fourier series - university of technology, iraq
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UNIVERSITY OF TECHNOLOGYElectromechanical Department
2009‐2010
Fourier series Dr.Eng Muhammad.A.R.Yass
Sultan
M O H D _ Y A S S 9 7 @ Y A H O O . C O M
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Advance Mathematics
Fourier series
3rd Class
Electromechanical Eng.
Dr. Eng
Muhammad. A. R. Yass
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Fourier series
Definition
The series
12 cos
∞
sin
Si the Fourier series of f on ( -L , L ) when the constant are chosen to be the Fourier coefficient of “f “on (-L , L)
Where
1 cos –
0 , 1 , 2
And
1 sin –
1 , 2 , 3
And
1
Example
Let f(x) = x for –π ≤ x ≤ π . we will write the Fourier series of “f” on [ -π , π] . the coefficient are
10
1cos
1cos sin 0
4
1 sin
1sin cos 0
2cos
21
Since cos ( n ) = (-1)n if n is an integer . the fourier series of “x” on [ - , ] is
2∞
1 sin 2 sin sin 223 sin 3
12 sin 4
25 sin 5 …
Example
Find the fourier series of the periodic function
1 0 10 1 2
Solution
d = 0
2 p = 2 p = 1
2 cos …… cos …… sin …… sin
1
11 1 0
| 1 1
1 cos
11
1 cos 0 cos
1sin | 0
1sin 0 0
f(t)
1 2
1
t
5
0
1 sin
1 1 sin 0 sin
1cos |
1cos
1cos 0
11 cos
1 1 cos 2/
12
1 cos 0
13 1 2
23
14 1 4 0
The fourier series become
12
2sin 0
23 sin 3 0
35 sin 5
Example
0 0
Solution
d = -
d + 2p =
p =
1
1
1
Always sin (n ) = 0
f(t)
t
-
6
1 cos–
1 –
cos cos
Integral by partial we get 0 if n = even and if n = odd
1 sin
1 sin
0
Fourier series will be
12
∞
cos 2 2 2
Example
Let f (x) = x for - ≤ x ≤
1
12 0
1cos
1cos
1cos sin 0
1cos
1cos
period
- 3 -3
7
1sin
1sin
sin cos
2cos
21
Fourier series will be
21 sin
∞
2 sin sin 223 sin 3
12 sin 4 ……
Example
Let
0 3 0 0 3
Solution
L = 3
13
13
32
13 cos 3
13 3
3cos 3 sin 3
31 1
13 sin 3
13 sin 3
3sin 3 cos 3
31
cos 1
period
3 -3
6
-6
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The Fourier series
34
3∞
1 1 cos 33
1 sin 3
Example
Let
2 2 find fourier series
12 dx
18
12 cos
nπx2 dx
8 164
12 sin 2
3sin 3 cos 3
3 1
The Fourier series will be
34
3 1 1
∞
cos 33
1 sin 3
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Even and odd function
Even Function
f is an even function on [-L , L ] if f(-x) = f(x) for -L ≤ x ≤ L
Odd Function
f is an odd function on [-L , L] if f(-x) = - f (x) -L≤ x ≤ L
Fig(1) Even function , cos
-3 +3
-1
Fig (2) Even function symmetrical a bout y-axis
Fig (3) Odd function , sin
Fig (4) Odd function symmetric through
the origin
+2 -2
+4
-4
+3 -3 -1 +1
-2 +2
+4
-1
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If the function is even then
/
/
2
/
While if is odd then
/
/
0
Also even function
2
2cos 0
Then the function will be
2 cos
0 0
2 sin
sin ∞
Symmetric a bout the origin
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Example
Find the fourier series of the function of the
2 1 01 0
2
Solution
Its an odd-function
0 0
2 sin
2 1 sin 2 sin
2 1cos
2cos
2 4cos
2cos 2
2 1 2 1
2cos 2
2
∞
1 2 1 cos 2 sin
Example
Find fourier series of the function
0 0
The function is odd then
0 ; 0
-
-2
+2
2 -
+1
-1
12
sin cos
1 10
∑ sin∞
Example
Let 1,1 find Fourier series
Solution
is an even function because f(-x) = f (x) ( i.e. example
f(-3) = f (3) on so on ) then 0
2 cos
2 cos8 6
1
the Fourier series
15 8
∞6 1 cos
Example
4 4
Solution
0
14 sin
4
12 sin 4 1 128 6/
-
h
-h
0
13
The Fourier series will be
1∞
128 6 sin 4
Conclusion
Even Function
Fourier series will be
12
∞
cos
2cos 0 , 1 , 2
Odd Function
Fourier series will be
∞
sin
2sin 1 , 2
Example
Find the Fourier series if f(x) = x2 0 < x < 2
Neither even nor odd
Period = 2L = 2 L =
0 2 -2
f(t)
4 4
14
1
83
1cos
1 sin 2
– cos2
sin
4 0
1sin
1
cos 2
cos 2
sin 4 0
1 sin
1
cos2
sin2cos 4
The Fourier series will be
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4∞
cos 44
sin
#Example
“a” odd f (-x) = - f (x)
2 0 32 3 0
-2
2
3 6 -3 -6
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“b” Neither EVEN nor odd
00 2 2
“C” EVEN f (-x) = f (x)
10 0 10 10
0
1
-1
2 -
-2
0 5 10 -10
2.5
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Half Range Expansions
Half range fourier series if function f (x) is defined only in the half fourier interval (0 ) the equation of such function can be problem into other half of period (- 0) infinite way .
a) An odd b) An even c) Neither odd nor even
Example
Give f(x) = x in the interval 0 0 < x <
a) Find the Fourier series an a even function ( cos function) b) Find the Fourier series an odd function (sin function)
a- Even function bn = 0
2
2cos
U = x du = dx
dv = cos (nx) dx v = sin (nx)
2 sin
1sin
2 sin sin 0
1cos
2 1 cos cos 0
21 1
4
24 cos∞
, ,
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b – odd function 0 0
2sin
sin 1cos
2 cos
1 cos 0
2 cos 0 cos 0
2 1
2 cos
21
2
2
2cos
2 cos∞
sin
Example
Find the sine and the cosine half range series of the function series
0
a. Even function
cos cos
cos 1
23
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∞
cos
0
f(x)
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b. Odd function
Solved Examples
The formula for a Fourier series on an interval [c,c+T] is:
Example (1)
Find the Fourier series for , .
Following the rules from the link above,
.
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.
So,
Example (2)
Find the Fourier series for .
FOURIER SERIES BOOKS
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Example (3)
Find the Fourier series for on
The general Fourier series on is:
The n = 0 case is not needed since the integrand in the formula for is .
In the present problem,
But since the right hand side is not defined if n = 0, the 0 index for a will have to be calculated seperately.
So the Fourier series is
for
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Example (4) Find the Fourier series for
on
The general Fourier series on is:
In the present problem,
22
So the Fourier series is:
Setting x = 0 gives
Example (5) Find the Fourier series for
on
So the Fourier series is:
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Example (6) Find the Fourier series for
on
The general Fourier series on is:
In the present problem,
So the Fourier series is:
on
Example (7) Find the Fourier series for a function
on .
Make the change of variables .
Now, look for the Fourier series of the function on
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Since ,
Example (8)
Find the Fourier series for on .
A general formula for the Fourier series of a function on an interval is:
In the current problem, and .
The function is odd, so the cosine coefficients will all equal zero. Nevertheless, should still be calculated separately.
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So the Fourier series for is
Example (9) Find the Fourier series for
This is the general Fourier Series:
So the given function can be replaced by its Fourier expansion:
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So the solution is
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Home Work
Problem (1) Find the Fourier series of the function
Answer.
Problem (2) . Find the Fourier series of the function
Answer. We have Therefore, the Fourier series of f(x) is
Problem (3) Find the Fourier series of
Answer.