Maulana's personal blog
Main feedMath/PhysicsSoftware Development

Fourier Series: part 2


Periodicity of a Fourier Series

As we already described in the previous article of part 1, if we consider a periodic function g(t)g(t) in the span of its period TT, we can decompose and break it down into its set of components CnC_n in which we can use it and reconstruct it back as a Fourier Series to represent the function.

Previously, we test it on a function cos(t)\cos(t) that has the same period as eite^{it}. What if the period is different? Or rather, a more important question, how do we even detect the period?

In the previous article, we found out the integral transform that lets us compute the coefficient (or component) for each basis in the Fourier Series.

Ck=1TTg(t)eiktdt(E1)\begin{align*} \tag{E1} C_k = \frac{1}{T} \int_T g(t) \, e^{-ikt} \, dt \end{align*}

As we can see, the span TT is directly tied with the unit of tt used by function g(t)g(t). Suppose that we want to define a complete period TT by having a value of 11, to denote its meaning as a complete cycle. Then we would have to redefine the parameter tt, or the transform itself. Suppose we don’t want to redefine tt because we want to have the function g(t)g(t) unaltered.

Then, if we want to keep the same inner product rule of Fourier basis, with symmetry around t=0t=0, then the span has to become Tinitial=12T_{initial}=-\frac{1}{2} and Tfinal=12T_{final}=\frac{1}{2}, such that the span T=1T=1. Consequently, we need to change the basis from eikte^{-ikt} into ei2πkte^{-i 2 \pi k t}. This is from a simple changes of unit from t=t2πt' = \frac{t}{2\pi}, with tt' the new parameter with ranges [12,12][-\frac{1}{2},\frac{1}{2}].

Writing 2π2\pi over and over again after this will be such a pain. So we introduce a corresponding constant called Tau τ=2π\tau = 2 \pi. One Tau τ\tau, means one complete cycle.

Then if we want to have the same and consistent coefficient value. We redefined the Fourier Series to use this following form.

g(t)=nNCneiτnt(E2) \begin{align*} \tag{E2} g(t) = \sum_n^N C_n \, e^{i \tau n t} \end{align*}

Now that we defined a standard “period” or “cycle” using unit τ\tau. We can redefine the series for arbitrary period, relative to this τ\tau unit.

Suppose that a function g(t)g(t) has period TT that is not 11. It can be less or greater. We actually want to find this period value.

The catch is, for any periodic function that has multiple harmonics, it will have multiple set of CnC_n value. Probably, it is not symmetric as well. Each CnC_n will have its corresponding period in this case. So let’s say we want to figure out the main or principal period. This has to be the period of C0C_0.

Fortunately, we have two ways of calculating C0C_0. One way, by setting the value k=0k=0, before doing the integration. So we have our regular integral under the curve g(t)g(t), for all tt:

C0=Tg(t)eiτktdt=Tg(t)dt\begin{align*} C_0 &= \int_T g(t) \, e^{-i \tau k t} \, dt \\ &= \int_T g(t) \, dt \\ \end{align*}

The second way, is by doing the integration first, then setting k=0k=0.

C0=Tg(t)eiτktdt=G(k,T)=G(0,T)\begin{align*} C_0 &= \int_T g(t) \, e^{-i \tau k t} \, dt \\ &= G(k, T) \\ &= G(0, T) \\ \end{align*}

From the integration, we will have an anti-derivative G(k,T)G(k, T) as a function of kk and TT, the span of the period. From the equality between the two results, we can calculate the principal period.

Since this relies on having C0C_0 defined as the total integral of g(t)g(t) over all values of tt. It is important that the integral actually converges for any tt.

A classic example would be a damping oscillator that takes form of g(t)=Aeλteiωtg(t) = A e^{-\lambda t} e^{i\omega t}.

Let’s calculate its principal coefficient C0C_0 using its total area integral under the curve, from t=0t=0 to t=t=\infty.

C0=Tg(t)eiτktdt=TAeλteiωtdt=[Aiωλeλteiωt]0=Aλiω\begin{align*} C_0 &= \int_T g(t) \, e^{-i \tau k t} \, dt \\ &= \int_T A e^{-\lambda t} e^{i\omega t}\, dt \\ &= \left[ \frac{A}{i\omega - \lambda} e^{-\lambda t} e^{i\omega t} \right]_{0}^\infty \\ &= \frac{A}{\lambda - i\omega} \end{align*}

Then we calculate using the integral transform while keeping kk as variable. Then we use the period TT, with initial value t=0t=0.

C0=Tg(t)eiτktdt=TAeλteiωteiτktdt=[Ai(ωτk)λeλtei(ωτk)t]0TAλiω=Aλi(ωτk)+Ai(ωτk)λeλTei(ωτk)T\begin{align*} C_0 &= \int_T g(t) \, e^{-i \tau k t} \, dt \\ &= \int_T A e^{-\lambda t} e^{i\omega t}\, e^{-i \tau k t} dt \\ &= \left[ \frac{A}{i(\omega-\tau k) - \lambda} e^{-\lambda t} e^{i(\omega - \tau k) t} \right]_{0}^T \\ \frac{A}{\lambda - i\omega} &= \frac{A}{\lambda - i(\omega -\tau k)} + \frac{A}{i(\omega-\tau k) - \lambda} e^{-\lambda T} e^{i(\omega - \tau k) T} \end{align*}

If we let any terms without TT be evaluated with k=0k=0, we will have:

Aλiω=Aλi(ωτk)+Ai(ωτk)λeλTei(ωτk)T0=eλTei(ωτk)T\begin{align*} \frac{A}{\lambda - i\omega} &= \frac{A}{\lambda - i(\omega -\tau k)} + \frac{A}{i(\omega-\tau k) - \lambda} e^{-\lambda T} e^{i(\omega - \tau k) T} \\ 0 &= e^{-\lambda T} e^{i(\omega - \tau k) T} \end{align*}

We are guessing that the above equation can be true for multiple of TT. The easiest trivial one is at the limit of T=T=\infty, in which the expression eλTe^{-\lambda T} will be so small, it can be evaluated as zero.

The other set is to solve the second expression: ei(ωτk)T=0e^{i(\omega - \tau k) T} = 0, which is not possible. So the principal period when k=0k=0 is actually T=T=\infty.

It was unfortunate that we actually contradict our original assumptions. But hey, that is called progress. So, even though we can predict that the damping oscillator will have an oscillating frequency for ω\omega it turns out the principal coefficient C0C_0 from the function can only be found when we assume T=T=\infty, the period to be really big.

Relearning what it means for a Period and Frequency in a Fourier Series

Since our assumption is wrong. We need to rethink what went wrong.

If we take another simple example, such as cos(t)\cos(t), we already know that its non-zero coefficient is in fact C1C_1 and C1C_{-1}. The C0C_0 is zero. So what does it mean for this kind of case?

From what we understand if we set k=0k=0, the inner product transform has become integral under the curve

C0=Tg(t)eiτktdt=Tcos(t)dt=sin(t)T\begin{align*} C_0 &= \int_T g(t) \, e^{-i \tau k t} \, dt \\ &= \int_T \cos(t) \, dt \\ &= \sin(t) |_T \end{align*}

Although the values were convergent, it is very dependent on TT. For example, if the ranges were [π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], the integral value is 2. But the function sin(t)\sin(t) is clearly periodic on T=2πT=2\pi, so why we can’t retrieve this value from the transform?

If we look at the definition of “periodic function”, it means any function that satisfies g(t)=g(t+T)g(t) = g(t+T) for a given TT. Important point is that TT is only affected by the “length” of the span, it is not affected by the starting point of the function. Consequently, any integer multiple of TT will also work. So that: g(t)=g(t+T)=g(t+mT)g(t) = g(t+T) = g(t+mT), when mm is an integer.

But, if we use the Fourier Series definition from E2, TT is a direct overlap from the period of the exponential function.

g(t)=g(t+mT)nNCneiτnt=nNCneiτn(t+mT)\begin{align*} g(t) &= g(t+mT) \\ \sum_n^N C_n \, e^{i \tau n t} &= \sum_n^N C_n \, e^{i \tau n \left(t + mT \right)} \end{align*}

Since the coefficient CnC_n will be the same between left hand side and right hand side, the only factor to make the equality is when:

1=eiτnmT\begin{align*} 1 &= e^{i \tau n m T } \end{align*}

Above equality only happens when nmTnmT is an integer, because τ=2π\tau = 2 \pi. If nmTnmT is not an integer, eiτnmTe^{i \tau n m T } becomes a complex number with real and imaginary component, clearly not equal to 11.

Suppose that TT is the fundamental period we are trying to find. The smallest non-zero period possible. Then mTmT is an integer multiple of the period. Let’s call it P=mTP = mT. Then, we have a reciprocal relationship between nn and PP. We also redefined connection between the index of the Fourier Series nn (as seen in E2), with the index of the integral transform kk (as seen in E1).

Our motivation is of course to make n=kn=k, so that we have a direct relationship between period used by the function g(t)g(t) in E2E2, with the span of the integration in E1E1.

In the case of n=kn=k, the connection is from the relation of the integral under the curve to retrieve C0C_0 from both E2 and E1.

Our problem is now the scaling for value C0C_0. If we use the transform and set k=0k=0, we got integral under the curve. The range of the integration PP can be broken down into mm partition of integrals with range TT each. So, essentially C0C_0 for range PP now becomes mm times the C0C_0 for range TT. It has different meaning, as illustrated below:

CP,0=Pg(t)eiτktdt=tt+Tg(t)eiτktdt+t+Tt+2Tg(t)eiτktdt+...=mTg(t)eiτktdt=mCT,0\begin{align*} C_{P,0} &= \int_P g(t) \, e^{i\tau k t } \, dt \\ &= \int_t^{t+T} g(t) \, e^{i\tau k t } \, dt + \int_{t+T}^{t+2T} g(t) \, e^{i\tau k t } \, dt + ... \\ &= m \int_T g(t) \, e^{i\tau k t } \, dt \\ &= m C_{T,0} \\ \end{align*}

Which is clearly inconsistent, since we have just defined that the Fourier Series representation will use the same coefficient C0C_0. To make it consistent for arbitrary period PP, the form of the integral transform need to be patched with normalization factor PP again.

Ck=1PPg(t)eiτktdt(E3)\begin{align*} \tag{E3} C_k = \frac{1}{P} \int_P g(t) \, e^{-i\tau k t} \, dt \end{align*}

This time, the meaning has slightly changed.

  • TT is the smallest non-zero period that served as the range of integration, the value can be arbitrary
  • PP is the span/length of the integration. It is like a test/window of evaluation to get the corresponding frequency kk
  • kk is an integer that corresponds to the “frequency”, and a reciprocal of set of values of mTmT.

This time, regardless with the arbitrary ranges PP, our Fourier series is now applicable for any linear combination of periodic function. If the coefficient exists for an integer frequency kk, that means a corresponding period Tk=mTT_k = m T exists.

Note that for the above convention to work, the description of the relation can’t be standalone. An integer kk would mean that PP is greater than TT. For example, suppose that we have a function cos(2πt)\cos(2\pi t). Using a range of integration PP from [5,5][-5,5] for parameter tt, we will have k=1k=1 results in C1=0C_1=0, if using formula E3E3.

However, from constructing the Fourier series directly, n=1n=1 means C1=12C_1=\frac{1}{2} if using formula E2E2. The index now doesn’t match.

Matching the index of Fourier Series with the index of the integral transform

If we want to have some kind of “inverse transform” relationship, the index nn and kk need to actually match the scale. Not just when both are 0.

Our new formula E3E3 now allows us to identify the smallest non-zero period TT. From there, we can match the scale of “frequency” kk to the index of the Fourier series term nn.

The smallest non-zero period TT should be able to be found, if we let kk to be a real number, not just an integer.

If a “cycle” in Fourier series term means nTnT, then a “cycle” in the integral transform corresponds to kPkP. This is because the following relation will be true for each basis used between formula E2E2 and E3E3

eiτnT=eiτkPe^{i \tau n T} = e^{i \tau k P}

Naturally, if we let nn to be integers, then kk has to be between 0 and 1 (a real number, not integer), because PP is much bigger than TT. The idea is that the right hand side uses integral transform as some kind of sampling process. So the data that going to be integrated has to be a much denser set than the smallest period the signal can have.

Rearranging things, we will have:

nP=kT\frac{n}{P} = \frac{k}{T}

Let us let the left hand side consists of only integers (both for nn and PP), and let the right hand side corresponds to real numbers (both for kk and TT). Since kT\frac{k}{T} is also a real number, we can even rename it as just one real number variable. We can call this the “continuous” version of “frequency”, and we can name it f=kTf=\frac{k}{T}.

Also, to make life simpler, we call the integral transform, a “Fourier Transform”. Which can take the following form in our current derivation:

The integer (nn and PP) parameter form:

Cn=1PPg(t)eiτnPtdtC_n = \frac{1}{P} \int_P g(t) \, e^{-i\tau \frac{n}{P} t} dt

The real number (kk and TT) parameter form:

Ck=1PPg(t)eiτkTtdtC_k = \frac{1}{P} \int_P g(t) \, e^{-i\tau \frac{k}{T} t} dt

The real number “continuous frequency” form:

Cf=1PPg(t)eiτftdtC_f = \frac{1}{P} \int_P g(t) \, e^{-i \tau f t} dt

We should be able to construct the corresponding Fourier Series, as long as we used the same span/length and index name for the terms.

These two operations is shown to be inverse of each other (proof maybe later on). The “Fourier Transform” is known as the “analysis” process, because it breaks down signals/functions into its corresponding set of coefficient of frequencies. Meanwhile, the “Fourier Series” is known as the “synthesis” process, because it uses the frequency sets to construct back the original signals/functions.

Extending Fourier Transform/Series pairs into non-periodic function

Previously, we had an accidental hint that Fourier Transform might work for non-periodic function, as long as the C0C_0 coefficient converges. If we think about it, the concept does make sense. C0C_0 means coefficient for the “Zero-th” frequency. So, the fundamental period/frequency is actually “no period/frequency applicable” for the previous decayed oscillation we discussed before. Meaning, as a whole, the function itself is not periodic, even though it oscillates (have other sets of frequencies).

While the Fourier Transform itself is computable for this kind of function. The “synthesis” part, the Fourier Series, doesn’t make any sense anymore. This is because there are no “fundamental frequency” to build the series. The TT just doesn’t exist because for such function, it can be set to an arbitrarily large length, since we never found the periodicity using previous method.

We will now try to understand what is the equivalent “synthesis” part of the duality for non-periodic function.

In this case, let PP be an infinite span. For a specific reason, this has to be an integer for now, so bear with me. Also, in this case, it is necessary to declare that T=PT=P, because the function is non-periodic (aperiodic).

We will use the “continuous” version of the Fourier Transform. This is because due to the reciprocity between period and frequency, “infinite period” means “infinitely small frequency”. So the frequency index can’t be an integer anymore.

Cf=1PPg(t)eiτftdt=limT1TT2T2g(t)eiτftdt(E4)\begin{align*} \tag{E4} C_f &= \frac{1}{P} \int_P g(t) \, e^{-i \tau f t} dt \\ &= \lim_{T \to \infty} \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} g(t) \, e^{-i \tau f t} dt \\ \end{align*}

To make sense of the above transform. First, imagine that the period TT is initially small. But we will gradually increase it. This, in turn will cause CfC_f to be increasingly small, since the normalization factor 1T\frac{1}{T} gets huge.

Now we want to construct the “synthesis” part. But first we began to construct it using integer for small value of TT. For each term, using integer index means the spacing between each successive value of nn is equal to Δn=1\Delta n = 1. If we now uses the “continuous” version of the index ff, spacing between successive terms becomes much denser, with each difference become Δf\Delta f, increasingly small.

Note that the relation nT=kPnT=kP has to be maintained.

nT=kPn=kTP=fPΔn=ΔfP\begin{align*} nT &= kP \\ n &= \frac{k}{T} P = fP \\ \Delta n &= \Delta f P \\ \end{align*}

If we write out the Fourier Series:

g(t)=nNCneiτnPt=nNCneiτnPtΔn=kCkeiτkTtΔn=fCfeiτftΔn=limΔf0fCfeiτftPΔf\begin{align*} g(t) &= \sum_n^N C_n \, e^{i\tau \frac{n}{P} t} = \sum_n^N C_n \, e^{i\tau \frac{n}{P} t} \Delta n \\ & = \sum_k C_k \, e^{i\tau \frac{k}{T} t} \, \Delta n = \sum_f C_f \, e^{i\tau f t} \, \Delta n \\ & = \lim_{\Delta f \to 0} \sum_f C_f \, e^{i\tau f t} \, P \, \Delta f \\ \end{align*}

Substituting CfC_f as TPT \to P and PP \to \infty.

g(t)=limΔf0,TP,PfPT[T2T2g(t)eiτftdt]eiτftΔf\begin{align*} g(t) & = \lim_{\Delta f \to 0, \, T \to P, \, P \to \infty} \sum_f \frac{P}{T} \left[ \int_{-\frac{T}{2}}^{\frac{T}{2}} g(t) \, e^{-i \tau f t} dt \right] \, e^{i\tau f t} \, \Delta f \\ \end{align*}

Under above condition, we can simplify things by taking into the limit:

Let’s introduce another function F(f)F(f).

Cf=limT1TT2T2g(t)eiτftdtF(f)=limTTCf=limTT2T2g(t)eiτftdtF(f)=g(t)eiτftdt(E5)\begin{align*} \tag{E5} C_f &= \lim_{T \to \infty} \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} g(t) \, e^{-i \tau f t} dt \\ F(f) &= \lim_{T \to \infty} T \, C_f = \lim_{T \to \infty} \int_{-\frac{T}{2}}^{\frac{T}{2}} g(t) \, e^{-i \tau f t} dt \\ F(f) &= \int_{-\infty}^{\infty} g(t) \, e^{-i \tau f t} dt \end{align*}

The Fourier Series/Sum now can be converted into an integral

g(t)=limΔf0,TP,PfPT[T2T2g(t)eiτftdt]eiτftΔf=limΔf0F(f)eiτftΔf=F(f)eiτftdf(E6)\begin{align*} \tag{E6} g(t) & = \lim_{\Delta f \to 0, \, T \to P, \, P \to \infty} \sum_f \frac{P}{T} \left[ \int_{-\frac{T}{2}}^{\frac{T}{2}} g(t) \, e^{-i \tau f t} dt \right] \, e^{i\tau f t} \, \Delta f \\ &= \lim_{\Delta f \to 0} F(f) \, e^{i\tau f t} \, \Delta f \\ &= \int_{-\infty}^{\infty} F(f) \, e^{i\tau f t} \, df \\ \end{align*}

Both the “analysis” and “synthesis” part now becomes integral transform and dramatically took a similar form.

The Fourier Transform now converts a function into another function with different domain. Since it depends on frequency ff, we usually call it the function on frequency domain.

The Fourier Series now takes form of another integral transform that can be called an Inverse Fourier Transform. The function is in the time domain.

Although we call it frequency and time domain, for non-periodic function, there is no canonical identity that lets us determine whether a function comes from a Fourier Transform (FT) or an Inverse Fourier Transform (IFT). This is because if both of them were complex function, the domain are the same.

Although, historically for physics related problem, a function in real number space is usually called function in the time domain. Meanwhile, a function that is in the complex number space (which has frequency and phase information) is called function in the frequency domain. For signal processing, it is very convenient to express a function in their frequency domain.

Constructing periodic Fourier Series from the non-periodic Fourier Transform

Our notion of the transformation now changes from finding coefficients of integer indices into finding a function in the frequency domain.

Surprisingly, the non-periodic formula of FT and IFT works really well for periodic function too. We just need to take into account the appropriate normalization scaling.

Let’s take an example of a function cos(ωt)\cos(\omega t).

First, we tried to determine whether this is periodic or non-periodic, from the FT at frequency 0. Suppose that the function is periodic, then there exists the smallest non-zero interval TT, in which the FT can be decomposed into a partition of nn integral with the same TT interval.

Its FT, the frequency-domain function:

F(f)=cos(ωt)eiτftdt=limnntt+Tcos(ωt)eiτftdtF(0)=limnntt+Tcos(ωt)dt=limnnω[sin(ωt+ωT)sin(ωt)]\begin{align*} F(f) &= \int_{-\infty}^{\infty} \cos(\omega t) \, e^{-i \tau ft} \, dt \\ &= \lim_{n \to \infty} n \, \int_{t}^{t+T} \cos(\omega t) \, e^{-i \tau ft} \, dt \\ F(0) &= \lim_{n \to \infty} n \, \int_{t}^{t+T} \cos(\omega t) \, dt \\ &= \lim_{n \to \infty} \frac{n}{\omega} \, \left[ \sin(\omega t + \omega T) - \sin(\omega t) \right] \\ \end{align*}

The FT with frequency 0 corresponds to the average/mean of the area of the integration. For periodic function, the average has to be the same for any integer multiple of TT, the smallest non-zero period. In this situation, the average area limnF(0)n\lim_{n\to \infty} \frac{F(0)}{n} has to be constant under changes of tt. So the derivative with respect to tt is zero.

Dt[limnF(0)n]=01ωDt[sin(ωt+ωT)sin(ωt)]=0cos(ωt+ωT)=cos(ωt)ωT=2πT=2πω\begin{align*} D_t \left[ \lim_{n\to \infty} \frac{F(0)}{n} \right] = 0 \\ \frac{1}{\omega} \, D_t \left[ \sin(\omega t + \omega T) - \sin(\omega t) \right] &= 0\\ \cos(\omega t + \omega T) = \cos(\omega t) \\ \omega T = 2 \pi \\ T = \frac{2 \pi}{\omega} \end{align*}

We can then reduce the range of the FT. Instead of integrating from -\infty to \infty. We can use πω-\frac{\pi}{\omega} to πω\frac{\pi}{\omega}

The FT then becomes:

F(f)=πωπωcos(ωt)eiτftdt=[1ωsin(ωt)eiτft]πωπω+iτfωπωπωsin(ωt)eiτftdt=0+[iτfω(1)ωcos(ωt)eiτft]πωπω+(iτf)2ω(1)ωπωπωcos(ωt)eiτftdt=iτfω2[eiτfπωeiτfπω]+(τf)2ω2F(f)=2τfω2sin(τfπω)+(τf)2ω2F(f)ω2F(f)(τf)2F(f)=2τfsin(τfπω)F(f)=2τfω2(τf)2sin(τfπω)\begin{align*} F(f) &= \int_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}} \cos(\omega t) \, e^{-i \tau ft} dt \\ &= \left[\frac{1}{\omega} \, \sin(\omega t) \, e^{-i \tau ft} \right]_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}} + \frac{i \tau f}{\omega} \int_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}} \sin(\omega t) \, e^{-i \tau ft} dt \\ &= 0 + \left[\frac{i \tau f}{\omega} \, \frac{(-1)}{\omega} \, \cos(\omega t) \, e^{-i \tau ft} \right]_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}} + \frac{(i \tau f)^2}{\omega} \, \frac{(-1)}{\omega} \, \int_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}} \cos(\omega t) \, e^{-i \tau ft} dt \\ &= - \frac{i \tau f}{\omega^2} \, \left[ e^{-i\tau f \frac{\pi}{\omega}} - e^{i\tau f \frac{\pi}{\omega} } \right] + \frac{(\tau f)^2}{\omega^2} \, F(f) \\ &= \frac{2 \tau f}{\omega^2}\, \sin(\tau f \frac{\pi}{ \omega}) + \frac{(\tau f)^2}{\omega^2} \, F(f) \\ \omega^2 \, F(f) - (\tau f)^2 \, F(f) &= 2 \tau f \, \sin(\tau f \frac{\pi}{ \omega}) \\ F(f) &= \frac{2 \tau f}{\omega^2 - (\tau f)^2} \, \sin(\tau f \frac{\pi}{\omega}) \end{align*}

If we see the above function F(f)F(f), it is immediately obvious that the function will have a peak frequency when ω2(τf)2=0\omega^2 - (\tau f)^2=0. Meaning that the function will have Fourier Series coefficient at fpeak=±ωτf_{peak}=\pm\frac{\omega}{\tau}.

F(fpeak)F(f_{peak}) will have indefinite form 00\frac{0}{0}, when fpeakf_{peak}. This would imply that we should use L’Hopital’s rule to get the limit of the value.

limffpeakF(f)=2τfω2(τf)2sin(τfπω)=2τ2τ2fsin(τfπω)+2τf2τ2fτπωcos(τfπω)=1τfsin(π)2τ2fπ2τ2fωcos(π)=πω\begin{align*} \lim_{f \to f_{peak}} F(f) &= \frac{2 \tau f}{\omega^2 - (\tau f)^2} \, \sin(\tau f \frac{\pi}{\omega}) \\ &= \frac{2 \tau}{- 2 \tau^2 f} \, \sin(\tau f \frac{\pi}{\omega}) + \frac{2 \tau f}{- 2 \tau^2 f} \, \tau \frac{\pi}{\omega} \cos(\tau f \frac{\pi}{\omega}) = -\frac{1}{\tau f} \, \sin(\pi) - \frac{2 \tau^2 f \pi}{2 \tau^2 f \omega} \, \cos(\pi) \\ &= \frac{\pi}{\omega} \end{align*}

From the same function of F(f)F(f), we can also check which frequency ff that has zero value of F(f)F(f). It’s easily be fn=nωτf_n = n \frac{\omega}{\tau}. We can then define the index nn that can be used in the Fourier Series approximation. So, the Fourier coefficient set only contains 2 values: F(fpeak)F(f_{-peak}) and F(fpeak)F(f_{peak})

The Fourier Series then will be like this:

g(t)=1TnNF(fn)eiτfnt=ω2π[πωeiωt+πωeiωt]=eiωt+eiωt2=cos(ωt)\begin{align*} g(t) &= \frac{1}{T} \sum_n^N F(f_n)\, e^{i\tau f_n t} \\ &= \frac{\omega}{2 \pi} \left[ \frac{\pi}{\omega}\, e^{i \omega t} + \frac{\pi}{\omega} \, e^{-i \omega t} \right] \\ &= \frac{e^{i \omega t} + e^{-i \omega t}}{2} \\ &= \cos(\omega t) \end{align*}

Which is indeed that we recover the original function using the Fourier Series.

As for why we are using the Fourier Series instead of the IFT integral transform, this is because for periodic function it will be essentially the same for finite range of integration TT. It is also easier to compute if the set of frequencies are finite.


The journey this time is quite long, but I think it is helpful as a simulation to build ideas from first principle.

As you may now, in engineering course, we often directly use Fourier Transform and Fourier Transform Inversion Theorem directly. At least when I first learn it, it feels un-intuitive and looks like hand-waving the theorem without intuitive grounds. Sure, it is easier to use because the theorem has been proven and reformed into a much more easier presentation. But sometimes it feels like it lacks fundamental understanding on why it has to be that way.

This illustration serves as a guide on how to view math as an exploration of ideas. We build it, test it, figure out what went wrong when it doesn’t do what we expect. Then finally, we condense the theory so that it can be easily used in a much more generic situation.

Let’s have a quick recap.

We start from building upon the ideas of Fourier Series, its alternative view as inner product orthogonal basis, and its integral transform to compute the coefficient. We then arrived at a much more generic formula that can be reduced into Fourier Series for a much more specific case.

Given a complex function with time parameter tg(t)t \to g(t). If the integral converges over its domain tt, then there exists a Fourier Transform FT:

F(f)=g(t)eiτftdt(T1)\begin{align*} \tag{T1} F(f) = \int_{-\infty}^{\infty} \, g(t) \, e^{-i\tau f t} \, dt \end{align*}

With F(f)F(f) another complex function, but on a frequency domain with parameter ff. The inverse transform: Inverse Fourier Transform IFT also exists:

g(t)=F(f)eiτftdf(T2)\begin{align*} \tag{T2} g(t) = \int_{-\infty}^{\infty} \, F(f) \, e^{i\tau f t} \, df \end{align*}

In the case of periodic function, we can find the smallest periodic interval TT, by finding the average value of the integration. Suppose F(f,n,T)F(f,n, T) is the function in frequency domain with boundary/interval parameter TT, and number of integer partitions nn.

F(f,n,T)=limnntt+Tg(t)eiτftdt(T3)\begin{align*} \tag{T3} F(f, n, T) = \lim_{n \to \infty} \, n \int_{t}^{t+T} \, g(t) \, e^{-i\tau f t} \, dt \end{align*}

If there exists TT as the smallest non-zero period. Then the zero-th frequency is the average/mean of integral under the curve. For periodic function, this value has to be constant under parameter tt. This can be expressed as:

ddt[limnF(0,n,T)n]=0(T4)\begin{align*} \tag{T4} \frac{d}{dt} \left[ \lim_{n\to\infty} \frac{F(0, n, T)}{n}\right] = 0 \end{align*}

Solve T4T4 to get the period TT. Use the boundary to perform FT. Construct discrete indices nn in which frequency fnf_n is defined. We can then approximate the Fourier Series for periodic function gT(t)g_T(t) using:

gT(t)=1TnNF(fn)eiτfnt(T5) \begin{align*} \tag{T5} g_T(t)=\frac{1}{T} \sum_n^N F(f_n)\, e^{i\tau f_n t} \end{align*}

For finite set of frequency spectrum F(fn)F(f_n), it can be shown that above approximation is equal to the original function, g(t)=gT(t)g(t)=g_T(t).

From above interpretation, we can then interpret the difference between Taylor approximation and Fourier approximation.

Taylor series is an approximation of the function at only the point tt of the evaluation. Fourier series is an approximation of the function at a given interval TT of the evaluation.

In this sense, Fourier Series can be thought of as a “compression” technique for an information about the function. For any periodic function with infinite domain, there exists a small subset of interval TT, in which we can sample the function to retrieve the frequency information. This discrete set of frequency is then enough to reconstruct the function for the whole intervals.

Rizky Maulana Nugraha

Written by Rizky Maulana Nugraha
Software Developer. Currently remotely working from Indonesia.
Twitter FollowGitHub followers