Fourier Analysis of Signals &
SystemsSystems
Dr. Sarmad Sohaib
Assistant Professor, UET Taxila
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.
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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.
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Fourier Analysis (cntd.)
• 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.
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Fourier Analysis (cntd.)
• 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.
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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
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0 0 0
1 1
( ) cos sink k
k k
x t A B k t C k tω ω∞ ∞
= =
= + +∑ ∑
0
2
T
πω =
The continuous trigonometric Fourier
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.
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The continuous trigonometric Fourier
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).
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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
−< <
=
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( )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).
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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,…
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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
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[ ]
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.
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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.
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The continuous trigonometric Fourier
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.
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Example-2
• Find the Fourier coefficients for the function
x(t)=2t over the range -½ to ½.
( )x t
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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
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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.
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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
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( ) [ ]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:
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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.
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Example-3 (home work)
• Determine the Fourier series of the
rectangular pulse train signal illustrated below
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( )x t
M
02
τ
2
τ−
tt−pTpT−
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