Fourier Series
Fourier Series Problems and Solution
Feb 17, 2016
Fourier series example problems with solution
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Fourier Series
Example 1
Find the fundamental frequency of the
following Fourier series:
ttttfb
tttfa
80cos40cos220cos5)(:)(
80cos40cos5)(:)(
Solution to Example 1
Hzf
f
f
SOLUTION
tttfa
20
402
402
:
80cos40cos5)(:)(
1
Hzf
f
f
SOLUTION
ttttfb
10
202
202
:
80cos40cos220cos5)(:)(
1
Example 2
Find the amplitude and phase of the
fundamental component of the function:
t
ttttf
1
111
3cos3...................
2sin5.3cos5.1sin5.0)(
Recall:
a
b
baR
where
tbtatR
1
22
tan
:
sincos)cos(
Solution to Example 2
• Fundamental component:
tt 11 cos5.1sin5.0
radiansa
b
baR
where
tbtatR
32.05.1
5.0tantan
58.15.05.1
:
sincos)cos(
11
2222
Example 3
Sketch the graph of the periodic function
defined by
1)(.......10..........)( Tperiodtttf
Solution to Example 3
-1 0 1 2 3 t
1
f(t)
Example 4
Write down a mathematical expression of
the function whose graph is:
-2 -1 0 1 2 3 4 t
1
f(t)
Solution to Example 4
21.............1
2..........10..............)(
t
Ttttf
Example 5:
Sketch the graph of the following periodic
functions:
2;11,)(:)( 2 Ttttfa
tt
Tt
tfb
2,sin
;2
0,0
)(:)(
10,
3;02,)(:)(
tt
Ttttfc
Solution to Example 5
(a)
-1 0 1 2 3 t
f(t)
Solution to Example 5
(b)
0 π/2 π 3π/2 t
f(t)
Solution to Example 5
(c)
-2 -1 0 1 2 t
Example 6
Show that f(t) is even
a)
t
f(t)
4
4
Solution to Example 6
(a) f(t) = f(-t)
cos t = cos (–t)
Example 6
Show that f(t) is even
b)
t
f(t)
Solution to Example 6
(b) f(t) = f(-t)
t2 = (-t)2
t2 = t2
Example 6
Show that f(t) is even
c)
t
f(t)
3
Solution to Example 6
(c.) f(t) = f(-t)
3 = 3
Example 7
Show that f(t) is odd
t
f(t)
4
4
Solution to Example 7
f(-t) = -f(t)
sin(-π/4) = -sin(π/4)
sin (-t) = -sin(t)
Example 8
State the product of the following
functions:
(a) f(t) = t3 sin wt
(b) f(t) = t cos 2t
(c) f(t) = t + t2
Solution to Example 8
f(t) = t3 sin wt
= (odd)(odd) = even
f(t) = t cos 2t
= (odd)(even) = odd
f(t) = t + t2
= odd + even = neither
Example
Find the Fourier series of the function
tonttf )(
1
0
sincos2
)(
cos2
0
0
n
n
n
ntnn
tf
nn
b
a
a
Answer:
Solution
0
)0(2
1
2
)(
2
)(
2
1
22
1
2
1
)(2
1
0
222
0
0
0
a
ta
tdta
dttfa
Solution
ntn
vdtdu
ntdtdvtu
ntdtta
ntdttfa
n
n
sin1
cos
cos1
cos)(1
Solution
0
cos1
cos1
sinsin1
)(cos1
cos1
)(sin)(
sin1
cos1
sin1
sin1
sin1
22
22
2
n
n
n
n
n
a
nn
nn
nn
nn
a
nn
nn
nn
nn
a
ntn
ntn
ta
ntdtn
ntn
ta
Solution
ntn
vdtdu
ntdtdvtu
ntdttb
ntdttfb
n
n
cos1
sin
sin1
sin)(1
Solution
nn
nn
b
nn
nn
nn
nn
b
nn
nn
nn
nn
b
ntn
ntn
tb
ntdtn
ntn
tb
n
n
n
n
n
cos2
cos21
sin1
sin1
coscos1
)(sin1
sin1
)(cos)(
cos1
sin1
cos1
cos1
cos1
22
22
2
Solution
1
sincos2
)(n
ntnn
tf
Solution using Half Range Sine
Series
Solution using Half Range Sine
Series
• Half range sine series
a0 = 0
an = 0
Solution using Half Range Sine
Series
bn:
00
0
0
cos1cos2
cos1
sin
sin)(2
sin)(2
ntdtnn
nttb
ntn
vdtdu
ntdtdvtu
ntdttb
ntdttfL
b
n
n
L
n
Solution using Half Range Sine
Series
bn:
nn
b
nn
b
nn
nn
nn
nn
b
ntnn
nttb
n
n
n
n
cos2
cos2
)0(sin1
sin1
)0(cos)0(
cos2
sin1cos2
22
0
2
0
Solution using Half Range Sine
Series
1
sincos2
)(n
ntnn
tf
Example
Expand the given function into a Fourier
series on the indicated interval.
05,4
50,4)(
t
ttf
Answer:
1
0
5sin)cos1(
8)(
)cos1(8
0
0
n
n
n
tnn
ntf
nn
b
a
a
Solution
0
202010
1
)0(4)5(4)5(4)0(410
1
4410
1
)4()4()5(2
1)(
2
1
0
0
0
5
0
0
50
5
0
0
50
a
a
a
tta
dtdtdttfL
aL
L
Solution
0
sin20
sin20
5
1
5
)0(sin
20
5
)5(sin
20
5
)5(sin
20
5
)0(sin
20
5
1
5sin
20
5sin
20
5
1
5sin
5)4(
5sin
5)4(
5
1
5cos)4(
5cos)4(
5
1
cos)(1
5
0
0
5
0
5
5
0
n
n
n
n
n
n
L
Ln
a
nn
nn
a
n
n
n
n
n
n
n
na
tn
n
tn
na
tn
n
tn
na
dttn
dttn
a
dtL
tntf
La
Solution
)cos1(8
)cos1)(2(20
5
1
20cos
20cos
2020
5
1
5
)0(cos
20
5
)5(cos
20
5
)5(cos
20
5
)0(cos
20
5
1
5cos
20
5cos
20
5
1
5cos
5)4(
5cos
5)4(
5
1
5sin)4(
5sin)4(
5
1
sin)(1
5
0
0
5
5
0
0
5
0
5
5
0
nn
b
nn
b
nn
nn
nnb
n
n
n
n
n
n
n
nb
tn
n
tn
nb
tn
n
tn
nb
dttn
dttn
b
dtL
tntf
Lb
n
n
n
n
n
n
n
L
Ln
Solution
1 5
sin)cos1(8
)(n
tnn
ntf
Solution using Half Range Sine
Series
Solution using Half Range Sine
Series
• Half range sine series
a0 = 0
an = 0
Solution using Half Range Sine
Series
5
0
5
0
5
0
0
5cos
20
5
2
5cos
5)4(
5
2
5sin)4(
5
2
sin)(2
tn
nb
tn
nb
dttn
b
dtL
tntf
Lb
n
n
n
L
n
Solution using Half Range Sine
Series
)cos1(8
)cos1(20
5
2
20cos
20
5
2
5
)0(cos
20
5
)5(cos
20
5
2
nn
b
nn
b
nn
nb
n
n
n
nb
n
n
n
n
Solution using Half Range Sine
Series
1 5
sin)cos1(8
)(n
tnn
ntf
Example
Find the Fourier series of the function
,)( 2 onttf
12
2
2
2
0
coscos4
3)(
0
cos4
3
n
n
n
ntnn
tf
b
nn
a
a
Answer:
Solution
3
3
2
2
1
3
)(
3
)(
2
1
32
1
2
1
)(2
1
2
0
3333
0
2
0
0
a
ta
dtta
dttfa
Solution
ntdttn
ntn
tntdt
n
tnt
n
ta
ntn
vtdtdu
ntdtdvtu
ntdtta
ntdttfa
n
n
n
sin2
sin1
sin2
sin1
sin1
2
cos
cos1
cos)(1
22
2
2
Solution
ntn
ntn
tnt
n
ta
ntn
ntn
t
nnt
n
ta
ntdtn
ntn
t
nnt
n
ta
ntn
vdtdu
ntdtdvtu
n
n
n
sin2
cos2
sin1
sin1
cos2
sin1
cos1
cos2
sin1
cos1
sin
32
2
2
2
2
Solution
nn
a
nn
a
nn
nn
nn
nn
nn
nn
a
nn
nn
nn
nn
nn
nn
a
n
n
n
n
cos4
cos41
sin2
sin2
cos2
cos2
sinsin1
)(sin2
sin2
)(cos)(2
cos2
)(sin)(
sin1
2
2
3322
22
3322
22
Solution
ntdttn
ntn
tntdt
n
tnt
n
tb
ntn
vtdtdu
ntdtdvtu
ntdttb
ntdttfb
n
n
n
cos2
cos1
cos2
cos1
cos1
2
sin
sin1
sin)(1
22
2
2
Solution
ntn
ntn
tnt
n
tb
ntn
ntn
t
nnt
n
tb
ntdtn
ntn
t
nnt
n
tb
ntn
vdtdu
ntdtdvtu
n
n
n
cos2
sin2
cos1
cos1
sin2
cos1
sin1
sin2
cos1
sin1
cos
32
2
2
2
2
Solution
0
cos2
cos2
sin2
sin2
coscos1
)(cos2
cos2
)(sin)(2
sin)(2
)(cos)(
cos1
3322
22
3322
22
n
n
n
b
nn
nn
nn
nn
nn
nn
b
nn
nn
nn
nn
nn
nn
b
1
2
2
coscos4
3)(
n
ntnn
tf
Solution using Half Range Cosine
Series
Solution using Half Range Cosine
Series
• Half range cosine series
bn = 0
Solution using Half Range Cosine
Series
3
3
1
3
)0(
3
)(1
3
1
1
)(1
2
0
333
0
3
0
0
2
0
00
a
ta
dtta
dttfa
Solution using Half Range Cosine
Series
00
2
00
2
2
0
2
0
sin2
sin2
sin2
sin2
sin1
2
cos
cos2
cos)(2
ntdttn
ntn
tntdt
n
tnt
n
ta
ntn
vtdtdu
ntdtdvtu
ntdtta
ntdttfa
n
n
n
Solution using Half Range Cosine
Series
0
3
0
2
0
2
0
2
00
2
000
2
sin2
cos2
sin2
sin1
cos2
sin2
cos1
cos2
sin2
cos1
sin
ntn
ntn
tnt
n
ta
ntn
ntn
t
nnt
n
ta
ntdtn
ntn
t
nnt
n
ta
ntn
vdtdu
ntdtdvtu
n
n
n
Solution using Half Range Cosine
Series
nn
a
nn
a
nn
nn
nn
nn
nn
nn
a
n
n
n
cos4
cos22
)0(sin2
sin2
)0(cos)0(2
cos2
)0(sin)0(
sin2
2
2
3322
22
Solution using Half Range Cosine
Series
1
2
2
coscos4
3)(
n
ntnn
tf
Example
Write the sine series of f(t) = 1 on [0,5]
1
0
5sin)cos1(
2)(
)cos1(2
0
0
n
n
n
tnn
ntf
nn
b
a
a
Answer:
Solution
Solution
• Half range sine series
a0 = 0
an = 0
Solution
bn:
5
0
5
0
0
5cos
5)1(
5
2
5sin)1(
5
2
sin)(2
tn
nb
dttn
b
dtL
tntf
Lb
n
n
L
n
Solution
bn:
)cos1(2
)cos1(5
5
2
5cos
5
5
2
5
)0(cos
5
5
)5(cos
5
5
2
nn
b
nn
b
nn
nb
n
n
n
nb
n
n
n
n
Solution
1 5
sin)cos1(2
)(n
tnn
ntf
Example
• Find the convergence of f(x) on [-2,2]
2,9
21,2
12,
)( 2
t
tx
te
tf
x
Solution
Solution
31.02
4.78
2
)2(2)2(
2
2)2(
2
)2()2()2(:2
)2(2
2
ef
exf
fffx
x
Solution
82.02
237.0
2
)1(2)1(
2
2)1(
2
)1()1()1(:1
2)1(
2
ef
xef
fffx
x
Solution
92
99
2
)2()2()2(:2
fffx
Example
• Express the function in terms of H(t) and find its
Fourier transform
0,
0,0)(
te
ttf
at
Solution
iaF
ia
e
ia
e
ia
eF
dtedteeF
dteedteF
dtetfF
iaiatia
tiatiat
tiatti
ti
1
2
1)(
2
1
)(2
1)(
2
1)(
2
1)(
)()0(2
1)(
)()(
)0)(()(
0
)(
0
)(
0
0
0
Seatwork
1. Find the Fourier series representation of the
function with period T= 1/50 given by:
02.001.0......0
01.00.......1)(
t
ttf
Seatwork
2. Find the Fourier series representation of
the function with period 2π defined by
20,)( 2 tttf
Seatwork
3. Find the half range sine series of
x
xxf 0;
)()(
2
Seatwork
4. Find the half range cosine series of
x
xxf 0;
)()(
2
Related Documents
