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Convolution and FourierTransform
Dr. Hariharan Muthusamy,
School of Mechatronic Engineering,
Universiti Malaysia Perlis.
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Convolution
Consider a linear system whose behaviour is specified by the impulse
response h(n). To find the output signal be y(n), resolve the signal x(n) into
a weighted sum of impulses, and the linearity and time shift property of the
LTI system.
The above expression gives the response y(n) of the LTI system as a function
of the input signal x(n) and impulse response h(n) is called the convolutionsum.
Steps involved in finding out the convolution sum
1. Folding: Fold the signal h(k) about the origin, ie at k=0;
2. Shifting: Shift h(-k) to the right by no,if no is positive or shift h(-k) to theleft by no, if no is negative to obtain h(no k).
3. Multiplication: Multiply x(k) by h(no k) to obtain the product sequence
yo(k) = x(k) h(no k)
k
k)x(k)h(ny(n)
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4. Summation: Sum all the values of
the product sequence yo(k) toobtain the value of the output at
time n = no
Find the convolution of two finite
duration sequences
otherwise
1n1
0
1x(n)
otherwise
1n1
0
1h(n)
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2. Find the convolution of the two signals x(n) = u(n) and h(n) =
anu(n), |a|=0.
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Fourier Transform When the repetition period T becomes infinity, the waveform f(t) become
non-periodic, the separation between two adjacent harmonics will be zero.
Assuming f(t) is initially periodic, we have
dtef(t)Tc)F(jn
Let
,ecTTf(t)
dtef(t)
T
1c
where
,ecf(t)
T/2
T/2
tjn
no
tjn
n
T/2
T/2
tjn
n
tjn
n
o
o
o
o
As T ->, the fundamental
frequency becomes infinitelysmall, hence o -> d
dtef(t))F(j
de)F(j
2
1f(t)
e)F(jn2
1
ecTT
1f(t)
tj
tj
o
tjn
o
tjn
n
o
o
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Properties of Fourier Transform
Linearity
FT obeys superposition and homogeneity principles.
According to this property FT of the sum of two signals is
equal to sum of FT of individual signals.F[f1(t)]= F1(j) F[f2(t)]= F2(j)
F[a1f1(t)+b1f2(t)] = a1 F1(j) + b1F2(j)
Where a1 and b1 are constants.
Time Shifting Time Reversal
F[f(t)]= F(j) F[f(t)]= F(j)
F[f(t-t0) = e-jt0 F(j) F[f(-t)]= F(-j)
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Frequency Shifting
According the this property multiplication of the signal ejot with f(t)shifts the frequency spectrum by o
F[f(t)]= F(j)
F[f(t) ejot ]= F[j(- o)]
Time Scaling:
F[f(t)]= F(j) F[f(at)]= 1/|a| F(j /a), when a>1 f(at) is a compressed
version of the signal f(t) by a factor a.When a < 1, f(at) is an expanded version of the signal x(t) by a factor
a
Differentiation in time
Differentiating a signal results in a multiplication of the Fourier
transform by j, F[f(t)]= F(j) then F[d/dt(f(t))]= j F(j)
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Differentiation in Frequency
Multiplication of signal x(t) with time t results in differentiation of
frequency spectrum F(j)
F[f(t)]= F(j) then F[t f(t)]= j d/d F(j)
Time Integration
Integration of a signal results in a division of Fourier transform by j.
However to account for the dc or average value that can result from
integration, we must add the term F(0)()
F[f(t)] = F(j), then
)F(j)(jdt
f(t)d
Fn
n
n
F(0)(F(j(j
1f(()F
t
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Conjugation
F[f(t)]= X(j), F[f*(t)] = F*(-j)
Fourier transform of complex and real functions
f(t) =fR(t)+jfI(t)Fourier transform of f(t) is given by
F[f(t)]=
Auto-correlation
F[Rff()] = F[f(t) * f*(-t)] = ff(j) = |F(j|2
Duality
F[f(t)]= F(j)
F[f(t)]= 2f(-j)
dtef(t))F(jtj
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Convolution
The convolution of two signals results in the multiplication of their Fourier
transforms in the frequency domain.
F[f(t)*h(t)]= F(j) H(j)
The output of a system can be obtained from the convolution of input
signal and the system impulse response
d))h(tf(y(t)
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Fourier transform of Single Gate Function
otherwise
T/2tT/2for
0
1,f(t)
2
TTsinc
2T
2
Tsin
T.
ej
1
dt.e1
dtef(t))F(j
T/2
T/2
tj
tj
T/2
T/2
tj
The amplitude spectrum is shown in Fig. At = 0, sin
c(T/2) = T, therefore F(j) at = 0 is equal to T.
At T/2 = n, sin c(T/2) = 0,
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Fourier Transform of Rectangular Pulse
otherwise
Tt0for
0
1,f(t)
2
T
2
Tsin
eT
2j
ee2e
eej
e
ej
1
dt.e1
dtef(t))F(j
T/2j-
T/2jT/2jT/2j
T/2jT/2jT/2j
T
0
tj
tj
T
0
tj
2
TsincTe
T/2j
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Fourier transform of a rectangular pulse 2 seconds
long with a magnitude of 10 volts
csine20
sine20
2j
eee20
eej
e10
j
1e
10
j
e10
dt0.e1
dtef(t))F(j
j
j
jjj-
jjj
j2-
tj
tj
T/2
T/2
tj
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Exponential Pulse
0tfor
0tfor
0
,ef(t)
-at
atan
a
1
ja
1
dte
0tfor0f(t)since,dtee0
dtef(t)dtef(t)
dtef(t))F(j
1
22
0
t)j(a
0
tjat
0
tj
0
tj
tj
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Fourier transform for the double exponential pulse
0tfor
0tfor
e
,ef(t)
at
-at
22
0
)(
ajF
a2
ja
1
ja
1
dtedte
dteedtee
dtef(t)dtef(t)
dtef(t))F(j
0
t)j(a
-
t)j(a
0
tjat
0
tjat
0
tj
0
tj
tj
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Fourier transform of triangular pulse
f(t) = A (1+ 2t/T), -T/2
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4
Tsi n2
T
4A
2
Tcos1
T
4A
dt
si ni
2Tsi n
2
T
T
4A
2Tsi n
2A
dt2tcostT
2Adtcoso2A
dtetdtetT2AdtedteAF j
obt ai nwei nt egral s,t hirdandfirstt hei n-t oChangi ng
2
2
2
T/ 2
0
T/ 2
0
T/ 2
0
T/ 2
0
j
T/ 2
0
j
T/ 2
0
T/ 2
0
j j
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Impulse Function (Unit Impulse)
The impulse function, which has an infinite amplitude and is infinitely narrow. This isdefined as (t) = 0 for all values except at t = 0
The Fourier transform or the impulse function (t) is obtained as
The frequency spectrum of the impulse function
(t) has a constant amplitude and extends
over positive and negative frequencies.
The inverse Fourier transform of the
unit impulse is given by
1dte(t))F(j tj
)(2
1.
Therefore2
1f(t)
t2
1
d)(2
ede)(
2
1f(t)
tjtj
Hence, the Fourier transform of the constant
function is an impulse at the origin with an area
equal to 2
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Signum Function
The signum function denoted by sgn(t) is defined by
0tif
0tif
0tif
1
0
1
f(t)
j
2
j
1
j
1
j
e
j
e
dte(1)dte1)(
dtesgn(t))F(j
0
tj0
tj
0
0
tjtj
tj
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Unit Step Function
The unit step function is obtained by suddenly closing a switch of a DC
circuit. For easier analysis, the waveform of the unit step function splitinto two component waveforms.
The first waveform is similar to the signum function with half amplitude.
Therefore, the Fourier transform function is given by
F1(j) = (2/j) = 1/j
The second waveform is related to the unit impulse function and hence its
Fourier transform is given by
F2(j) = [2 ()] = ()
Therefore, the Fourier transform of the step function becomes
F(j) = F1(j) + F2(j)
=1/j + ()
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Sinusoidal Functions
The Fourier transforms of the sinusoidal functions cos ot and sin ot areobtained as given below.
F[cos ot] =
Using the transform pair
F[cos ot] = [( o) + (+o)]
Similarly
F[sin ot] = [( + o) + ( - o)]
2
eeF
tjtj oo
getwe),(2e otj o
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FOURIER TRANSFORM OF POWER AND ENERGY
SIGNALS The average power of a signal x(t) over a single period (t1, t1+T)
is given by
Where x(t) is a complex periodic signal.
A signal f(t) is called a power signal, if the average power
expressed by
If x(t) is bounded, P
is finite. Every bounded and periodic signal is
a power signal. But it is true that a power signal is not
necessarily a bounded and periodic signal.
dtx(t)
T
1P
2Tt
t
av
1
i
T
T
2
tx x(t)
2T
1LtP
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Energy Signal
A signal x(t) is called an Energy signal, if its total energy over the interval
(- , ) is finite, that is
For a digital signal, the energy is defined by
As n ->, the energy of period signals becomes infinite, whereas the energy
of aperiodic pulse-signals have a finite value.
dtx(t)LtE
T
T
2
Tx
2
n
x(n)E
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Determine the signal energy and signal power for (a). f(t) = e-3|t|, (b).
f(t) = e-3t
31e62
dte2
dtedte
dteE
06t
0
6t
0
6t
0
6t
2
|t|3
The signal power P
= 0,
since Eis finite. Hence,
the signal f(t) is an energysignal
TT
T
T
T
e6
2
6T
T
6t-
3t-
e6
1
dte
dteE
As T -> , ET approaches infinity.
Its average power is
Hence, e-3t is neither an energy signal nor a power
signal.
T
eeLtE
TLtP
TT
TT
T 82
166
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Compute the signal energy for x(t) = e-4t u(t)
Find the Fourier transform of the signal f(t) shown in Fig.
f(t) = (A/T) t, for 0 < t < T, A , for T < t < 2T
Tj2T/2jT/2jT/2j2
Tj2Tj
2
2T
T
tjT
0
2
tjtj
2T
T
tj
T
0
tj-
eA
jeeeT
A
eA
j1e
T
A
j
eA
)j(
e
j
et
T
A
dteAdtetT
AE
Tj2T/2j
Tj2T/2j
Tj2T/2j
2
e2Tsince
jA
eA
j2
Tsince
Aj
e
A
j2
T
sineT
2A
j
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Obtain the Fourier transform of the trapezoidal pulse is shown in Fig.
f(t) = At/(tp ta) + Atp/(tp ta), for tp < t < -ta
= A, for ta < t < ta
= Atp/(tp ta) - At/(tp ta), for tp < t < -ta
Therefore,
p
a
p
a
a
p
a
p
a
p
t
t ap
tjt
t ap
tj
p
t
t ap
tj
t
t ap
tjp
t
t ap
tj
dt
tt
Atedt
tt
etAdt
tt
Ate
dttt
eAtdt
tt
Ate)F(j
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j
eeA
j
ee
tt
At
ee1
ee1
j
eet
tt
A
jeA
je
je
ttAt
e
j
tee
j
te
tt
A
dteAdtedtett
At
dtetdtettt
A
aaaa
ppaa
aa
a
a
p
a
a
p
p
a
a
p
a
a
a
p
p
a
a
p
p
a
tjtjtjtj
ap
p
tjtj
2
tjtj
2
tjtj
a
ap
t
t
tjt
t
tjt
t
tj
ap
p
t
t
2
tjtjt
t
2
tjtj
ap
t
t
tj
t
t
t
t
tjtj
ap
p
t
t
t
t
tjtj
ap
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pa2
pa2
ap
p
ap
p
a
aa
ap
p
p2a2a
a
ap
tcostcos2
tcostcos2
tt
t
tt
t1tsin
2A
t2sinA
t2sin1
tt
At
t2cos1
t2cos1
tsin2t
tt
A