Module -7
Properties of Fourier Series and Complex Fourier Spectrum.
Objective:To understand the change in Fourier series coefficients due to different signal
operations and to plot complex Fourier spectrum.
Introduction:
The Continuous Time Fourier Series is a good analysis tool for systems with
periodicexcitation. Understanding properties of Fourier series makes the work simple in
calculating the Fourier series coefficients in the case when signals modified by some basic
operations.
Graphical representation of a periodic signal in frequency domain represents Complex
Fourier Spectrum.
Description:
Properties of continuous- time Fourier series
The Fourier series representation possesses a number of important properties
that are useful for various purposes during the transformation of signals from one form to other .
Some of the properties are listed below.
[x1(t) and x2(t)] are two periodic signals with period T and with Fourier series
coefficients Cn and Dn respectively.
1) Linearity property
The linearity property states that, if
𝐹𝑠
x1(t) Cn and x2 (𝑡)
then
proof:
𝐹𝑠
𝐹𝑠
Dn
Ax1(t)+Bx2(t) ACn+BDn
From the definition of Fourier series, we have
1 𝑡 +𝑇
FS[Ax1(t)+Bx2(t)]=𝑇 𝑡 0 [𝐴𝑥1 𝑡 + 𝐵𝑥2 𝑡 ]𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡
0
1
𝑡 +𝑇
1
=𝐴(𝑇 𝑡 0 𝑥1 𝑡 𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡)+ 𝐵(𝑇
0
=ACn+BDn
𝐹𝑠
or
Ax1(t)+Bx2(t) ACn+BDn
proved
2) Time shifting property
The time shifting property states that,if
𝐹𝑠
x(t) Cn
𝐹𝑠
then
x(t-t0) 𝑒 −𝑗𝑛 𝜔 0 𝑡0 Cn
proof: From the definition of Fourier series, we have
𝑗𝑛 𝜔 0 𝑡
x(t)= ∞
𝑛=−∞ 𝐶𝑛 𝑒
𝑗𝑛 𝜔 0 (𝑡−𝑡 0 )
x(t-t0)= ∞
𝑛=−∞ 𝐶𝑛 𝑒
∞
−𝑗𝑛 𝜔 0 𝑡 𝑗𝑛 𝜔 0 𝑡
= 𝑛=−∞ [𝐶𝑛 𝑒
]𝑒
-1
−𝑗𝑛 𝜔 0 𝑡
=FS [Cn𝑒
]
x(t-t0)
𝐹𝑠
𝑒 −𝑗𝑛 𝜔 0 𝑡0 Cn
𝑡 0 +𝑇
𝑥2
𝑡0
𝑡 𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡)
Or
3) Time reversal property
The time reversal property states that,if
proved
𝐹𝑠
x(t) Cn
𝐹𝑠
then
x(-t) C-n
proof: From the definition of Fourier series, we have
𝑗𝑛 𝜔 0 𝑡
x(t)= ∞
𝑛=−∞ 𝐶𝑛 𝑒
∞
x(-t)= 𝑛=−∞ 𝐶𝑛 𝑒 𝑗𝑛 𝜔 0 (−𝑡)
substituting n = -p in the right hand side, we get
𝑗 (−𝑝)𝜔 0 (−𝑡)
𝑗𝑝 𝜔 0 𝑡
x(-t)= −∞
= ∞
𝑝=∞ 𝐶−𝑝 𝑒
𝑝=−∞ 𝐶−𝑝 𝑒
substituting p = n , we get
𝑗𝑛 𝜔 0 𝑡
x(-t)= ∞
= FS-1[C-n]
𝑛=−∞ 𝐶−𝑛 𝑒
𝐹𝑠
x(-t) C-n
proved
4) Time scaling property
The time scaling property states that , if
𝐹𝑠
x(t) Cn
𝐹𝑠
x(𝛼t) Cn with ω0→𝛼ω0
From the definition of Fourier series, we have
𝑗𝑛 𝜔 0 𝑡
x(t)= ∞
𝑛=−∞ 𝐶𝑛 𝑒
𝑗𝑛 𝜔 0 𝛼𝑡
𝑗𝑛 (𝜔 0 𝛼)𝑡
x(𝛼t)= ∞
= ∞
= FS-1[Cn]
𝑛=−∞ 𝐶𝑛 𝑒
𝑛=−∞ 𝐶𝑛 𝑒
where ω0→𝛼ω0.
then
proof:
𝐹𝑠
x(𝛼t) Cn with fundamental frequency of 𝛼ω0
proved.
5) Time differential property
The time differential property states that, if
𝐹𝑠
x(t) Cn
𝑑𝑥 (𝑡) 𝐹𝑠
jnω0Cn
From the definition of Fourier series, we have
𝑗𝑛 𝜔 0 𝑡
x(t)= ∞
𝑛=−∞ 𝐶𝑛 𝑒
Differentiating both sides with respect to t, we get
then
proof:
𝑑𝑥 (𝑡)
𝑑𝑡
=
𝑑𝑡
∞
𝑛=−∞
𝐶𝑛
𝑑(𝑒 𝑗𝑛 𝜔 0 𝑡 )
𝑑𝑡
=
∞
𝑛=−∞
𝑑𝑥 (𝑡) 𝐹𝑠
𝑑𝑡
𝐶𝑛 (jnω0 ) 𝑒 𝑗𝑛 𝜔 0 𝑡
𝑗𝑛 𝜔 0 𝑡
= ∞
= FS-1[jnω0 Cn]
𝑛=−∞ 𝐶𝑛 (jnω0 ) 𝑒
jnω0Cn
6) Time integration property
Proved.
The time integration property states that,
𝐹𝑠
if
x(t) Cn
𝑡
𝑥
−∞
then
𝜏 𝑑𝜏
𝐹𝑆
𝐶𝑛
(if C0 = 0)
𝑗𝑛 𝜔 0
proof:
From the definition of Fourier series, we have
𝑗𝑛 𝜔 0 𝑡
x(t)= ∞
𝑛=−∞ 𝐶𝑛 𝑒
𝑡
𝑡
𝑗𝑛 𝜔 0 𝜏
𝑥 𝜏 𝑑𝜏 = −∞ [ ∞
]𝑑𝜏
𝑛=−∞ 𝐶𝑛 𝑒
−∞
Interchanging the order of integration and summation, we get
𝑡
𝑥
−∞
𝜏 𝑑𝜏 =
𝑡
∞
𝑗𝑛 𝜔 0 𝜏
𝑑𝜏
𝑛=−∞ 𝐶𝑛 −∞ 𝑒
=
=
𝑡
𝑥
−∞
𝜏 𝑑𝜏
𝐹𝑆
∞
𝑛=−∞
𝑒 𝑗𝑛 𝜔 0 𝜏
] 𝑡−∞
𝑗𝑛 𝜔 0
𝐶𝑛
∞
𝑗𝑛 𝜔 0 𝑡
=
𝑛=−∞ (𝑗𝑛 𝜔 ) 𝑒
0
𝐶𝑛
𝑗𝑛 𝜔 0
𝐶𝑛 [
𝐶
FS-1[𝑗𝑛 𝜔𝑛 ]
0
Proved.
, C0 = 0
7) Convolution theorem or property
The convolution theorem or property states that , The Fourier series of the
convolution of two time domain functions x1(t) and x2(t) is equal to the multiplication of their
Fourier series coefficients, i.e.”Convolution of two functions in time domain is equivalent to
multiplication of their Fourier coefficients in frequency domain”.
The convolution property states that,if
𝐹𝑠
x1(t) Cn and x2 (𝑡)
𝐹𝑠
𝐹𝑠
Dn
then
x1(t)*x2(t) TCnDn
proof: From the definition of Fourier series, we have
1 𝑡 +𝑇
1 𝑇
FS[x1(t)*x2(t)]=𝑇 𝑡 0 [x1 (t) ∗ x2 (t)] 𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡 =𝑇 0 [x1 (t) ∗ x2 (t)] 𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡
0
But from the convolution integral for a periodic signal, we have
𝑇
𝑇
x1 (t) ∗ x2 (t)= 0 x1 (τ) ∗ x2 (t − τ) 𝑑𝜏 or x1 (t) ∗ x2 (t)= 0 x1 (t − τ) ∗ x2 (τ) 𝑑𝜏
Substituting back, we get
1 𝑇 𝑇
FS[x1(t)*x2(t)] =𝑇 0 ( 0 x1 (τ) ∗ x2 (t − τ) 𝑑𝜏) 𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡
Interchanging the order of integration, we get
𝑇
1 𝑇
FS[x1(t)*x2(t)] = 0 x1 (τ)[ 0 x2 (t − τ) 𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡] 𝑑𝜏
𝑇
Substituting t-𝛕 =t' in RHS, we have dt=dt'. Substituting back, we get
′
𝑇−τ
1 𝑇
FS[x1(t)*x2(t)] =𝑇 0 x1 (τ)[ −τ x2 (t′) 𝑒 −𝑗𝑛 𝜔 0 (𝑡 +τ) 𝑑𝑡′] 𝑑𝜏
1
𝑇
1
𝑇
=T(𝑇 0 x1 (τ)𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝛕) (𝑇 0 x1 (t′)𝑒 −𝑗𝑛 𝜔 0 𝑡′ 𝑑t′)
=T(Cn)(Dn)
0 to T or –𝛕 to T-𝛕 or t0 to t0+T will have the same period.
𝐹𝑠
x1(t)*x2(t) TCnDn
Proved.
8) Modulation or Multiplication property
The Modulation or Multiplication property states that,if
𝐹𝑠
x1(t) Cn and x2 (𝑡)
then
proof:
𝐹𝑠
Dn
𝐹𝑠
∞
x1(t)x2(t)
𝑙=−∞ 𝐶𝑙 𝐷𝑛−𝑙
From the definition of Fourier series, we have
1 𝑡 +𝑇
FS[x1(t)x2(t)]= 𝑇 𝑡 0 x1 t x2 t 𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡
0
𝑡 +𝑇
=𝑇 𝑡 0 x1
0
1 𝑡 0 +𝑇
=𝑇 𝑡
x1
0
1
t (
∞
𝑗𝑛 𝜔 0 𝑡
) 𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡
𝑙=−∞ 𝐶𝑙 𝑒
−𝑗 (𝑛−𝑙)𝜔 0 𝑡
t ( ∞
) 𝑑𝑡
𝑙=−∞ 𝐶𝑙 𝑒
Interchanging the order of integration and summation, we have
1 𝑡 0 +𝑇
FS[x1(t)x2(t)]= ∞
x1 t 𝑒 −𝑗 (𝑛−𝑙)𝜔 0 𝑡 𝑑𝑡) = ∞
𝑙=−∞ 𝐶𝑙 (𝑇 𝑡
𝑙=−∞ 𝐶𝑙 𝐷𝑛−𝑙
0
x1(t)x2(t)
𝐹𝑠
Proved.
∞
𝑙=−∞ 𝐶𝑙 𝐷𝑛−𝑙
9) Conjugation and Conjugate symmetry property
𝐹𝑠
If
Then the Conjugation property states that
x*(t)
𝐹𝑠
x(t) Cn
∗
𝐶−𝑛
[for complex x(t)]
proof: Conjugation property
From the definition of Fourier series, we have
1 𝑡 +𝑇
1
FS[x*(t)]= 𝑇 𝑡 0 x ∗ (t) 𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡=(𝑇
1
=(𝑇
x*(t)
𝐹𝑠
0
𝑡 0 +𝑇
x(t) 𝑒 𝑗𝑛 𝜔 0 𝑡 𝑑𝑡)*
𝑡0
𝑡 0 +𝑇
x(t) 𝑒 −𝑗 (−𝑛)𝜔 0 𝑡 𝑑𝑡)*=[C-n]*
𝑡0
∗
𝐶−𝑛
Proved.
10) Parseval’s Relation or Theorem or property
𝐹𝑠
If
x1(t) Cn
𝐹𝑠
[for complex x1(t)]
And
x2 (𝑡) Dn [for complex x2(t)]
Then, the parsevals relation states that
1 𝑡 0 +𝑇
∗
x1 t x2∗ t 𝑑𝑡= ∞
𝑛=−∞ Cn 𝐷𝑛 [for complex x1(t) and x2(t)]
𝑇 𝑡0
And parseval’s identity states that
1 𝑡 0 +𝑇
2
⎹x(t)⎸2 𝑑𝑡 = ∞
if x1(t) = x2(t) = x(t)
𝑛=−∞ ⎹Cn ⎸
𝑇 𝑡0
Proof: parsevals relation
1 𝑡 +𝑇
1 𝑡 +𝑇
𝑗𝑛 𝜔 0 𝑡
LHS =𝑇 𝑡 0 x1 t x2∗ t 𝑑𝑡 = 𝑇 𝑡 0 ( ∞
) 𝑥2∗ 𝑑𝑡
𝑛=−∞ 𝐶𝑛 𝑒
0
0
Interchanging the order of integration and summation in the RHS , we have
1 𝑡 0 +𝑇
1 𝑡 0 +𝑇 ∗
x1 t x2∗ t 𝑑𝑡 = ∞
x2 t 𝑒 𝑗𝑛 𝜔 0 𝑡 𝑑𝑡)
𝑛=−∞ 𝐶𝑛 (𝑇 𝑡
𝑇 𝑡
0
0
=
∞
𝑛=−∞
1
𝐶𝑛 (𝑇
𝑡 0 +𝑇
x2
𝑡0
t 𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡)*=
∞
𝑛=−∞ Cn
[Dn ]∗
Parseval’s identity
𝑡 0 +𝑇
𝑇 𝑡0
1
Proved
x1 t
x2∗
t 𝑑𝑡=
∞
𝑛=−∞ Cn
𝐷𝑛∗
If x1(t) = x2(t) = x(t), then the relation changes to
1 𝑡 0 +𝑇
𝑇 𝑡0
x t x ∗ t 𝑑𝑡=
∞
𝑛=−∞ Cn
𝐶𝑛∗
Since ⎹x(t)⎸2 = x t x ∗ t and ⎹Cn ⎸2 = Cn𝐶𝑛∗ , substituting these values in above equation, we get
1 𝑡 0 +𝑇
⎹x(t)⎸2
𝑇 𝑡0
𝑑𝑡 =
∞
2
𝑛=−∞ ⎹Cn ⎸
Proved.
Complex Fourier Spectrum
The Fourier spectrum of a periodic signal x(t) is a plot of its Fourier coefficients versus
frequency ω. It is in two parts : (a) Amplitude spectrum and (b) phase spectrum. The plot of the
amplitude of Fourier coefficients verses frequency is known as the amplitude spectra, and the
plot of the phase of Fourier coefficients verses frequency is known as phase spectra. The two
plots together are known as Fourier frequency spectra of x(t).This type of representation is also
called frequency domain representation. The Fourier spectrum exists only at discrete frequencies
nωo, where n=0,1,2,….. Hence it is known as discrete spectrum or line spectrum. The envelope
of the spectrum depends only upon the pulse shape, but not upon the period of repetition.
The trigonometric representation of a periodic signal x(t) contains both sine and cosine
terms with positive and negative amplitude coefficients (an and bn) but with no phase angles.
The cosine representation of a periodic signal contains only positive amplitude
coefficients with phase angle θn. Therefore, we can plot amplitude spectra(An versus ω)and phase
spectra (θn versus ω). Since, in this representation, Fourier coefficients exist only for positive
frequencies, this spectra is called single-sided spectra.
The exponential representation of a periodic signal x(t) contains amplitude coefficients
Cn Which are complex. Hence, they can be represented by magnitude and phase. Therefore, we
can plot two spectra, the magnitude spectrum(⎹Cn⎸versus ω) and phase spectrum ( 𝐶𝑛 versus ω).
The spectra can be plotted for both positive and negative frequencies.Henceit is called two-sided
spectra.
The below figure (a) represents the spectrum of a trigonometric Fourier series extending
from 0 to∞,producing a one-sided spectrum as no negative frequencies exist here. The figure (b)
represents the spectrum of a complex exponential Fourier series extending from ∞𝑡𝑜∞,producing a two-sided spectrum.
The amplitude spectrum of the exponential Fourier series is symmetrical Fourier series is
symmetrical about the vertical axis. This is true for all periodic functions.
Fig: Complex frequency spectrum for (a) Trigonometric Fourier series and (b) complex
exponential Fourier series.
If Cn is a general complex number, then
Cn = ⎹Cn⎸𝑒 𝑗 𝜃𝑛
C-n = ⎹Cn⎸𝑒 −𝑗 𝜃𝑛
Cn = ⎹C-n⎸
The magnitude spectrum is symmetrical about the vertical axis passing
through the origin, and the phase spectrum is antisymmetrical about the vertical axis passing
through the origin. So the magnitude spectrum exhibits even symmetry and phase spectrum
exhibits odd symmetry.
When x(t) is real , then C-n=𝐶𝑛∗ , the complex conjugate of Cn.
Solved Problems:
Problem 1: Show that the magnitude spectrum of every periodic function is symmetrical about
the vertical axis passing through the origin and the phase spectrum is antisymmetrical about the
vertical axis passing through the origin.
Solution:
The coefficient Cn of exponential Fourier series is given by
1
Cn = 𝑇
T
2
𝑇
−
2
1
And
C-n = 𝑇
𝑥 𝑡 𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡)
T
2
𝑇
−
2
𝑥 𝑡 𝑒 𝑗𝑛 𝜔 0 𝑡 𝑑𝑡)
It is evident from these equations that the coefficients Cn and C-n are complex conjugate of each
other , that is
Cn= 𝐶𝑛∗
Hence
⎹Cn ⎸=⎹C−n ⎸
It, therefore, follows that the magnitude spectrum is symmetrical about the vertical axis
passing through the origin, and hence is an even function of ω.
If Cn is real, then C-n is also real and Cn is equal to C-n. is complex, let
Cn=⎹Cn ⎸ejθ n
C-n=⎹Cn ⎸e−jθ n
then
The phase of Cn is θn ;however , the phase of C-n is−θn .Hence , it is obvious that the phase
spectrum is antisymmetrical, and the magnitude spectrum is symmetrical about the vertical axis
passing through the origin.
Problem 2: With regard to Fourier series representation, justify the following statement: odd
functions have only sine terms.
Solution: We know that the trigonometric Fourier series of a periodic function x(t) in any
𝑇
𝑇
interval t0 to t0+T or 0 to T or -2 to 2 is given by
∞
𝑛=1(𝑎𝑛
x(t)= 𝑎0 +
T
2
𝑇
−
2
2
𝑎𝑛 = 𝑇
and𝑏𝑛 =
T
2
𝑇
−
2
1
𝑎0 =
where
𝑇
cos 𝑛𝜔0 𝑡 + 𝑏𝑛 sin 𝑛ω0 t)
𝑥 𝑡 𝑑𝑡
𝑥 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡
T
2
𝑇
−
2
2
𝑇
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡
For an odd function
x(t) = -x(-t)
Also even part is zero, i.e. xe(t) = 0 and x(t) = x0(t)
1
𝑎0 =
𝑇
T
2
𝑇
−
2
𝑥 𝑡 𝑑𝑡 =
1
𝑇
T
2
𝑇
−
2
𝑥0 𝑡 𝑑𝑡 = 0
Since the integration of an odd function,over one cycle is always zero.
2
𝑎𝑛 = 𝑇
T
2
𝑇
−
2
𝑥 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡 =
2
𝑇
T
2
𝑇
−
2
𝑥0 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡
Here 𝑥0 (t) is odd and cos 𝑛𝜔0 𝑡 is even. So 𝑥0 𝑡 cos 𝑛𝜔0 𝑡, i.e. the product if an odd function
and even function is odd. So integration over one cycle is zero. Therefore,
𝑎𝑛 = 0
𝑏𝑛 =
Now,
2
𝑇
T
2
𝑇
−
2
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡 =
2
𝑇
T
2
𝑇
−
2
𝑥0 𝑡 sin 𝑛ω0 t 𝑑𝑡
Now, 𝑥0 (t) is odd sin 𝑛ω0 t is also odd. Therefore, the product 𝑥0 (t)sin 𝑛ω0 t is even. So we
have to evaluate the integral:
𝑏𝑛 =
2
𝑇
T
2
𝑇
−
2
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡 =
x(t)=
4
T
2
𝑇 0
∞
𝑛=1 𝑏𝑛
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡
sin 𝑛ω0 t
Thus, the Fourier series of odd functions contains only sine terms.
Problem3: With regard to Fourier series representation, justify the following statement: Even
functions have no sine terms.
Solution: We know that the trigonometric Fourier series of a periodic function x(t) in any
𝑇
𝑇
interval t0 to t0+T or 0 to T or -2 to 2 is given by
x(t)= 𝑎0 +
1
𝑎0 =
where
T
2
𝑇
−
2
2
𝑎𝑛 = 𝑇
𝑇
∞
𝑛=1(𝑎𝑛
T
2
𝑇
−
2
cos 𝑛𝜔0 𝑡 + 𝑏𝑛 sin 𝑛ω0 t)
𝑥 𝑡 𝑑𝑡
𝑥 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡
𝑏𝑛 =
and
T
2
𝑇
−
2
2
𝑇
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡
For an even function
x(t) = x(-t)
Also odd part is zero, i.e. x(t) = xe(t)
𝑎0 =
𝑎𝑛 =
T
2
𝑇 −𝑇
2
T
2 2
𝑇 −𝑇
2
1
𝑥 𝑡 𝑑𝑡 =
1
𝑇
T
2
𝑇
−
2
2
𝑥𝑒 𝑡 𝑑𝑡 = 𝑇
𝑥 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡 =
2
𝑇
T
2
𝑇
−
2
T
2
0
𝑥 𝑡 𝑑𝑡
𝑥𝑒 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡
𝑥𝑒 (t) is even and also cos 𝑛𝜔0 𝑡 is even. So, the product 𝑥𝑒 (t)cos 𝑛𝜔0 𝑡 is even. So we have to
evaluate the integral:
2
𝑎𝑛 = 𝑇
𝑏𝑛 =
2
𝑇
T
2
𝑇
−
2
T
2
𝑇
−
2
4
𝑥𝑒 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡 = 𝑇
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡 =
2
𝑇
T
2
0
T
2
𝑇
−
2
𝑥 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡
𝑥𝑒 𝑡 sin 𝑛ω0 t 𝑑𝑡
Here 𝑥𝑒 (t) is even sin 𝑛ω0 tand is odd. So, the product 𝑥𝑒 𝑡 sin 𝑛ω0 t is odd. Thus, the integral
over one complete cycle is zero.
𝑏𝑛 = 0
x(t)= 𝑎0 + ∞
𝑛=1 𝑎𝑛 cos 𝑛𝜔0 𝑡
Thus, the Fourier series of even functions contains no sine terms.
Problem 4: With regard to Fourier series representation, justify the following statement:
Functions with half wave symmetry have only odd harmonics.
Solution: We know that the trigonometric Fourier series of a periodic function x(t) in any
𝑇
𝑇
interval t0 to t0+T or 0 to T or -2 to 2 is given by
x(t)= 𝑎0 +
∞
𝑛=1(𝑎𝑛
cos 𝑛𝜔0 𝑡 + 𝑏𝑛 sin 𝑛ω0 t)
𝑎0 =
where
2
𝑎𝑛 = 𝑇
T
2
𝑇
−
2
T
2
𝑇
−
2
1
𝑇
𝑥 𝑡 𝑑𝑡
𝑥 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡
T
2
𝑇
−
2
2
𝑏𝑛 =
and
𝑇
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡
For a half wave symmetric function,
𝑇
x(t) = -x(t±2 )
𝑎0 = 0
Therefore,
2
𝑎𝑛 = 𝑇
T
𝑥
0
𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡
T
2
2
=𝑇[
0
T
𝑥 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡 +
𝑇
2
𝑥 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡 ]
To have the limits of second integration also from 0 to T/2 , change the variable t by t + (T/2) in
the second integration.
T
2
2
𝑎𝑛 =𝑇 [
0
T
2
𝑥 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡 +
T
T
𝑥 𝑡 + 2 cos 𝑛𝜔0 (𝑡 + 2 ) 𝑑𝑡 ]
0
𝑇
But
x(t) = -x(t±2 )
T
2
2
𝑎𝑛 = 𝑇 [
𝑥 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡 +
0
T
2
−𝑥 𝑡 cos(𝑛𝜔0 𝑡 + 𝑛π) 𝑑𝑡 ]
0
2
=𝑇[
0
𝑎𝑛 =
T
2
0
𝑥 𝑡 cos 𝑛𝜔0 𝑡 − 𝑥 𝑡 cos(𝑛𝜔0 𝑡 + 𝑛π)] 𝑑𝑡
𝑓𝑜𝑟 𝑒𝑣𝑒𝑛 𝑛
T
2
4
𝑇 0
𝑥 𝑡 cos 𝑛𝜔0 𝑡
𝑓𝑜𝑟 𝑜𝑑𝑑 𝑛
Now,
𝑏𝑛 =
2
T
2
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡
𝑇 0
T
2
2
=𝑇[
0
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡 +
T
𝑇
2
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡]
To have the limits of second integration also from 0 to T/2 change the variable t by t + (T/2) in
the second integration.
2
𝑏𝑛 = 𝑇 [
But
T
2
0
T
2
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡 +
0
𝑇
x(t) = -x(t±2 )
𝑇
T
𝑥 𝑡 + 2 sin 𝑛ω0 (t + 2 ) 𝑑𝑡]
2
𝑏𝑛 = 𝑇 [
T
2
0
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡 +
2
=𝑇[
T
2
0
0
𝑏𝑛 =
4
T
2
0
−𝑥 𝑡 sin( 𝑛ω0 t + nπ) 𝑑𝑡]
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡 -
T
2
0
𝑥 𝑡 sin 𝑛ω0 t cos nπ) 𝑑𝑡]
𝑓𝑜𝑟 𝑒𝑣𝑒𝑛 𝑛
T
2
𝑥 𝑡 sin 𝑛ω0 t
𝑓𝑜𝑟 𝑜𝑑𝑑 𝑛
Thus, the Fourier series of half wave symmetric function consists of only odd harmonics.
𝑇 0
Problem 5: Draw the complex fourier spectrum of given figure.
Solution: It is a plot between Cn and nω0 = ω. As Cn is real, only amplitude plot is sufficient.
The amplitude versus frequency plot, called the amplitude spectrum.
To draw the frequency spectrum,
x(t) =
2𝐴
𝜋
+
A −1 n
∞
𝑛=−∞ [
𝑛≠0 π 2n+1
exponential Fourier series coefficients
2𝐴
C0 = 𝜋
2𝐴
C1= 3𝜋
2𝐴
C2= 15𝜋
2𝐴
C3= 35𝜋
+
−1 n +1
2n−1
]ej2nt
Problem 6: For the periodic gate function shown in below figure, plot the magnitude and phase
spectra.
Solution:
Amplitude and phase spectra are shown in below figures respectively.
Example 7: Find the complex exponential Fourier series and the trigonometric Fourier series of
unit impulse train δT (t) shown in bellow figure.
Solution:
Let
The periodic waveform shown in above figure with period T can be expressed as
x(t) = δT (t) = ∞
𝑛=−∞ 𝛿(𝑡 − 𝑛𝑇)
𝑇
t0= - 2
𝑇
So
t0+T = - 2
In one period, only 𝛿(t) exists.
Fundamental frequency ω0 = (2𝜋/T)
1
Cn =
𝑇
1
=𝑇
𝑡 0 +𝑇
𝑥(𝑡)𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡
𝑡0
T
2
𝑇
−
2
1
𝛿(t) 𝑒 −𝑗𝑛 𝜔 0 𝑡 𝑑𝑡
𝛿(t)= 1
=0
at t = 0
elsewhere
1
= 𝑇 e0 = 𝑇
1
Cn = 𝑇
The exponential Fourier series is
x(t)=
∞
𝑛=−∞ 𝐶𝑛
𝑒 𝑗𝑛 𝜔 0 𝑡 =
1 𝑗𝑛 (2π )𝑡
∞
T
𝑛=−∞ 𝑇 𝑒
1
=𝑇
∞
𝑛=−∞
𝑒 𝑗𝑛
2π
T
𝑡
To get the trigonometric Fourier coeifficients , we have
1
𝑎0 = C0 = 𝑇
2
𝑎𝑛 = 𝑇
T
𝑥 𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡
0
2 T
𝑥 𝑡 sin 𝑛ω0 t 𝑑𝑡
𝑇 0
=
T
δ
𝑇 0
2 T
𝑥
𝑇 0
2
2
𝑡 cos 𝑛𝜔0 𝑡 𝑑𝑡 = 𝑇
𝑏𝑛 =
=
𝑡 sin 𝑛ω0 t 𝑑𝑡 = 0
Therefore , the trigonometric Fourier series is
x(t) +𝑎0 + ∞
𝑛=1 𝑎𝑛 cos 𝑛𝜔0 𝑡 + 𝑏𝑛 sin 𝑛ω0 t =
The complex Fourier series coefficients are
1
Cn = 𝑇
for all n
The magnitude and phase spectrum are
1
⎹Cn ⎸= ⎹𝑇 ⎸for all n
𝐶𝑛 = 0o
for all n
The frequency spectra are plotted in bellow figure.
1
𝑇
2
+𝑇
∞
𝑛=1 cos 𝑛𝜔0 𝑡
Problem 8: Find the Fourier series of the signal x(t)=e-t with T = 1 sec as shown in bellow
figure. Draw its magnitude and phase spectra.
x(t)=e-t
Period T = 1 sec
2𝜋
2𝜋
Fundamental frequency ω0 = 𝑇 = 1 =2𝜋
The exponential Fourier series is
Solution:
Given signal
1
Cn = 𝑇
=
=
x(t)=
𝑇
𝑥(𝑡)𝑒 −𝑗𝑛 𝜔 0 𝑡
0
1
𝑑𝑡 = 1
1 −t −𝑗𝑛 2𝜋𝑡
e 𝑒
0
𝑑𝑡
1 −(1+𝑗𝑛 2𝜋)𝑡
𝑒 −(1+𝑗𝑛 2𝜋 )𝑡 1
𝑒
𝑑𝑡
=
[
]
0
−(1+𝑗𝑛 2𝜋) 0
𝑒 −(1+𝑗𝑛 2𝜋 ) −1
1−𝑒 −1 𝑒 −𝑗𝑛 2𝜋
1−𝑒 −1
−(1+𝑗𝑛 2𝜋)
∞
𝑛=−∞ 𝐶𝑛
=
𝑒 𝑗𝑛 𝜔 0 𝑡 =
=
(1+𝑗𝑛 2𝜋)
(1+𝑗𝑛 2𝜋)
1−𝑒 −1
∞
𝑗𝑛 2𝜋𝑡
𝑛=−∞ (1+𝑗𝑛 2𝜋 ) 𝑒
The spectra are shown in bellow figure.
Problem 9: Obtain the exponential Fourier Series for the wave form shown in below figure.
Also draw the frequency spectrum.
Solution: The periodic waveform shown in fig with a period T= 2π can be expressed as:
A
0≤t≤π
x(t) =
−A
π ≤ t ≤ 2π
Let
t 0 = 0, t 0 +T = 2π
2π
2π
and
Fundamental frequency ⍵0 = T = 2π =1
Exponential Fourier series
1
C0 = T
Cn =
T
T 0
= 2π
x(t)e
=
Cn =
1
+ 2π
2π
π
−A dt = 0
dt
π
2π 0
A
1
=−
∴
x(t) dt
π
A dt
0
−jn ⍵0 t
1
1
T
0
1
Ae−jnt dt + 2π
−Ae−jnt dt
(−1)n − 1 − 1 − (−1)n = −j
j2nπ
2A
−j πn
A
2nπ
for odd n
0
x(t) = C0 +
2π
π
for even n
∞
jn ⍵0 t
n=−∞ Cn e
n≠0
2𝐴
=
A
∞
jnt
n=−∞ −j 2nπ e
2𝐴
2𝐴
C0 = 0, C-1 = C1 = 𝜋 , 𝐶−3 = C3 = 3𝜋 , 𝐶−5 = C5= 5𝜋
The frequency spectrum is shown in the below figure.
Problem10: Find the exponential Fourier series and plot the frequency spectrum for the full
wave rectified sine wave given in below figure
Solution: The waveform shown in fig can be expressed over one period(0 to π) as:
2π
x(t) = A sin ⍵t where ⍵= 2π =1
because it is part of a sine wave with period = 2π
x(t) = A sin ⍵t 0≤ t ≤ π
The full wave rectified sine wave is periodic with period T = π
Let
t 0 = 0, t 0 +T = 0+π = π
2π
2π
and
Fundamental frequency ⍵0 = T = π = 2
The exponential Fourier series is
jn ⍵0 t
j2nt
x(t) = ∞
= ∞
n=−∞ Cn e
n=−∞ Cn e
1 T
where
Cn = T 0 x(t)e−jn ⍵0 t dt
π
A π
A sin t e−j2nt dt = π 0 sin t e−j2nt
0
A e j 1−2n t −e 0
e −j 1−2n t −e 0
1
=π
= j2π
j 1−2n
−
dt
−j 1−2n
2A
∴Cn = π(1−4n 2 )
T
0
1 π
π 0
1
C0 = T
x(t) dt
A
=
A sin t dt = π − cos 𝑡 𝜋0 =
The exponential Fourier series is given by
2A
π
x(t) =
2A
∞
j2nt 2A
= π
n=−∞ π(1−4n 2 ) e
2𝐴
2𝐴
2A
+π
∞
n=−∞
n≠0
2𝐴
e j2nt
1−4n 2
2𝐴
C0 = 𝜋 , C-1 = C1 = 3𝜋 , 𝐶−2 = C2 = − 15𝜋 , 𝐶−3 = C3 = − 35𝜋
The frequency spectrum is shown in the below figure.
Assignment Problems:
1. Find the trigonometric Fourier series for the periodic signal x(t) shown in figure below. Also
draw the frequency spectrum.
2. Find the trigonometric Fourier series for the periodic signal x(t) shown in figure below. Also
draw the frequency spectrum.
3. Compute the exponential Fourier series of the signal shown in figure below. Also draw the
frequency spectrum.
4. Find the exponential series of the signal shown in the figure below. Also draw the frequency
spectrum.
5. Find the trigonometric Fourier series of x(t) =t2 over the interval (-1, 1).
6. Find the Fourier series for the periodic signal x(t) =t2 for 0 ≤ 𝑡 ≤ 1, so that it repeats every 1
sec.
7. Determine the time signal corresponding to the magnitude and phase spectra shown in below
figure with ⍵o = 𝜋.
Figure (a) Magnitude and (b) Phase spectra
8. Determine the Fourier series representation for x(t) = 2 sin(2𝜋𝑡 − 3)+sin6𝜋𝑡.
9. Find the complex exponential Fourier series coefficient of the signal x(t) = 3cos 4⍵0 t. Also
draw the frequency spectrum.
10. Find the complex exponential Fourier series coefficient of the signal x(t) = 2cos 3⍵0 t. Also
draw the frequency spectrum.
Simulation:
MATLAB can be used to find and plot the exponential Fourier series of a periodic function.
Using the symbolic math toolbox, we can perform the integrations necessary to find the
coefficients Fn.
Exponential Fourier Series
For the waveform, f(t), shown below:
we can write the exponential Fourier Series expansion as:
∞
𝐹𝑛 𝑒 𝑗𝑛 𝜔 𝑜 𝑡
𝑓 𝑡 =
𝑛 =−∞
Where
The Fn are complex coefficients. Thus, when we plot them we need to plot the magnitude
andphase separately. In Matlab we use the “abs” function to find the magnitude and “angle” to
findthe phase angle. The following Matlab statements show one method for obtaining these
plots:
Program
clc;clear all; close all;
n=-10:10;
Fn=(1./(j*n*2*pi)).*(4*exp(j*n*2*pi/7)+ 2*exp(-j*n*12*pi/7)-6*exp(j*n*6*pi/7));
Fn(find(n==0))=10/7;
stem(n,abs(Fn))
xlabel('n-->');
ylabel('|Fn|-->');
title('Magnitude spectrum');
figure,
stem(n,angle(Fn))
xlabel('n-->');
ylabel('L Fn -->');
title('Phase spectrum');
Since the value of Fn is different at n = 0, we assign that value using the find command.
Thestatement: find(n==0) will locate the index of the n vector that contains a zero. In this case
thatindex is 6, since the zero is found in the sixth position in n. That value is then used to index
Fn,so the value of Fn(6) is set to 10/7. This corresponds to the value for n = 0. These
commandsresult in the following magnitude and phase plots:
Note that the magnitude spectrum is even and the phase spectrum is odd as expected.
References:
[1] Alan V.Oppenheim, Alan S.Willsky and S.Hamind Nawab, “Signals & Systems”, Second
edition, Pearson Education, 8th Indian Reprint, 2005.
[2] M.J.Roberts, “Signals and Systems, Analysis using Transform methods and MATLAB”,
Second edition,McGraw-Hill Education,2011
[3] John R Buck, Michael M Daniel and Andrew C.Singer, “Computer explorations in Signals
and Systems using MATLAB”,Prentice Hall Signal Processing Series
[4] P Ramakrishna rao, “Signals and Systems”, Tata McGraw-Hill, 2008
[5] Tarun Kumar Rawat, “Signals and Systems”, Oxford University Press,2011
[6] A.Anand Kumar, “Signals and Systems” , PHI Learning Private Limited ,2011