fourier series problems and solution
DESCRIPTION
Fourier series example problems with solutionTRANSCRIPT
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