MM Vercelli.

L5: Fourier Transform.

Contents: • History and some motivation. • Signal are tempered distributions. • Fourier Transform. • Convolution. • The sampling theorem.

1 History and some motivation

In the 18 century Euler and d’Alembert observed that many interesting functions can be written as a sum of sines and cosines and applied them to solving important problems.

But was Fourier that developed the theory and applied it to the heat equation in his important memoir (1807) On the Propagation of Heat in Solid Bodies.

The idea is that signals can be described from two points of view. Namely, from the obvious point of view of time t but also from the point of view of frequencies ξ or ω.

Let see and example of the kind of mathematical trick we have to learn to use.

Since signals are composed by frequencies we would like to eliminate some frequencies. This is done we means of a so called filter. So let Hb = 1 for the desired frequencies and Hb = 0 for the undesirable frequencies. Then when a signal fb is fed into the filter, an output signal g is obtained in time. We will see that there is a mathematical operation called convolution and denoted by ∗ such that g = f ∗ H. The function H is called impulse function because is the output generated when the input is an impulse at time 0, namely a delta δ.

Very simpler filters are constructed by means of RLC circuit. A RLC circuit is mathematically modeled as a second order ordinary differential equation: LCg00(t) + RCg0(t) + g(t) = f(t)

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we will see that in this case 1 H(ω) = . −LCω2 + i RCω + 1

So given f(t) (input signal) we have to solve the above differential equation to obtain the output signal g. As explained we will use convolutions, the dirac delta, and other mathematical tools to do this.

2 Signals are tempered distributions.

Classically a signal is any function of f(t) where t ∈ R is regarded as time. Usually f(t) is also a real number but we will allow f(t) to be a complex number because the periodic or undulatory phenomena are better understood by means of complex numbers.

Actually, functions f(t) do not give an adequate mathematical model or setting for the physical phenomena of signals or spectral analysis. Tempered distributions are the natural setting to study such signals. Here is another example showing the necessity of the concept of distribution.

The signal f(t) = sin(ω0t) is obviously composed by a single frequency ω0 . Then we ˆ expect that f(ω) = (Ff)(ω) should be zero unless ω = ω0 . Unfortunately, the improper R − i ωt integral of the Fourier transform of e sin(ω0t)dt fails to exist due to the oscillatory R nature of sin function. It is inside the theory of distributions that the Fourier transform of sin(ω0t), regarded as a distribution, do exist and have the properties we expect.

Why distributions and not just functions?. Consider the function f(t) = sin(ω0t). It is clear that f(t) contains a single frequency. So we expect its Fourier transform fˆ(ω) to be zero unless ω = ω0 . Unfortunately, the integral Z ∞ − i ωt sin(ω0t)e dt −∞ fails to exist for any ω due to the oscillations (Check it!). But as we will see the dis- tribution Lf associated to the function f(t) = sin(ω0t) has a Fourier transform with a property similar to that we expect, i.e. its spectrum is concentrated in ±ω0 . For more see [Smith03].

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3 Fourier Transform

If f(x) is in S its Fourier transform Ff is a function of another variable, usually denoted by ξ defined as Z Ff(ξ) = fˆ(ξ) := f(x)e− i ξxdx R and it is possible to prove that Ff ∈ S 1 [Stri94, page 34]. The inverse Fourier transform or anti-transform of g(ξ) ∈ S is given by the integral

1 Z F −1g(x) =g ˇ(x) := g(ξ)ei ξxdξ 2π R Warning: There is no general agreement with the defini- tion of Fourier transform. Indeed, there are small differences 1 with the minus sign in the exponential, the factor 2π , etc. Therefore, in using Fourier transform formulas from different sources, always check which definition is being used [Stri94, page 29].

3.1 Two important formulas Fourier inversion formula

F(F −1f) = F −1(Ff) = f

and Parseval-Plancherel formula Z 1 Z |f(x)|2dx = |fˆ(ξ)|2dξ 2π R R Actually, this is the content of Appendix A.

As consequence notice that if Ff is identically zero then f is identically zero. That is to say the Fourier transform is a bijection of S .

1This is a fundamental property of the Schwartz class S , i.e. FS = S . This property is a consequence of the duality under Fourier transform of differentiability and decrease at infinity (see [Stri94, page 35] for a detailed discussion).

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3.2 Linearity of F

(αf\+ βg)(ξ) = αfb(ξ) + βgb(ξ) and the same identity hold for F −1 .

3.3 Properties of the Fourier transform in S . Duality. Since gˆ(x) gˇ(−x) = 2π we get gˆ(x) = 2πg(−x) which is known as duality.

By iteration we get ˆ gˆ = 4π2g

Shiftings and scalings. Given a function f we denote with τyf its shifting or trans- lation f(x + y) we have

i ξy F(τyf)(ξ) = e Ff(ξ) and if g(ξ) is another function we have

−1 i ξy −1 −1 F (e g)(x) = τy(F g)(x) = F g(x + y) For the scaling f(αx) we have:

1 ξ f\(αx)(ξ) = fb( ) |α| α

df Derivatives and linear operators. Here is the relation between F and dx . df F (ξ) = i ξ(Ff)(ξ) dx or

dcf (ξ) = i ξfb(ξ) . dx

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We also have

dFf = − i F(xf) dξ By iteration by get

dnf (ξ) = (i ξ)nfb(ξ) dxn Recall that a linear differential operator L maps functions to functions by

dnf dnf df L(f) := a + a + ··· + a + a f n dxn n−1 dxn−1 1 dx 0 n n−1 Let L(X) = anX + an−1X + ··· + a1X + a0 be the polynomial associated to L. d Namely L(f) = L( dx )(f). Then FL(f) = L(i ξ)Ff or L([f) = L(i ξ)fb We also have d L( )Ff(ξ) = F(L(− i x)f)(ξ) dξ

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3.4 The Gaussian The Gaussian is the function e−tx2 , where t > 0 is a parameter. Here is its Fourier transform r 2 −tx2 π − ξ Fe (ξ) = e 4t t 1 Notice that for t = 2 we get √ Ff(ξ) = 2πf(ξ)

2 − x where f(x) = e 2 . This is a consequence of the residue calculus or by solving differential equation. As an application we will see that for P > 0 there exists ϕ ∈ S such that

2π 2π F ϕ(ξP ) = ϕ( ξ) (1) P P

Assume that for some P0 we have a solution ϕP0 . Let us see that ϕP0 (αx), for α > 0, is also a solution for P = αP0 . Indeed, 1 ξ F ϕ (αx)(ξ) = F ϕ ( ) P0 α P0 α 1 P0ξ = F ϕP0 ( ) α P0α 1 2π 2π ξ = ϕP0 ( ) α P0 P0 P0α 1 2π 2π ξ = ϕ (α ) P0 2 α P0 P0 P0α so 2π 2π F ϕP0 (αx)(ξαP0) = ϕP0 (α ξ) αP0 αP0 which shows that ϕP0 (αx) is a solution of equation (1) with P = αP0 .

Thus, to show the existence of a solution of equation (1) for any P > 0 it is√ enough to find some special value of P0 for which there exists a solution. Now if P0 = 2π and 2 − x ϕ(x) = e 2 we get a solution of equation (1). Indeed,

2π 2 2 √ ( √ ξ) − x 2π − 2π Fe 2 ( 2πξ) = √ e 2 2π

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4 Convolution

The convolution (f ∗ g)(x), f, g ∈ S is a function which solve the problem of finding the anti-transform of the product fb(ξ)gb(ξ). Namely,

f[∗ g(ξ) = fb(ξ)gb(ξ) Here is the formula for the computation of (f ∗ g)(x): Z (f ∗ g)(x) = f(s)g(x − s)ds R Let us compute f[∗ g(ξ): Z Z  f[∗ g(ξ) = f(s)g(x − s)ds e− i xξdx R R regarding the above double integral as an integral we have: Z Z f[∗ g(ξ) = f(s)g(x − s)e− i xξds dx R R by changing the variables s = u and x − s = v we get we get Z Z f[∗ g(ξ) = f(u)g(v)e− i(v+u)ξdu dv ZR ZR = f(u)g(v)e− i vξe− i uξdu dv ZR ZR = f(u)e− i uξ
g(v)e− i vξ
du dv R R = fb(ξ)gb(ξ) By the way this shows that the convolution is commutative and associative since the product between functions enjoy such properties.

4.1 Convolution and approximations of the identity

Let Kn(x) be a approximation of the identity of function in S and let f ∈ S . Then the convolution f ∗ Kn is given by Z f ∗ Kn(x) = f(s)Kn(x − s)ds R

Mathematical Methods, L5 7 Mathematical Methods MM Vercelli. performing a change of variable we get Z f ∗ Kn(x) = f(x − u)Kn(u)du R

Since Kn converges weakly to δ we have extend the definition of the convolution allowing δ to be also convolutioned. Namely,

f ∗ δ(x) = f(x) That is to say the δ is the identity of the convolution which gives an explanation of the name ’approximations of the identity’. More in general if T ∈ S0 is a tempered distribution and f ∈ S we define their convolution to be

(T ∗ g)(x) = hT, rx(g)i

where rx(g)(s) := g(x − s). So if Tf is a tempered distribution associated to the function f we get: Z (Tf ∗ g)(x) = hTf , rx(g)i = f(s)g(x − s)ds =: f ∗ g(x) . R 5 Fourier transform of tempered distributions

If T ∈ S0 is a tempered distribution we define its Fourier transform FT as follows

hFT, ϕi := hT, F ϕi for all ϕ ∈ S0 .

Since a function f ∈ S gives a distribution Tf we wonder if TFf = FTf ?? The answer is yes and actually this is the motivation of the definition of FT . Indeed, if f, ϕ ∈ S the double integral of f(x) ϕ(ξ)e− i xξ can be computed on R×R in any order and this is exactly the meaning of

TFf = FTf

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5.1 Examples Example 5.1. (translations) − i ξa FTa = e FT and i xa F(e T ) = (FT )a

Example 5.2. (delta) Fδ = T1 and FT1 = 2πδ. Example 5.3. (single frequency) The classical Fourier transform of the function f(t) = ei ω0t does not exist, due to the oscillatory nature of this function. But

FTei ω0t = 2πδω0 .

i ω0t i ω0t Indeed, Tei ω0t = e T1 and the above examples imply FTei ω0t = Fe T1 = (FT1)ω0 =

(2πδ)ω0

ei ω0t−e− i ω0t Example 5.4. (sine) Since sin(ω0t) = 2 i it follows

FTsin(ω0t) = i π(δω0 − δ−ω0 ) Example 5.5. (Heaviside) To compute FH recall that H0 = δ. So

0 FH = i ξFH = Fδ = T1 .

Since i ξP V solves the equation i ξX = T1 we get FH = i ξP V + c δ

−x2 for some constant ’c’ to be determined. To obtain ’c’ we test with Gaussian e 2 . Namely, −ξ2 −x2 Z e 2 −x2 hFH, e 2 i = lim dξ + ce 2 δ a→0 |ξ|>a i ξ = 0 + c

−x2 = hH, Fe 2 i

√ −x2 = hH, 2πe 2 i +∞ √ Z −x2 = 2π e 2 dx √0 √ 2π = 2π 2 = π .

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So PV FH = + πδ. i Example 5.6. (sign function) The function sgn(x) is defined as follow:  −1 if x < 0,  sgn(x) = 0 if x = 0, 1 if x > 0.

To compute Fsgn just notice that sgn = 2H − 1. So PV PV Fsgn = 2FH − FT = 2 + 2πδ − 2πδ = 2 1 i i

5.2 Convolution theorem Here is the convolution theorem.

Theorem 5.7. For T ∈ S0 and g ∈ S the following holds

T[∗ g = Tbgb

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6 The sampling theorem (ST).

The sampling theorem says that sometimes we can reconstruct a continuous signal from discrete samples of the continuous signal. B. Kosko wrote in his beautiful book Noise see [Ko06, page 115]:

The sampling theorem reflects the artist’s intuition that we can approximate the image of a face if we use enough small dots of color.

The history of the sampling theorem is the object of the very readable and broad survey [Hi85].

We suggest the reader to have a look to: http://en.wikipedia.org/wiki/Nyquist-Shannon_sampling_theorem.

For the history and origins of the sampling theorem see [Lu99] and [Hi85].

6.1 Mathematical statement of the ST. Here is the mathematical statement of the sampling theorem: Theorem 6.1. (Whittaker-Nyquist-Shannon’s sampling theorem) Suppose f ∈ S and F (ω) = Ff(ω) is band limited, i.e. assume that there is a frequency ωm > 0 such that F (ω) is supported on the interval [−ωm, ωm]. Let Ts be the distance between the consecutive samples f(nTs) , n ∈ Z. Then f can be recovered from its samples {f(nTs)} according to the formula2

X sin(ωm(t − nTs)) f(t) = f(nTs) . (2) ωm(t − nTs) n∈Z if the Nyquist’s condition ω ≥ 2ω is satisfied, where ω = 2π . s m s Ts Besides the mathematical statement in concrete applications there are four technical conditions to be satisfied.

The first condition is that the signal f(t) must NOT have an infinite frequency spectrum. This is exactly the meaning of the existence of ωm > 0 such that the support of the Fourier transform F is contained in [−ωm, ωm].

2The infinite series in equation (2) is sometimes called cardinal series see [Hi85] for more details.

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The second condition is that the number of samples f(nTs) must be infinite, i.e. n run over Z. In practice this is of course impossible since we can not take infinitely many samples. Another problem here is that the sampling theorem requires the exact value of the sample but this is also impossible since we use rational numbers represented as binary numbers inside a computer. So this introduce another kind of error to be controlled in practical applications.

The third condition is the most important for digital sampling. It states that the sampling rate Ts must exceed twice the largest frequency. This is expressed in mathe- matical terms by the condition 0 < T ≤ π . s ωm

The fourth condition is the practical problem of the infinite summation in equa- tion (2) involving infinite sinc’s functions sin(ωm(t−nTs)) . ωm(t−nTs)

The above fourth conditions are better explained in [Ko06, page 115].

Here is the original version published by C. E. Shannon in [Sh48, Theorem 13]: Let f(t) contains no frequencies over W. Then

∞ X sin π(2Wt − n) f(t) = X n π(2Wt − n) −∞

n where Xn = f( 2W ). Here is the idea of the proof of the sampling theorem.

The main ingredient is Poisson’s summation formula which is the contain of Lemma 6.3 below. Namely,

FfS = TFS (3) P P where fS = f(nTs)δnT = f(t).( δnT ) is the so called impulse function con- n∈Z s n∈Z s structed from the sample sequence S = {f(nTs), n ∈ Z} and FS is the ’periodic’ spec- trum of a signal constructed from the original spectrum of the signal f , namely 1 X FS(ξ) = (Ff)nωs . Ts n∈Z

It is clear that if Ff is supported in [−ωm, ωm] and Nyquist’s condition 2π ω = ≥ 2ω s T m

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is satisfied then we can recover F(f) from FfS by filtering the higher frequencies. That is to say, by multiplication of FfS by H(ξ) which equals Ts on the interval [−ωm, ωm] and 0 elsewhere, i.e. H is called an ideal low-pass filter.

Set h := F −1H . Then the above argument proves

f = fS ∗ h . (4)

Indeed, Ff = H.FfS = Fh.FfS and we get (4) from the convolution formula and the ω uniqueness of the Fourier transform. Finally, since h(t) = Ts R m ei ωtdω we get 2π −ωm X f = fS ∗ h = f(nTs)δnTs ∗ h n∈Z X = f(nTs)h(t − nTs) n∈Z X sin(ωm(t − nTs)) = f(nTs) ωm(t − nTs) n∈Z Actually the above proof is not 100% correct since the h(t) is not in the Schwartz’ class S and the above convolution has no meaning. To fix the problem is necessary to approximate h(t) by the so called mesa functions, see [Smith03, page 220] for the details.

6.2 Transform of the impulse train. Here we show the computations of a special case of Poisson’s summation formula.

2π Lemma 6.2. (impulse train or Dirac’s comb) Let P > 0 and ω0 = P . The Fourier transform of the tempered distribution X T = δnP n∈Z is X FT = ω0 δnω0 . n∈Z Proof. First of all notice that T ∈ D0 . Moreover if ϕ ∈ S then hT, ϕi < ∞ so T is a tempered function. Obviously T has a period P . Since the Fourier transform is continuous we have

X X X − i ξnP FT = F δnP = FδnP = e n∈Z n∈Z n∈Z

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Then we notice that FT has the following two properties: ( ei P ξFT = FT, 2π FT (ξ + P ) = FT (ξ) i P ξ P Then the first property imply (e −1)FT = 0 which implies that FT = z∈Zeros czδz , where Zeros := {ξ | ei P ξ − 1 = 0}. Namely, X FT = cnδ 2πn . P n∈Z

Now the periodicity of FT imply that all cn are equals to some number c. So we get X FT = c δ 2πn . P n∈Z

2 2π (αx) − 2 To see that c = P = ω0 it is enough to test against the function ϕ(x) = e , where α = √P . Recall that ϕ is a solution of 2π 2π 2π F ϕ(ξP ) = ϕ( ξ) P P and so

X 2πn X 2π X 2π hFT, ϕi = c ϕ( ) = hT, F ϕi = F ϕ(nP ) = ϕ( n) P P P n∈Z n∈Z n∈Z 2π then c = P since ϕ > 0 and the sum is convergent. 2

Here is a generalization of the above result discovered by D.S.Poisson.

2π Lemma 6.3. (impulse function & Poisson summation formula) Let T0 > 0 and ω0 = . P T0 For f ∈ S define fS = f(nT0)δnT . Then n∈Z 0

FfS = TFS

1 P where FS(ξ) = (Ff)nω . Moreover, the series defining FS(ω) converges abso- T0 n∈Z 0 ∞ lutely for each ω, and FS is C .

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The interested reader can read the proof in [Smith03, Proof of Lemma 17, page 231].

Here we want to underline the relation of the above statement with the most classical:

∞ ∞ X 1 X g(x + 2πk) = gˆ(n)ei nx (5) 2π k=−∞ n=−∞

Indeed, here g(x) play the role of (Ff)(ξ) andg ˆ(n) can be regarded as the sample 3 f(nT0). This suggest that perhaps in 1820 D.S.Poisson may have thought about the sampling theorem. This seems not to be the case as explained in [Hi85, page 52].

P∞ Here is the proof of the above equation (5). Set h(x) = k=−∞ g(x + 2πk). Then h is 2π-periodic. Let us compute its Fourier coefficients hˆ(n). Namely,

1 Z 2π hˆ(n) = h(x)e− i nxdx = 2π 0 ∞ 1 X Z 2π = g(x + 2πk)e− i nxdx = 2π k=−∞ 0 ∞ 1 X Z 2π(k+1) = g(x)e− i nxdx = 2π k=−∞ 2πk 1 Z ∞ 1 = g(x)e− i nxdx = gˆ(n) 2π −∞ 2π

3http://www-history.mcs.st-and.ac.uk/history/Mathematicians/Poisson.html

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References

[Hi85] Higgins, J. R.: Five short stories about the cardinal series. Bull. Amer. Math. Soc. (N.S.) 12 (1985), no. 1, 4589. projecteuclid.org/euclid.bams/1183552334

[Ko06] Kosko, B.: Noise. Viking Penguin Group 2006.

[Lu99] Luke,¨ H.D.: The Origins of the Sampling Theorem. IEEE Communications Magazine, April 1999, pp.106-108. http://www.hit.bme.hu/~papay/edu/Conv/pdf/origins.pdf [Sh48] Shannon, C.E.: A mathematical theory of communication Bell Sys. Tech. J., vol. 27, 1948, pp.379-423,623-56. http://cm.bell-labs.com/cm/ms/what/shannonday/shannon1948.pdf

[Smith03] Smith, D.C.: An introduction to distribution theory for signals analysis. Digital Signal Processing 13 (2003) 201-232. http://www.sciencedirect.com/science/article/pii/ S1051200402000350

[Stri94] Strichartz, R.S.: A guide to distribution theory and Fourier transforms. Studies in Advanced Mathematics. CRC Press, Boca Raton, FL, 1994.

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