Fourier Analysis

Fourier Analysis of Signals &

SystemsSystems

Dr. Sarmad Sohaib

Assistant Professor, UET Taxila

[email protected]

Fourier Analysis

• Decompose the signal in terms of sinusoidal

(or complex exponential) components.

• With such decomposition, the signal is said to • With such decomposition, the signal is said to

be represented in the frequency domain.

• Fourier analysis applies to both periodic and

aperiodic signals.

12/7/2010 2Dr. Sarmad Sohaib (Signals & Systems)

Fourier Analysis (cntd.)

• When analyzing periodic waveforms, the FOURIER SERIES applies.

• When analyzing aperiodic signals, we apply the FOURIER TRANSFORM.

• Furthermore, signal may be continuous or discrete, so • Furthermore, signal may be continuous or discrete, so there are FOUR varieties of the analysis technique, i.e.:

– The continuous Fourier series

– The continuous Fourier transform

– The discrete Fourier series, also called the discrete Fourier Transform

– The discrete time Fourier transform.

12/7/2010 3Dr. Sarmad Sohaib (Signals & Systems)


• Fourier analysis allows a signal to be expressed as

a series of sinusoidal in ascending frequency.

• These sinusoids are represented as a set of cosine

and sine functions.and sine functions.

• But this is not strictly necessary: it is quite

permissible to express the series as a set of sine

function only (or cosine function only), since each

is simply a phase shifted version of the other.

12/7/2010 Dr. Sarmad Sohaib (Signals & Systems) 4


• Fourier transformation is used in

• Signal restoration

• Signal enhancement and degradation

• Signal compression and decompression• Signal compression and decompression

• Signal modulation and demodulation

• Phase shifting

• Digital wave synthesis

• Interpolation and so on …

• Every time you use your mobile phone, DVD, CD or MP3

player, Fourier is at the heart of the matter.


The continuous trigonometric Fourier

series for periodic signals

• A periodic waveform may be represented as a series

involving cosine and sine coefficients: in general,

although not always, the series in infinite. Thus the

synthesis equation may be written as∞ ∞

where ω0 represents the fundamental frequency of

the system, given by


0 0 0

1 1

( ) cos sink k

k k

x t A B k t C k tω ω∞ ∞

= =

= + +∑ ∑

0

2

T

πω =


series for periodic signals (cntd.)

where

• T represents the time duration of a single period of the waveform.

• A0 coefficient represents the mean signal level (also known a the DC level or zeroth harmonic),

• A0 coefficient represents the mean signal level (also known a the DC level or zeroth harmonic),

• Bk is the coefficient representing the magnitude of the cosine wave content of the kth harmonic

• Ck is the coefficient representing the magnitude of the sine wave content of the kth harmonic.



series for periodic signals (cntd.)• In order to calculate A0, Bk, and Ck, we employ the analysis

equations as follows:

the limits 0 to Tmeans single timeperiod, the limit

0

0

1( )

2

T

T

A x t dtT

= ∫period, the limit does not have to start specifically

from 0.

• The Equation for calculating Bk and Ck allows us to calculate “how much” of a given sinusoid, at a particular frequency, is presented in the waveform x(t).


0

0

0

0

2( )cos( )

2( )sin( )

T

k

T

k

B x t k t dtT

C x t k t dtT

ω

ω

=

=

∫

∫

Example-1

• Calculate the Fourier components of the square wave pulse

train shown below over the range t=-½ to ½, which is defined

as1

0, 02

( )

t

x t

−< <

=


( )1

1, 02

x t

t

= < <

( )x t

1

01/ 21/ 2− tt−

Example-1 (cntd.)

• In this example, t ranges from -½ to ½, T=1. The mean signal level is given by

0

0

1/2

1( )

1

T

A x t dtT

= ∫

• Clearly, we could have obtained this by inspection (DC level =1/2).


1/2

1/2

0 1/2

1/2 0

1( )

1

1(0) (1)

2

x t dt

dt dt

−

−

=

= + =

∫

∫ ∫

Example-1 (cntd.)

• The cosine terms are given by

1/2

0 0

0 1/2

0 1/2

0 0

2 2( ) cos( ) ( ) cos( )

1

2 (0)cos( ) (1)cos( )

T

kB x t k t dt x t k t dtT

k t dt k t dt

ω ω

ω ω

−

= =

= +

∫ ∫

∫ ∫

• From the final line of the above equation we deduce that whatever the

value of k, Bk is always zero:

Bk=0, k=1,2,3,…


0 0

1/2 0

1/2 1/2

0

00 0

2 (0)cos( ) (1)cos( )

sin( ) sin( 2 )2 2

2

k t dt k t dt

k t k t

k k

ω ω

ω π

ω π

−

= +

= =

∫ ∫

Example-1 (cntd.)

• The sine terms are given by

1/2

0 0

0 1/2

0 1/2

0 0

2 2( )sin( ) ( )sin( )

1

2 (0)sin( ) (1)sin( )

T

kC x t k t dt x t k t dtT

k t dt k t dt

ω ω

ω ω

−

= =

= +

∫ ∫

∫ ∫

• In this case whenever k is even, Ck = 0. However, whenever k is odd, we

have:

For k = odd


[ ]

0 0

1/2 0

1/2 1/2

0

00 0

2 (0)sin( ) (1)sin( )

cos( ) cos( 2 ) 12 2 cos(0) cos( )

2

k t dt k t dt

k t k tk

k k k

ω ω

ω ππ

ω π π

−

= +

= − = − = −

∫ ∫

2k

Ckπ

=

Example-1 (cntd.) + Observations

• So the Fourier series for this square wave pulse train is

2 1 1 1( ) 0.5 sin sin3 sin5 sin 7x t t t t tω ω ω ω

= + + + + + ⋯

• Therefore we need infinitely large number of harmonics to recreate the ideal square wave pulse train.

• This is because the waveform exhibits discontinuities where it changes state from zero to one and vice versa.


0 0 0 0

2 1 1 1( ) 0.5 sin sin3 sin5 sin 7

3 5 7x t t t t tω ω ω ω

π

= + + + + +

⋯

Observations

• The signal in example-1 is odd, therefore has

only sine terms (as Bk=0).

• Therefore,

– Odd signal �all Fourier coefficients would be sine – Odd signal �all Fourier coefficients would be sine

terms

– Even signal �all Fourier coefficients would be

cosine terms

– Neither even nor odd �Fourier series would

contain both cosine and sine harmonics.



series for aperiodic signals

• If we have a signal that does not repeat, but

we want to find the Fourier coefficients within

some range, say -½T to ½T, we pretend that it

is cyclical outside of this range.is cyclical outside of this range.

• We then use exactly the same analysis

equations as for continuous periodic signals.


Example-2

• Find the Fourier coefficients for the function

x(t)=2t over the range -½ to ½.

( )x t


t

( )x t

0

1/ 2− 1/ 2

11

1−

Example-2 (contd.)

• Construct a series of repeating ramps outside

of the range of interest, shown by dotted

lines.

• Function has odd symmetry, which means: • Function has odd symmetry, which means:

– a) the DC term, A0, is zero

– b) there are no cosine terms, Bk, in the Fourier

expansion


Example-2 (contd.)

• The sine terms are thus:

• Now the integral here is product of two functions, so we use the

1/2

0 0

0 1/2

2( )sin( ) 2 2 sin( )

T

kC x t k t dt t k t dt

Tω ω

−

= =∫ ∫

• Now the integral here is product of two functions, so we use the

integration by parts formula, i.e.


0 0 0

0

where, 2 , hence 2, and 2

1Similarly, sin( ) , so sin( ) cos( ).

u dv uv v du

duu t du dt

dt

dv k t dt v k t dt k tk

ω ω ωω

= −

= = =

= = = −

∫ ∫

∫

Example-2 (contd.)

• Inserting these into formula, we get

( )

( ) [ ]

1/2 1/2

0 0

0 0 1/21/2

1/2

1/2

4 4cos cos( )

4 4cos sin( )

k

tC k t k t dt

k k

tk t k t

ω ωω ω

ω ω

−−

= − +

= − +

∫

• Fortunately, the final sine term in the above equation cancels out, leaving


( ) [ ]0 02 2 1/20 01/2

cos sin( )k t k tk k

ω ωω ω −

−

= − +

2 2 2cos( ) cos( ) cos( )

2 2k

C k k kk k k

π π ππ π π

= − − − = −

Example-2 (contd.)

• Hence when k is odd, we have

• And when k is even, we have

2k

Ckπ

=

2k

Ckπ

= −

• So the Fourier series is:


0 0 0 0 0

2 1 1 1 1( ) 0 sin( ) sin(2 ) sin(3 ) sin(4 ) sin(5 )

2 3 4 5x t t t t t tω ω ω ω ω

π

= + − + − + +

⋯

Practice more complex Practice more complex

signals at home.


Example-3 (home work)

• Determine the Fourier series of the

rectangular pulse train signal illustrated below


( )x t

M

02

τ

2

τ−

tt−pTpT−

Fourier Analysis

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