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“Fourier series and fourier Transform”
Civil Engineering Department
JD College off Engineering and Management, Nagpur
•Vikas s. Rathod
•Vijay b. Mujmule
•Sonu chichghre
•Dipika s. Kathoute
•Achal r. Wadhai
UNDER GUIDANCE OF
PROF. ROHIT PATANE SIR
Civil Engineering Department
JD College of Engineering and management, Nagpur
HEAD OF THE DEPARTMENT (HOD)
PROF. ABDUL GAFFAR SIR
UNIT : 3 : - FOURIER TRANSFORM
Definition :-
Fourier transform of the function of f(x) defined in the interval (−∞ ,∞) is defined as , 𝑓
(ℷ)= ∫𝑓(𝑢)𝑒−𝑖ℷ𝑢.𝑑𝑢∞−∞
……….(a)
And the inverse fourier transform of 𝑓(ℷ) or fourier integral representation of f(x) in
interval (−∞ ,∞) is given by ,
F(x) = 12𝜋 𝑓(ℷ)𝑒𝑖ℷ𝑢.𝑑ℷ
……………(b)
To find inverse fourier transform of 𝑓(ℷ) or fourier integral representation of any function
of f(x) , first and the value of 𝑓(ℷ) by using equation (a) and then put the value of 𝑓(ℷ) in
equation (b)
Fourier integral :-
Important formulaes : - 𝑓(𝑥)𝑏𝑒 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑡ℎ𝑒𝑛,
𝑓(𝑥)= ∫[𝑎(ℷ)𝑐𝑜𝑠ℷ𝑥+𝐵(ℷ)𝑠𝑖𝑛ℷ𝑥 ]∞0.𝑑𝑥
………………….(1)
Now,
Where,
A(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑐𝑜𝑠∞−∞ℷ𝑡.𝑑𝑡 ………………..(2)
And
B(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑠𝑖𝑛∞−∞ℷ𝑡.𝑑𝑡…………(3)
Equation (1) is the fourier integral representation of the functions f(x) with coefficient.
Defined by equaqtion (2) and (3)
Example – Find fourier integral representation of the rectangle function.
F(x) = h ( 1 - |𝑥| ) , where it is a Heaviside unit step function .
STEP (1) –
Let
F(x) = h ( 1 - |𝑥| )
…………………..(1)
Where h is a Heaviside unit step function
STEP(2) –
We have to find Fourier integral representation of f(x) given inn equation (1)
STEP(3) –
We know that the fourier integral representation of f(x) is –
𝑓(𝑥)= ∫[𝑎(ℷ)𝑐𝑜𝑠ℷ𝑥+𝐵(ℷ)𝑠𝑖𝑛ℷ𝑥 ]∞0.𝑑𝑥
………………….(2)
Where
A(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑐𝑜𝑠∞−∞ℷ𝑡.𝑑𝑡 ………………..(3)
And
B(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑠𝑖𝑛∞−∞ℷ𝑡.𝑑𝑡
………………..(4)
STEP(4)-
From euation(1) F(x) = h ( 1 - |𝑥| )
Now,
h ( 1 - |𝑥| ) = 1 ,−1≥x≥1
0 , Otherwise
………(5)
STEP(5) –
To find A(ℷ) and B(ℷ)
A(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑐𝑜𝑠∞−∞ℷ𝑡.𝑑𝑡
F(x) = 1 ,−1≥x≥1
0 , Otherwise
Now,
A(ℷ)=1𝜋 {∫𝑓(𝑡)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡+ ∫𝑓(𝑡)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡+ ∫𝑓(𝑡)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡∞11−1−1−∞}
A(ℷ)=1𝜋 {∫(0)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡+ ∫(1)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡+ ∫(0)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡∞11−1−1−∞}
A(ℷ)=1𝜋 {0+ ∫(1)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡+0 }1−1
A(ℷ)=1𝜋{ [ 𝑠𝑖𝑛ℷ𝑡𝑡]}
A(ℷ)=1𝜋{ [ 𝑠𝑖𝑛ℷ(−1)𝑡 - 𝑠𝑖𝑛ℷ(−1)𝑡] }
A(ℷ)=1𝜋{ [ 𝑠𝑖𝑛ℷ𝑡 + 𝑠𝑖𝑛ℷ𝑡] }
A(ℷ)= 𝟐𝒔𝒊𝒏𝒏ℷ𝝅ℷ
…………………..(6)
To findB(ℷ)
B(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑐𝑜𝑠∞−∞ℷ𝑡.𝑑𝑡
F(x) = 1 ,−1≥x≥1
0 , Otherwise
Now,
B(ℷ)=1𝜋 {∫𝑓(𝑡)𝑠𝑖𝑛ℷ𝑡.𝑑𝑡+ ∫𝑓(𝑡)𝑠𝑖𝑛ℷ𝑡.𝑑𝑡+ ∫𝑓(𝑡)𝑠𝑖𝑛ℷ𝑡.𝑑𝑡∞11−1−1−∞}
A(ℷ)=1𝜋 {∫(0)𝑠𝑖𝑛𝑛ℷ𝑡.𝑑𝑡+ ∫(1)𝑠𝑖𝑛𝑛ℷ𝑡.𝑑𝑡+ ∫(0)𝑠𝑖𝑛ℷ𝑡.𝑑𝑡∞11−1−1−∞}
A(ℷ)=1𝜋 {0+ ∫(1)𝑠𝑖𝑛ℷ𝑡.𝑑𝑡+0 }1−1
A(ℷ)=1𝜋{ [𝑐𝑜𝑠ℷ𝑡𝑡]}
A(ℷ)=1𝜋{ [1−𝑐𝑜𝑠ℷ(1)𝑡 - −𝑐𝑜𝑠ℷ(−1)𝑡] }
A(ℷ)=1𝜋{ [−𝑐𝑜𝑠ℷ𝑡 + 𝑐𝑜𝑠ℷ𝑡] }
A(ℷ)=1𝜋(0) = 0
…………………….(7)
Now
Put equation 6 amd 7 in equation (2) , we get 𝑓(𝑥)= ∫[𝑎(ℷ)𝑐𝑜𝑠ℷ𝑥+𝐵(ℷ)𝑠𝑖𝑛ℷ𝑥
]∞0.𝑑ℷ
= ∫[ 𝟐𝒔𝒊𝒏𝒏ℷ𝝅ℷ 𝒄𝒐𝒔ℷ𝒙+(𝟎)𝒔𝒊𝒏ℷ𝒙 ∞0].𝑑ℷ = ∫[ 𝟐𝒔𝒊𝒏𝒏ℷ𝝅ℷ 𝒄𝒐𝒔ℷ𝒙.∞0 𝑑ℷ
Which is required Fourier integral of f(x)
Fourier sine and cosine integrals :-
Case-(1) :-
If the function f(x) is an given function then fourier cosine integral of f(x) is given by ,
𝑓(𝑥)= ∫𝐴(ℷ∞0)cosℷx.dℷ
…………..(1)
Where ,
A(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑐𝑜𝑠∞−∞ℷ𝑥.𝑑𝑥
= 2𝜋 ∫𝑓(𝑥)∞0 𝑐𝑜𝑠ℷ𝑥.𝑑𝑥
………….(2)
Case-(2):-
If the function f(x) is an given function then fourier sine integral of f(x) is
given by ,
𝑓(𝑥)= ∫𝐴(ℷ∞0)sinℷx.dℷ
………….(3)
Where,
B(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑠𝑖𝑛∞−∞ℷ𝑥.𝑑𝑥
= 2𝜋 ∫𝑓(𝑥)∞0 𝑠𝑖𝑛ℷ𝑥.𝑑𝑥
…………(4)
PROPERTIES OF FOURIER TRANSFORM : -
Modulation Theorem :-
If F(s) is The complex Fourier Transform of the Function f(x) then,
F{f(x).cosax} = 12{𝐹(𝑠+𝑎)+ 𝐹(𝑠−𝑎)}
Let,
F(s) is the complex Fourier Transform of F(x)
i.e; F(s) = F {f(x)} ……..(1)
We have to show
F{f(x).cosax} = 12{𝐹(𝑠+𝑎)+ 𝐹(𝑠−𝑎)}
…………..(2)
Then ,
We know that the fourier transform of f(x) is
F{f(x)} = ∫𝑒𝑖𝑠𝑥𝑓(𝑥).𝑑𝑥∞−∞
…………..(3)
F{f(x).cosax} = ∫𝑒𝑖𝑠𝑥 𝑓(𝑥),𝑐𝑜𝑠𝑎𝑥.𝑑𝑥∞−∞
F{f(x).cosax} = ∫𝑒𝑖𝑠𝑥 𝑓(𝑥).( 𝑒𝑖𝑎𝑥+ 𝑒−𝑖𝑎𝑥2).𝑑𝑥∞−∞
F{f(x).cosax} = 12 [∫𝑒𝑖𝑠𝑥∞−∞ 𝑓(𝑥).𝑒𝑖𝑎𝑥.𝑑𝑥]
F{f(x).cosax} = 12 [∫𝑒𝑖𝑠𝑥 𝑓(𝑥)𝑒𝑖𝑎𝑥∞−∞.𝑑𝑥+ ∫𝑒𝑖𝑠𝑥∞−∞ 𝑓(𝑥).𝑒−𝑖𝑎𝑥.𝑑𝑥]
F{f(x).cosax} = 12 [∫𝑒𝑖𝑠𝑥+𝑖𝑎𝑥 𝑓(𝑥)∞−∞ 𝑑𝑥+ ∫𝑒𝑖𝑠𝑥−𝑖𝑎𝑥 𝑓(𝑥)∞−∞.𝑑𝑥]
Now,
F{f(x).cosax} = 12{𝐹(𝑠+𝑎)+ 𝐹(𝑠−𝑎)}
F{f(x).cosax} = 𝟏𝟐{𝑭(𝒔+𝒂)+ 𝑭(𝒔−𝒂)}
Hence proved .
Convolution Theorem For Fourier Transform :-
The convolution of the function f(x) and g(x) over the interval ,( ∞
,−∞) is defined as
F*g = ∫𝑓(𝑢).𝑔(𝑥−𝑢).𝑑𝑢∞−∞
= h(x)
According to convolution theorem , the fourier transform of the
convolution of f(x) and g(x) is the product of their fourie
transforms ,
i.e;
F{f(x)*g(x) } = f{f(x)} . F{g(x)}
Now,
Let,F{f(x)*g(x) } = F {∫𝑓(𝑢).𝑔(𝑥−𝑢).𝑑𝑢}∞−∞
F{f(x)*g(x) } = ∫𝑒𝑖𝑠𝑥 {∫𝑓(𝑢).𝑔(𝑡−𝑢).𝑑𝑢}.𝑑𝑥∞−∞∞−∞
[ ( F{f(x)} = ∫𝑒𝑖𝑠𝑥∞−∞.f(x).dx ) ]
Thank you

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fourier series and fourier transform

  • 1. . . “Fourier series and fourier Transform” Civil Engineering Department JD College off Engineering and Management, Nagpur •Vikas s. Rathod •Vijay b. Mujmule •Sonu chichghre •Dipika s. Kathoute •Achal r. Wadhai UNDER GUIDANCE OF PROF. ROHIT PATANE SIR Civil Engineering Department JD College of Engineering and management, Nagpur HEAD OF THE DEPARTMENT (HOD) PROF. ABDUL GAFFAR SIR
  • 2. UNIT : 3 : - FOURIER TRANSFORM Definition :- Fourier transform of the function of f(x) defined in the interval (−∞ ,∞) is defined as , 𝑓 (ℷ)= ∫𝑓(𝑢)𝑒−𝑖ℷ𝑢.𝑑𝑢∞−∞ ……….(a) And the inverse fourier transform of 𝑓(ℷ) or fourier integral representation of f(x) in interval (−∞ ,∞) is given by , F(x) = 12𝜋 𝑓(ℷ)𝑒𝑖ℷ𝑢.𝑑ℷ ……………(b) To find inverse fourier transform of 𝑓(ℷ) or fourier integral representation of any function of f(x) , first and the value of 𝑓(ℷ) by using equation (a) and then put the value of 𝑓(ℷ) in equation (b)
  • 3. Fourier integral :- Important formulaes : - 𝑓(𝑥)𝑏𝑒 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑡ℎ𝑒𝑛, 𝑓(𝑥)= ∫[𝑎(ℷ)𝑐𝑜𝑠ℷ𝑥+𝐵(ℷ)𝑠𝑖𝑛ℷ𝑥 ]∞0.𝑑𝑥 ………………….(1) Now, Where, A(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑐𝑜𝑠∞−∞ℷ𝑡.𝑑𝑡 ………………..(2) And B(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑠𝑖𝑛∞−∞ℷ𝑡.𝑑𝑡…………(3) Equation (1) is the fourier integral representation of the functions f(x) with coefficient. Defined by equaqtion (2) and (3)
  • 4. Example – Find fourier integral representation of the rectangle function. F(x) = h ( 1 - |𝑥| ) , where it is a Heaviside unit step function . STEP (1) – Let F(x) = h ( 1 - |𝑥| ) …………………..(1) Where h is a Heaviside unit step function STEP(2) – We have to find Fourier integral representation of f(x) given inn equation (1) STEP(3) –
  • 5. We know that the fourier integral representation of f(x) is – 𝑓(𝑥)= ∫[𝑎(ℷ)𝑐𝑜𝑠ℷ𝑥+𝐵(ℷ)𝑠𝑖𝑛ℷ𝑥 ]∞0.𝑑𝑥 ………………….(2) Where A(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑐𝑜𝑠∞−∞ℷ𝑡.𝑑𝑡 ………………..(3) And B(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑠𝑖𝑛∞−∞ℷ𝑡.𝑑𝑡 ………………..(4) STEP(4)- From euation(1) F(x) = h ( 1 - |𝑥| ) Now, h ( 1 - |𝑥| ) = 1 ,−1≥x≥1
  • 6. 0 , Otherwise ………(5) STEP(5) – To find A(ℷ) and B(ℷ) A(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑐𝑜𝑠∞−∞ℷ𝑡.𝑑𝑡 F(x) = 1 ,−1≥x≥1 0 , Otherwise Now, A(ℷ)=1𝜋 {∫𝑓(𝑡)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡+ ∫𝑓(𝑡)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡+ ∫𝑓(𝑡)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡∞11−1−1−∞} A(ℷ)=1𝜋 {∫(0)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡+ ∫(1)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡+ ∫(0)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡∞11−1−1−∞} A(ℷ)=1𝜋 {0+ ∫(1)𝑐𝑜𝑠ℷ𝑡.𝑑𝑡+0 }1−1
  • 7. A(ℷ)=1𝜋{ [ 𝑠𝑖𝑛ℷ𝑡𝑡]} A(ℷ)=1𝜋{ [ 𝑠𝑖𝑛ℷ(−1)𝑡 - 𝑠𝑖𝑛ℷ(−1)𝑡] } A(ℷ)=1𝜋{ [ 𝑠𝑖𝑛ℷ𝑡 + 𝑠𝑖𝑛ℷ𝑡] } A(ℷ)= 𝟐𝒔𝒊𝒏𝒏ℷ𝝅ℷ …………………..(6) To findB(ℷ) B(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑐𝑜𝑠∞−∞ℷ𝑡.𝑑𝑡 F(x) = 1 ,−1≥x≥1 0 , Otherwise Now, B(ℷ)=1𝜋 {∫𝑓(𝑡)𝑠𝑖𝑛ℷ𝑡.𝑑𝑡+ ∫𝑓(𝑡)𝑠𝑖𝑛ℷ𝑡.𝑑𝑡+ ∫𝑓(𝑡)𝑠𝑖𝑛ℷ𝑡.𝑑𝑡∞11−1−1−∞}
  • 8. A(ℷ)=1𝜋 {∫(0)𝑠𝑖𝑛𝑛ℷ𝑡.𝑑𝑡+ ∫(1)𝑠𝑖𝑛𝑛ℷ𝑡.𝑑𝑡+ ∫(0)𝑠𝑖𝑛ℷ𝑡.𝑑𝑡∞11−1−1−∞} A(ℷ)=1𝜋 {0+ ∫(1)𝑠𝑖𝑛ℷ𝑡.𝑑𝑡+0 }1−1 A(ℷ)=1𝜋{ [𝑐𝑜𝑠ℷ𝑡𝑡]} A(ℷ)=1𝜋{ [1−𝑐𝑜𝑠ℷ(1)𝑡 - −𝑐𝑜𝑠ℷ(−1)𝑡] } A(ℷ)=1𝜋{ [−𝑐𝑜𝑠ℷ𝑡 + 𝑐𝑜𝑠ℷ𝑡] } A(ℷ)=1𝜋(0) = 0 …………………….(7) Now Put equation 6 amd 7 in equation (2) , we get 𝑓(𝑥)= ∫[𝑎(ℷ)𝑐𝑜𝑠ℷ𝑥+𝐵(ℷ)𝑠𝑖𝑛ℷ𝑥 ]∞0.𝑑ℷ = ∫[ 𝟐𝒔𝒊𝒏𝒏ℷ𝝅ℷ 𝒄𝒐𝒔ℷ𝒙+(𝟎)𝒔𝒊𝒏ℷ𝒙 ∞0].𝑑ℷ = ∫[ 𝟐𝒔𝒊𝒏𝒏ℷ𝝅ℷ 𝒄𝒐𝒔ℷ𝒙.∞0 𝑑ℷ Which is required Fourier integral of f(x)
  • 9. Fourier sine and cosine integrals :- Case-(1) :- If the function f(x) is an given function then fourier cosine integral of f(x) is given by , 𝑓(𝑥)= ∫𝐴(ℷ∞0)cosℷx.dℷ …………..(1) Where , A(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑐𝑜𝑠∞−∞ℷ𝑥.𝑑𝑥 = 2𝜋 ∫𝑓(𝑥)∞0 𝑐𝑜𝑠ℷ𝑥.𝑑𝑥 ………….(2)
  • 10. Case-(2):- If the function f(x) is an given function then fourier sine integral of f(x) is given by , 𝑓(𝑥)= ∫𝐴(ℷ∞0)sinℷx.dℷ ………….(3) Where, B(ℷ)= 1𝜋 ∫𝑓(𝑡).𝑠𝑖𝑛∞−∞ℷ𝑥.𝑑𝑥 = 2𝜋 ∫𝑓(𝑥)∞0 𝑠𝑖𝑛ℷ𝑥.𝑑𝑥 …………(4)
  • 11. PROPERTIES OF FOURIER TRANSFORM : - Modulation Theorem :- If F(s) is The complex Fourier Transform of the Function f(x) then, F{f(x).cosax} = 12{𝐹(𝑠+𝑎)+ 𝐹(𝑠−𝑎)} Let, F(s) is the complex Fourier Transform of F(x) i.e; F(s) = F {f(x)} ……..(1) We have to show F{f(x).cosax} = 12{𝐹(𝑠+𝑎)+ 𝐹(𝑠−𝑎)} …………..(2) Then , We know that the fourier transform of f(x) is F{f(x)} = ∫𝑒𝑖𝑠𝑥𝑓(𝑥).𝑑𝑥∞−∞ …………..(3) F{f(x).cosax} = ∫𝑒𝑖𝑠𝑥 𝑓(𝑥),𝑐𝑜𝑠𝑎𝑥.𝑑𝑥∞−∞
  • 12. F{f(x).cosax} = ∫𝑒𝑖𝑠𝑥 𝑓(𝑥).( 𝑒𝑖𝑎𝑥+ 𝑒−𝑖𝑎𝑥2).𝑑𝑥∞−∞ F{f(x).cosax} = 12 [∫𝑒𝑖𝑠𝑥∞−∞ 𝑓(𝑥).𝑒𝑖𝑎𝑥.𝑑𝑥] F{f(x).cosax} = 12 [∫𝑒𝑖𝑠𝑥 𝑓(𝑥)𝑒𝑖𝑎𝑥∞−∞.𝑑𝑥+ ∫𝑒𝑖𝑠𝑥∞−∞ 𝑓(𝑥).𝑒−𝑖𝑎𝑥.𝑑𝑥] F{f(x).cosax} = 12 [∫𝑒𝑖𝑠𝑥+𝑖𝑎𝑥 𝑓(𝑥)∞−∞ 𝑑𝑥+ ∫𝑒𝑖𝑠𝑥−𝑖𝑎𝑥 𝑓(𝑥)∞−∞.𝑑𝑥] Now, F{f(x).cosax} = 12{𝐹(𝑠+𝑎)+ 𝐹(𝑠−𝑎)} F{f(x).cosax} = 𝟏𝟐{𝑭(𝒔+𝒂)+ 𝑭(𝒔−𝒂)} Hence proved . Convolution Theorem For Fourier Transform :- The convolution of the function f(x) and g(x) over the interval ,( ∞ ,−∞) is defined as F*g = ∫𝑓(𝑢).𝑔(𝑥−𝑢).𝑑𝑢∞−∞
  • 13. = h(x) According to convolution theorem , the fourier transform of the convolution of f(x) and g(x) is the product of their fourie transforms , i.e; F{f(x)*g(x) } = f{f(x)} . F{g(x)} Now, Let,F{f(x)*g(x) } = F {∫𝑓(𝑢).𝑔(𝑥−𝑢).𝑑𝑢}∞−∞ F{f(x)*g(x) } = ∫𝑒𝑖𝑠𝑥 {∫𝑓(𝑢).𝑔(𝑡−𝑢).𝑑𝑢}.𝑑𝑥∞−∞∞−∞ [ ( F{f(x)} = ∫𝑒𝑖𝑠𝑥∞−∞.f(x).dx ) ]