0.5)

Then taking inverse Laplace Transform of Y (s) we get y(t) = 1.5e−0.5t u(t).

2. (20 points) What is the Nyquist frequency for the signal x1 (t) = A + eit sinc(3π(5 − t))?

Solution:

To find the Nyquist frequency we need to find a Fourier transform of the signal x1 (t). Let
x(t) = sinc(t) = sinπtπt . We know that Fourier transform of x(t) = sinc(t) is X(ω) = u(ω +
π) − u(ω − π) = Π(ω/2π). Since sinc is even function then sinc(3π(5 − t)) = sinc(3π(t − 5)).
Using scaling property we have,
1 1 ω
sinc(3πt) ←→ [u(ω + 3π 2 ) − u(ω − 3π 2 )] = Π( 2 )
3π 3π 6π
Then, using delay property we have,

x(t − td ) ←→ e−jωtd X(ω)

1 ω
sinc(3π(t − 5)) ←→ Π( 2 )e−j5ω
3π 6π
and
x(t)ejω0 t ←→ X(ω − ω0 )
1 ω − 1 −j5(ω−1)
eit sinc(3π(t − 5)) ←→ Π( )e
3π 6π 2
Finally,
1 ω − 1 −j5(ω−1)
x1 (t) = A + eit sinc(3π(t − 5)) ←→ 2πAδ(ω) + Π( )e
3π 6π 2
The highest frequency component of the signal x1 (t) is ωc = 3π 2 + 1 rad. Thus, the Nyquist
frequency is ωs = 2ωc = 6π 2 + 2 rad.

9
3. (20 points) Assume you are given a causal LTI system

(a) Find the Laplace transform of the triangular signal Λ(t) = r(t + 1) − 2r(t) + r(t − 1) and
its ROC. Hint. Be careful for zero-pole cancelations.

Solution:
1
Laplace transform of r(t) is s2
. Using the linearity and time shift property we have:

1
L[Λ(t)] = Λ(s) = (es − 2 + e−s )
s2
Now, we find ROC.
The zeros of Λ(s) are the values of s that make es − 2 + e−s = 0. Multiplying this
equation with e−s we get (1 − e−s )2 = 0. Therefore we have double zeros at sk = j2πk,
k = 0, ±1, ±2, .... In particular when k = 0 there are two zeros at 0 which cancel the
two poles at 0 resulting from the denominator s2 . Thus, Λ(s) has an infinite number of
zeros but no poles given this pole-zero cancellation. Therefore, Λ(s), as a signal of finite
support, has the whole s− plane as its region of convergence and can be calculated at
s = jω.
(b) Assume the output the LTI system equals y(t) = r(t) − 2r(t − 1) + r(t − 2) when the
input is x(t) = u(t) − u(t − 1). Find the impulse response of this system.

Solution: We can find the impulse response by using Laplace transform as follows,

Y (s) L(r(t) − 2r(t − 1) + r(t − 2)) 1

s2
(1 − 2e−s + e−2s ) 1
H(s) = = = 1 = (1 − e−s )
X(s) L(u(t) − u(t − 1) s (1 − e−s ) s

Thus,

h(t) = u(t) − u(t − 1)

10
Problem 5 [Extra Points Problem] (50 points)

1. (10 points) Calculate the Fourier Transform for the signal r(t) = tu(t).

Solution:
d
Let X3 (ω) denote Fourier transform of r(t). We know that u(t) = dt r(t). Using the Fourier
transform derivative property we have :
d 1
F {u(t)} = F { r(t)} = jωX3 (ω) =⇒ X3 (ω) = F {u(t)}
dt jω
Therefore,
 
1 1 −jπδ(ω) 1
X3 (ω) = πδ(ω) + = − 2
jω jω ω ω

2. (10 points) Find the energy of the signal x(t) = sinc(3t+2).

Solution: The energy of x(t) is given by:

Z ∞ Z ∞
1
|x(t)|2 dt = |X(jω)|2 dω
−∞ 2π −∞

Now,
1 1 ω
X(jω) = ej2ω/3 X(jω/3) = ej2ω/3 Π( )
3 3 6π
Therefore,
Z ∞ Z 3π
1 2 1 1 6π 1
|X(jω)| dω = 1dω = = .
2π −∞ 2π 9 −3π 18π 3

3. (30 points) For a periodic signal x(t), of some period T0 , assume we can compute the Laplace
transform of a period of x(t), namely

x1 (t) = x(t)(u(t − t0 ) − u(t − t0 − T0 )) (13)

for any constant t0 (since it does not matter from where we start counting a period). Your
friend claims that we can then calculate the FS coefficients of x(t) using the formula:
1
ck = L[x1 (t)]s=ikω0 (14)
T0
(a) (15 points) Provide an argument on why your friend is correct.

Solution:
The claim above can be verified by comparing Xk coefficient with the Laplace transform
of a period x1 (t) = x(t)(u(t − t0 ) − u(t − t0 − T0 )) of x(t):
Z t0 +T0
1
ck = x(t)e−ikω0 t dt
T0 t0
Z t0 +T0
1
= x1 (t)e−st dt|s=ikω0
T0 t0
1
= L[x1 (t)]s=ikω0
T0

11
(b) (15 points) Use this method to calculate the FS coefficients for the signal x(t) = | cos(πt)|.

Solution:
A period x1 (t) of x(t) = | cos(πt)| can be expressed as x1 (t − 0.5) = sin πtu(t) + sin(π(t −
1))u(t − 1). Using the Laplace transform we have
π
X1 (s)e−0.5s = (1 + e−s )
s2 + π2
π
X1 (s) = (e0.5s + e−0.5s)
s2 + π2
The Fourier coefficients are then
1
Xk = X1 (s)|s=iω0 k
T0
where T0 = 1 and ω0 = 2π, giving
π
Xk = 2 cos(2πk/2)
(i2πk)2 + π 2
2(−1)k
=
π(1 − 4k 2 )

since cos πk = (−1)k . The DC value is X0 = 2/π. Notice that the Fourier coefficients
are real given that the signal is even.

12
 Fourier Transform Properties

R∞ −jωt dt.
1. The Fourier Transform: F (ω) = −∞ f (t)e
1
R∞ jωt dω
2. The Inverse Fourier Transform: f (t) = 2π −∞ F (ω)e
R∞
3. Sufficient condition (not necessary) for FT to exist: −∞ |x(t)|dt <∞

4. Properties of Fourier Transform:

1) Linearity F {ax1 (t) + bx2 (t)} = aF {x1 (t)} + bF {x2 (t)}, ∀a, b constants
1 ω
2) Scaling x(at) ←
→ |a| X( a ) ∀a ̸= 0, a is a real constant
Special case: x(−t) ←
→ X(−ω)
3) Delay → e−jωtd X(ω)
x(t − td ) ←
x(t)ejω0 t ←
→ X(ω − ω0 )
4) Symmetry x∗ (t) ←
→ X(−ω)
even and real ←→ even and real
even and imaginary ← → even and imaginary
odd and real ←→ odd and imaginary
odd and imaginary ←→ odd and real
5) Convolution y(t) = x1 (t) ∗ x2 (t) ←
→ Y (ω) = X1 (ω)X2 (ω)
1
6) Multiplication y(t) = x1 (t)x2 (t) ←
→ Y (ω) = 2π X1 (ω) ∗ X2 (ω)
dk x(t)
7) Derivative Property dtk
←
→ (jω)k X(ω)
R∞ 1
R∞
8) Parseval’s Identity −∞ |x(t)|2 dt = 2π
2
−∞ |X(ω)| dω
9) Duality if x(t) ←
→ X(ω) then
X(t) ←→ 2πx(−ω)

5. Fourier Transform Pairs:

Time Frequency
δ(t) 1
1 2πδ(ω)
e−at u(t) 1
jω+a
1
eat u(−t) a−jω
e−|t| 2a
a2 +ω 2
ejω0 t 2πδ(ω − ω0 )
A cos(ω0 t + ϕ) πAejϕ δ(ω − ω0 ) + πAe−jϕ δ(ω + ω0 )
sin(ω0 t) −jπδ(ω − ω0 ) + jπδ(ω + ω0 )
cos(ω0 t) πδ(ω − ω0 ) + πδ(ω + ω0 )
sin(ωb t)
πt u(ω + ωb ) − u(ω − ωb )
sin(ωT /2)
u(t + T /2) − u(t − T /2) ω/2 = T sinc( ωT
2π )
1
u(t) πδ(ω) + jω

13
1. Laplace Transform Pairs

2. Laplace Transform Properties

Laplace Transform
1. Convolution Property: f1 (t) ∗ f2 (t) ←−−−−−−−−−−→ F1 (s)F2 (s)
dn f (t) Laplace Transform Pn n−i f i−1 (0− )
2. n-th Derivative: dtn ←−−−−−−−−−−→ sn F (s) − i=1 s

14