Tài liệu Electronics and Circuit Analysis Using MATLAB P8 ppt
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Attia, John Okyere. “Fourier Analysis.”
Electronics and Circuit Analysis using MATLAB.
Ed. John Okyere Attia
Boca Raton: CRC Press LLC, 1999
© 1999 by CRC PRESS LLC
CHAPTER EIGHT
FOURIER ANALYSIS
In this chapter, Fourier analysis will be discussed. Topics covered are Fou-
rier series expansion, Fourier transform, discrete Fourier transform, and fast
Fourier transform. Some applications of Fourier analysis, using MATLAB,
will also be discussed.
8.1 FOURIER SERIES
If a function
gt
()
is periodic with period
T
p
, i.e.,
gt gt T
p
() ( )
=±
(8.1)
and in any finite interval
gt
()
has at most a finite number of discontinuities
and a finite number of maxima and minima (Dirichlets conditions), and in
addition,
gtdt
T
p
()
<∞
∫
0
(8.2)
then
gt
()
can be expressed with series of sinusoids. That is,
gt
a
a nwt b nwt
nn
n
() cos( ) sin( )
=+ +
=
∞
∑
0
00
1
2
(8.3)
where
w
T
p
0
2
=
π
(8.4)
and the Fourier coefficients
a
n
and
b
n
are determined by the following equa-
tions.
a
T
g t nw t dt
n
p
t
tT
o
op
=
+
∫
2
0
()cos( )
n
= 0, 1,2, … (8.5)
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b
T
g t nw t dt
n
p
t
tT
o
op
=
+
∫
2
0
( )sin( )
n
= 0, 1, 2 … (8.6)
Equation (8.3) is called the trigonometric Fourier series. The term
a
0
2
in
Equation (8.3) is the dc component of the series and is the average value of
gt
()
over a period. The term
anwtbnwt
nn
cos( ) sin( )
00
+
is called the
n
-
th harmonic. The first harmonic is obtained when
n
= 1. The latter is also
called the fundamental with the fundamental frequency of ω
o
. When n = 2,
we have the second harmonic and so on.
Equation (8.3) can be rewritten as
gt
a
Anwt
nn
n
() cos( )
=+ +
=
∞
∑
0
0
1
2
Θ
(8.7)
where
Aab
nnn
=+
22
(8.8)
and
Θ
n
n
n
b
a
=−
−
tan
1
(8.9)
The total power in
gt
()
is given by the Parseval’s equation:
P
T
g t dt A
A
p
t
tT
dc
n
n
o
op
==+
+
=
∞
∫
∑
1
2
22
2
1
()
(8.10)
where
A
a
dc
2
0
2
2
=
(8.11)
The following example shows the synthesis of a square wave using Fourier
series expansion.
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© 1999 CRC Press LLC
Example 8.1
Using Fourier series expansion, a square wave with a period of 2 ms, peak-to-
peak value of 2 volts and average value of zero volt can be expressed as
gt
n
nft
n
()
()
sin[( ) ]
=
−
−
=
∞
∑
41
21
212
0
1
π
π
(8.12)
where
f
0
500
=
Hz
if
at
()
is given as
at
n
nft
n
()
()
sin[( ) ]
=
−
−
=
∑
41
21
212
0
1
12
π
π
(8.13)
Write a MATLAB program to plot
at
()
from 0 to 4 ms at intervals of 0.05
ms and to show that
at
()
is a good approximation of
g(t
).
Solution
MATLAB Script
% fourier series expansion
f = 500; c = 4/pi; dt = 5.0e-05;
tpts = (4.0e-3/5.0e-5) + 1;
for n = 1: 12
for m = 1: tpts
s1(n,m) = (4/pi)*(1/(2*n - 1))*sin((2*n - 1)*2*pi*f*dt*(m-1));
end
end
for m = 1:tpts
a1 = s1(:,m);
a2(m) = sum(a1);
end
f1 = a2';
t = 0.0:5.0e-5:4.0e-3;
clg
plot(t,f1)
xlabel('Time, s')
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© 1999 CRC Press LLC
ylabel('Amplitude, V')
title('Fourier series expansion')
Figure 8.1 shows the plot of
at
()
.
Figure 8.1 Approximation to Square Wave
By using the Euler’s identity, the cosine and sine functions of Equation (8.3)
can be replaced by exponential equivalents, yielding the expression
g t c jnw t
n
n
( ) exp( )
=
=−∞
∞
∑
0
(8.14)
where
c
T
gt jnwtdt
n
p
t
T
p
p
=−
−
∫
1
2
2
0
( ) exp( )
/
/
(8.15)
and
w
T
p
0
2
=
π
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© 1999 CRC Press LLC
Equation (8.14) is termed the exponential Fourier series expansion. The coeffi-
cient
c
n
is related to the coefficients
a
n
and
b
n
of Equations (8.5) and (8.6)
by the expression
cab
b
a
nnn
n
n
=+∠−
−
1
2
22 1
tan ( )
(8.16)
In addition,
c
n
relates to
A
n
and
φ
n
of Equations (8.8) and (8.9) by the rela-
tion
c
A
n
n
n
=∠Θ
2
(8.17)
The plot of
c
n
versus frequency is termed the discrete amplitude spectrum or
the line spectrum. It provides information on the amplitude spectral compo-
nents of
gt
().
A similar plot of
∠c
n
versus frequency is called the dis-
crete phase spectrum and the latter gives information on the phase components
with respect to the frequency of
gt
()
.
If an input signal
xt
n
()
x t c jnw t
nn o
( ) exp( )
=
(8.18)
passes through a system with transfer function
Hw
()
, then the output of the
system
yt
n
()
is
y t H jnw c jnw t
nono
( ) ( ) exp( )
=
(8.19)
The block diagram of the input/output relation is shown in Figure 8.2.
H(s)x
n
(t) y
n
(t)
Figure 8.2 Input/Output Relationship
However, with an input
xt
()
consisting of a linear combination of complex
excitations,
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x t c jnw t
n
n
no
( ) exp( )
=
=−∞
∞
∑
(8.20)
the response at the output of the system is
y t H jnw c jnw t
n
n
on o
( ) ( ) exp( )
=
=−∞
∞
∑
(8.21)
The following two examples show how to use MATLAB to obtain the coeffi-
cients of Fourier series expansion.
Example 8.2
For the full-wave rectifier waveform shown in Figure 8.3, the period is 0.0333s
and the amplitude is 169.71 Volts.
(a) Write a MATLAB program to obtain the exponential Fourier series
coefficients
c
n
for
n
= 0,1, 2, .. , 19
(b) Find the dc value.
(c) Plot the amplitude and phase spectrum.
Figure 8.3 Full-wave Rectifier Waveform
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Solution
diary ex8_2.dat
% generate the full-wave rectifier waveform
f1 = 60;
inv = 1/f1; inc = 1/(80*f1); tnum = 3*inv;
t = 0:inc:tnum;
g1 = 120*sqrt(2)*sin(2*pi*f1*t);
g = abs(g1);
N = length(g);
%
% obtain the exponential Fourier series coefficients
num = 20;
for i = 1:num
for m = 1:N
cint(m) = exp(-j*2*pi*(i-1)*m/N)*g(m);
end
c(i) = sum(cint)/N;
end
cmag = abs(c);
cphase = angle(c);
%print dc value
disp('dc value of g(t)'); cmag(1)
% plot the magnitude and phase spectrum
f = (0:num-1)*60;
subplot(121), stem(f(1:5),cmag(1:5))
title('Amplitude spectrum')
xlabel('Frequency, Hz')
subplot(122), stem(f(1:5),cphase(1:5))
title('Phase spectrum')
xlabel('Frequency, Hz')
diary
dc value of g(t)
ans =
107.5344
Figure 8.4 shows the magnitude and phase spectra of Figure 8.3.
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Figure 8.4 Magnitude and Phase Spectra of a Full-wave
Rectification Waveform
Example 8.3
The periodic signal shown in Figure 8.5 can be expressed as
gt e t
gt gt
t
()
()()
=−≤<
+=
−
2
11
2
(i) Show that its exponential Fourier series expansion can be expressed as
gt
ee
jn
jn t
n
n
()
()( )
()
exp( )
=
−−
+
−
=−∞
∞
∑
1
22
22
π
π
(8.22)
(ii) Using a MATLAB program, synthesize
gt
()
using 20 terms, i.e.,
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gt
ee
jn
jn t
n
n
()
()( )
()
exp( )
∧
−
=−
=
−−
+
∑
1
22
22
10
10
π
π
024
t(s)
g(t)
1
Figure 8.5 Periodic Exponential Signal
Solution
(i)
g t c jnw t
no
n
( ) exp( )
=
=−∞
∞
∑
where
c
T
gt jnwtdt
n
p
T
T
o
p
p
=−
−
∫
1
2
2
( ) exp( )
/
/
and
w
T
o
p
===
22
2
ππ
π
ctjntdt
n
=−−
−
∫
1
2
2
1
1
exp( ) exp( )
π
c
ee
jn
n
n
=
−−
+
−
()( )
()
1
22
22
π
thus
© 1999 CRC Press LLC
© 1999 CRC Press LLC
