chapter3_fourier_series. on signal and system

Fourier Series
Spring 2024 EE220 Signals and Systems
Dr. Muhammad Rehan Chaudhry
Biomedical Engineering Department
UET Lahore Narowal Campus
April 15, 2024
Dr. Muhammad Rehan Chaudhry Fourier Series Spring 2024 EE220 Signals and Systems 1 / 59

The Fourier Series Representation - Synthesis and Analysis
Intuition behind the Fourier Series
You likely have many colorful pictures stored on your laptop or smartphone.
In a colored image, each pixel is composed of specific values for its red,
green, and blue components, often denoted as r, g, and b respectively.
When these components have their maximum value, the pixel appears white
on the screen, while at their minimum value, the pixel appears black.
Each of these colors can be thought of as analogous to exponential functions
in the Fourier series. When combining these three primary colors in varying
intensities, we can create a wide range of colors. This additive color mixing
is a fundamental principle in the RGB color model used in digital displays
and image processing.
The concept of combining different wavelengths of light to create colors can
be likened to the principle of combining complex exponential functions of
different frequencies in Fourier series to represent complex periodic signals.

Figure: Color Wheel: Intersection Producing White

In linear algebra, basis vectors are a set of vectors in a vector space that
are linearly independent and span the entire space. This means that any
vector in the space can be expressed as a unique linear combination of the
basis vectors
Figure: Every vector a in three dimensions is a linear combination of the standard basis vectors î, ĵ and k̂.
1
https://en.wikipedia.org/wiki/Standard basis#/media/File:3D Vector.svg

The Fourier series is a way to represent a periodic function as an infinite
sum of complex exponentials. These complex exponentials can be thought
of as the “basis functions” for the space of periodic functions, because just
like basis vectors, they can be used to construct any function in that space.
Just like in linear algebra, where basis vectors are orthogonal to each other,
the basis functions in Fourier series are also orthogonal over one period.
This orthogonality property is crucial for decomposing a function into its
Fourier series representation.
Just as any vector in a vector space can be represented as a linear
combination of basis vectors in linear algebra, any periodic function can be
represented as a linear combination of the orthogonal basis functions (sine
and cosine waves) in Fourier series.

The Fourier Series Representation - Synthesis and Analys is
Let’s consider a complex exponential, which is periodic with period T.
x(t) = ejωot
, T =
2π
ωo
(1)
Let’s consider signals x1(t) and x2(t), that are formed by adding complex
exponentials. Are they periodic? What is their fundamental period?
x1(t) = 2ejωot
+ 3ej2ωot
, (2)
Fundamental Period T1 =?
x2(t) = e−jωot
+ 2 + ejωot
+ 2ej2ωot
+ 3ej3ωot
, (3)
Fundamental Period T2 =?

Let’s consider a signal x(t) that can be expressed as,
x(t) =
∞
X
k=−∞
akejkωot
, (4)
where k is an integer and we are summing over k from negative infinity to
positive infinity.
Is x(t) periodic?
What is the fundamental period?

Let’s consider a signal x(t) that can be expressed as,
x(t) =
∞
X
k=−∞
akejkωot
, (4)
where k is an integer and we are summing over k from negative infinity to
positive infinity.
Is x(t) periodic?
What is the fundamental period?
Yes, x(t) is periodic.
Fundamental period is T = 2π
ωo

x(t) =
∞
X
k=−∞
akejkωot
(4)
−→ Equation (4) is the Fourier Series representation.
−→ Term corresponding to k = ±1 : fundamental / first harmonics
−→ Term corresponding to k = ±2 : Second harmonics
−→ Term corresponding to k = ±n : nth harmonics
Fourier claimed that any periodic signal can be written in the form
of equation (4). Really?

Example(1) : x(t) = sinωot. Can we write this in the form of (4)
.
ejωot
= cosωot + jsinωot (i)
e−jωot
= cosωot − jsinωot (ii)
Adding (i) and (ii),
cosωot =
ejωot + e−jωot
2
Subtracting (i) and (ii),
sinωot =
ejωot − e−jωot
2j

Example(1) : x(t) = sinωot. Can we write this in the form of (4)
. So,
sinωot =
−1
2j
e−jωot
+
1
2j
ejωot
=
∞
X
k=−∞
akejkωot
with,
ak =





−1
2j , k = −1
1
2j , k = +1
0, otherwise

Example(2) : x(t) = 1 + sinωot + 2cosωot + cos(2ωot + π
4
).
Represent this as FS : (4).
x(t) = 1 +
ejωot − e−jωot
2j
+ 2(
ejωot + e−jωot
2
) +
ej(ωot+π
4
)
+ e−j(ωot+π
4
)
2
x(t) = 1 + ejωot
(1 +
1
2j
) + e−jωot
(1 −
1
2j
) +
1
2
ej π
4 ej2ωot
+
1
2
e−j π
4 e−j2ωot
x(t) =
∞
X
k=−∞
akejkωot
with ak =





















1, k = 0
1 + 1
2j , k = +1
1 − 1
2j , k = −1
1
2ej π
4 , k = +2
1
2e−j π
4 , k = −2
0, otherwise

Determining the FS coefficient {ak}
Assuming the FS representation of x(t) exists, meaning x(t) can be
written in the form
x(t) =
∞
X
k=−∞
akejkωot
, x(t) = x(t + T), T =
2π
ωo
Is there a way to obtain ak using mathematical analysis?
Let us multiply x(t) with e−jnωot and integrate over one period.
Z T
0
x(t)e−jnωot
dt (5)
=
Z T
0
(
∞
X
k=−∞
akejkωot
)e−jnωot
dt, replaced x(t) in (5) with (4).

=
∞
X
k=−∞
ak
Z T
0
ej(k−n)ωot
dt (6)
solving the integral,
Z T
0
(ej(k−n)ωot
)dt =
1
j(k − n)ωo
ej(k−n)ωot
T
0
=
1
j(k − n)ωo
(ej(k−n)ωoT
− 1)
=
1
j(k − n)ωo
(ej(k−n)ωo
2π
ωo − 1)
=
1
j(k − n)ωo
(ej(k−n)2π
− 1) = 0 k ̸= n

when k = n, then
Z T
0
(ej(k−n)ωot
)dt =
Z T
0
(1)dt = T
Therefore,
Z T
0
ej(k−n)ωot
dt =
(
T, k = n
0, k ̸= n
From(5),
Z T
0
x(t)e−jnωot
dt =
∞
X
k=−∞
ak
Z T
0
ej(k−n)ωot
dt
Z T
0
x(t)e−jnωot
dt = anT

Z T
0
x(t)e−jnωot
dt = anT
=⇒ an =
1
T
Z T
0
x(t)e−jnωot
dt
In conclusion, if the FS representation of a periodic signal x(t) with period T = 2π
ωo
exists, then
x(t) =
∞
X
k=−∞
akejkωot
←− (Synthesis Equation)
ak =
1
T
Z T
0
x(t)e−jkωot
dt ≜
1
T
Z
T
x(t)e−jkωot
dt ←− (Analysis Equation)
{ak} ←− (Fourier Seies/ Spectral coefficients)

The Fourier Series Examples
Example 1 : Square Wave
x(t) =
(
1, |t| < T1
0, T1 < |t| < T
2
x(t) is periodic with period T.
−T −T
2
−T1 0 T1 T
2
T
· · · · · ·
1
x(t)
t

Solution:
−T −T
2
−T1 0 T1 T
2
T
T
· · · · · ·
1
x(t)
t
ak =
1
T
Z
T
x(t)e−jkωot
dt
ak =
1
T
Z T
2
−T
2
x(t)e−jkωot
dt =
1
T
Z T1
−T1
(1)e−jkωot
dt =
1
T
Z T1
−T1
e−jkωot
dt
=
1
T
1
−jkωo
e−jkωot
T1
−T1
=
1
T
1
−jkωo
(e−jkωoT1
− ejkωoT1
)

=
1
T
1
jkωo
(ejkωoT1
− e−jkωoT1
) =
1
T
2
kωo
(
ejkωoT1 − e−jkωoT1
2j
)
=
2sin(kωoT1)
kωoT
=
2sin(kωoT1)
kωo
2π
ωo
=
sin(kωoT1)
kπ
, k ∈ −∞, ..., −2, −1, 0, 1, 2, ..., ∞
x(t) =
∞
X
−∞
akeejkωot
ak =
sin(kωoT1)
kπ

Let’s consider a square wave with T = 0.02 sec, and T1 = 0.005 sec.
−0.02 −0.01−0.005 0 0.005 0.01 0.02
· · · · · ·
1
x(t)
t(sec)

The Fourier Series Examples (Example 1 : Square Wave)
ωo 2ωo
3ωo
4ωo 5ωo
6ωo
7ωo
−ωo
−2ωo
−3ωo
−4ωo
−5ωo
−6ωo
−7ωo

Example 2 : x(t) is periodic with period T = 2.
x(t) = e−t
, −1 < t < 1

x(t) = e−t
, −1 < t < 1
Solution:
ak =
1
T
Z
T
x(t)e−jkωot
dt
ak =
1
2
Z 1
−1
e−t
e−jkωot
dt =
1
2
Z 1
−1
e(−1+jkωo)t
dt

x(t) = e−t
, −1 < t < 1
ak =
1
2
e(−1+jkωo)t
−(1 + jkωo)
+1
−1
=
1
2
e(1+jkωo) − e−(1+jkωo)
(1 + jkωo)
ωo =
2π
T
=
2π
2
= πrad/s
ak =
1
2
e1ejkπ − e−1e−jkπ
(1 + jkπ)
ak =
1
2
(−1)k
(1 + jkπ)
(e1
− e−1
)

x(t) = e−t
, −1 < t < 1
x(t)=
P∞
−∞ akejkω0t
ak= 1
2
(−1)k
(1+jkπ)
(e1−e−1)

The Fourier Series Examples (Example 2 : Periodic
Exponential Signal)

Gibb’s Phenomenon
The Gibbs phenomenon is a common occurrence in Fourier analysis.
It refers to the overshoot and oscillations that occur near jump
discontinuities of a function.
The phenomenon is observed when approximating a discontinuous
function using its Fourier series.

Gibb’s Phenomenon : Mathematical Explanation
The Gibbs phenomenon arises due to the inability of the Fourier series
to accurately represent discontinuous functions.
Near a jump discontinuity, the Fourier series overshoots the function’s
value and produces oscillations that persist even as the number of
terms in the series increases.
The overshoot near the discontinuity is typically about 9% of the
height of the jump. The exact proportion is given by the
Wilbraham-Gibbs Constant.
1
π
Z π
0
sint
t
dt −
1
2
= 0.089489. . .
It may also be noted that as more number of terms in the series are
added, the frequency increases and the overshoots become sharper,
but the amplitude of the adjoining oscillation reduces, i.e., the error
between the original signal x(t) and the approximated signal in red
reduces except at edges as the n increases.

Gibb’s Phenomenon : Implications
In biomedical signals (such as ECG, EEG, etc.) and images (such as MRI,
CT scans, etc.), Gibbs’ phenomenon can lead to:
Ringing artifacts: Oscillations near abrupt changes (e.g., edges,
boundaries) in the signal or image.
False features: Spurious oscillations that may be misinterpreted as
actual features.
Image quality degradation: The overshoots and oscillations can
affect the accuracy of measurements and diagnoses.

Properties of FS
Linearity
x(t)
FS
←→ ak
y(t)
FS
←→ bk
both x(t) and y(t) have the same period T.
z(t) = Ax(t) + By(t)
z(t)
FS
←→ ck
ck =
1
T
Z
T
z(t)e−jkωot
dt =
1
T
Z
T
[Ax(t) + By(t)]e−jkωot
dt
ck =
A
T
Z
T
x(t)e−jkωot
dt +
B
T
Z
T
y(t)e−jkωot
dt
ck = Aak + Bbk

Properties of FS
Time Shifting
x(t)
FS
←→ ak
x(t − to)
FS
←→ bk =?
bk =
1
T
Z
T
x(t − to)e−jkωot
dt
Let τ = t − to =⇒ dt = dτ
t = 0 → τ = −to
t = T → τ = T − to
bk =
1
T
Z T−to
to
x(τ)e−jkωo(τ+to)
dτ =
e−jkωoto
T
Z T−to
to
x(τ)e−jkωoτ
dτ
bk = e−jkωoto
ak

Properties of FS
Time Shifting
x(t)
FS
←→ ak
x(t − to)
FS
←→ bk = e−jkωoto
ak
e−jkωoto represents the change in phase.
|bk| = |e−jkωoto
||ak|
|bk| = |ak|, ∵ |e−jkωoto
| = 1

Properties of FS
Time Reversal
x(t)
FS
←→ ak
x(−t)
FS
←→ bk =?
bk =
1
T
Z
T
x(−t)e−jkωot
dt
Let τ = −t =⇒ −dt = dτ
t = 0 → τ = 0
t = T → τ = −T
bk =
−1
T
Z −T
0
x(τ)ejkωoτ
dτ
=
1
T
Z 0
−T
x(τ)ejkωoτ
dτ

Properties of FS
Time Reversal
bk =
1
T
Z
T
x(τ)e−j(−k)ωoτ
dτ
bk = a−k
x(−t) =
∞
X
k=−∞
bkejkωot
=
∞
X
k=−∞
a−kejkωot
What if the signal is even?
if x(t) = x(−t)
x(t)
FS
←→ ak
x(−t)
FS
←→ a−k



=⇒ ak = a−k
The FS coefficients are symmetric, i.e.,ak = a−k.

Properties of FS
Time Reversal
What if the signal is odd?
if x(t) = −x(−t)
x(t)
FS
←→ ak
x(−t)
FS
←→ a−k



=⇒ ak = −a−k
The FS coefficients are anti-symmetric, i.e.,ak = −a−k.
What can we say about a0?
a0 = −a0 =⇒ a0 = 0

Properties of FS
Time Scaling
x(t)
FS
←→ ak
x(αt) is periodic with a period T
α .
if x(t) =
∞
X
k=−∞
akejkωot
x(αt) =
∞
X
k=−∞
akejkωoαt
=
∞
X
k=−∞
akejk(αωo)t
With time scaling, FS coefficients do not change, but the frequency/ time
period does.

Properties of FS
Conjugation
x(t)
FS
←→ ak
x∗
(t)
FS
←→ bk =?
bk =
1
T
Z
T
x∗
(t)e−jkωot
dt
=

1
T
Z
T
x∗
(t)e−jkωot
dt
∗∗
=

1
T
Z
T
x(t)ejkωot
dt
| {z }
a−k
∗
bk = (a−k)∗
= a∗
−k

Properties of FS
Conjugation
If x(t) is real. x(t) = x∗(t)
x(t)
FS
←→ ak
x∗
(t)
FS
←→ a∗
−k



=⇒ ak = a∗
−k
FS coefficients are conjugate symmetric. a0 = a∗
−0
=⇒ a0 = a∗
0 =⇒ a0 is real.

Properties of FS
Conjugation
If x(t) is real and even.
Real: ak = a∗
−k (i)
Even: ak = a−k (ii)
=⇒ a∗
k = a−k (iii)
comparing (ii) and (iii): ak = a∗
k.
=⇒ ak is real.
If x(t) is real and even, then FS coefficients must be real.

Properties of FS
Conjugation
If x(t) is real and odd.
Real: ak = a∗
−k (iv)
Odd: ak = −a−k (v)
comparing (iv) and (v): ak = −a∗
k.
=⇒ ak is imaginary. Real part of ak’s must be zero.
If x(t) is real and even, then FS coefficients must be purely imaginary.

Properties of FS : Example
Determine the FS representation for x(t).
T = 4, ωo =
2π
T
=
π
2

Recall we have already solved a square wave signal provided above and it’s
FS representation was calculated to be:
g(t)
FS
←→ ak =
(sin(k π
2
)
kπ k ̸= 0
1
2 k = 0

By looking at the two signals we can write x(t) in terms of g(t) as,
x(t) = g(t − 1) −
1
2
g(t − 1)
FS
←→ ake−jk π
2
.1
y(t) =
1
2
FS
←→ ck =
(
1
2 k = 0
0 k ̸= 0
x(t) = g(t − 1) −
1
2
FS
←→ bk = ake−jk π
2
.1
− ck
bk =
(sink π
2
kπ ejk π
2 k ̸= 0
0 k = 0
bk =
(sink π
2
kπ (j)k k ̸= 0
0 k = 0

Properties of FS
Parseval’s Relation
Power in one time period =
1
T
Z
T
|x(t)|2
dt
x(t) =
∞
X
k=−∞
akejkωot
P =
1
T
Z
T
|x(t)|2
dt =
1
T
Z
T
x(t)x∗
(t)dt
=
1
T
Z
T
|x(t)|2
dt =
1
T
Z
T
∞
X
k=−∞
akejkωot
| {z }
x(t)
∞
X
m=−∞
a∗
me−jmωot
| {z }
x∗(t)

dt
=
1
T
∞
X
k=−∞
∞
X
m=−∞
Z
T
aka∗
mejkωot
e−jmωot
dt

Properties of FS
Parseval’s Relation
=
1
T
∞
X
k=−∞
∞
X
m=−∞
aka∗
m
Z
T
ej(k−m)ωot
dt
Z
T
ej(k−m)ωot
dt =
(
T, m = k
0, m ̸= k
=
1
T
∞
X
k=−∞
ak
∞
X
m=−∞
a∗
m
Z
T
ej(k−m)ωot
dt
| {z }
a∗
kT

=
1
T
∞
X
k=−∞
aka∗
kT =
∞
X
k=−∞
aka∗
k =
∞
X
k=−∞
|ak|2
1
T
Z
T
|x(t)|2
dt =
∞
X
k=−∞
|ak|2

Properties of FS
Example:
1 x(t) is real.
2 x(t) is periodic with period T = 4.
3 The FS coefficients ak are zero for |k| 1.
4 The signal with FS given by bk = a−kejk π
2 is odd.
5 1
4
R
T |x(t)|2dt = 1
2.
Determine x(t).
Solution:
Since ak = 0 for |k| 1, then only possible non-zero FS coefficients
are a−1, a0, a+1.
T = 4, ωo = 2π
4 = π
2 .
Using the synthesis expression : x(t) =
P+∞
k=−∞ akejkωot.
=⇒ x(t) = a−1e−j π
2
t
+ a0 + a+1ej π
2
t
(A)
Only need to determine a+1, a0, a+1.

Since the signal is real. a+1 = a∗
−1, a0 is real.
The signal corresponding to bk is a shifted and flipped version of x(t)
and therefore must be real.Since, the signal is odd,
b0 = 0, ∵ bk = ejk π
2 a−k we have a0 = 0.
Since the signal is real and odd,bk’s are imaginary.
bk = ejk π
2 a−k =⇒ b1 = ej π
2 a−1 = ja−1
=⇒ b−1 = ej π
2 a+1 = −ja+1
=⇒ a−1 and a+1 must be real.
Since a−1 = a∗
+1 and a−1 and a+1 are real, a−1 = a+1

1
4
Z
4
|x(t)|2
dt =
1
2
=⇒ |a−1|2
+

0
|a0|2
+ |a+1|2
=
1
2
=⇒ |a1|2
=
1
4
=⇒ |a1| =
1
2
=⇒ x(t) = a+1ej π
2
t
+ a−1e−j π
2
t
=
1
2
ej π
2
t
+
1
2
e−j π
2
t
or = −
1
2
ej π
2
t
−
1
2
e−j π
2
t
x(t) = ±cos(
π
2
t)

Fourier Series and LTI Systems
Response of LTI Systems to Complex Exponentials
Complex Exponential : x(t) = sst
where, s = σ
|{z}
ℜ(s)
+j ω
|{z}
ℑ(s)
LTI
h(t)
y(t) =?
x(t) = est
y(t) =
Z ∞
−∞
h(τ)x(t − τ)dτ =
Z ∞
−∞
h(τ)es(t−τ)
dτ
= est
Z ∞
−∞
h(τ)e−sτ
dτ
| {z }
H(s)

y(t) = est
Z ∞
−∞
h(τ)e−sτ
dτ
| {z }
H(s)
y(t) = est
H(s) where, H(s) =
Z ∞
−∞
h(τ)e−sτ
dτ
By looking at the y(t) we can say if the input to an LTI system is a
complex exponential then output is also a complex exponential multiplied
by H(s).
In linear algebra we have learned about eigen vectors and eigen values.

y(t) = est
Z ∞
−∞
h(τ)e−sτ
dτ
| {z }
H(s)
y(t) = est
H(s) where, H(s) =
Z ∞
−∞
h(τ)e−sτ
dτ
By looking at the y(t) we can say if the input to an LTI system is a
complex exponential then output is also a complex exponential multiplied
by H(s).
In linear algebra we have learned about eigen vectors and eigen values.
Ax̄ = λ
|{z}
Eigen value
x̄
|{z}
Eigen vector
We say that x(t) = est is an eigenfunction for an LTI system with a
corresponding eigenvalue H(s).

LTI
h(t)
y(t) =?
x(t) =
P∞
k=−∞ akejkωot
LTI
h(t)
x(t) = ejkωot y(t) =?

LTI
h(t)
y(t) =?
x(t) =
P∞
k=−∞ akejkωot
LTI
h(t)
x(t) = ejkωot y(t) = ejkωotH(jkωo)

LTI
h(t)
LTI
h(t)
x(t) = akejkωot y(t) =?

LTI
h(t)
LTI
h(t)
x(t) = akejkωot y(t) = akejkωotH(jkωo)

LTI
h(t)
LTI
h(t)
x(t) =
P∞
k=−∞ akejkωot y(t) =?

LTI
h(t)
LTI
h(t)
x(t) =
P∞
k=−∞ akejkωot
y(t) =
P∞
k=−∞ akejkωotH(jkωo)
y(t) =
∞
X
k=−∞
akH(jkωo)ejkωot

y(t) =
∞
X
k=−∞
akH(jkωo)ejkωot
What can we say about the periodicity of y(t)?

y(t) =
∞
X
k=−∞
akH(jkωo)ejkωot
What can we say about the periodicity of y(t)? Indeed y(t) is periodic
with period T = 2π
ωo
.
y(t) =
∞
X
k=−∞
akH(jkωo)
| {z }
FS coefficients bk
ejkωot
y(t) =
∞
X
k=−∞
bkejkωot
where bk = akH(jkωo)

chapter3_fourier_series. on signal and system

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