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Fourier Series: part 4

23-Feb-2024

Fourier Transform Identities and Properties

We covered the basics of FT properties in previous article of part 3

In this article, we are going to cover some more properties, in a more abstract way.

Fourier Transform of a derivative

If we have function g(t)g(t), and its derivative g(t)g'(t). Then there is a relation between the fourier transform of g(t)g(t) and g(t)g'(t).

Suppose that the FT of g(t)g(t) is G(f)=g^(f)G(f)=\widehat{g}(f).

F{g(t)}=g^(f)=G(f)\begin{align*} \mathcal{F}\left\{ g(t)\right\} &= \widehat{g}(f) = G(f) \end{align*}

Then the FT of the derivative g(t)g'(t) with respect to tt is:

F{g(t)}=g(t)eiτftdt=[g(t)eiτft]+iτfg(t)eiτftdt=0+iτfg^(f)=iτfg^(f)=iτfG(f)\begin{align*} \mathcal{F}\left\{ g'(t)\right\} &= \int_{-\infty}^\infty g'(t) \, e^{-i\tau f t} dt \\ &= \left[ g(t)\, e^{-i\tau f t} \right]_{-\infty}^\infty + i \tau f \, \int_{-\infty}^\infty g(t) \, e^{-i\tau f t} dt \\ &= 0 + i \tau f \, \widehat{g}(f) \\ &= i \tau f \, \widehat{g}(f) = i \tau f \, G(f) \\ \end{align*}

From step (1) to (2), we use integration by parts. From step (2) to (3), we rely on the assumption that if g(t)g(t) is periodic, then the first term will vanish to zero. If it is not periodic, then we are going to assume that the average also vanish to zero as limtg(t)=0\lim_{t\to\infty}g(t) = 0. Finally we got result (4).

So interestingly, an FT of a derivative wrt to tt, is the same as the FT of the original function, times iτfi \tau f, with ff being the dual variable of tt (its frequency domain).

Basically FT turns derivative into a multiplication in the frequency domain.

The other interesting aspect is that you will get the same result, even if you tried to derive it using the inverse FT.

g(t)=F1{G(f)}=G(f)eiτftdfddtg(t)=ddtG(f)eiτftdfg(t)=G(f)[ddteiτft]df=[iτfG(f)]eiτftdfg(t)=F1{iτfG(f)}F{g(t)}=iτfG(f)\begin{align*} g(t) &= \mathcal{F}^{-1}\left\{ G(f) \right\} \\ &= \int_{-\infty}^\infty G(f) \, e^{i\tau f t} df \\ \frac{d}{dt} g(t) &= \frac{d}{dt} \int_{-\infty}^\infty G(f) \, e^{i\tau f t} df \\ g'(t) &= \int_{-\infty}^\infty G(f) \, \left[ \frac{d}{dt} e^{i\tau f t} \right] df \\ &= \int_{-\infty}^\infty \left[ i \tau f G(f) \right] \, e^{i\tau f t} \, df \\ g'(t) &= \mathcal{F}^{-1}\left\{ i \tau f G(f) \right\} \\ \mathcal{F}\left\{ g'(t) \right\} &= i \tau f G(f) \\ \end{align*}

Step (1) and (2) is using the definition of inverse Fourier Transform. Step (3) to (4) is assuming the convergence of the integrals, then by linearity, we can swap the order of integrals and derivative, because it is against a different variable. In step (4), we took derivative of eiτfte^{i\tau f t} because it is the only term that contains variable tt. Step (5) is using the definition of inverse Fourier Transform, but backwards. The term in the bracket must be the Fourier Transform of g(t)g'(t).

In summary, we conclude:

F{g(t)}=G(f)F{g(t)}=iτfG(f)F{Dtng(t)}=(iτf)nG(f)(P7)\begin{align*} \tag{P7} \mathcal{F}\left\{ g(t) \right\} &= G(f) \\ \mathcal{F}\left\{ g'(t) \right\} &= i \tau f G(f) \\ \mathcal{F}\left\{ D_t^n g(t) \right\} &= (i \tau f)^n G(f) \\ \end{align*}

The last row is a direct result if we took the derivative of g(t)g(t), as much as nn times.

Fourier Transform of a constant function

The next interesting thing to observe after figuring out derivative, is how we model a Fourier Transform of a constant function.

We can have two different approach.

Approach A: use the fact that a constant function can be thought of as a periodic function with finite smallest non-zero period T. Then, we can use Fourier Series approximation to construct back the original function.

Approach B: use the fact that a constant function can also be thought of as a non-periodic function with infinite period T. Then, we can use Inverse Fourier Transform integral to construct back the original function.

Constant function, interpreted from periodic function

A constant function is tricky. It expands to both \infty and -\infty, so the integral (area under the curve), is definitely \infty as well.

In terms of periodicity, we can think of constant function as a function that has “arbitrarily small” fundamental period. We can set it as small as possible, and it can still be considered as a periodic function, since the output is constant.

This made us wondering. Suppose that there exists a smallest interval possible (but still non-zero), that can’t be divided again. Can we use this as the fundamental period TT?

Let’s challenge this idea. But first, let us use a constant hh instead of TT to represent this smallest interval. Also, remember, it can’t be zero. In this interval, the value of g(t)g(t) is decidedly 1 (a constant).

We are going to use periodic approximation from article 2-T3

G(f,N,h)=Ntt+hg(t)eiτftdt=Ntt+h1eiτftdt=Nth2t+h21eiτftdt=N[eiτftiτf]th2t+h2=N[eiτftiτf]t+h2th2=N[eiτft+iτfh2eiτftiτfh2iτf]=Neiτft[eiτfh2eiτfh2iτf]=Neiτfthsin(τfh2)τfh2=Nheiτftsinc(τfh2)\begin{align*} G(f, N, h) &= N \int_{t}^{t+h} g(t) \, e^{-i\tau f t} \, dt \\ &= N \int_{t}^{t+h} 1 \, e^{-i\tau f t} \, dt \\ &= N \int_{t-\frac{h}{2}}^{t+\frac{h}{2}} 1 \, e^{-i\tau f t} \, dt \\ &= N \left[ \frac{e^{-i\tau f t}}{-i \tau f} \right]_{t-\frac{h}{2}}^{t+\frac{h}{2}} \\ &= N \left[ \frac{e^{-i\tau f t}}{i \tau f} \right]_{t+\frac{h}{2}}^{t-\frac{h}{2}} \\ &= N \left[ \frac{e^{-i\tau f t + i \tau f \frac{h}{2}}-e^{-i\tau f t - i \tau f \frac{h}{2}}}{i \tau f} \right] \\ &= N \, e^{-i\tau f t} \left[ \frac{e^{i \tau f \frac{h}{2}}-e^{-i \tau f \frac{h}{2}}}{i \tau f} \right] \\ &= N \, e^{-i\tau f t} \, h \, \frac{\sin(\tau f \frac{h}{2})}{\tau f \frac{h}{2}} \\ &= N \, h \, e^{-i\tau f t} \operatorname{sinc}(\tau f \frac{h}{2}) \\ \end{align*}

From step (1) to (2), we evaluate the value of g(t)=1g(t)=1 in this interval.

From step (2) to (3), notice that for a constant value, we can shift the interval so that tt is the center. The range is shifted into t±h2t\pm\frac{h}{2} because the length needs to be the same hh.

From step (7) to (8), we use the definition of sin(ϕ)=eiϕeiϕ2i\sin(\phi) = \frac{e^{i\phi} - e^{-i\phi}}{2i}.

From step (8) to (9), we use the definition of sine cardinal function: sinc=sin(ϕ)ϕ\operatorname{sinc} = \frac{\sin(\phi)}{\phi}.

Looking at the last expression, we know that the whole interval of domain tt is approximately NhNh. This is because the integral under the curve had to match (means the frequency f=0f=0).

For a constant function g(t)g(t), we can pick any tt and it should behave the same way as if t=0t=0, the center of the function. So, evaluating t=0t=0 would cause eiτft=1e^{-i \tau f t} = 1. We will then being left with the final expression:

G(f,N,h)=Nhsinc(τfh2)G(f,N,h) = N \, h \, \operatorname{sinc}(\tau f \frac{h}{2})

But, it’s not over yet. Ideally the function should be independent from any variable other than ff. So, we need to figure out how to eliminate NN and hh. We will use the Fourier Series form to do that. Taken from article 2-T5

gh(t)=1Nf=G(f,N,h)eiτft=1Nf=Nhsinc(τfh2)eiτft=hf=sin(τfh2)τfh2eiτft=f=eiτfh2eiτfh2iτfeiτft\begin{align*} g_h(t) &= \frac{1}{N} \sum_{f=-\infty}^\infty G(f,N,h) \, e^{i\tau f t} \\ &= \frac{1}{N} \sum_{f=-\infty}^\infty N \, h \, \operatorname{sinc}(\tau f \frac{h}{2}) \, e^{i\tau f t} \\ &= h \sum_{f=-\infty}^\infty \frac{\sin(\tau f \frac{h}{2})}{\tau f \frac{h}{2}} \, e^{i\tau f t} \\ &= \sum_{f=-\infty}^\infty \frac{e^{i \tau f \frac{h}{2}} - e^{-i \tau f \frac{h}{2}}}{i \tau f} \, e^{i\tau f t} \\ \end{align*}

Notice that, since gh(t)g_h(t) should be a constant function, then its derivative with respect to tt is zero. However, this can only mean, term inside the sum evaluates to zero (the right hand side).

gh(t)=f=eiτfh2eiτfh2iτfeiτftddtgh(t)=f=eiτfh2eiτfh2iτfddt[eiτft]0=f=(eiτfh2eiτfh2)eiτfteiτfh2=eiτfh2\begin{align*} g_h(t) &= \sum_{f=-\infty}^\infty \frac{e^{i \tau f \frac{h}{2}} - e^{-i \tau f \frac{h}{2}}}{i \tau f} \, e^{i\tau f t} \\ \frac{d}{dt} g_h(t) &= \sum_{f=-\infty}^\infty \frac{e^{i \tau f \frac{h}{2}} - e^{-i \tau f \frac{h}{2}}}{i \tau f} \, \frac{d}{dt} \left[ e^{i\tau f t} \right] \\ 0 &= \sum_{f=-\infty}^\infty \left( e^{i \tau f \frac{h}{2}} - e^{-i \tau f \frac{h}{2}} \right) \, e^{i\tau f t} \\ e^{i \tau f \frac{h}{2}} &= e^{-i \tau f \frac{h}{2}} \\ \end{align*}

In step (2) to (3) from derivation above, you will notice that the factor iτfi \tau f cancels out with the derivative of eiτfte^{i\tau f t}. This allows a very neat equation under the summation signs. Also, a peculiar one at that.

Suppose that we want to check the equality before integration, then for any tt, and any ff it seems that:

1=eiτfh\begin{align*} 1 = e^{i \tau f h} \end{align*}

This would contradict our assumption in a strange way. The consequences: one of the following needs to be true:

  1. hh is actually dependent on the value of ff that is currently being evaluated in the Fourier Transform.
  2. hh can be the smallest non-zero value possible (doesn’t matter what it is), but fhfh needs to be related with an integer nn such that fh=nf h = n. So that eiτn=ei2πn=ei2π=1e^{i\tau n}= e^{i 2 \pi n} = e^{i 2\pi} = 1

The condition for 1. will be covered automatically if we are dealing with signal scaling. But this is not the case for constant function, since stretching the function horizontally will have no effect. So, we have to accept point 2.

But if we accept point 2, then sin(τfh2)=sin(nπ)\sin(\tau f \frac{h}{2}) = \sin(n \pi) which is zero for all nn. Then, the function G(f,N,h)G(f,N,h) is zero for all nn, which means the sums also zero.

There is one weird catch, though. When f=0f=0, G(f,N,h)=NhG(f,N,h)=Nh like what we concluded before (because sinc(0)=1\operatorname{sinc}(0)=1).

Concluding what we have found now, we know that G(f,N,h)G(f,N,h) is one peculiar “function” that is zero everywhere, except when f=0f=0. Now, we need to figure out the value of NhNh and check if this has limit of some sorts.

From the Fourier Series representation, but taking t=0t=0 (the center), the value of gh(t)=0g_h(t)=0 must be zero. We also substitute f=nhf=\frac{n}{h}.

gh(t)=1Nf=G(f)eiτft=hf=sinc(τfh2)=hn=sinc(nπ)\begin{align*} g_h(t) &= \frac{1}{N} \sum_{f=-\infty}^{\infty} G(f) \, e^{i\tau f t} \\ &= h \sum_{f=-\infty}^{\infty} \operatorname{sinc}(\tau f \frac{h}{2}) \\ &= h \sum_{n=-\infty}^{\infty} \operatorname{sinc}(n \pi) \\ \end{align*}

The last sum is easy to solve. Since we already decided that nn is an integers, sinc(nπ)\operatorname{sinc}(n \pi) is zero everywhere, except when n=0n=0, where the value of sinc(0)=1\operatorname{sinc}(0)=1.

It would immediately follows that gh(t)=h=1g_h(t)=h=1.

Whaaaat? So, in fact h=1h=1 is the smallest interval that we can use.

Our G(f,N,h)=G(f,N)=Nsinc(fπ)G(f,N,h)=G(f,N)=N \, \operatorname{sinc}(f \pi).

To make it consistent with our convention that integer frequency should be named kk, we uses G(k,N)=Nsinc(kπ)G(k,N) = N \, \operatorname{sinc}(k \pi).

In our particular case here, NN really depends on the choice of the span of domain tt. So if tt ranges from -\infty to \infty, then limN=\lim_{N\to\infty}=\infty, so the series has a very tall spike when k=0k=0, but zero in everywhere else.

G(k,N)=Nsinc(kπ)(P8)\begin{align*} \tag{P8} G(k,N)= N \operatorname{sinc}(k \pi) \end{align*}

Constant function, interpreted from non-periodic function

As we have discussed before, we can also thought of constant function as a non-periodic function with infinite periods.

This line of thinking makes it easy to generalize the concept, but it is a much more challenging ideas to constructs.

Our starting line is the same. Take a segment of a constant function g(t)g(t) with the span TT. Its Fourier Transform is a function in frequency domain that can represent the function, up to this span TT. This step is similar with Approach A, where we thought that there is a periodic slices with interval hh.

However, this time, instead of multiplying it by NN (the count of the interval). We instead want to enlarge the slice by taking the limit of TT to infinity. This is analoguous to stretching the function horizontally to positive and negative x-axis direcation.

Our G(f)G(f) is a representation of the interval TT, now. We are going to use the Fourier Transform definition in article 2-T1.

G(f)=limTTTg(t)eiτftdtGT(f)=TTg(t)eiτftdt=TT1eiτftdt=2Tsinc(τfT)\begin{align*} G(f) &= \lim_{T\to\infty} \int_{-T}^{T} \, g(t) \, e^{-i\tau f t} \, dt \\ G_T(f) &= \int_{-T}^{T} \, g(t) \, e^{-i\tau f t} \, dt \\ &= \int_{-T}^{T} \, 1 \, e^{-i\tau f t} \, dt \\ &= 2 T \operatorname{sinc}(\tau f T) \\ \end{align*}

Step (1) to (2), is just a notational convenience. GT(f)G_T(f) means G(f)G(f) in the limit that limT\lim_{T\to\infty}.

This way, the end result GT(f)G_T(f) is a function that took the form of a function that its limit hasn’t been taken yet.

Let’s evaluate it for a moment. From this form, it is clear that when f=0f=0, the function GT(0)G_T(0) evaluates to \infty and becomes really huge. This is because sinc(0)=1\operatorname{sinc}(0)=1 and the value 2T2T approaches a really huge number.

What we are not sure yet, is how this function behaves on continuous ff. Suppose that fT=nfT=n an integer, just like what we concluded in Approach A. It means that the function evaluates to zero. But what about when fTfT not equal to an integer? Especially when we are going to set TT as a really huge number.

Let’s calculate the Inverse Fourier Transform, and we are going to see from there.

g(t)=G(f)eiτftdfgT(t)=GT(f)eiτftdf=2Tsinc(τfT)eiτftdf=2Tsin(τfT)τfTeiτftdf=2eiτfTeiτfTiτfeiτftdf\begin{align*} g(t) &= \int_{-\infty}^{\infty} G(f) \, e^{i\tau f t} \, df \\ g_T(t) &= \int_{-\infty}^{\infty} G_T(f) \, e^{i\tau f t} \, df \\ &= 2 T \int_{-\infty}^{\infty} \operatorname{sinc}(\tau f T) \, e^{i\tau f t} \, df \\ &= 2 T \int_{-\infty}^{\infty} \frac{\sin(\tau f T)}{\tau f T} \, e^{i\tau f t} \, df \\ &= 2 \int_{-\infty}^{\infty} \frac{e^{i \tau f T} - e^{-i \tau f T}}{i \tau f} \, e^{i\tau f t} \, df \\ \end{align*}

Just like what we do before, we are going to take derivative with respect to tt. Since this is a constant function, the derivative should be zero.

gT(t)=2eiτfTeiτfTiτfeiτftdfddtg(t)=2eiτfTeiτfTiτfddt[eiτft]df0=(eiτfTeiτfT)eiτftdfeiτf(t+T)df=eiτf(tT)df\begin{align*} g_T(t) &= 2 \int_{-\infty}^{\infty} \frac{e^{i \tau f T} - e^{-i \tau f T}}{i \tau f} \, e^{i\tau f t} \, df \\ \frac{d}{dt} g(t) &= 2 \int_{-\infty}^{\infty} \frac{e^{i \tau f T} - e^{-i \tau f T}}{i \tau f} \, \frac{d}{dt} \left[ e^{i\tau f t} \right] \, df \\ 0 &= \int_{-\infty}^{\infty} \left( e^{i \tau f T} - e^{-i \tau f T} \right) \, e^{i\tau f t} \, df \\ \int_{-\infty}^{\infty} e^{i \tau f (t+T)} \, df &= \int_{-\infty}^{\infty} e^{i \tau f (t-T)} \, df \\ \end{align*}

Just like before, it would imply that for any tt and any TT, it seems that:

1=eiτ2fT=ei4πfT\begin{align*} 1 = e^{i \tau 2 f T} = e^{i 4\pi f T} \end{align*}

We arrived at the same contradiction. But, the situation is different now. Previously, the only resolution makes sense is statement 2, where hh needs to be the smallest non-zero value possible, with fh=nfh=n as an integer.

This time, ff is continuous, and TT is in the limit of some large number approaching to infinity (not a small number). There is a specific relation needs to be hold such that both ff and TT were dependent from each other in such a way that fT=n2fT=\frac{n}{2}.

It implies a natural constraint, such that we can set ff to be as small as possible, as long as fT=n2fT=\frac{n}{2}. If we stretch the signal to TT approaching infinity, for example, in the case of our constant function. Then its Fourier Transform domain ff is so localized that G(f)G(f) only has non-zero values around f=0f=0.

It has similar (if not the same) consequences as Approach A, albeit with entirely different reasoning.

In conclusion, GT(f)G_T(f) behaves the same way as its Fourier Series counterpart. It is zero everywhere when fT=n2fT=\frac{n}{2}, except when f=0f=0, where its value spikes to infinity.

Next, from the FT to IFT relationship, we will try to find another condition such that the transform is invertible. Suppose that at t=0t=0, the value g(t)=1g(t)=1 is a constant. We also substitute f=n2Tf=\frac{n}{2T}

gT(t)=GT(f)eiτftdf=2Tsinc(τfT)df=sinc(nπ)dn\begin{align*} g_T(t) &= \int_{-\infty}^{\infty} G_T(f) \, e^{i\tau f t} \, df \\ &= 2T \int_{-\infty}^{\infty} \operatorname{sinc}(\tau f T ) \, df \\ &= \int_{-\infty}^{\infty} \operatorname{sinc}(n \pi) \, dn \\ \end{align*}

Since there are no variable left on the right hand side. We can guess that the right hand side evaluates to 1, in order for the limit to converge with the left hand side g(t)=1g(t)=1. But we had to make sure.

Previously we evaluate a sum, and it was easy to do because for any value of nn, the function evaluates to zero, except when n=0n=0. This time nn represents a continuous variable. This last integral is a little bit tricky to solve. But we will use several steps to change the form.

In order to evaluate the integral, notice that sinc(ϕ)\operatorname{sinc}(\phi) is an even symmetric function. So we can change the integral into this:

sinc(nπ)dn=20sinc(nπ)dn\int_{-\infty}^{\infty} \operatorname{sinc}(n \pi) \, dn = 2 \int_{0}^{\infty} \operatorname{sinc}(n \pi) \, dn \\

We expand sinc(nπ)=sin(nπ)nπ\operatorname{sinc}(n \pi)=\frac{\sin(n \pi) }{n \pi}

20sinc(nπ)dn=20sin(nπ)nπdn2 \int_{0}^{\infty} \operatorname{sinc}(n \pi) \, dn = 2 \int_{0}^{\infty} \frac{\sin(n\pi)}{n\pi} \, dn

We are going to decompose 1n\frac{1}{n} as an integral: 1n=0ensds\frac{1}{n} = \int_0^\infty e^{-n s} \, ds

20sin(nπ)nπdn=2π00sin(nπ)ensdsdn2 \int_{0}^{\infty} \frac{\sin(n\pi)}{n\pi} \, dn = \frac{2}{\pi} \int_{0}^{\infty} \int_{0}^{\infty} \sin(n\pi) \,e^{-n s} \, ds \, dn

Using Fubini’s Theorem, assuming the integral converges, we swap the order of the integration.

2π00sin(nπ)ensdsdn=2π00sin(nπ)ensdnds\frac{2}{\pi} \int_{0}^{\infty} \int_{0}^{\infty} \sin(n\pi) \,e^{-n s} \, ds \, dn = \frac{2}{\pi} \int_{0}^{\infty} \int_{0}^{\infty} \sin(n\pi) \,e^{-n s} \, dn \, ds

Let’s break apart into smaller terms. Suppose that I(n)=0sin(nπ)ensdnI(n)=\int_0^\infty \sin(n\pi) \, e^{-ns} \, dn, then we solve it.

I(n)=0sin(nπ)ensdn=1s[sin(nπ)ens]0+πs0cos(nπ)ensdn=0+πs0cos(nπ)ensdnsπI(n)=1s[cos(nπ)ens]0πs0sin(nπ)ensdn=1sπsI(n)s2+π2π2I(n)=1πI(n)=πs2+π2\begin{align*} I(n) &= \int_0^\infty \sin(n\pi) \, e^{-ns} \, dn \\ &= \frac{1}{-s}\left[ \sin(n\pi) \, e^{-ns} \right]_0^\infty + \frac{\pi}{s} \int_0^\infty \cos(n\pi) \, e^{-ns} \, dn \\ &= 0 + \frac{\pi}{s} \int_0^\infty \cos(n\pi) \, e^{-ns} \, dn\\ \frac{s}{\pi} I(n) &= \frac{1}{-s}\left[ \cos(n\pi) \, e^{-ns} \right]_0^\infty - \frac{\pi}{s} \int_0^\infty \sin(n\pi) \, e^{-ns} \, dn \\ &= \frac{1}{s} - \frac{\pi}{s} I(n) \\ \frac{s^2+\pi^2}{\pi^2}I(n)&=\frac{1}{\pi} \\ I(n)&=\frac{\pi}{s^2+\pi^2} \\ \end{align*}

Substituting this result back, we have:

2π00sin(nπ)ensdnds=201s2+π2ds\frac{2}{\pi} \int_{0}^{\infty} \int_{0}^{\infty} \sin(n\pi) \,e^{-n s} \, dn \, ds = 2 \int_0^\infty \frac{1}{s^2+\pi^2} \, ds \\

That last integral is just a trigonometric integral, that can be solved by choosing s=πtan(ϕ)s=\pi \tan(\phi). So that dsdϕ=πsec2(ϕ)\frac{ds}{d\phi}=\pi \sec^2(\phi). With the integration boundaries becomes ϕ[0,π2]\phi \to [0,\frac{\pi}{2}]

201s2+π2ds=20π2πsec2(ϕ)π2(tan2(ϕ)+1)dϕ=2π0π2dϕ=2ππ2=1\begin{align*} 2 \int_0^\infty \frac{1}{s^2+\pi^2} \, ds &= 2 \int_0^\frac{\pi}{2} \frac{\pi \sec^2(\phi)}{\pi^2 (\tan^2(\phi) + 1)} \, d\phi \\ &= \frac{2}{\pi} \int_0^\frac{\pi}{2} d\phi \\ &= \frac{2}{\pi} \frac{\pi}{2} \\ &= 1 \end{align*}

With this, we concluded that indeed g(t)=gT(t)=1g(t)=g_T(t)=1. In the limit of large TT, turns out the integral converges, even without us having to specify that nn has to be integers.

So the Fourier Transform of a constant signal with arbitrarily large span TT is:

F{g(t)}=limTGT(f)F{1}=limT2Tsinc(τfT)(P9)\begin{align*} \tag{P9} \mathcal{F}\left\{ g(t) \right\} &= \lim_{T\to\infty} G_T(f) \\ \mathcal{F}\left\{1\right\} &= \lim_{T\to\infty} 2T \operatorname{sinc}(\tau f T) \\ \end{align*}

Note that, although the concept is different, when T=12T=\frac{1}{2}, we recover the Fourier Series representation of a constant function: G(k)=sinc(kπ)G(k)= \operatorname{sinc}(k \pi)

Constant function, interpreted as distribution

Originally, we have two approaches A and B. We now realized that both is similar but not equal. It appears that one approach is most suited when the time interval is very small. The other one is most suited when the time interval is very large.

Can’t we have both?

For instance, both representation of the FT, G(f)G(f) seems to behave like these:

  1. The value G(f)G(f) at f=0f=0, is either very large, dependent on the span of the domain tt, or just \infty.
  2. The value G(f)G(f) at f0f\ne0, is zero everywhere.
  3. The integral over all domain ff for G(f)G(f) is 1.

This seems to be a very weird function. How come it has value \infty, but the integral area under the curve is normalized at 11? Doesn’t look like a function to me.

In the case of Fourier Transform, this looks like an identity basis.

In the previous two approach, we started from a constant function, then we try to find its FT. However, if we go the other way around: “What is a function that its FT is a constant function?”, then we might understand something.

Suppose that there exists a function (or property, whatever), called the delta function δ(t)\delta(t). Delta function has a property in such a way that its FT D(f)D(f) is a constant 1. Can this function exists?

The reason why we called it “Delta function” is probably special in its way. It is called “Delta” because it represent a small slice of interval.

Note that it is the inverse/reverse direction of approach A & B where we assume that constant function is in the time domain. Now we assume the constant function is in the frequency domain. The only benefit that we have, is that we can immediately construct a Fourier Series δ(t)\delta(t) just by composing the frequencies.

From a Fourier Series definition if a function δ(t)\delta(t) has a Fourier Transform D(f)=1D(f)=1. Then it immediately follows:

δ(t)=1Nf=1eiτft=1Nf=eiτft\begin{align*} \delta(t) &= \frac{1}{N} \sum_{f=-\infty}^{\infty} 1 \, e^{i\tau f t} \\ &= \frac{1}{N} \sum_{f=-\infty}^{\infty} e^{i\tau f t} \\ \end{align*}

Using this kind of representation. We can figure out 3 criterias of δ(t)\delta(t) to be the same with the function G(f)G(f) we are trying to find before.

  1. When t=0t=0, δ(t)\delta(t) values approach infinity. This is because the sum happens to sum 1 from negative infinity to infinity. So this becomes a huge number.
  2. When t0t\ne0, there is a corresponding indices ff, in such a way that τft=nπ\tau f t = n\pi where nn is an integer.
  3. The integral over the domain tt must equal to 1.

This last property can be proven using integral.

t=δ(t)dt=1t=1Nf=eiτftdt=1Nf=t=eiτftdt=1Nf=limTTTeiτftdt=1Nf=limT2sin(τfT)τf=1NlimTf=2sin(τfT)τf=1NlimTn=2Tsin(nπ)nπ=1NlimT2Tn=sin(nπ)nπ=limT2TN=1limT2T=N\begin{align*} \int_{t=-\infty}^{\infty} \delta(t) \, dt &= 1 \\ \int_{t=-\infty}^{\infty} \frac{1}{N}\sum_{f=-\infty}^{\infty} e^{i\tau f t} \, dt &= \\ \frac{1}{N} \sum_{f=-\infty}^{\infty} \int_{t=-\infty}^{\infty} e^{i\tau f t} \, dt &= \\ \frac{1}{N} \sum_{f=-\infty}^{\infty} \lim_{T\to\infty} \int_{-T}^{T} e^{i\tau f t} \, dt &= \\ \frac{1}{N} \sum_{f=-\infty}^{\infty} \lim_{T\to\infty} \frac{2 \sin(\tau f T)} {\tau f}&= \\ \frac{1}{N} \lim_{T\to\infty} \sum_{f=-\infty}^{\infty} \frac{2 \sin(\tau f T)} {\tau f}&= \\ \frac{1}{N} \lim_{T\to\infty} \sum_{n=-\infty}^{\infty} \frac{2 T \sin(n \pi)} {n \pi}&= \\ \frac{1}{N} \lim_{T\to\infty} 2T \sum_{n=-\infty}^{\infty} \frac{\sin(n \pi)} {n \pi}&= \\ \lim_{T\to\infty} \frac{2T}{N} &= 1 \\ \lim_{T\to\infty} 2T &= N \end{align*}

The end result only makes sense if the equality is true. Which means the normalization factor of this delta function depends from the span of the integration, the domain of tt.

Now, here’s the connection we are trying to make.

Since δ(t)\delta(t) and D(f)=1D(f)=1 is a Fourier Dual, with g(t)=1g(t)=1 and G(f)G(f) is a Fourier Dual… It would imply that δ(t)=G(f)\delta(t)=G(f).

But…, the representation is different:

δ(x)=G(x)limT12Tf=eiτfx=limT2Tsinc(πx)\begin{align*} \delta(x) &= G(x) \\ \lim_{T\to\infty} \frac{1}{2T} \sum_{f=-\infty}^{\infty} e^{i\tau f x} &= \lim_{T\to\infty} 2T \operatorname{sinc}(\pi x ) \end{align*}

When x=0x=0:

limT12Tf=eiτfx=limT2Tsinc(πx)\begin{align*} \lim_{T\to\infty} \frac{1}{2T} \sum_{f=-\infty}^{\infty} e^{i\tau f x} &=\lim_{T\to\infty} 2T\operatorname{sinc}(\pi x ) \\ \end{align*}

These symbols must have some correspondence. Since all the values for x0x\ne 0 is the same. We can only care when limT\lim_{T\to \infty}. It should increase the same way in the case x=0x=0. Now, since both of TT depends on the span of the integration (the domain of x). We should see some relationship.

We can see that 2T2T is affected by the normalization factors. So in a sense, δ(x)\delta(x) is some sort of distribution. This distribution, can be approximated by sinc(ax)\operatorname{sinc}(ax) for any value of aa.

Suppose that this distribution exists, then it will be easier for us to define Fourier Transform pair, as a function.

Let’s explore the ideas further.

For non-periodic function, suppose there exists a distribution called δ(t)\delta(t). Because it is a distribution, its integral over its domain is normalized:

δ(t)dt=1(E1)\begin{align*} \tag{E1} \int_{-\infty}^{\infty} \delta(t) \, dt &= 1 \end{align*}

Suppose that its Fourier Transform is 11. Meaning:

F{δ(t)}=δ^(f)=1(E2)\begin{align*} \tag{E2} \mathcal{F}\left\{ \delta(t) \right\} &= \widehat{\delta}(f) = 1 \end{align*}

Now, in parallel, suppose there exists a Fourier Transform dual g(t)g(t) and G(f)G(f). Meaning, we can write g(t)g(t) like this, using inverse Fourier Transform:

g(t)=G(f)eiτftdf(E3)\begin{align*} \tag{E3} g(t) &= \int_{-\infty}^{\infty} G(f) \, e^{i\tau f t} df \end{align*}

Using the same trick like we do by finding a Fourier Transform of a derivative, now we try to find a Fourier Transform of a total integral over all of its domain. Integrate left side and right side over all values of tt.

g(t)=G(f)eiτftdfg(t)dt=G(f)eiτftdfdt\begin{align*} g(t) &= \int_{-\infty}^{\infty} G(f) \, e^{i\tau f t} \, df \\ \int_{-\infty}^{\infty} g(t) \, dt &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} G(f) \, e^{i\tau f t} \, df \, dt \\ \end{align*}