Chapter 15 - Fourier Series and Fourier Transform

8/9/2019 Chapter 15 - Fourier Series and Fourier Transform

1/41

Problems

Section 15.2: The Fourier Series

P15.2-1

2

022 s = rad/s and ( ) for 0 22

T f t t t = = = . The coefficients of the Fourier

series are given by:

( )

2 2

0 0

2 2

20

2 2

0

1 432

2 4cos

2

2 4sin

2

n

n

a t dt

a t n t dt n

b t n t dt n

= =

= =

= =

( ) 2 21 1

4 4 1 4 1cos sin

3 n n f t n t

n nn t

= = = +

P15.2-2

4 24 2

04

04

2 2 2 1 2 2cos 2 cos sin 2sin

1(sin 0) 2 (sin ) sin

2 2( 1)

1sin

2

T TT T

n TT

n

a n t dt n t dt n t nT T T n T T

n nn

n

n

n

+

t

= + = +

= +

= =

( )1 2

odd

0 even

nn

n

( )4 2

4 2

04

04

2 2 1 22sin 2 sin cos 2 cos

1(2cos ( ) 1) cos

23

is odd

22,6,10,

0 4,8,12,

T T

T T

TnT

b n t dt n t dt n t nTT T n T

nn

n

nn

nn

n

2t

T

= + = +

=

= =

=

1


2/41

P15.2-3

( )0 average value of2

Aa f t= =

( ) 1 for 0t

f t A t T T

=

( ) ( ) ( )

0 0 0

2

0

2 2

2 2 2 2 1 21 cos cos cos

2 2 2cos sin

2 10

2

cos 2 cos 0 2 sin 2 0 02

T T Tn

T

t Aa A n t dt n t dt t n t dt T T T T T T T

n t n t n t A T T T

T Tn

T

An n n

n

= =

+

=

= +

=

( ) ( )( ) ( )( )

0 0 0

2

0

2 2

2 2 2 2 1 21 sin sin sin

2 2 2sin cos

2 10

2

sin 2 sin 0 2 cos 2 02

T T Tn

T

t Ab A n t dt n t dt t n t dt T T T T T T T

n t n t n t A T T T

T Tn

T

A An n n

n n

= =

=

=

=

( )1

2sin

2 n

A A f t n

n Tt

=

= +

2


3/41

P15.2-4

0

22 s, rad/s

2T

= = = , ( )0 average value of 1a f t= = ,

( ) for 0 2 f t t t =

( )( ) ( ) ( )

( )

( ) ( ) ( )

2

2

20

0

2 2

cos sin2cos

2

1cos 2 cos 0 2 sin 2 0

0

n

n t n t n t a t n t dt

n

n n nn

+= =

= +

=

( )( ) ( ) ( )

( )( ) ( )( ) ( )( )

2

2

20

0

2 2

sin cos2sin

2

1sin 2 sin 0 2 cos 2 0

2

n

n t n t n t b t n t dt

n

n n nn

n

= =

=

=

( )1

2 21 sin

n

f t nn T

t

=

=

Use Matlab to check this answer:

% P15.2-4

pi=3.14159;

A=2; % input waveform parameters

T=2; % period

w0=2*pi/T; % fundamental frequency, rad/s

tf=2*T; % final time

dt=tf/200; % time increment

t=0:dt:tf; % time, s

a0=A/2; % avarage value of input

v1=0*t+a0; % initialize input as vector

for n=1:1:51 % for each term in the Fourier series ...an=0; % specify coefficients of the input

series

bn=-A/pi/n;

cn=sqrt(an*an + bn*bn); % convert to magnitude and angle form

thetan=-atan2(bn,an);

v1=v1+cn*cos(n*w0*t+thetan); % add the next term of the input

Fourier series

end

3


4/41

plot(t, v1,'black') % plot the Fourier series

grid

xlabel('t, s')

ylabel('f(t)')

title('P15.3-4')

4


5/41

Section 15-3: Symmetry of the Function f(t)

P15.3-1

o

24 s rad/s

4 2

T

= = = .

The coefficients of the Fourier series are:

( )0 average value of 0da v= =t

0 because vd(t) is an odd function oft.na =

( )

( )( ) ( )( )

4

0

4 4

0 0

4 4

2 2

00

2 2

16 3 sin

2 2

33 sin sin

2 2 2

cos3 12

3 sin cos2 2 2 2

2 4

6 61 cos 2 sin 2 0 2 co

nb t n t dt

n t dt t n t dt

n t

n t n t n t n

n

n n nn n

=

=

=

=

( )( )( )12

s 2nn

=

The Fourier series is:

( )d1

12sin

2nv t n t

n

=

=

P15.3-2

( ) ( ) ( )1 1

12 121 6 6 sin 1 6 sin

2 2c d

n n

v t v t n t n t nn n 2

= =

= = + = +

1


6/41

P15.3-3

o

2 10006 ms 0.006 s rad/s krad/s

.006 3 3T

= = = = =

The coefficients of the Fourier series are:

( )0

3 2

12average value of V6 2

aa v t

= = =

because va(t) is an even function oft.0nb =

( )

( )

0.001

0

0.001 0.0016

0 0

2 6 2

2 10002 3 3000 cos

0.006 3

1000 10002000 cos 2 10 cos

3 3

1000sin

1000 1000 1000 100032000 cos sin1000 10 3 3 3

3 9

na t n t dt

n t dt t n t dt

n t

n t n t n t n

n

=

=

= +

0.001

0

2 3 2

2 2

3 92000 sin 0 cos 1 sin 0

1000 3 10 3 3 3

6 18 6sin cos 1 sin

3 3 3

18

n n n nn n

n n nn n n

= +

=

=

2 2cos 1

3

n

n

The Fourier series is

( )a 2 21

1 18 10001 cos cos

2 3n

nv t n t

n 3

=

= +

P15.3-4

( ) ( ) ( )b a 2 21

2 21

1 18 10000.002 1 1 1 cos cos 0.002

2 3 3

1 18 1000 21 cos cos

2 3 3 3

n

n

nv t v t n t

n

nn t n

n

=

=

= = + +

= +

2


7/41

P15.3-5

( ) f t t t = < <

( )

0

0

00

22 , 1

2

average value: 0

2sin

0 since have odd function

T

n

n

T

a

b f t n t dt T

a

= = =

=

=

=

( )2

1 2 3

2 12 1 sin cossin

2

2, 1, and 2 3

n

nnt t nt

b t nt dt n n n

b b b

= = =

= = =

P15.3-6

08 s, 4 rad/sT = =

( )0 because is an even functionnb f t=

( )0

2 2 2 1average 1 4

8a

= = =

( ) ( )2 00

1 2

0 1

4cos

42 cos cos

4 48

23 sin sin

4 2

T

na f t n t dt T

n t dt n t dt

n n

n

=

=

=

1 2 3.714, .955, .662a a a= = =

3


8/41

P15.3-7

0

2

0

2

2 ,

2cos

T

Aa A t dt

= =

= =

( ) ( ) ( )( ) ( )( )

( )( )

( )( )

( ) ( )

( )( ) ( ) ( ) ( )

2 2

2 2

2

2

2

2 2cos cos 2 cos 2 1 cos 2 1

2

sin 2 1 sin 2 12

2 2 1 2 2 1

sin 2 1 sin 2 12 2 2

2 1 2 1

22 1 sin 2 1 2 1 sin 2 1

2 24 1

n

Aa A t n t dt n t n n t dt

n t n t A

n n

n nA

n n

An n n n

n

= = +

+= +

+

+

= + +

= +

( )( )

( )

( )

2

2

4cos

4 1

4 1

4 1

n

An

n

A

n

=

=

0nb = since ( )f t is an even function.

P15.3-8

0

20.4 s, 5 rad sT

T

= = =

0

0

cos 0 .1

( ) 0 .1 .3

cos .3 .4

A t t

f t t

A t t

=


9/41

[ ]

.1 2

1 0.1

.1

0 0.1

.1

.1

2

5 cos2

5 cos cos

15 cos 5 (1 ) cos 5 (1 )

22 cos ( / 2)

11

n

Aa A t dt

a A t n t dt

A n t

A nn

n

= =

=

= + +

=

n t dt

0 because the function is even.nb =

P15.3-9

0 0 because the average value is zero

0 because the function is odd

10 for even due to wave symmetry

4

n

n

a

a

b

=

=

=

Next:

( )2 24

0 2 24

2 2

88 sin 4 cos 1,5,9, ...

2 2 sin

83,7,11, ...

T

n T

n nn for n

nb t n t dt

nfor n

n

= = = = =

5


10/41

Section 15.5: Exponential Form of the Fourier Series

P15.5-1

21

1

oT 2

= = = , the coefficients of the complex Fourier series are given by:

( )

( ) ( )( )( )

( )

( )

( )

1 12 2

0 0

1 2 1 2 1

0

12 1 2 1

2

0

1sin

1 2

2

2

2 2 1 2 1 (4 1)

j t j t j n t j n t

n

j n t j n t

j n t j n t

e e A t e dt A e dt

j

Ae e dt

j

A e e A

j j n j n n

+

+

= =

=

= =

+

C

where we have used and2 1j ne = j je e = .

P15.5-22 2

20 0

1 j nt j nt T TT T

n

A At e dt t e dt

T T T

= =

C

Recall the formula for integrating by parts:2 22

11 1

t tt

tt tu dv u v v du= . Take u andt=

2 j nt

Tdv e dt

= . When , we get0n

2 22 2

2 20

00

2 2

2

1

2 2 2 2

1

2 22

TT

j nt j nt j nT T j nt T

Tn

j n j n

A t e A T e ee dt

T j nT j n j n nT T T

A T e e Aj

T j nn

T

n

= + = +

= + =

C

Now for n = 0 we have

0 0

1

2

T A AC t dt

T T= =

Finally,

465


11/41

( )2

0

1

2 2

n j n t T

nn

A A f t j e

n

=

=

= +

P15.5-3/ 2

22

/ 2

/ 2

/ 2

2 2

2

sin

sin

d j n t j n d j n d

T T j n t dT

nd

d

2

T

j n d j n d T T

A A e A e ee dt

T T T j n j n j n

T T

A e e

n j

A n d n T

n d

Ad T

n dT

T

T

= = =

=

=

=

C

P15.5-4

( )( )

0

0

1o

t T j n t

n dta f t t b e dt

T

+ = +

C

Let dt t = , then dt t= + .

( )( ) ( )

( )( )

( )( )

( ) ( ) ( )

0

0

0

0

0

0

0 0

0 0

1

1

1 1

do d

d

do o d

d

o dd

o

d

d do d o o d o

d d

t T t j n t n

t t

t T t j n j n t

t t

j n t t T t j n

t t

t T t t T t j n t j n j n t j n

t t t t

a f b e d T

a f b e e d T

ea f b e d

T

a e f e d e b e d T T

+ +

+

+

+ +

= +

= +

= +

= +

C

But

0

0

0

0

0 0

0

do

do

d

d

t T tj n

t T t j n

t to t t

neb e d b

bj n

+ +

= =

= so

466


12/41

0 0C a C b= +

and

0o d j n t n na e n= C C

P15.5-5

( )0 0 2 2 2 1 12 18 s, rad/s, average value4 8

T CT

= = = = =

4=

The coefficients of the exponential Fourier series are calculated as

1 1 24 4 4

n2 1 1

1 1 2

4 4 4

2 1 1

4 2 4 4 2 4

12

8

12

8

4 4 4

22

32

n n n j t j t j t

n n n j t j t j t

n n n n n n j j j j j j

j

e dt e dt e dt

e e e

n n n

j j j

je e e e e e

n

je

n

= + +

= + +

= +

=

C

4 4 2 2

13 2 sin 2 sin 3sin sin

2 4 2 4

n n n n j j j

e e e

j n n nj j

n n

2

n

= =

and

( )

1 1 24 4

n2 1 1

1 1 2

4 4 4

2 1 1

4 2 4 4 2 4

11 2 1

8

11 2 1

8

4 4 4

2

2

1

n n j t j t j t

n n n j t j t j t

n n n n n n j j j j j j

e dt e dt e d

e e e

n n n j j j

je e e e e e

n

= + +

= + +

= +

=

C 4n

t

nsin 3sin

2 4

n n

n

= C

The function is represented as

467


13/41

( ) 4 40 n n1 1

4 4

1 1

4

1

1 1 13sin sin sin 3sin

4 4 2 2 4

1 23sin sin

4 4 2

j n t j n t

n n

j n t j n t

n n

j n t

n

f t C e e

n n n ne e

n n

n ne

n

= =

= =

=

= + +

= + +

= +

C C

This result can be checked using MATLAB:

pi = 3.14159;

N=100;

T = 8; % period

t = linspace(0,2*T,200); % time

c0 = 1/4; % average value

w0 = 2*pi/T; % fundamental frequency

for n = 1: N

C(n) = -j*((exp(+j*n*pi/4)-exp(+j*n*pi/2))-2*(exp(-j*n*pi/4)-

exp(+j*n*pi/4))+(exp(-j*n*pi/2)-exp(-j*n*pi/4)))/(2*pi*n);

end

for i=1:length(t)

f(i)=c0;

for n=1:length(C)

f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+C(n)*exp(-j*n*w0*t(i));

end

end

plot(t,f,'black');

xlabel('t, sec');

ylabel('f(t)');

468


14/41

469


15/41

Alternately, this result can be checked using Mathcad:

N 15:= n 1 2, N..:= m 1 2, N..:=

T 8:= 2

T:=

d T200

:= i 1 2, 400..:= ti d i:=

Cn

2

1

t1 exp j n t( )

d

1

1

t2 exp j n t( )

d+1

2

t1 exp j n t( )

d+

T:=

Cm

2

1

t1 exp j m t( )

d

1

1

t2 exp j m t( )

d+1

2

t1 exp j m t( )

d+

T:=

f i( )

1

N

n

Cn

exp j n ti

( )= 1

N

m

Cm

exp 1 j m ti

( )=

+:=

f i( )

.643

.685

.745

.807

.856

1.88

.872

.831

.767

.693

.628

.589

.589

.633

.717

.825

=C

n

0.357

0.477

0.331

0

0.199

0.159

0.051

0

0.04

0.095

0.09

0

0.076

0.068

0.024

= Cm

0.357

0.477

0.331

0

0.199

0.159

0.051

0

0.04

0.095

0.09

0

0.076

0.068

0.024

=

100 200 300 4002

0

2

f i( )

i

470


16/41

P15.5-6

The function shown at right is related to the

given function by

( ) ( )1 1 6v t v t = +

(Multiply by 1 to flip v1 upside-down; subtract

6 to fix the average value; replace tby t+1 toshift to the left by 1 s.)

From Table 15.5-1

( )( ) ( )

0 21

1 6 1n n

j n t j n t

n n

j A jv t e e

n n

= =

= =

Therefore

( )

( ) ( ) ( )12 2

6 1 6 1

6 6

n n j n t j n j n t

n n

j j

v t e e en n2

+

= =

= =

The coefficients of this series are:

( )2

0 n

6 16 and

nj nj

C en

= = C

This result can be checked using Matlab:

pi = 3.14159;

N=100;

A = 6; % amplitude

T = 4; % periodt = linspace(0,2*T,200); % time

c0 = -6; % average value

w0 = 2*pi/T; % fundamental frequency

for n = 1: N

C(n) = (-j*A*(-1)^n/n/pi)*exp(+j*n*pi/2);

D(n) = (+j*A*(-1)^n/n/pi)*exp(-j*n*pi/2);

end

for i=1:length(t)

f(i)=c0;

for n=1:length(C)

f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+D(n)*exp(-j*n*w0*t(i));

end

end

plot(t,f,'black');

xlabel('t, sec');

ylabel('f(t)');

title('p15.5-6')

471


17/41

472


18/41

P15.5-7

Represent the function as

( )( )

5

5 1 5

1 0

1 2

t

t

e tf t

e e t

1 =

(Check: ( ) ( ) ( )5 50 0, 1 1 1, 2 f f e f e e = = = 5 0 = )

0 0

2 12 s, , also average value

2 2T C

= = = = =

The coefficients of the exponential Fourier series are calculated as

( ) ( )( )

( )

( )

( )( )( )

( )

( )

1 2 5 15 5

n0 1

1 1 2 255 5 5

0 0 1 1

1 21 25 5

5 5

0 10 1

11

2

1

2

1

2 5 5

tt j n t j n t

j n t j n t t j n t j n t

j n t j n t j n t j n t

e e dt e e e dt

e dt e e dt e e dt e e dt

e e e ee e

j n j n j n j n

+

+ +

= +

= +

= + + +

C

( )

( ) ( )

( )

( ) ( )

5 2 55 25 5

5 5 2 25

1 1 1

2 5 5

1 1 1

2 5 5

j n j n j n j n j n j n

j n j n j n j n j n j n

e e e e e e ee e

j n j n j n j n

e e e e e e e ee

j n j n j n j n

+ +

= + + +

= + + +

( ) ( )( )

( )( )

( )5 5 51 1 1 1 1 1 112 5 5

n n n n

e e e j n j n j n j n

= + + +

The terms that include the factor are small and can be ignored.5 0.00674e =

( )( )

( )( )

( ) ( )

n

1 1 11 1

2 5 5

1 1odd

50 even

5odd

5

0 even

n n

j n j n j n

n

j n j nn

n j n j n

n

= + + +

+=

+=

C

473


19/41

This result can be checked using Matlab:

pi = 3.14159;

N=101;

T = 2; % period

t = linspace(0,2*T,200); % time

c0 = 0.5; % average valuew0 = 2*pi/T; % fundamental frequency

for n = 1:2:N

if n == 2*(n/2)

C(n) = 5/((+j*pi*n)*(5+j*pi*n));

D(n) = 5/((-j*pi*n)*(5-j*pi*n));

else

C(n)=0;

D(n)=0

end

end

for i=1:length(t)

f(i)=c0;

for n=1:length(C)

f(i)=f(i)+C(n)*exp(j*n*w0*t(i))+D(n)*exp(-j*n*w0*t(i));

end

end

plot(t,f,'black');

xlabel('t, sec');

ylabel('f(t)');

title('p15.5-7')

474


20/41

Section 15-6: The Fourier Spectrum

P15.6-1

( )

40 2

42

2

A Tt tT

f tA T

A t t T

T

=

+

Documents

Chapter 15 - Fourier Series and Fourier Transform