laplace ppt

1. LAPLACE AND FOURIER TRANSFORMS
Class test -2
Presented By:
Name :- Tangudu SRAVANI
Reg.No:-200101120010
Branch:-CSE
Semester:-2ND
Group:- 1
Guided by :
Prof. Balaji Padhy
(Asst . Professor)
Topic:- Half range expansions with examples.

2. content
• ✍Introduction to half range
• ✍ Half range sine series
• ✍Half range cosine series
• ✍Examples of solving sine and cosine
series
• ✍conclusion

3. Introduction:-
• Suppose a function is defined in the range(0,L),
instead of the full range (- L,L).Then the
expansion f(x) contains in a series of sine or cosine
terms only .The series is termed as half range sine
series or half range cosine series.

4. Half range Fourier sine series
• If f(x) is taken to be an odd function, its
Fourier series expansion will consists of only
sine terms. Hence the Fourier series expansion
of f(x) represents “Half range expansion of
Fourier sine series”.

5. Half range cosine series
• If f(x) is taken to be an even function , then its
Fourier series expansion will consists of
constant and cosine terms. Therefore the
Fourier series expansion of f(x) represents
“Half range expansion of Fourier series”

6. Example 1: If f (x) = 1 −
𝑥
𝐿
in 0 < x < L find (a) Fourier cosine series (b) Fourier
sine series of f (x). Graph the corresponding periodic continuation of f (x).
• Fourier cosine series of a non periodic, not even function f (x). Define (or construct) a new
function g(x) such that (i) g(x) = f (x) in (0, L) (ii) g(x) is even periodic function in (−L, L). Now
obtain the Fourier cosine series of g(x) in (−L, L)
• which is the required Fourier cosine series of f (x) in (0, L) since f (x) and g(x) coincide in (0, L).
Define
• g(x) = f (x) = 1 −
𝑥
𝐿
in 0 < x < L
• g(x) = 1 +
𝑥
𝐿
in − L < x < 0f(-x)
• Now g(x) is even in (−L, L) and is periodic with period 2L. The graph in Fig. 17.9 is the even

7. • g(x)=
𝑎𝑜
2
+ 𝑛=1
∞
𝑎𝑛 cos
𝑛𝜋𝑥
𝐿
• Where
• 𝑎𝑜=
2
𝐿 0
𝐿
𝑓 𝑥 𝑑𝑥 =
2
𝐿 0
L
1 −
𝑥
L
𝑑𝑥
• 𝑎𝑜=
2
𝐿
𝑥 −
𝑥2
2𝐿
𝐿
0
=
2
𝐿
𝐿 −
𝐿2
2𝐿
=1
• 𝑎𝑛=
2
𝐿 0
𝐿
𝑓 𝑥 cos
𝑛𝜋𝑥
𝐿
𝑑𝑥
• =
2
𝐿 0
𝐿
1 −
𝑥
L
cos
𝑛𝜋𝑥
𝐿
𝑑𝑥
• =
2
𝐿
1 −
𝑥
L
𝐿
𝑛𝜋
sin
𝑛𝜋𝑥
𝐿
−
−1
𝐿
𝐿2
𝑛2𝜋2 −𝑐𝑜𝑠
𝑛𝜋𝑥
𝐿
𝐿
0
• =
2
𝑛2𝜋2 [1 − (−1)𝑛
]
• Thus the required Fourier cosine series of f (x) in (0, L) is
• g(x)=
1
2
+ 𝑛=1
∞ 2
𝑛2𝜋2 [1 − (−1)𝑛
] cos
𝑛𝜋𝑥
𝐿

8. • Fourier sine series of f (x) in (0, L): On similar lines, define a new function h(x) such that (i) h(x) = f (x) in (0,
L) and (ii) h(x) is odd periodic function. Define.
• h(x) = f (x) 1 −
𝑥
𝐿
in 0 < x < L
• h(x) = − (1 −
𝑥
𝐿
)in -L < x < 0 h(x)=-(f(-x))
• Thus h(x) is odd periodic function in (−L, L) with period 2L. The graph in Fig. 17.11 is the odd periodic
continuation (or extension) of f (x) in fig. Now the Fourier sine series expansion of g(x) in the interval (−L,
L) is

11. • Conclusion- Through half range expansion, the
expansion of any function can be possible
either odd or its even ,,can be able to find any
of the extension.