F1.3YF2 Mathematical Techniques 1

EXAMPLES 1: FOURIER SERIES

1. Find the Fourier series of each of the following functions

(i) f (x) = 1−x2, −1< x< 1.

(ii) g(x) = |x|, −π< x< π.

(iii) h(x) ={

0 if −2< x< 01 if 0≤ x< 2.

In each case sketch the graph of the function to which the Fourier series convergesover anx- range of three periods of the Fourier series.

2. Find the Fourier series forf (x) = x2

4 , −π< x< π. Hence deduce that

(i) π2

6 = 1+ 122 + 1

32 + 142 + 1

52 + . . . .

(ii) π2

12 = 1− 122 + 1

32 − 142 + 1

52 − . . ..

(iii) π2

8 = 1+ 132 + 1

52 + 172 + 1

92 + . . . .

3. Find the Fourier cosine series and the Fourier sine series for the function

f (x) ={

1 if 0< x< 10 if 1≤ x< 2.

4. Find the Fourier cosine series for the functionf (x) = sin(x), 0< x< π.What is the Fourier sine series forf ?

5. Supposef : R→ R is a periodic function of period 2L with Fourier series

a0 +∞

∑n=1

ancos(nπxL

)+bnsin(nπxL

).

(i) Using integration by parts, show that, iff ′ exists and is a bounded function onR, then there exists a constantk such that|an| ≤ k

n and|bn| ≤ kn for all n≥ 1.

(ii) Show that, if f ′′ exists and is a bounded function onR, then the Fourier seriesfor f is absolutely convergent for allx.

F1.3YF2 Fourier Series – Solutions 1

EXAMPLES 1: FOURIER SERIES – SOLUTIONS1. (i) We must calculate the Fourier coefficients.a0 =

12

R 1�1(1�x2)dx= 1

2(x�13x3j1�1 =

23.

an =11

R 1�1(1�x2)cos(nπx)dx=

R 1�1cos(nπx)dx�

R 1�1 x2cos(nπx)dx

= 0� 1nπ x2sin(nπx)j1

�1+2nπ

R 1�1xsin(nπx)dx

= 0�0� 2n2π2 xcos(nπx)j1�1+

2n2π2

R 1�1cos(nπx)dx

=�4

n2π2 cos(nπ)+0= (�1)n+1 4n2π2 :

Also bn =11

R 1�1(1�x2)sin(nπx)dx= 0 as integrand is odd.

Hence

f (x) =23+

4π2fcos(πx)�

14

cos(2πx)+19

cos(3πx)� : : :g:

(ii) a0 =12π

R π�π jxjdx= 2

2πR π

0 xdx= 1π

12x2jπ0 = π

2.

an =1πR π�π jxjcos(nx)dx= 2

πR π

0 xcos(nx)dx

= 2nπxsin(nx)jπ0�

2nπ

R π0 sin(nx)dx= 0+ 2

n2π cos(nx)jπ0

= 2n2π [(�1)n�1] =

��

4n2π if n is odd

0 if n is even.

Also bn =1πR π�π jxjsin(nx)dx= 0 as integrand is odd.

Hence

g(x) =π2�

4πfcos(x)+

19

cos(3x)+125

cos(5x)+ : : :g:

(iii ) a0 =14

R 2�2 h(x)dx= 1

4

R 20 dx= 1

2.

an =12

R 2�2 h(x)cos(nπx

2 )dx= 12

R 20 cos(nπx

2 )dx= 12

2nπ sin(nπx

2 )j20 = 0.

bn =12

R 2�2 h(x)sin(nπx

2 )dx= 12

R 20 sin(nπx

2 )dx=�12

2nπ cos(nπx

2 )j20

= 1nπ [1�cos(nπ)] =

�0 if n is even2nπ if n is odd.

Hence

h(x) =12+

2πfsin(

πx2)+

13

sin(3πx2

)+15

sin(5πx2

)+ : : :g:

In (i) and(ii), if functions are extended as periodic functions, they are continuous at all points andso the Fourier series converge to the function value at all points.

Thus the Fourier series forf converges to the following function

1− 1

F1.3YF2 Fourier Series – Solutions 2

and the Fourier series forg converges to

− π π

In (iii ), if function is extended as a periodic function, it is discontinuous atx= 0;�2;�4; thus theFourier series converges to12 at these points and converges to the value of the function at all otherpoints.

42 6

x x x x

2. Again calculating the Fourier coefficients we havea0 =

12π

R π�π

x2

4 dx= 12π

14

13x3jπ�π = π2

12.

an =1πR π�π

x2

4 cos(nx)dx= 1π

1n

x2

4 sin(nx)jπ�π�1

2πn

R π�π xsin(nx)dx

= 0+ 12πn2 xcos(nx)jπ�π�

12πn2

R π�π cos(nx)dx= 1

n2 cos(nπ).

Also bn =1πR π�π

x2

4 sin(nx)dx= 0 as the integrand is odd.

Since f is continuous when extended as a periodic function with period 2π, the Fourier series forfconverges tof (x) for all x2 R.

− π π

Hencex2

4=

π2

12�cos(x)+

122 cos(2x)�

132 cos(3x)+

142 cos(4x)� : : :

Settingx= π, we haveπ2

4=

π2

12+1+

122 +

132 +

142 + : : : :

F1.3YF2 Fourier Series – Solutions 3

i.e.,

1+122 +

132 +

142 + : : :=

π2

4�

π2

12=

π2

6(1)

Settingx= 0, we have

0=π2

12�1+

122 �

132 +

142 �

152 + : : :

i.e.,

1�122 +

132 �

142 +

152 � : : :=

π2

12(2)

(1)+(2) gives

2(1+132 +

152 + : : :) =

π2

6+

π2

12=

π2

4:

Hence

1+132 +

152 + : : :=

π2

8:

3. The Fourier cosine series off is given bya0+∑∞n=1ancos(nπx

2 ) where

a0 =14

Z 2

�2f (x)dx=

12

Z 2

0f (x)dx=

12

Z 1

0dx=

12

andan = 1

2

R 2�2 f (x)cos(nπx

2 )dx=R 1

0 cos(nπx2 )dx= 2

nπ sin(nπx2 )j10

= 2nπ sin(nπ

2 ) =

8<:

0 if n is even2nπ if n= 4k+1�

2nπ if n= 4k+3:

Hence

f (x) =12+

2πfcos(

πx2)�

13

cos(3πx2

)+15

cos(5πx2

)� : : :g:

The Fourier sine series off is given by∑∞n=1bnsin(nπx

2 ) where

bn = 12

R 2�2 f (x)sin(nπx

2 )dx=R 1

0 sin(nπx2 )dx=�

2nπ cos(nπx

2 )j10

= 2nπ(1�cos(nπ

2 )) =

8<:

2=nπ if n is odd0 if n= 4k4=nπ if n= 4k+2:

Hence

f (x) =2πfsin(

πx2)+

22

sin(2πx2

)+13

sin(3πx2

)+sin(5πx2

)+26

sin(6πx2

)+ : : :g:

F1.3YF2 Fourier Series – Solutions 4

4. The Fourier cosine series off is given bya0+∑∞n=1ancos(nx) where

a0 =12π

Z π

�πf (x)dx=

1π

Z π

0sin(x)dx= (

1π) :(�cos(x))jπ0 =

2π

and

an =1π

Z π

�πf (x)cos(nx)dx=

2π

Z π

0sin(x)cos(nx)dx:

If n= 1, a1 =2πR π

0 sin(x)cos(x)dx= 1πR π

0 sin(2x)dx= 0.

In n 6= 1, an =1πR π

0 fsin[(n+1)x]+sin[(1�n)x]gdx

= 1πf�

1n+1 cos[(n+1)x]+ 1

n�1 cos[(n�1)x]gjπ0= 1

πf1

n+1(1�cos[(n+1)π])+ 1n�1(cos[(n�1)π]�1)g:

Thusan = 0 if n is odd andan =2π(

1n+1�

1n�1) =�

4π(n2�1) if n is even.

Hence

sin(x) =2π�

4πf

13

cos(2x)+115

cos(4x)+135

cos(6x)+ : : :g:

The Fourier sine series is simplyf (x) = sin(x)

5(i) Since f 0 is a bounded function onR, there exists a constantc> 0 such thatj f 0(x)j � c for allx2 R. Now

an =1L

Z L

�Lf (x)cos(

nπxL

)dx=1nπ

f (x)sin(nπxL

)jL�L�1nπ

Z L

�Lf 0(x)sin(

nπxL

)dx:

As x! f (x)sin(nπxL ) has period 2L (or alternatively as sin(�nπ) = 0), f (x)sin(nπx

L jL�L = 0.

Hencean =�1nπ

R L�L f 0(x)sin(nπx

L )dx and so

janj �2Lnπ

max�L�x�Lj f0(x)sin(

nπxL

)j �2Lcnπ

=kn:

A similar argument shows thatjbnj �kn.

(ii) Since f 00 is a bounded function onR, there exists a constantc1 such thatj f 00(x)j � c1 for allx2 R. By abovean =�

1nπ

R L�L f 0(x)sin(nπx

L )dx

= Ln2π2 f 0(x)cos(nπx

L )jL�L�L

n2π2

R L�L f 00(x)cos(nπx

L )dx

=�L

n2π2

R L�L f 00(x)cos(nπx

L )dx (asx! f 0(x)cos(nπxL ) has period 2L).

Hence

janj �2L2

n2π2max�L�x�Lj f00(x)cos(

nπxL

)j �2L2c1

n2π2 =k1

n2 :

A similar argument shows thatjbnj �k1n2 .

Thus

jancos(nπxL

)+bnsin(nπxL

)j � janj+ jbnj �2k1

n2 :

Since∑∞n=1

1n2 converges, it follows from the comparison test that∑∞

n=0 jancos(nπxL )+bnsin(nπx

L )jconverges.

Hence∑∞n=0(ancos(nπx

L )+bnsin(nπxL )) converges absolutely.

Fourier series for output voltages of inverter waveforms.

The Fourier series for a periodic function vo(ωt) can be expressed as

∑∞

=

++=1

)sin()cos()(n

nnoo tnbtnaatv ωωω

For an odd quarter-wave symmetry waveform,

00 == no aa and

= ∫n

ntdtnvb on

evenfor0

oddfor)()sin(4 2

0

π

ωωπ

Therefore, vo(ωt) can be written as

∑∞

=

=oddn

no tnbtv )sin()( ωω

(i) Square-wave π

π/2 2π π

+Vdc

-Vdc

(ii) Quasi square-wave

π/2 α 2π π

+Vdc

-Vdc

[ ]

π

ωπ

ωωπ

ωωπ

π

π

nV

tnconnV

tdtnV

tdtnvb

dc

dc

dc

on

4

(4

)()sin(4

)()sin(4

20

2

0

2

0

=

−=

=

=

∫

∫

π

[ ]

)cos(4

(4

)()sin(4

)()sin(4

2

2

2

απ

ωπ

ωωπ

ωωπ

π

α

π

α

α

nnV

tnconnV

tdtnV

tdtnvb

dc

dc

dc

on

=

−=

=

=

∫

∫

(iii) Notched waveform (Harmonics Elimination PWM)

α2 α1 π π/2 α3

Vdc Vdc

π (iv) Si

[ ] [ ]

[ ])cos()cos()cos(4

)2

cos()cos()cos()cos(4

)cos(4

((4

)()sin()()sin(4

)()sin(4

231

231

,

2,

2

2

2

0

31

2

3

2

1

3

2

1

αααπ

παααπ

ωπ

ωωπ

ωωωωπ

ωωπ

ααπ

α

π

ααα

π

α

α

α

nnnnV

nnnnnV

tnnV

tncontnconnV

tdtnVtdtnV

tdtnvb

dc

dc

dc

dc

dcdc

on

−+=

−−+=

=

−+−=

+=

=

∫∫

∫

nusoidal PWM (unipolar and bipolar)

Unipolar SPWM waveform as an example

[ ]

termsharmonicforFunctionBessel)sin(

)cos()2(2sin4)sin()(

1

+=

++= ∑

∞

=

tVMn

tknVtVMtv

dca

n

dcdcao

ω

ωπ

πωω

The tables are required to resolve for the Bessel function for harmonic terms. The harmonics in the inverter output appear as sidebands, centered around the switching frequency, that is, around mf, 2mf, 3mf and so on. This general pattern hold true for all values of ma in the range 0 – 1 and mf>9. The unipolar SPWM switching scheme has the advantage of “effectively” doubling the switching frequency as far as the output harmonics are concerned, compared to the bipolar SPWM switching scheme. Because of that the harmonics in the inverter output of unipolar SPWM are centered around 2mf, 4mf, 6mf and so on.

TABLE 8.3 NORMALIZED FOURIER COEFFICIENTS Vn/Vdc FOR BIPOLAR SPWM Ma = 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 n = 1 1.00 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 n = mf 0.60 0.71 0.82 0.92 1.01 1.08 1.15 1.20 1.24 1.27 n = mf±1 0.32 0.27 0.22 0.17 0.13 0.09 0.06 0.03 0.02 0.00

Table 8.3 shows the first harmonic frequencies in the output spectrum at and around mf for the bipolar SPWM switching scheme. The harmonics at and around 2mf, 3mf, 4mf and so on are not indicated. TABLE 8.5 NORMALIZED FOURIER COEFFICIENTS Vn/Vdc FOR UNIPOLAR SPWM Ma = 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 n = 1 1.00 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 n = 2mf ±1 0.18 0.25 0.31 035 0.37 0.36 0.33 0.27 0.19 0.10 n = 2mf ±3 0.21 0.18 0.14 0.10 0.07 0.04 0.02 0.01 0.00 0.00

Table 8.5 shows the first harmonic frequencies in the output spectrum at and around 2mf for the uipolar SPWM switching scheme. The harmonics at and around 4mf, 6mf, 8mf and so on are not indicated. Table 8.3 and 8.5 can be used to predict the THD for the ouput current of the inverter connected to RL load. Higher order harmonics are assumed to contribute little power and effect, so they can be neglected. To evaluate the THD for the output voltage of the inverter, higher order harmonics should be taken into account.

-tammat-