17 août 2020 · Proof We demonstrate how to prove some of these properties because the other proofs are similar (Property 1—4) These properties follow
Previous PDF | Next PDF |
[PDF] 1 Fourier Transform - Mathtorontoedu
17 août 2020 · Proof We demonstrate how to prove some of these properties because the other proofs are similar (Property 1—4) These properties follow
[PDF] Chapter 1 The Fourier Transform - Math User Home Pages
1 mar 2010 · we will consider the transform as being defined as a suitable limit of Fourier series, and will prove the results stated here Definition 1 Let f : R
[PDF] 19 Fourier transform - NDSU
1 Since the Fourier transform plays a somewhat auxiliary role in my course, I will the duality principle or by a direct proof, the Fourier transform of xf(x) is i d ˆf
[PDF] Chapter 4 Fourier Transform
n1 e¡inx dx=Z¡1 1 f(x) e¡ix dx= f^(); and this completes the proof D The following de nition gives an inverse relation to the Fourier transform De nition 4 5
[PDF] Chapter 3 Fourier Transforms of Distributions
1) How do we transform a function f /∈ L1(R), f /∈ L2(R), for example Weierstrass Proof If ψ = the inverse Fourier transform of ϕ, then ϕ = ˆψ and the formula
[PDF] Lecture 15 Fourier Transforms (contd)
Fourier transform of a scaled function (“Scaling Theorem”): For a b > 0, F(f(bt)) = 1 bF (ωb) (15) Once again, because of its importance, we provide a proof,
[PDF] Fourier Transform Theorems - CS-UNM
Fourier Transform Theorems • Addition Theorem Shift Theorem (variation) F −1 {F(s−s0)}(t) = e j2πs0 t f(t) Proof: F −1 {F(s−s0)}(t) = / ∞ −∞ F(s−s0)e
[PDF] 1 Properties and Inverse of Fourier Transform - Department of
28 août 2016 · In the study of Fourier Transforms, one function which takes a niche (1) for all points of continuity for any integrable function x(t) (proof done in
[PDF] The Fourier transform - UBC Math
2 mai 2020 · Proof I do this in several steps Step 1 I first show that the inversion formula is valid for one particular function
[PDF] fourier transform of 1/2
[PDF] fourier transform of 1/k^2
[PDF] fourier transform of 1/t
[PDF] fourier transform of 1/x
[PDF] fourier transform of 1/x^2
[PDF] fourier transform of a constant derivation
[PDF] fourier transform of a constant signal
[PDF] fourier transform of a signal
[PDF] fourier transform of constant function proof
[PDF] fourier transform of cos(wt + theta)
[PDF] fourier transform of cos(wt) in matlab
[PDF] fourier transform of cos(wt+phi)
[PDF] fourier transform of cosine matlab
[PDF] fourier transform of cosine plot
August 17, 2020 APM346 { Week 12 Justin Ko
1 Fourier Transform
We introduce the concept of Fourier transforms. This extends the Fourier method for nite intervals to innite domains. In this section, we will derive the Fourier transform and its basic properties.1.1 Heuristic Derivation of Fourier Transforms
1.1.1 Complex Full Fourier Series
Recall thatDeMoivre formulaimplies that
sin() =eiei2iand cos() =ei+ei2 This implies that the set of eigenfunctions for the full Fourier series on [L;L] n1;cosxL
;cos2xL ;:::;sinxL ;sin2xL ;:::o is generated by the set of complex exponentialsfeinxL gn2Z. Consider the inner product for complex valued functions hf;gi=Z LLf(x)g(x)dx:
Since the complex conjugatee
inxL =einxL , it is also easy to check forn6=m, that heinxL ;eimxL i=Z LLeinxL
eimxL dx=(1)nm(1)mni(nm)= 0; so the complex exponentials are an orthogonal set. We have the following reformulation of the fullFourier series using complex variables.
Denition 1.Thecomplex form of the full Fourier seriesis given by f(x) =1X n=1c neinxL (1) where the (complex valued) Fourier coecients are given by c n=hf(x);einxL iheinxL ;einxL i=R LLf(x)einxL
dxR LLeinxL
einxL dx=12LZ LLf(x)einxL
dx:(2) The proof of Parseval's equality also implies that 1 X n=1jcnj2=12LZ LLjf(x)j2dx:(3)
1.1.2 Fourier Transform
We now formally extend the Fourier series to the entire line by takingL! 1. If we substitute (2) into ( 1 ), then f(x) =12L1 X n=1 ZLLf(x)einxL
dx e inxLPage1of14
August 17, 2020 APM346 { Week 12 Justin Ko
We denekn=nL
and k=knkn1=L then this simplies to f(x) =121 X n=1 ZLLf(x)eiknxdx
e iknxk=1p21 X n=1C(kn)eiknxk: whereC(k) =1p2Z
LLf(x)eikxdx:
If we takeL! 1, then
C(k) =1p2Z
LLf(x)eikxdx!1p2Z
1 1 f(x)eikxdx(4) and interpreting the sum as a right Riemann sum, f(x) = limL!11p21 X n=1C(kn)eiknxk!1p2Z 1 1C(k)eikxdk:(5)
Similarly, Parseval's equality (
3 ) becomes Z LLjf(x)j2dx= 2L1X
n=1jcnj2= 2L1X n=124L2 1p2Z LLf(x)einxL
dx2 =1X n=1jC(kn)j2k so takingL! 1impliesZ1 1 jf(x)j2dx=Z 1 1 jC(k)j2dk:(6) Remark 1.The step where we tookL! 1was not rigorous, because the bounds of integration and the function depend onL. A rigorous proof of this extension is much trickier.1.2 Denition of the Fourier Transform
The Fourier transformFis an operator on the space of complex valued functions to complex valued functions. The coecientC(k) dened in (4) is called the Fourier transform. Denition 2.Letf:R!C. TheFourier transformoffis denoted byF[f] =^fwhere f(k) =1p2Z 1 1 f(x)eikxdx(7) Similarly, theinverse Fourier transformoffis denoted byF1[f] =fwhere f(x) =1p2Z 1 1 f(k)eikxdk:(8)The follows from (
5 ) thatFandF1are indeed inverse operations.Theorem 1 (Fourier Inversion Formula)Iffandf0are piecewise continuous, thenF1[Ff] =fandF[F1f] =f. In particular,
f(x) =1p2Z 11^f(k)eikxdkandf(k) =1p2Z
11f(x)eikxdx:Remark 2.Technically the Fourier inversion theorem holds for almost everywhere iffis discontinuous.
In fact, one can show thatF1[Ff] =f(x)+f(x+)2
, similarly to the pointwise convergence theorem. The Fourier transforms of integrable and square integrable functions are also square integrable ( 6Page2of14
August 17, 2020 APM346 { Week 12 Justin Ko
Theorem 2 (Plancherel's Theorem)Iffis both integrable and square integrable, thenkfkL2=k^fkL2=kfkL2, i.e.
Z 1 1 jf(x)j2dx=Z 1 1 j^f(k)j2dk=Z 1 1jf(x)j2dx:(9)Remark 3.In the Denition2 , we also assume thatfis an integrable function, so that that its
Fourier transform and inverse Fourier transforms are convergent. Remark 4.Our choice of the symmetric normalizationp2in the Fourier transform makes it a linear unitary operator fromL2(R;C)!L2(R;C), the space of square integrable functionsf:R!C. Dierent books use dierent normalizations conventions.1.3 Properties of Fourier Transforms
The Fourier transform behaves very nicely under several operations of functions. We have already seen that the formulas for the solutions of several PDEs we have encountered can be expressed as convolutions. Denition 3.Theconvolutionoffandgis a functionfgdened by (fg)(x) =Z 1 1 f(xy)g(y)dy=Z 1 1 f(y)g(xy)dy:(10) We now state several properties satised by Fourier transforms. Theorem 3 (Properties of Fourier Transforms)Iffandgare integrable, then1.F[f(xa)] =eika^f(k)
2.F[f(x)eibx] =^f(kb)
3.F[f(x)] =jj1^f(1k)
4.F[^f(x)] =f(k)5.F[f0(x)] =ik^f(k)
6.F[xf(x)] =i^f0(k)
7.F[(fg)(x)] =p2^f(k)^g(k)
8.F[f(x)g(x)] =1p2(^f^g)(k)Proof.We demonstrate how to prove some of these properties because the other proofs are similar.
(Property 1|4)These properties follow immediately from a change of variables. For example, property 1 follows becauseF[f(xa)] =1p2Z
1 1 f(xa)eikxdxy=xa=1p2Z 1 1 f(y)eik(y+a)dy eikap2Z 1 1 f(y)eikydy =eika^f(k): (Property 5|6)These properties follow immediately by interchanging dierentiation and integra- tion. For example, if we writef(x) as the inverse Fourier transform of^f f(x) =1p2Z 11^f(k)eikxdk:
Page3of14
August 17, 2020 APM346 { Week 12 Justin Ko
then computing the derivative implies that f0(x) =1p2ddx
Z 11^f(k)eikxdk=1p2Z
1 1 ik^f(k)eikxdk=F1[ikf(k)] =) F[f0(x)] =ik^f(k):We are able to pass the derivative inside the integral by Leibiz's rule provided thatik^f(k)eikxis in-
tegrable. This integrability condition is implicitly satised because we assumed that the functions in
the theorem have convergent Fourier and inverse Fourier transforms. (Property 7|8)These properties are the hardest to prove, so we will show it in detail. The proof of property 7 follows from a change of variables and Fubini's theorem