07 periodic functions and fourier series
DESCRIPTION
engineering mathsTRANSCRIPT
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Periodic Functions and Fourier Series
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Periodic Functions
A function f is periodic
if it is defined for all real and if there is some positive number,
T such that fTf .
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0
T
f
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0
T
f
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0
T
f
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Fourier Series
f be a periodic function with period 2
The function can be represented by a trigonometric series as:
11
0 sincosn
nn
n nbnaaf
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11
0 sincosn
nn
n nbnaaf
What kind of trigonometric (series) functions are we talking about?
32
and32
sin,sin,sin
cos,cos,cos
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0
0
2
cos cos 2 cos 3
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0
0 2
sin sin 2 sin 3
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We want to determine the coefficients,
na and nb .
Let us first remember some useful integrations.
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dmndmn
dmn
cos2
1cos
2
1
coscos
0
dmn coscos mn
dmn coscos mn
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dmndmn
dmn
sin2
1sin
2
1
cossin
0
dmn cossin
for all values of m.
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dmndmn
dmn
cos2
1cos
2
1
sinsin
0
dmn sinsin mn
dmn sinsin mn
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Determine 0aIntegrate both sides of (1) from
to
dnbnaa
df
nn
nn
110 sincos
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dnb
dnada
df
nn
nn
1
10
sin
cos
000
dadf
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002 0
adf
dfa
2
10
0a is the average (dc) value of the
function, f .
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dnbnaa
df
nn
nn
2
011
0
2
0
sincos
It is alright as long as the integration is performed over one period.
You may integrate both sides of (1) from
0 to 2 instead.
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dnb
dnada
df
nn
nn
2
01
2
01
2
0 0
2
0
sin
cos
002
0 0
2
0
dadf
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002 0
2
0 adf
dfa
2
00 2
1
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Determine naMultiply (1) by mcosand then Integrate both sides from
to
dmnbnaa
dmf
nn
nn
cossincos
cos
110
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Let us do the integration on the right-hand-side one term at a time.
First term,
00 dma cos
Second term,
dmnan
n coscos1
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m
nn admna
coscos1
Second term,
Third term,
01
dmcosnsinb
nn
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Therefore,
madmf cos
,2,1cos1
mdmfam
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Determine nbMultiply (1) by msinand then Integrate both sides from
to
dmsinnsinbncosaa
dmsinf
nn
nn
110
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Let us do the integration on the right-hand-side one term at a time.
First term,
00 dmsina
Second term,
dmsinncosa
nn
1
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01
dmsinncosa
nn
Second term,
Third term,
m
nn bdmnb
sinsin1
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Therefore,
mbdmsinf
,,mdmsinfbm 211
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dfa
2
10
The coefficients are:
,2,1cos1
mdmfam
,2,1sin1
mdmfbm
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dfa
2
10
We can write n in place of m:
,,ndncosfan 211
,,ndnsinfbn 211
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The integrations can be performed from
0 to 2 instead.
dfa
2
00 2
1
,,ndncosfan 211 2
0
,,ndnsinfbn 211 2
0
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Example 1. Find the Fourier series of the following periodic function.
0
f
2 3 4 5
A
-A
2
0
whenA
whenAf
ff 2
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02
12
12
1
2
0
2
0
2
00
dAdA
dfdf
dfa
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011
1
1
2
0
2
0
2
0
n
nsinA
n
nsinA
dncosAdncosA
dncosfan
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ncosncoscosncosn
A
n
ncosA
n
ncosA
dnsinAdnsinA
dnsinfbn
20
11
1
1
2
0
2
0
2
0
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oddisnwhen4
1111
20
n
An
A
ncosncoscosncosn
Abn
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evenisnwhen0
1111
20
n
A
ncosncoscosncosn
Abn
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Therefore, the corresponding Fourier series is
7sin
7
15sin
5
13sin
3
1sin
4A
In writing the Fourier series we may not be able to consider infinite number of terms for practical reasons. The question therefore, is – how many terms to consider?
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When we consider 4 terms as shown in the previous slide, the function looks like the following.
1.5
1
0.5
0
0.5
1
1.5
f ( )
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When we consider 6 terms, the function looks like the following.
1.5
1
0.5
0
0.5
1
1.5
f ( )
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When we consider 8 terms, the function looks like the following.
1.5
1
0.5
0
0.5
1
1.5
f ( )
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When we consider 12 terms, the function looks like the following.
1.5
1
0.5
0
0.5
1
1.5
f ( )
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The red curve was drawn with 12 terms and the blue curve was drawn with 4 terms.
1.5
1
0.5
0
0.5
1
1.5
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The red curve was drawn with 12 terms and the blue curve was drawn with 4 terms.
0 2 4 6 8 101.5
1
0.5
0
0.5
1
1.5
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The red curve was drawn with 20 terms and the blue curve was drawn with 4 terms.
0 2 4 6 8 101.5
1
0.5
0
0.5
1
1.5
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Even and Odd Functions
(We are not talking about even or odd numbers.)
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Even Functions
f(
The value of the function would be the same when we walk equal distances along the X-axis in opposite directions.
ff Mathematically speaking -
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Odd Functions The value of the function would change its sign but with the same magnitude when we walk equal distances along the X-axis in opposite directions.
ff Mathematically speaking -
f(
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Even functions can solely be represented by cosine waves because, cosine waves are even functions. A sum of even functions is another even function.
10 0 10
5
0
5
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Odd functions can solely be represented by sine waves because, sine waves are odd functions. A sum of odd functions is another odd function.
10 0 10
5
0
5
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The Fourier series of an even function fis expressed in terms of a cosine series.
1
0 cosn
n naaf
The Fourier series of an odd function fis expressed in terms of a sine series.
1
sinn
n nbf
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Example 2. Find the Fourier series of the following periodic function.
ff 2
x
f(x)
3 5 7 90
xwhenxxf 2
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332
1
2
1
2
1
23
20
x
x
x
dxxdxxfa
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nxdxx
dxnxxfan
cos1
cos1
2
Use integration by parts. Details are shown in your class note.
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nn
an cos4
2
odd is n when4
2nan
even is n when4
2nan
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This is an even function.
Therefore, 0nbThe corresponding Fourier series is
222
2
4
4cos
3
3cos
2
2coscos4
3
xxxx
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Functions Having Arbitrary Period
Assume that a function tf has period, T . We can relate angle ( ) with time ( t ) in the following manner.
t
is the angular velocity in radians per second.
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f 2f is the frequency of the periodic function,
tf
tf 2 Tf
1where
tT
2Therefore,
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tT
2 dt
Td
2
Now change the limits of integration.
2
Tt t
T
2
2
Tt t
T
2
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dfa
2
10
dttfT
a
T
T
2
2
0
1
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,,ndncosfan 211
,2,12
cos2 2
2
ndttT
ntf
Ta
T
Tn
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,,ndnsinfbn 211
,2,12
sin2 2
2
ndttT
ntf
Tb
T
Tn
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Example 4. Find the Fourier series of the following periodic function.
0 t
T/2
f(t)
T/4
3T/4
T-T/2 2T
4
3
42
44T
tT
whenT
t
Tt
Twhenttf
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tfTtf
This is an odd function. Therefore, 0na
dttT
ntf
T
dttT
ntf
Tb
T
T
n
2sin
4
2sin
2
2
0
0
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dttT
nTt
T
dttT
nt
Tb
T
T
T
n
2sin
2
4
2sin
4
2
4
4
0
Use integration by parts.
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2sin
2
2sin
2.2
4
22
2
n
n
T
n
n
T
Tbn
0nb when n is even.
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Therefore, the Fourier series is
t
Tt
Tt
T
T 10
5
16
3
122222sinsinsin
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The Complex Form of Fourier Series
11
0 sincosn
nn
n nbnaaf
Let us utilize the Euler formulae.
2
jj ee
cos
i
eesin
jj
2
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The nth harmonic component of (1) can beexpressed as:
i
eeb
eea
nbnajnjn
n
jnjn
n
nn
22
sincos
22
jnjn
n
jnjn
n
eeib
eea
![Page 69: 07 periodic functions and fourier series](https://reader033.vdocument.in/reader033/viewer/2022061206/5482f1f75806b5d9048b4731/html5/thumbnails/69.jpg)
jnnnjnnn
nn
ejba
ejba
nbna
22
sincos
Denoting
2nn
n
jbac
2
nnn
jbac,
and 00 ac
![Page 70: 07 periodic functions and fourier series](https://reader033.vdocument.in/reader033/viewer/2022061206/5482f1f75806b5d9048b4731/html5/thumbnails/70.jpg)
jn
njn
n
nn
ecec
nsinbncosa
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The Fourier series for f can be expressed as:
n
jnn
n
jnn
jnn
ec
ececcf
1
0
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The coefficients can be evaluated in the following manner.
dnf
jdnf
jbac nnn
sin2
cos2
1
2
dnjnf sincos
2
1
def jn
2
1
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dnf
jdnf
jbac nnn
sin2
cos2
1
2
dnjnf sincos
2
1
def jn
2
1
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.
2
nnn
jbac
2nn
n
jbac
nc nc
Note that is the complex conjugate of
. Hence we may write that
defc jnn 2
1
,,,n 210
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The complex form of the Fourier series of
f 2 with period is:
n
jnnecf
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Example 1. Find the Fourier series of the following periodic function.
0
f
2 3 4 5
A
-A
2
0
whenA
whenAf
ff 2
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A 5
f x( ) A 0 x if
A x 2 if
0 otherwise
A01
2 0
2
xf x( )
d
A0 0
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n 1 8
An1
0
2
xf x( ) cos n x( )
d
A1 0 A2 0 A3 0 A4 0
A5 0 A6 0 A7 0 A8 0
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Bn1
0
2
xf x( ) sin n x( )
d
B1 6.366 B2 0 B3 2.122 B4 0
B5 1.273 B6 0 B7 0.909 B8 0
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Complex Form
n
jnnecf
defc jnn 2
1
,,, 210 n
C n( )1
2 0
2
xf x( ) e1i n x
d
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C 0( ) 0 C 1( ) 3.183i C 2( ) 0 C 3( ) 1.061i
C 4( ) 0 C 5( ) 0.637i C 6( ) 0 C 7( ) 0.455i
C 1( ) 3.183i C 2( ) 0 C 3( ) 1.061i
C 4( ) 0 C 5( ) 0.637i C 6( ) 0 C 7( ) 0.455i
C n( )1
2 0
2
xf x( ) e1i n x
d