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

x1(t)𝐹𝑠 Cn and x2 (𝑡)

𝐹𝑠 Dn

then Ax1(t)+Bx2(t) 𝐹𝑠 ACn+BDn

proof: From the definition of Fourier series, we have

FS[Ax1(t)+Bx2(t)]=1

𝑇 [𝐴𝑥1 𝑡 + 𝐵𝑥2 𝑡 ]𝑒−𝑗𝑛𝜔0𝑡𝑑𝑡𝑡0+𝑇

𝑡0

=𝐴(1

𝑇 𝑥1 𝑡 𝑒

−𝑗𝑛𝜔0𝑡𝑑𝑡𝑡0+𝑇

𝑡0)+ 𝐵(

1

𝑇 𝑥2 𝑡 𝑒

−𝑗𝑛𝜔0𝑡𝑑𝑡)𝑡0+𝑇

𝑡0

=ACn+BDn

or proved

2) Time shifting property

The time shifting property states that,if

x(t)𝐹𝑠 Cn

then x(t-t0)𝐹𝑠 𝑒−𝑗𝑛𝜔0𝑡0Cn

proof: From the definition of Fourier series, we have

x(t)= 𝐶𝑛∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝑡

x(t-t0)= 𝐶𝑛∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0(𝑡−𝑡0)

= [𝐶𝑛∞𝑛=−∞ 𝑒−𝑗𝑛𝜔0𝑡]𝑒𝑗𝑛 𝜔0𝑡

=FS-1

[Cn𝑒−𝑗𝑛𝜔0𝑡]

Ax1(t)+Bx2(t)𝐹𝑠 ACn+BDn

x(t-t0)𝐹𝑠 𝑒−𝑗𝑛𝜔0𝑡0Cn

Or proved

3) Time reversal property

The time reversal property states that,if

x(t)𝐹𝑠 Cn

then x(-t)𝐹𝑠 C-n

proof: From the definition of Fourier series, we have

x(t)= 𝐶𝑛∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝑡

x(-t)= 𝐶𝑛∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0(−𝑡)

substituting n = -p in the right hand side, we get

x(-t)= 𝐶−𝑝−∞𝑝=∞ 𝑒𝑗 (−𝑝)𝜔0(−𝑡) = 𝐶−𝑝

∞𝑝=−∞ 𝑒𝑗𝑝 𝜔0𝑡

substituting p = n , we get

x(-t)= 𝐶−𝑛∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝑡 = FS

-1[C-n]

proved

4) Time scaling property

The time scaling property states that , if

x(t)𝐹𝑠 Cn

then x(𝛼t)𝐹𝑠 Cn with ω0→𝛼ω0

proof: From the definition of Fourier series, we have

x(t)= 𝐶𝑛∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝑡

x(𝛼t)= 𝐶𝑛∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝛼𝑡= 𝐶𝑛

∞𝑛=−∞ 𝑒𝑗𝑛 (𝜔0𝛼)𝑡= FS

-1[Cn]

where ω0→𝛼ω0.

proved.

5) Time differential property

The time differential property states that, if

x(t)𝐹𝑠 Cn

then 𝑑𝑥 (𝑡)

𝑑𝑡

𝐹𝑠 jnω0Cn

proof: From the definition of Fourier series, we have

x(t)= 𝐶𝑛∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝑡

Differentiating both sides with respect to t, we get 𝑑𝑥 (𝑡)

𝑑𝑡= 𝐶𝑛

𝑑(𝑒 𝑗𝑛 𝜔0𝑡)

𝑑𝑡∞𝑛=−∞ = 𝐶𝑛(jnω0)∞

𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝑡

= 𝐶𝑛(jnω0)∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝑡 = FS

-1[jnω0Cn]

Proved.

6) Time integration property

x(-t)𝐹𝑠 C-n

x(𝛼t)𝐹𝑠 Cn with fundamental frequency of 𝛼ω0

𝑑𝑥 (𝑡)

𝑑𝑡

𝐹𝑠 jnω0Cn

The time integration property states that,

if x(t)𝐹𝑠 Cn

then 𝑥 𝜏 𝑑𝜏𝑡

−∞

𝐹𝑆

𝐶𝑛

𝑗𝑛 𝜔0 (if C0 = 0)

proof: From the definition of Fourier series, we have

x(t)= 𝐶𝑛∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝑡

𝑥 𝜏 𝑑𝜏𝑡

−∞ = [ 𝐶𝑛

∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝜏]𝑑𝜏

𝑡

−∞

Interchanging the order of integration and summation, we get

𝑥 𝜏 𝑑𝜏𝑡

−∞ = 𝐶𝑛

∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝜏𝑑𝜏

𝑡

−∞

= 𝐶𝑛∞𝑛=−∞ [

𝑒 𝑗𝑛 𝜔0𝜏

𝑗𝑛 𝜔0 ] −∞

𝑡

= (𝐶𝑛

𝑗𝑛 𝜔0)∞

𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝑡 = FS-1

[𝐶𝑛

𝑗𝑛 𝜔0]

Proved.

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

x1(t)𝐹𝑠 Cn and x2 (𝑡)

𝐹𝑠 Dn

then x1(t)*x2(t) 𝐹𝑠 TCnDn

proof: From the definition of Fourier series, we have

FS[x1(t)*x2(t)]=1

𝑇 [x1(t) ∗ x2(t)]𝑡0+𝑇

𝑡0𝑒−𝑗𝑛𝜔0𝑡𝑑𝑡 =

1

𝑇 [x1(t) ∗ x2(t)]𝑇

0𝑒−𝑗𝑛𝜔0𝑡𝑑𝑡

But from the convolution integral for a periodic signal, we have

x1(t) ∗ x2(t)= x1(τ) ∗ x2(t − τ)𝑇

0𝑑𝜏 or x1(t) ∗ x2(t)= x1(t − τ) ∗ x2(τ)

𝑇

0𝑑𝜏

Substituting back, we get

FS[x1(t)*x2(t)] =1

𝑇 ( x1(τ) ∗ x2(t − τ)

𝑇

0𝑑𝜏)

𝑇

0𝑒−𝑗𝑛𝜔0𝑡𝑑𝑡

Interchanging the order of integration, we get

FS[x1(t)*x2(t)] =1

𝑇 x1(τ)[ x2(t − τ)

𝑇

0

𝑇

0𝑒−𝑗𝑛𝜔0𝑡𝑑𝑡] 𝑑𝜏

Substituting t-𝛕 =t' in RHS, we have dt=dt'. Substituting back, we get

FS[x1(t)*x2(t)] =1

𝑇 x1(τ)[ x2(t′)

𝑇−τ

−τ

𝑇

0𝑒−𝑗𝑛 𝜔0(𝑡 ′ +τ)𝑑𝑡′] 𝑑𝜏

=T(1

𝑇 x1(τ)𝑒−𝑗𝑛𝜔0𝑡𝑑𝛕𝑇

0) (

1

𝑇 x1(t′)𝑒−𝑗𝑛𝜔0𝑡′𝑑t′𝑇

0)

=T(Cn)(Dn)

0 to T or –𝛕 to T-𝛕 or t0 to t0+T will have the same period.

Proved.

𝑥 𝜏 𝑑𝜏𝑡

−∞

𝐹𝑆

𝐶𝑛

𝑗𝑛 𝜔0 , C0 = 0

x1(t)*x2(t)𝐹𝑠 TCnDn

8) Modulation or Multiplication property

The Modulation or Multiplication property states that,if

x1(t)𝐹𝑠 Cn and x2 (𝑡)

𝐹𝑠 Dn

then x1(t)x2(t) 𝐹𝑠 𝐶𝑙𝐷𝑛−𝑙

∞𝑙=−∞

proof: From the definition of Fourier series, we have

FS[x1(t)x2(t)]= 1

𝑇 x1 t x2 t 𝑡0+𝑇

𝑡0𝑒−𝑗𝑛𝜔0𝑡𝑑𝑡

=1

𝑇 x1 t ( 𝐶𝑙𝑒

𝑗𝑛 𝜔0𝑡∞𝑙=−∞ )

𝑡0+𝑇

𝑡0𝑒−𝑗𝑛𝜔0𝑡𝑑𝑡

=1

𝑇 x1 t ( 𝐶𝑙𝑒

−𝑗 (𝑛−𝑙)𝜔0𝑡∞𝑙=−∞ )

𝑡0+𝑇

𝑡0𝑑𝑡

Interchanging the order of integration and summation, we have

FS[x1(t)x2(t)]= 𝐶𝑙∞𝑙=−∞ (

1

𝑇 x1 t 𝑒−𝑗 (𝑛−𝑙)𝜔0𝑡𝑡0+𝑇

𝑡0𝑑𝑡) = 𝐶𝑙𝐷𝑛−𝑙

∞𝑙=−∞

Proved.

9) Conjugation and Conjugate symmetry property

If x(t)𝐹𝑠 Cn

Then the Conjugation property states that

x*(t)

𝐹𝑠 𝐶−𝑛

∗ [for complex x(t)]

proof: Conjugation property

From the definition of Fourier series, we have

FS[x*(t)]=

1

𝑇 x ∗ (t)𝑡0+𝑇

𝑡0𝑒−𝑗𝑛𝜔0𝑡𝑑𝑡=(

1

𝑇 x(t)𝑡0+𝑇

𝑡0𝑒𝑗𝑛 𝜔0𝑡𝑑𝑡)*

=(1

𝑇 x(t)𝑡0+𝑇

𝑡0𝑒−𝑗 (−𝑛)𝜔0𝑡𝑑𝑡)*

=[C-n]*

Proved.

10) Parseval’s Relation or Theorem or property

If x1(t)𝐹𝑠 Cn [for complex x1(t)]

And x2 (𝑡)𝐹𝑠 Dn [for complex x2(t)]

Then, the parsevals relation states that 1

𝑇 x1 t x2

∗ t 𝑡0+𝑇

𝑡0𝑑𝑡= Cn

∞𝑛=−∞ 𝐷𝑛

∗ [for complex x1(t) and x2(t)]

And parseval’s identity states that 1

𝑇 ⎹x(t)⎸2𝑡0+𝑇

𝑡0𝑑𝑡 = ⎹Cn⎸

2∞𝑛=−∞ if x1(t) = x2(t) = x(t)

Proof: parsevals relation

LHS =1

𝑇 x1 t x2

∗ t 𝑡0+𝑇

𝑡0𝑑𝑡 =

1

𝑇 ( 𝐶𝑛𝑒

𝑗𝑛 𝜔0𝑡∞𝑛=−∞ )

𝑡0+𝑇

𝑡0𝑥2∗𝑑𝑡

Interchanging the order of integration and summation in the RHS , we have 1

𝑇 x1 t x2

∗ t 𝑡0+𝑇

𝑡0𝑑𝑡 = 𝐶𝑛

∞𝑛=−∞ (

1

𝑇 x2

∗ t 𝑒𝑗𝑛 𝜔0𝑡𝑡0+𝑇

𝑡0𝑑𝑡)

= 𝐶𝑛∞𝑛=−∞ (

1

𝑇 x2 t 𝑒−𝑗𝑛𝜔0𝑡𝑡0+𝑇

𝑡0𝑑𝑡)*

= Cn∞𝑛=−∞ [Dn]∗

x1(t)x2(t)𝐹𝑠 𝐶𝑙𝐷𝑛−𝑙

∞𝑙=−∞

x*(t)

𝐹𝑠 𝐶−𝑛

∗

Proved

Parseval’s identity

If x1(t) = x2(t) = x(t), then the relation changes to

1

𝑇 x t x∗ t 𝑡0+𝑇

𝑡0𝑑𝑡= Cn

∞𝑛=−∞ 𝐶𝑛

∗

Since ⎹x(t)⎸2= x t x∗ t and ⎹Cn⎸2= Cn𝐶𝑛

∗ , substituting these values in above equation, we get

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.

1

𝑇 x1 t x2

∗ t 𝑡0+𝑇

𝑡0𝑑𝑡= Cn

∞𝑛=−∞ 𝐷𝑛

∗

1

𝑇 ⎹x(t)⎸2𝑡0+𝑇

𝑡0𝑑𝑡 = ⎹Cn⎸

2∞𝑛=−∞

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⎸𝑒𝑗 𝜃𝑛

C-n = ⎹Cn⎸𝑒−𝑗𝜃𝑛

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

Cn = 1

𝑇 𝑥 𝑡 𝑒−𝑗𝑛𝜔0𝑡𝑑𝑡

T

2

−𝑇

2

)

And C-n = 1

𝑇 𝑥 𝑡 𝑒𝑗𝑛 𝜔0𝑡𝑑𝑡

T

2

−𝑇

2

)

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⎸ejθn

then C-n=⎹Cn⎸e−jθn

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

interval t0 to t0+T or 0 to T or -𝑇

2 to

𝑇

2 is given by

x(t)= 𝑎0+ (𝑎𝑛 cos𝑛𝜔0𝑡∞𝑛=1 + 𝑏𝑛 sin𝑛ω0t)

where 𝑎0 = 1

𝑇 𝑥 𝑡 𝑑𝑡

T

2

−𝑇

2

𝑎𝑛 = 2

𝑇 𝑥 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

2

−𝑇

2

and𝑏𝑛 = 2

𝑇 𝑥 𝑡 sin 𝑛ω0t𝑑𝑡

T

2

−𝑇

2

For an odd function

x(t) = -x(-t)

Also even part is zero, i.e. xe(t) = 0 and x(t) = x0(t)

𝑎0 = 1

𝑇 𝑥 𝑡 𝑑𝑡

T

2

−𝑇

2

= 1

𝑇 𝑥0 𝑡 𝑑𝑡

T

2

−𝑇

2

= 0

Since the integration of an odd function,over one cycle is always zero.

𝑎𝑛 = 2

𝑇 𝑥 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

2

−𝑇

2

= 2

𝑇 𝑥0 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

2

−𝑇

2

Here 𝑥0(t) is odd and cos 𝑛𝜔0𝑡 is even. So 𝑥0 𝑡 cos𝑛𝜔0𝑡, i.e. the product if an odd function

and even function is odd. So integration over one cycle is zero. Therefore,

𝑎𝑛 = 0

Now, 𝑏𝑛 = 2

𝑇 𝑥 𝑡 sin 𝑛ω0t𝑑𝑡

T

2

−𝑇

2

= 2

𝑇 𝑥0 𝑡 sin 𝑛ω0t𝑑𝑡

T

2

−𝑇

2

Now, 𝑥0(t) is odd sin𝑛ω0t is also odd. Therefore, the product 𝑥0(t)sin𝑛ω0t is even. So we

have to evaluate the integral:

𝑏𝑛 = 2

𝑇 𝑥 𝑡 sin𝑛ω0t𝑑𝑡

T

2

−𝑇

2

= 4

𝑇 𝑥 𝑡 sin𝑛ω0t𝑑𝑡

T

20

x(t)= 𝑏𝑛 sin𝑛ω0t∞𝑛=1

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

interval t0 to t0+T or 0 to T or -𝑇

2 to

𝑇

2 is given by

x(t)= 𝑎0+ (𝑎𝑛 cos𝑛𝜔0𝑡∞𝑛=1 + 𝑏𝑛 sin𝑛ω0t)

where 𝑎0 = 1

𝑇 𝑥 𝑡 𝑑𝑡

T

2

−𝑇

2

𝑎𝑛 = 2

𝑇 𝑥 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

2

−𝑇

2

and 𝑏𝑛 = 2

𝑇 𝑥 𝑡 sin 𝑛ω0t𝑑𝑡

T

2

−𝑇

2

For an even function

x(t) = x(-t)

Also odd part is zero, i.e. x(t) = xe(t)

𝑎0 = 1

𝑇 𝑥 𝑡 𝑑𝑡

T

2

−𝑇

2

= 1

𝑇 𝑥𝑒 𝑡 𝑑𝑡

T

2

−𝑇

2

= 2

𝑇 𝑥 𝑡 𝑑𝑡

T

20

𝑎𝑛 = 2

𝑇 𝑥 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

2

−𝑇

2

= 2

𝑇 𝑥𝑒 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

2

−𝑇

2

𝑥𝑒(t) is even and also cos 𝑛𝜔0𝑡 is even. So, the product 𝑥𝑒(t)cos𝑛𝜔0𝑡 is even. So we have to

evaluate the integral:

𝑎𝑛 = 2

𝑇 𝑥𝑒 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

2

−𝑇

2

= 4

𝑇 𝑥 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

20

𝑏𝑛 = 2

𝑇 𝑥 𝑡 sin𝑛ω0t𝑑𝑡

T

2

−𝑇

2

= 2

𝑇 𝑥𝑒 𝑡 sin 𝑛ω0t𝑑𝑡

T

2

−𝑇

2

Here 𝑥𝑒(t) is even sin𝑛ω0tand is odd. So, the product 𝑥𝑒 𝑡 sin 𝑛ω0t is odd. Thus, the integral

over one complete cycle is zero.

𝑏𝑛 = 0

x(t)= 𝑎0+ 𝑎𝑛 cos𝑛𝜔0𝑡∞𝑛=1

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

interval t0 to t0+T or 0 to T or -𝑇

2 to

𝑇

2 is given by

x(t)= 𝑎0+ (𝑎𝑛 cos𝑛𝜔0𝑡∞𝑛=1 + 𝑏𝑛 sin𝑛ω0t)

where 𝑎0 = 1

𝑇 𝑥 𝑡 𝑑𝑡

T

2

−𝑇

2

𝑎𝑛 = 2

𝑇 𝑥 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

2

−𝑇

2

and 𝑏𝑛 = 2

𝑇 𝑥 𝑡 sin 𝑛ω0t𝑑𝑡

T

2

−𝑇

2

For a half wave symmetric function,

x(t) = -x(t±𝑇

2)

Therefore, 𝑎0 = 0

𝑎𝑛 = 2

𝑇 𝑥 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

0

= 2

𝑇[ 𝑥 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

20

+ 𝑥 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡T𝑇

2

]

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.

𝑎𝑛 =2

𝑇[ 𝑥 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

20

+ 𝑥 𝑡 +T

2 cos𝑛𝜔0(𝑡 +

T

2) 𝑑𝑡

T

20

]

But x(t) = -x(t±𝑇

2)

𝑎𝑛 = 2

𝑇[ 𝑥 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

20

+ −𝑥 𝑡 cos(𝑛𝜔0𝑡 + 𝑛π) 𝑑𝑡T

20

]

= 2

𝑇[ 𝑥 𝑡 cos𝑛𝜔0𝑡 − 𝑥 𝑡 cos(𝑛𝜔0𝑡 + 𝑛π)] 𝑑𝑡

T

20

𝑎𝑛 = 0 𝑓𝑜𝑟 𝑒𝑣𝑒𝑛 𝑛

4

𝑇 𝑥 𝑡 cos𝑛𝜔0𝑡

T

20

𝑓𝑜𝑟 𝑜𝑑𝑑 𝑛

Now,

𝑏𝑛 = 2

𝑇 𝑥 𝑡 sin𝑛ω0t𝑑𝑡

T

20

= 2

𝑇[ 𝑥 𝑡 sin𝑛ω0t𝑑𝑡

T

20

+ 𝑥 𝑡 sin 𝑛ω0t𝑑𝑡T𝑇

2

]

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.

𝑏𝑛 = 2

𝑇[ 𝑥 𝑡 sin𝑛ω0t𝑑𝑡

T

20

+ 𝑥 𝑡 +𝑇

2 sin𝑛ω0(t +

T

2)𝑑𝑡

T

20

]

But x(t) = -x(t±𝑇

2)

𝑏𝑛 = 2

𝑇[ 𝑥 𝑡 sin𝑛ω0t𝑑𝑡

T

20

+ −𝑥 𝑡 sin(𝑛ω0t + nπ)𝑑𝑡T

20

]

= 2

𝑇[ 𝑥 𝑡 sin𝑛ω0t𝑑𝑡

T

20

- 𝑥 𝑡 sin𝑛ω0t cos nπ) 𝑑𝑡T

20

]

𝑏𝑛 = 0 𝑓𝑜𝑟 𝑒𝑣𝑒𝑛 𝑛

4

𝑇 𝑥 𝑡 sin𝑛ω0t

T

20

𝑓𝑜𝑟 𝑜𝑑𝑑 𝑛

Thus, the Fourier series of half wave symmetric function consists of only odd harmonics.

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

π[ −1 n

2n+1+

−1 n +1

2n−1]ej2nt∞

𝑛=−∞𝑛≠0

exponential Fourier series coefficients

C0 = 2𝐴

𝜋

C1= 2𝐴

3𝜋

C2= 2𝐴

15𝜋

C3= 2𝐴

35𝜋

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: The periodic waveform shown in above figure with period T can be expressed as

x(t) = δT(t) = 𝛿(𝑡 − 𝑛𝑇)∞𝑛=−∞

Let t0= -𝑇

2

So t0+T = - 𝑇

2

In one period, only 𝛿(t) exists.

Fundamental frequency ω0 = (2𝜋/T)

Cn =1

𝑇 𝑥(𝑡)𝑒−𝑗𝑛𝜔0𝑡

𝑡0+𝑇

𝑡0

𝑑𝑡

=1

𝑇 𝛿(t) 𝑒−𝑗𝑛 𝜔0𝑡𝑑𝑡

T

2

−𝑇

2

= 1

𝑇e0 =

1

𝑇

Cn= 1

𝑇

The exponential Fourier series is

x(t)= 𝐶𝑛∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝑡 =

1

𝑇∞𝑛=−∞ 𝑒𝑗𝑛 (

2π

T)𝑡

=1

𝑇 𝑒𝑗𝑛

2π

T 𝑡∞

𝑛=−∞

To get the trigonometric Fourier coeifficients , we have

𝑎0 = C0= 1

𝑇

𝑎𝑛 = 2

𝑇 𝑥 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

0 =

2

𝑇 δ 𝑡 cos𝑛𝜔0𝑡 𝑑𝑡

T

0 =

2

𝑇

𝑏𝑛 = 2

𝑇 𝑥 𝑡 sin𝑛ω0t𝑑𝑡

T

0 =

2

𝑇 𝑥 𝑡 sin𝑛ω0t𝑑𝑡

T

0 = 0

Therefore , the trigonometric Fourier series is

x(t) +𝑎0 + 𝑎𝑛cos𝑛𝜔0𝑡∞𝑛=1 + 𝑏𝑛 sin𝑛ω0t =

1

𝑇 +

2

𝑇 cos𝑛𝜔0𝑡

∞𝑛=1

The complex Fourier series coefficients are

Cn= 1

𝑇 for all n

The magnitude and phase spectrum are

⎹Cn⎸= ⎹1

𝑇⎸for all n

𝐶𝑛 = 0o for all n

The frequency spectra are plotted in bellow figure.

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.

𝛿(t)= 1 at t = 0

= 0 elsewhere

Solution: Given signal x(t)=e

-t

Period T = 1 sec

Fundamental frequency ω0 =2𝜋

𝑇 =

2𝜋

1 =2𝜋

The exponential Fourier series is

Cn =1

𝑇 𝑥(𝑡)𝑒−𝑗𝑛𝜔0𝑡𝑇

0𝑑𝑡 =

1

1 e−t𝑒−𝑗𝑛 2𝜋𝑡1

0𝑑𝑡

= 𝑒−(1+𝑗𝑛 2𝜋)𝑡1

0𝑑𝑡 = [

𝑒−(1+𝑗𝑛 2𝜋 )𝑡

−(1+𝑗𝑛 2𝜋) ]0

1

= 𝑒−(1+𝑗𝑛 2𝜋 )−1

−(1+𝑗𝑛 2𝜋) =

1−𝑒−1𝑒−𝑗𝑛 2𝜋

(1+𝑗𝑛 2𝜋) =

1−𝑒−1

(1+𝑗𝑛 2𝜋)

x(t)= 𝐶𝑛∞𝑛=−∞ 𝑒𝑗𝑛 𝜔0𝑡 = (

1−𝑒−1

1+𝑗𝑛 2𝜋)∞

𝑛=−∞ 𝑒𝑗𝑛 2𝜋𝑡

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:

x(t) = A 0 ≤ t ≤ π

−A π ≤ t ≤ 2π

Let

t0 = 0, t0+T = 2π

and Fundamental frequency ⍵0=2π

T =

2π

2π =1

Exponential Fourier series

C0 = 1

T x(t)

T

0 dt

= 1

2π Aπ

0 dt +

1

2π −A

2π

π dt = 0

Cn = 1

T x(t)e−jn⍵0tT

0 dt

= 1

2π Ae−jntπ

0 dt +

1

2π −Ae−jnt2π

π dt

= −A

j2nπ (−1)n − 1 − 1 − (−1)n = −j

A

2nπ

Cn = −j

2A

πn for odd n

0 for even n

∴ x(t) = C0+ Cnejn⍵0t∞n=−∞

n≠0 = −j

A

2nπejnt∞

n=−∞

C0 = 0, C-1 = C1 = 2𝐴

𝜋, 𝐶−3 = C3 =

2𝐴

3𝜋, 𝐶−5 = C5=

2𝐴

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:

x(t) = A sin ⍵t where ⍵= 2π

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

t0 = 0, t0+T = 0+π = π

and Fundamental frequency ⍵0=2π

T =

2π

π = 2

The exponential Fourier series is

x(t) = Cnejn⍵0t∞n=−∞ = Cnej2nt∞

n=−∞

where Cn = 1

T x(t)e−jn⍵0tT

0 dt

= 1

π A sin t e−j2ntπ

0 dt =

A

π sin t e−j2ntπ

0 dt

= A

j2π

ej 1−2n t−e0

j 1−2n −

e−j 1−2n t−e0

−j 1−2n

∴Cn = 2A

π(1−4n2)

C0 = 1

T x(t)

T

0 dt

= 1

π Aπ

0 sin t dt =

A

π − cos 𝑡 0

𝜋 = 2A

π

The exponential Fourier series is given by

x(t) = 2A

π(1−4n2)ej2nt∞

n=−∞ = 2A

π +

2A

π ∞

n=−∞n≠0

ej2nt

1−4n2

C0 = 2𝐴

𝜋, C-1 = C1 =

2𝐴

3𝜋, 𝐶−2 = C2 = −

2𝐴

15𝜋, 𝐶−3 = C3 = −

2𝐴

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⍵0t. Also

draw the frequency spectrum.

10. Find the complex exponential Fourier series coefficient of the signal x(t) = 2cos 3⍵0t. 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:

𝑓 𝑡 = 𝐹𝑛𝑒𝑗𝑛 𝜔𝑜 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