Fourier Analysis Review for engineering.

Review on Fourier Analysis
© Samy S. Soliman
Faculty of Engineering - Cairo University, Egypt
University of Science and Technology - Zewail City, Egypt
American University in Cairo, Egypt
Email: samy.soliman@ualberta.net
Website: http://scholar.cu.edu.eg/samysoliman
© Samy S. Soliman Fourier Analysis 1 / 61

Review on Fourier Analysis
1 Introduction to Fourier Analysis
2 Fourier Series of Periodic Continuous Signals
3 Periodic Rectangular Signal
4 Periodic Train of Impulses
5 Properties of CT Fourier Series
6 Continuous Time Fourier Transform
7 Fourier Transform of Periodic CT Signals
8 Basic Fourier Transform Pairs
9 Properties of Continuous Time Fourier Transform

Introduction to Fourier
Analysis

Introduction to Fourier Analysis
1 It is advantageous in the study of LTI systems to decompose a signal
into a linear combination of basic signals
2 Basic signals should possess the following two properties:
Can construct a broad class of signals.
Generate simple response by LTI systems to provide a convenient
representation for the response of the system to any signal constructed
as a linear combination of the basic signals
3 Complex exponential signals provide both properties in continuous
and discrete time

Transform Analysis

Introduction to Fourier Analysis: CT Signals
Assume a CT signal x(t) = est.
Assume x(t) is the input of an LTI systems characterized by h(t).
Then the system’s output is:
y(t) =
Z ∞
−∞
h(τ)x(t − τ)dτ
=
Z ∞
−∞
h(τ)es(t−τ)
dτ
= est
Z ∞
−∞
h(τ)e−sτ
dτ
= H(s)est
, where H(s) =
Z ∞
−∞
h(τ)e−sτ
dτ

For an input signal x(t) =
P
k akesk t, the system’s output is:
y(t) =
X
k
akH(sk)esk t
Conclusion
If the input of an LTI system is a linear component of exponential signals,
the output will be a linear combination of scaled exponential.
Note: H(s) is a characteristic to the system related to its impulse
response h(t).
If we can represent input signals in terms of exponential signals, the
system’s output can be easily obtained.

Example: System y(t) = x(t − 3)
Impulse Response: h(t) = δ(t − 3)
Then, H(s) =
Z ∞
−∞
δ(τ − 3)e−sτ
dτ
= e−3s
For an input signal x(t) = ej2t, the output can be evaluated as
(s = 2j) ⇒ y(t) = ej2t
H(j2) = ej2t
e−j6
Note:
H(s) = e−j6 is an Eigenvalue of the Eigenfunction est = ej2t

Fourier Series Analysis

Fourier Series of Periodic Continuous Signals
For a periodic signal x(t) with period T = 2π
ω0
,
Theorem (FS of Continuous-Time Signals)
Synthesis Equation
x(t) =
∞
X
k=−∞
akejkω0t
=
∞
X
k=−∞
akejk 2π
T
t
Analysis Equation
ak =
1
T
Z
T
x(t)e−jkω0t
dt =
1
T
Z
T
x(t)e−jk 2π
T
t
dt
T is the Fundamental Period, ω0 is the Fundamental Frequency

Example on Fourier Series
x(t) = cos(4πt) + 2 sin(8πt)
x(t) =
1
2

ej4πt
− e−j4πt

+ 2
1
2j

ej8πt
− e−j8πt

ω1 = 4π, ω2 = 8π ⇒ T1 =
1
2
, T2 =
1
4
⇒ T =
1
2
and ω0 = 4π
x(t) =
∞
X
k=−∞
akejkω0t
⇒ ar =















1
2, r = 1
−1
2 , r = −1
1
j , r = 2
−1
j , r = −2
0, otherwise

Periodic CT Rectangular Signal
Find the Fourier series representation of the periodic rectangular signal,
where one period from −T/2 to T/2 is defined as
x(t) =
(
1 , |t| T1
0 , T − 1 |t| T/2

ak =
1
T
Z T/2
−T/2
x(t)e−jkω0t
dt
=
1
T
Z T1
−T1
(1)e−jkω0t
dt
=
1
−jkω0T
h
e−jkω0T1
− e+jkω0T1
i
, k 6= 0
= 2
sin(kω0T1)
kω0T
=
2T1
T
sinc

kω0T1
π

=
2T1
T
sinc

k
2T1
T

a0 =
2T1
T

FS of Periodic CT Rectangular Signal
ak =
2T1
T
sinc

k
2T1
T


Periodic Train of Impulses
Train of Impulses
x(t) =
∞
X
−∞
δ(t − kT)
The Fourier series coefficients are obtained as
ak =
1
T
Z T/2
−T/2
δ(t)e−jkω0t
dt
=
1
T
∀k

Properties of FS of CT Signals
Assuming two periodic signals, with the same period, such that
x(t) FS
←
→ ak
y(t) FS
←
→ bk
1- Linearity
z(t) = Ax(t) + By(t) FS
←
→ ck = Aak + Bbk
2- Time Shifting
x(t − t0) FS
←
→ ake−jkω0t0

3- Time Reversal
x(−t) FS
←
→ a−k
Activity: Think (If x(t) is even ⇒ ak is · · · )
Activity: Think (If x(t) is odd ⇒ ak is · · · )
4- Time Scaling
x(αt) FS
←
→ ak
Activity: Think (What is the period of x(αt)?)
Activity: Think (What is the fundamental frequency of x(αt)?)
Activity: Analyze (Comment on the signals x(t) and x(αt), as well as
their FS representations

5- Multiplication
z(t) = x(t) × y(t) FS
←
→ ck =
∞
X
m=−∞
ambk−m
Activity: Think (How to prove this property?)
6- Conjugation
x∗
(t) FS
←
→ a∗
−k
Activity: Identify (If x(t) is real and even ⇒ ak is · · · )
Activity: Think (If x(t) is real and odd ⇒ ak is · · · )

7- Frequency Shifting
ejmω0t
x(t) FS
←
→ ak−m
8- Periodic Convolution
Z
T
x(τ)y(t − τ)dτ FS
←
→ Takbk
9- Even-Odd Decomposition of Real Signals
E{x(t)} FS
←
→ R{ak}
O{x(t)} FS
←
→ jI{ak}

10- Differentiation
d
dt
x(t) FS
←
→ jkω0ak
11- Integration
Z t
−∞
x(τ)dτ FS
←
→
1
jkω0
ak

Fourier Series: Parseval’s Theorem
Theorem (Parseval’s Theorem)
The total average power in a periodic signal equals to the sum of the
average powers in all its harmonic components
Parseval’s Relation
1
T
Z
T
|x(t)|2
dt =
∞
X
k=−∞
|ak|2
Activity: Analyze (Can you analyze Parseval’s Theorem and prove it?)

Find the Fourier series representation of the periodic rectangular signal,
where one period from −T/2 to T/2 is defined as
x(t) =
(
1 , |t| T1
0 , T − 1 |t| T/2

y(t) =
dx(t)
dt
, x(t) =
Z t
−∞
y(τ)dτ, ak =
1
jkω0
bk
y(t) = z(t + T1) − z(t − T1), z(t) is an Impulse Train

bk = ejkω0T1
ck − e−jkω0T1
ck, ck =
1
T
ak =
1
jkω0
bk =
1
jkω0T
h
ejkω0T1
− e−jkω0T1
i
=
2T1
T
sinc

k
2T1
T


Continuous Time Fourier
Transform

Continuous Time Fourier Transform
Fourier Transform is used to represent Aperiodic signals
Theorem (FT of Continuous-Time Signals)
Synthesis Equation - IFT Equation
x(t) =
1
2π
Z ∞
−∞
X(jω)ejωt
dω
Spectrum Equation - FT Equation
X(jω) =
Z ∞
−∞
x(t)e−jωt
dt

x(t) = x̃(t), as T → ∞
x(t) =
(
x̃(t), |t| T/2
0, otherwise

Fourier Transform
Tak =
Z T/2
−T/2
x̃(t)e−jkω0t
dt
=
Z ∞
−∞
x(t)e−jkω0t
dt
Letting Tak = X(jkω0)
X(jω) =
Z ∞
−∞
x(t)e−jωt
dt ⇒ FT

Inverse Fourier Transform
x̃(t) =
∞
X
−∞
akejkω0t
=
∞
X
−∞
1
T
X(jkω0)ejkω0t
=
∞
X
−∞
ω0
2π
X(jkω0)ejkω0t
As T → ∞, x(t) =
1
2π
Z ∞
−∞
X(jω)ejωt
dω ⇒ IFT

Fourier Transform is used to represent Aperiodic signals
Theorem (FT of Continuous-Time Signals)
Synthesis Equation - IFT Equation
x(t) =
1
2π
Z ∞
−∞
X(jω)ejωt
dω
Spectrum Equation - FT Equation
X(jω) =
Z ∞
−∞
x(t)e−jωt
dt
Note:
The signal must be absolutely integrable, i.e.
Z ∞
−∞
|x(t)|dt ∞

CTFT Examples
Delta Function
x(t) = δ(t)
X(jω) =
Z ∞
−∞
δ(t)e−jωt
dt = 1

CTFT Examples
Aperiodic Rectangular Signal
x(t) = 1, |t| T1
X(jω) =
Z T1
−T1
e−jωt
dt = 2
sin(ωT1)
ω

CTFT Examples
Sinc Signal: Duality
X(jω) = 1, |ω| W
x(t) =
1
2π
Z W
−W
ejωt
dω =
sin(Wt)
πt

CTFT of Periodic Signals
For X(jω) = 2πδ(ω − kω0), the IFT is obtained as
x(t) =
1
2π
Z ∞
−∞
X(jω)ejωt
dω = ejkω0t
Then, applying linearity,
For a periodic signal, x(t), expressed in terms of its Fourier series
coefficients, ak,
x(t) =
∞
X
k=−∞
akejkω0t
The Fourier transform can be obtained as
X(jω) =
∞
X
k=−∞
2πakδ(ω − kω0)

Periodic Rectangular Signal
X(jω) =
∞
X
k=−∞
2π
sin(kω0t)
πt
δ(ω − kω0)

x(t) = sin(ω0t)
X(jω) =
π
j
δ(ω−ω0)−
π
j
δ(ω+ω0)
x(t) = cos(ω0t)
X(jω) = πδ(ω−ω0)+πδ(ω+ω0)

Train of Impulses
x(t) =
∞
X
k=−∞
δ(t − kT) ⇔ X(jω) =
∞
X
k=−∞
2π
T
δ(ω − k
2π
T
)

Basic Fourier Transform Pairs

Fourier Transform: Properties
Assuming two signals such that
x(t) FT
←
→ X(jω)
y(t) FT
←
→ Y (jω)
1- Linearity
z(t) = Ax(t) + By(t) FT
←
→ ck = AX(jω) + BY (jω)
2- Time Shifting
x(t − t0) FT
←
→ e−jωt0
X(jω)

3- Time Reversal
x(−t) FT
←
→ X(−jω)
4- Time and Frequency Scaling
x(αt) FT
←
→
1
|α|
X

jω
α


Example
x(t) = x2(t − 2.5) + 0.5x1(1 − 2.5)
X(jω) = e−j2.5ω
[X2(jω) + 0.5X1(jω)]
= e−j2.5ω

2 sin(3ω/2)
ω
+ 0.5
2 sin(ω/2)
ω


5- Frequency Shifting
ejω0t
x(t) FT
←
→ X (j(ω − ω0))
6- Conjugation
x∗
(t) FT
←
→ X∗
(−jω)
E{x(t)} FT
←
→ R{X(jω)}
O{x(t)} FT
←
→ jI{X(jω)}
Activity: Identify (If x(t) is real and even ⇒ X(jω) is · · · )
Activity: Identify (If x(t) is real and odd ⇒ X(jω) is · · · )

Example
x(t) = e−a|t|
, where a 0
x(t) = eat
u(−t) + e−at
u(t)
= 2E{e−at
u(t)}
X(jω) = 2R{
1
a + jω
}
=
2a
a2 + ω2

7- Multiplication
z(t) = x(t) × y(t) FT
←
→ Z(jω) =
1
2π
Z ∞
−∞
X(jθ)Y (j(ω − θ))dθ
Note: This property is sometimes called the Modulation Property
8- Convolution
x(t) ∗ y(t) =
Z t
−∞
x(τ)y(t − τ)dτ FT
←
→ X(jω) × Y (jω)

Example
A signal s(t), whose FT is S(jω), is modulated by p(t) = cos(ω0t). Find
the FT of the resulting signal, R(jω)
r(t) = s(t) × p(t)
R(jω) =
1
2π
Z ∞
−∞
S(jθ)P(j(ω − θ))dθ
=
1
2π
Z ∞
−∞
S(jθ)π [δ(ω − ω0 − θ) + δ(ω + ω0 − θ)] dθ

R(jω) =
1
2
Z ∞
−∞
S(jθ)δ(ω − ω0 − θ)dθ +
Z ∞
−∞
S(jθ)δ(ω + ω0 − θ)dθ

=
1
2
[S(j(ω − ω0)) + S(j(ω + ω0))]

Example
Find the output, y(t), of a system whose impulse response, h(t), and
input, x(t), are defined as
h(t) = e−at
u(t), x(t) = e−bt
u(t), where a, b 0, a 6= b
y(t) = x(t) ∗ h(t)
Y (jω) = X(jω)H(jω)
=
1
(a + jω)
1
(b + jω)
=
1/(b − a)
a + jω
+
1/(a − b)
b + jω
y(t) =
1
b − a
h
e−at
− e−bt
i
u(t)

9- Differentiation
d
dt
x(t) FT
←
→ jωX(jω)
10- Integration
Z t
−∞
x(τ)dτ FT
←
→
1
jω
X(jω) + πX(0)δ(ω)

Example
y(t) =
d
dt
x(t)
Y (jω) =
2 sin(ω)
ω
− (ejω
+ e−jω
) =
2 sin(ω)
ω
− 2 cos(ω)
X(jω) =
1
jω
Y (jω) + πY (0)δ(ω) =
2 sin(ω)
jω2
−
2 cos(ω)
jω
+ 0

11- Duality
For any transform pair, there is a dual pair with the time and frequency
variables interchanged
X(t) FT
←
→ 2πx(−ω)
This property can be used to drive other properties such as:
Differentiation and Integration in Frequency Domain
−jtx(t) FT
←
→
d
dω
X(jω)
1
−jt
x(t) + πx(0)δ(t) FT
←
→
Z ω
−∞
X(jθ)dθ

Example
Find the output, y(t), of a systems whose impulse response, h(t), and
input, x(t), are defined as
h(t) = e−at
u(t), x(t) = e−at
u(t), where a 0
y(t) = x(t) ∗ h(t)
Y (jω) = X(jω)H(jω)
=
1
(a + jω)
1
(a + jω)
=
1
(a + jω)2
y(t) = te−at
u(t)

Fourier Transform: Parseval’s Theorem
Parseval’s Relation
Z ∞
−∞
|x(t)|2
dt =
1
2π
Z ∞
−∞
|X(jω)|2
dω

Summary of CTFT Properties
Properties of CTFT
z(t) = Ax(t) + By(t) FT
←
→ Z(jω) = AX(jω) + BY (jω)
x(t − t0) FT
←
→ e−jωt0
X(jω)
x(−t) FT
←
→ X(−jω)
x (αt) FT
←
→
1
|α|
X

j
ω
α

z(t) = x(t) × y(t) FT
←
→ Z(jω) =
1
2π
X(jω) ∗ Y (jω)
x∗
(t) FT
←
→ X∗
(−jω)
E{x(t)} FT
←
→ R{X(jω)}
O{x(t)} FT
←
→ jI{X(jω)}

Summary of CTFT Properties
Properties of CTFT
ejω0t
x(t) FT
←
→ X (j(ω − ω0))
x(t) ∗ y(t) FT
←
→ X(jω) × Y (jω)
dx(t)
dt
FT
←
→ jωX(jω)
Z t
−∞
x(τ)dτ FT
←
→
1
jω
X(jω) + πX(0)δ(ω)
tx(t) FT
←
→ j
d
dω
X(jω)
Z ∞
−∞
|x(t)|2
dt =
1
2π
Z ∞
−∞
|X(jω)|2
dω

References
Alan V. Oppenheim, and Alan S. Willsky (1997)
Signals and Systems, 2nd Edition.
Prentice Hall.

Thank You!
samy.soliman@ualberta.net
https://www.youtube.com/c/SamySSoliman

Fourier Analysis Review for engineering.

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