chapter 15 - fourier series and fourier transform
TRANSCRIPT
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8/9/2019 Chapter 15 - Fourier Series and Fourier Transform
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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
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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
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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
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plot(t, v1,'black') % plot the Fourier series
grid
xlabel('t, s')
ylabel('f(t)')
title('P15.3-4')
4
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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
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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
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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
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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
=
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[ ]
.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
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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,
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( )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
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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
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( ) 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)');
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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
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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')
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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
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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')
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Section 15-6: The Fourier Spectrum
P15.6-1
( )
40 2
42
2
A Tt tT
f tA T
A t t T
T
=
+