comlpex fourier series
TRANSCRIPT
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Complex Fourier Series
Muhammad Nadeem
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In many cases, the complex Fourier series is easier to obtain ratherthan the trigonometric Fourier series
Recall that, Eulers identity,
xixeix
sincos =
yields
2cos
ixix ee
x
+
=i
ee
x
ixix
2sin
=and
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Then the Fourier series representation becomes
=
=
+=
+
++=
++=
1
0
1
0
22
)sincos()(
xinxinxinxin
n
xinxin
n
xinxin
n
n
nn
eeee
i
eeb
eeaa
xnbxnaaxf
=
=
=
=
++
+=
++
+=
11
0
1
0
1
22
22
22
n
xinnn
n
xinnn
n
xinnnxinnn
n
nn
eiba
eiba
a
eiba
eiba
a
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Here, let we name
=
=
++
+=
11
022
)(n
xinnn
n
xinnn eiba
eiba
axf
2
nnn
ibac
= ,
2
nnn
ibac
+=
Hence,
=
=
++=
11
0
n
xin
n
n
xin
n ececc
and 00 ac =
c0 cncn
=
=
=
=
=
+=
++=
++=
0
0
1
1
0
11
0
nn
xin
n
n
xin
n
n
xin
n
n
xinn
n
xinn
ecc
ececc
ececc
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Then, the coefficient cn can be derived from
=
=
=
TT
TT
nnn
xdxnxfixdxnxf
xdxnxfT
ixdxnxf
T
ibac
00
sin)(cos)(1
sin)(2
2
cos)(2
2
1
2
=
=
Txin
T
dxexfT
dxxnixnxfT
0
0
00
)(1
]sin)[cos(1
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In summary, Complex Fourier series of
function f(x) is:
=
+=
0
)(
nn
xin
neccxf
o
Where complex Fourier coefficients are:
=
Txin
n dxexfT
c0
)(1
T
2=
=T
dxxfT
c0
)(1
o
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Obtain the complex Fourier series of the followingfunction
2e
)(xf
Example
Solution
xifexf x 20)(
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So complex Fourier series of function f(x) is:
=
2
0
)(2
1 dxexfcinx
n
=+=
0
)(nn
inx
neccxf o
Where complex Fourier coefficients are:
=
2
0
)(2
1 dxxfco
[ ]
21
21
2
1)(
1
22
0
2
00
0
==
==
ee
dxexdxfT
c
x
xT
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1
2
1
2
1
)(
1
2)1(
2
0
)1(
2
00
e
dxe
dxeedxexfTc
xin
xin
inxx
T
xin
n
=
=
==
)1(2
1
)1(2
1
)1(2
1222)1(2
0
in
e
in
ee
in
en
niin
=
=
=
1012sin2cos2
===
nineni
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inx
nn
nn
inx
n
ein
ee
eccxf
=
=
+
=
+=
0
22
0
)1(2
1
2
1
)(
o
Therefore, the complex Fourier series off(x) is
0
2
0
2
0 2
1
)1(2
1c
e
in
ec
n
nn=
=
=
=
=
*Notes: Even though c0 can be found by substituting cn with n = 0,sometimes it doesnt works (as shown in the next example). Therefore, itis always better to calculate c0 alone.
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Obtain the complex Fourier series of the followingfunction
Example
SolutionThe function is described by the following graph:
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2
1012
1)(2
1)(11
0
2
1
2
00
0 =
+=== dtdtdttfdttfT
cT
)(21)(1
2
00
== tin
T
tinn dtetfdtetf
Tc
]1)1[(2
)1(22
1
012
1
0
10
==
=
+=
nin
tin
tintin
nie
ni
ine
dtedte
nin
nnine )1(cossincos===
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Thus,
== even,0
odd,/
]1)1[(2 n
nni
n
i
cn
n
*Here notice that 00 cc nn =
+=)(tin
necctf
o
Therefore, the complex
=
=
=
=
+=
odd0
0
0
2
1
]1)1[(22
1
nn
n
tjn
xin
n
n
n
nn
en
i
en
i
our er ser es o t s