AEM (2130002) Topic: Fourier Series ( Periodic Function to change of interval)

2. Prepared By:
Himal Desai 140120109008
Abhishek Choksi 140120109005
Soham Davara 140120109007

3. Fourier series
• Periodic function that occur in many physical and
engineering problems for example, in conduction
of heat and mechanical vibration are complicated
and it is desirable to represent them in a series of
sine and cosines. Most of the single valued
function can be expressed in a trigonometric series
of the form
𝑎0
2
+ 𝑎1 𝑐𝑜𝑠𝑥 + 𝑎2 cos 2𝑥 + 𝑎3 cos 3𝑥 + … .
+ 𝑏1 sin 𝑥 + 𝑏2 sin 2𝑥 + 𝑏3 sin 3𝑥 + … .
---------- (1)

4. • Within a desired range of value of the variable such
as a series is known as Fourier Series
• Thus the function f(x) defined in the interval
c ≤ x ≤ c + 2𝜋 can be expressed in the Fourier Series
f(x) =
𝑎0
2
+ 𝑛=1
∞
(𝑎 𝑛 cos 𝑛𝑥 + 𝑏 𝑛 sin 𝑛𝑥) -------- (2)
• Where 𝑎0, 𝑎 𝑛, 𝑏 𝑛 ( n = 1,2,3,…) are constant, called
the Fourier coefficients of f(x) are required to be
determined.

5. Change Of Interval
• In many engineering problems, it is required to
expand a function in a Fourier series over an
interval of length 2l instead of 2 𝜋.
• The transformation from the function of period
p = 2 𝜋 to those of period p = 2l is quite simple.
• This can be achieved by transformation of the
variable.
• Consider a periodic function f(x) defined in the
interval c ≤ x ≤ c + 2l.
• To change the interval into length 2 𝜋.

6. • Put z =
𝜋𝑥
𝑙
So that when x = c, z =
𝜋𝑐
𝑙
= d
and when x = c + 2l, z =
𝜋(𝑐+2𝑙)
𝑙
=
𝜋𝑐
𝑙
+ 2 𝜋 = d + 2 𝜋
• Thus the function f(x) of period 2l in c to c + 2l is
transformed to the function.
• f(
𝑙𝑧
𝜋
) = f(z) of the period 2 𝜋 in d to d + 2 𝜋 and f(z)
can be expressed as the Fourier series.

7. PROOF:
• F(z) =
𝑎0
2
+ 𝑛=1
∞
(𝑎 𝑛 cos 𝑛𝑧 + 𝑏 𝑛 sin 𝑛𝑧) ---------- (1)
• Where,
𝑎0 =
1
𝜋 𝑑
𝑑+2𝜋
𝐹 𝑧 𝑑𝑧
𝑎 𝑛=
1
𝜋 𝑑
𝑑+2𝜋
𝐹 𝑧 cos 𝑛𝑧 𝑑𝑧, n= 1,2,3…
𝑏 𝑛 =
1
𝜋 𝑑
𝑑+2𝜋
𝐹 𝑧 sin 𝑛𝑧 𝑑𝑧, n= 1,2,3…
Now making the inverse substitution
z =
𝜋𝑥
𝑙
, dz =
𝜋
𝑙
dx
When, z = d, x = c
and when, z = d + 2 𝜋, x = c + 2l

8. • The expression 1 becomes
f(z) = f(
𝜋𝑥
𝑙
) = f(x) =
𝑎0
2
+ 𝑛=1
∞
(𝑎 𝑛 cos
𝑛𝜋𝑥
𝑙
+ 𝑏 𝑛 sin
𝑛𝜋𝑥
𝑙
)
• Thus the Fourier series for f(x) in the interval c to
c + 2l is given by,
f(x) =
𝑎0
2
+ 𝑛=1
∞
(𝑎 𝑛 cos
𝑛𝜋𝑥
𝑙
+ 𝑏 𝑛 sin
𝑛𝜋𝑥
𝑙
) ------ (2)
Where,
𝑎0 =
1
𝑙 𝑑
𝑑+2𝑙
𝐹 𝑥 𝑑𝑧
𝑎 𝑛=
1
𝑙 𝑑
𝑑+2𝑙
𝐹 𝑥 cos
𝑛𝜋𝑥
𝑙
𝑑𝑥, n= 1,2,3…
𝑏 𝑛 =
1
𝑙 𝑑
𝑑+2𝑙
𝐹 𝑥 sin
𝑛𝜋𝑥
𝑙
𝑑𝑥, n= 1,2,3…

9. Example: Obtain Fourier series for the function
f(x) = 𝜋x, 0 ≤ x ≤ 1 p = 2l = 2
= 𝜋 (2 - x) 1 ≤ x ≤ 2
• Solution:
Let f(x) =
𝑎0
2
+ 𝑛=1
∞
(𝑎 𝑛 cos
𝑛𝜋𝑥
𝑙
+ 𝑏 𝑛 sin
𝑛𝜋𝑥
𝑙
)
Where,
𝑎0 =
1
𝑙
0
2𝑙
𝑓 𝑥 𝑑𝑥 =
1
1
0
2
𝑓 𝑥 𝑑𝑥
= 0
1
𝜋𝑥 𝑑𝑥 + 1
2
𝜋(2 − 𝑥)𝑑𝑥
= 𝜋
𝑥2
2
1
0
+ 𝜋 2𝑥 −
𝑥2
2
2
1
= 𝜋
1
2
+ 𝜋 4 − 2 − (2 −
1
2
)
= 𝜋

10. 𝑎 𝑛 =
1
𝑙
0
2𝑙
f x cos
𝑛𝜋𝑥
𝑙
𝑑𝑥 =
1
1
0
2
𝑓 𝑥 cos 𝑛𝜋𝑥 𝑑𝑥
= 𝜋 0
1
𝑥 cos 𝑛𝜋𝑥 𝑑𝑥 + 𝜋 1
2
2 − 𝑥 cos 𝑛𝜋𝑥 𝑑𝑥
= 𝜋 𝑥
1
𝑛𝜋
sin 𝑛𝜋𝑥 − 1 −
1
𝑛2 𝜋2 cos 𝑛𝜋𝑥
1
0
+
𝜋 2 − 𝑥
1
𝑛𝜋
sin 𝑛𝜋𝑥 − −1 −
1
𝑛2 𝜋2 cos 𝑛𝜋𝑥
2
1
• Since sin n 𝜋 = sin 2n 𝜋 = 0, cos 2n 𝜋 = 1 for all n = 1,2,3….
=
2
𝑛2 𝜋
cos 𝑛𝜋 − 1
=
2
𝑛2 𝜋
(−1) 𝑛
−1
= 0 if n is even
= −
4
𝑛2 𝜋
if n is odd

11. 𝑏 𝑛 =
1
𝑙
0
2𝑙
f x sin
𝑛𝜋𝑥
𝑙
𝑑𝑥 =
0
2
𝑓 𝑥 sin 𝑛𝜋𝑥 𝑑𝑥
= 𝜋 0
1
𝑥 sin 𝑛𝜋𝑥 𝑑𝑥 + 𝜋 1
2
2 − 𝑥 sin 𝑛𝜋𝑥 𝑑𝑥
= 𝜋 𝑥 −
1
𝑛𝜋
cos 𝑛𝜋𝑥 − 1 −
1
𝑛2 𝜋2 sin 𝑛𝜋𝑥
1
0
+
𝜋 2 − 𝑥 −
1
𝑛𝜋
cos 𝑛𝜋𝑥 − −1 −
1
𝑛2 𝜋2 sin 𝑛𝜋𝑥
2
1
= 𝜋 −
cos 𝑛𝜋
𝑛𝜋
+
cos 𝑛𝜋
𝑛𝜋
= 0
ANS:
f(x) =
𝜋
2
−
4
𝜋
cos 𝜋𝑥
12 +
cos 3𝜋𝑥
32 +
cos 5𝜋𝑥
52 + ⋯

12. Example: Find the Fourier series with period 3 to
represent f(x) = 2𝑥 − 𝑥2
, in the range (0,3)
• Solution:
Here p = 2l = 3
So, l =
3
2
For this period 2l = 3, we have
f(x) =
𝑎0
2
+ 𝑛=1
∞
(𝑎 𝑛 cos
2𝑛𝜋𝑥
3
+ 𝑏 𝑛 sin
2𝑛𝜋𝑥
3
)
Where,
𝑎0 =
1
𝑙
0
2𝑙
𝑓 𝑥 𝑑𝑥 =
2
3
0
3
2𝑥 − 𝑥2 𝑑𝑥
=
2
3
𝑥2 −
𝑥3
2
3
0
= 0

13. 𝑎 𝑛 =
1
𝑙
0
2𝑙
f x cos
𝑛𝜋𝑥
𝑙
𝑑𝑥 =
2
3
0
3
2𝑥 − 𝑥2
cos
2𝑛𝜋𝑥
3
𝑑𝑥
=
2
3
2𝑥 − 𝑥2 3
2𝑛𝜋
sin
2𝑛𝜋𝑥
3
− 2 − 2𝑥 −
9
4𝑛2 𝜋2 cos
2𝑛𝜋𝑥
3
+ −2 −
27
8𝑛3 𝜋3 sin
2𝜋𝑛𝑥
3
3
0
• Since sin 2n 𝜋 = sin 0 = 0 cos 2n 𝜋 = cos 0 = 1 for all n =
1,2,3….
=
2
3
−
9
4𝑛2 𝜋2 4 −
9
4𝑛2 𝜋2 2
= −
9
𝑛2 𝜋

14. 𝑏 𝑛 =
1
𝑙
0
2𝑙
f x sin
𝑛𝜋𝑥
𝑙
𝑑𝑥 =
2
3
0
3
2𝑥 − 𝑥2
sin
2𝑛𝜋𝑥
3
𝑑𝑥
=
2
3
2𝑥 − 𝑥2
−
3
2𝑛𝜋
cos
2𝑛𝜋𝑥
3
− 2 − 2𝑥 −
9
4𝑛2 𝜋2 sin
2𝑛𝜋𝑥
3
+
−2
27
8𝑛3 𝜋3 cos
2𝜋𝑛𝑥
3
3
0
=
3
𝑛𝜋
ANS:
f(x) = −
9
𝜋2 𝑛=1
∞ 1
𝑛2 cos
2𝑛𝜋𝑥
3
+
3
𝜋 𝑛=1
∞ 1
𝑛
sin
2𝑛𝜋𝑥
3

15. Example : Find the corresponding Fourier series
F(x) =
8, 0 < 𝑥 < 2
−8, 2 < 𝑥 < 4
p = 2l = 4
• Solution:
Here 2l = 4
So, l = 2
Let,
f(x) =
𝑎0
2
+ 𝑛=1
∞
(𝑎 𝑛 cos
𝑛𝜋𝑥
2
+ 𝑏 𝑛 sin
𝑛𝜋𝑥
2
)
Where,
𝑎0 =
1
𝑙
0
2𝑙
𝑓 𝑥 𝑑𝑥 =
1
2
0
4
𝑓 𝑥 𝑑𝑥
=
1
2 0
2
8 𝑑𝑥 + 2
4
−8 𝑑𝑥
=
8
2
𝑥
2
0
− 𝑥
4
2

16. 𝑎 𝑛 =
1
𝑙
0
2𝑙
f x cos
𝑛𝜋𝑥
𝑙
𝑑𝑥 =
1
2
0
4
𝑓(𝑥) cos
𝑛𝜋𝑥
2
𝑑𝑥
=
1
2
8 0
2
cos
𝑛𝜋𝑥
2
𝑑𝑥 − 8 2
4
cos
𝑛𝜋𝑥
2
𝑑𝑥
= 4
2
𝑛𝜋
sin
𝑛𝜋𝑥
2
2
0
−
2
𝑛𝜋
sin
𝑛𝜋𝑥
2
4
2
= 0
𝑏 𝑛 =
1
𝑙
0
2𝑙
f x sin
𝑛𝜋𝑥
𝑙
𝑑𝑥 =
1
2
0
4
𝑓(𝑥) sin
𝑛𝜋𝑥
2
𝑑𝑥
=
1
2
8 0
2
sin
𝑛𝜋𝑥
2
𝑑𝑥 − 8 2
4
sin
𝑛𝜋𝑥
2
𝑑𝑥

17. = 4 −
2
𝑛𝜋
cos
𝑛𝜋𝑥
2
2
0
+
2
𝑛𝜋
cos
𝑛𝜋𝑥
2
4
2
=
8
𝑛𝜋
− cos 𝑛𝜋 + 1 + 1 − cos 𝑛𝜋
=
16
𝑛𝜋
1 − cos 𝑛𝜋
=
16
𝑛𝜋
1 − (−1) 𝑛
=
32
𝑛𝜋
, if n is odd
= 0 , if n is even
ANS:
f(x) =
32
𝜋
1
1
sin
𝜋𝑥
2
+
1
3
sin
3𝜋𝑥
2
+
1
5
sin
5𝜋𝑥
2
+ ⋯