1
$\begingroup$

Why exactly continious Fourier-series's coefficients aren't periodic like coefficients of discrete Fourier series (DFS)? $e^{-j2\pi}$ is periodic in both sequences.

$\endgroup$
  • $\begingroup$ Why do you think they should be periodic? The FS of a sine wave has just two coefficients, so it' cant be periodic. The DTF applies to a discrete signal, whose spectrum is by definition periodic. $\endgroup$ – MBaz Oct 4 '16 at 16:38
  • $\begingroup$ I understand in general, that continious fourier-series's coefficients aren't periodic. But i'm asking a little bit deeper question. What exactly makes DTF coefficients poriodic and continious FS not periodic? Where is the reason? $\endgroup$ – Дмитрий Oct 4 '16 at 16:44
1
$\begingroup$

The reason why Fourier series coefficients of continuous functions are generally not periodic is because of the continuity of the function. The discrete Fourier series coefficients are periodic because the analyzed signals are discrete. Note the duality relationship of the Fourier transform

$$\text{periodicity}\Longleftrightarrow\text{discreteness} $$

Periodicity in one domain leads to discreteness in the other domain. E.g., a periodic function has only discrete frequency components (as shown by its Fourier series). If these discrete frequency components are periodic, then, consequently, the periodic function must be discrete, i.e., it must be non-zero only at discrete times.

In his answer, Fat32 gave an example of a periodic signal with periodic Fourier series coefficients. The most general form of such a signal is

$$x(t)=\sum_{m=-\infty}^{\infty}b_m\delta\left(t-m\frac{T}{N}\right)\tag{1}$$

where the coefficients $b_m$ are essentially the discrete Fourier series coefficients of the $N$-periodic Fourier coefficients $a_k$ of the $T$-periodic signal $x(t)$:

$$b_m=\frac{T}{N}\sum_{k=0}^{N-1}a_ke^{j2\pi km/N}\tag{2}$$ $$x(t)=\sum_{k=-\infty}^{\infty}a_ke^{j2\pi kt/T}\tag{3}$$

As mentioned above, the signal $x(t)$ is of course discrete in the sense that it is non-zero only at discrete time instances $t_m=mT/N$. Any periodic signal $x(t)$ with $N$-periodic Fourier coefficients $a_k$ must be of the form $(1)$.

A derivation of Equations $(1)$ and $(2)$ can be found in this answer.

$\endgroup$
  • 1
    $\begingroup$ I disagree with $(3)$ as a "discrete in the sense that it is nonzero only at discrete time instances...." because it suggests that the values of $x(t)$ at those discrete time instances are finite real numbers; they are not. If the Fourier coefficients are a periodic sequence, then in terms of Fourier transforms, the transform is a periodic impulse train and the signal is also an impulse train. $\endgroup$ – Dilip Sarwate Oct 5 '16 at 1:11
  • $\begingroup$ i up-arrowed both Matt and @DilipSarwate. the way i would put it is in this: Uniform sampling of a signal in one domain (say, the "time domain") corresponds to periodic extension of the Fourier Transform of that signal in the reciprocal domain (say, the "frequency domain"). And, because of the symmetry of the Fourier Transform and its inverse, the converse is also just as true. $\endgroup$ – robert bristow-johnson Oct 5 '16 at 2:48
  • $\begingroup$ well, my up-arrow is locked in unless the answer is edited. Matt, you didn't quote Fat's answer verbatim, and modifications to a verbatim quote are good if they increase simplicity, conciseness, and understanding and not-so-good if the modifications to the quote decrease such. i would modify the answer to fix it, but usually Matt reverses such. so this answer gets a free and poorly-considered endorsement from me. $\endgroup$ – robert bristow-johnson Oct 5 '16 at 3:00
  • $\begingroup$ @DilipSarwate: I wouldn't say that Eq. $(3)$ implies that the function $x(t)$ is finite everywhere. If we accept the equation $$\sum_k\delta(t-kT)=\frac{1}{T}\sum_ke^{j2\pi kt/T}$$ (which is just the Fourier series representation of an impulse train) then we should also accept the more general Eq. $(3)$ for the signal $x(t)$ given by $(1)$. $\endgroup$ – Matt L. Oct 5 '16 at 7:09
  • $\begingroup$ @robertbristow-johnson: I didn't at all intend to quote or paraphrase Fat's answer. Which part of my answer would you like to modify and why? Also have a look at my response to Dilip's comment (which you seem to agree with). An impulse train does have a Fourier series, and a Fourier series representation by no means implies that the function is finite everywhere. $\endgroup$ – Matt L. Oct 5 '16 at 7:13
1
$\begingroup$

In fact here is an example of a family of signals whose continuous time Fourier series (CTFS) coefficients is periodic. The signal whose CTFS cefficients being periodic is: $$ x(t) = \sum_{k=-\infty}^{\infty} {\delta(t - k T)} $$ for which the CTFS coefficients are found as $$ c_n = \frac 1T $$ for all n. Hence the CTFS can be considered to be periodic with the period being one.

That being said however, looking at the analysis equation of CTFS $$ c_n = \frac 1T \int_{<T>} {x(t) e^{-j {n\omega}_0 t} dt } $$

The condition of periodicity on $c_n$ can be translated into the following: $$ c_n = c_{n+M} = \frac 1T \int_{<T>} {x(t) e^{-j {(n+M)\omega}_0 t} dt } $$ $$ e^{-j M{\omega}_0 t } = 1 = e^{-j2\pi m }$$ for all $t \in R$, and for $M \in Z$, which translates to: $$ M {\omega}_0 t = 2\pi m \rightarrow M = \frac {2 \pi m }{{\omega}_0 t} $$

As can be inferred, for the CTFS to be periodic with an integer period $M$, (as requested by the fact that the CTFS coefficients $c_n$ can be considered as a discrete sequence $c[n]$ for which periodicity naturally implies an integer period) a number of conditions must be met loosely stated as:

1- ${\omega}_0$ being a rational multiple of $\pi$

2- $t$ being a rational multipe of integer m.

(actually it is ${\omega}_0 t$ which should be a rational multiple of $\pi$, so that $M$ can be integer)

The example I have provided (continuous time periodic impulse train) thanksfully satisfies the two conditions as: $$ {\omega}_0 = \frac {2\pi}{T} $$ and $$ t = k T $$ due to the presence of impulse sampling on the time argument t. From which the period can be deduced as: $$ M = \frac{2\pi m}{ \frac {2\pi}{T} kT } = \frac mk \rightarrow M= 1 $$ , as the smallest integer to satisfy it for any arbitrary integers $k$ and $m$.

However for most other functions, such will not be the case and hence their CTFS coefficients will not be periodic. And indeed the given example of the CT periodic impulse train is a mathematical convenience to model (describe) the behaviour of sampling of continuous time signals and hence is closely associated with the discrete time signals, whose DFS are always periodic !

$\endgroup$
  • $\begingroup$ You correctly gave an example of a signal with periodic Fourier series coefficients. The last paragraph of your answer seems to imply that that signal is the only signal with that property. However, it is just a special case. The most general form of such a signal is given by Eq. (1) in my answer. Also note that $\omega_0$ (and, consequently, the period of $x(t)$) can be chosen arbitrarily. You seem to imply that there's a restriction on $\omega_0$, but I might be misinterpreting your answer here. $\endgroup$ – Matt L. Oct 4 '16 at 21:43
  • $\begingroup$ @MattL. Thanks for the comment. I have editted the answer according to your last statement. My initial aim was just to give a counter example, therefore it's a little loose on the assertion of conditions that is required for the periodicity of CTFS... Your answer apparently yields a broader perspective. $\endgroup$ – Fat32 Oct 4 '16 at 22:33
  • $\begingroup$ @Fat32 I think your answer for continuous time FT + mine for discrete time FT form a complete and sufficient answer to the question being asked, in the sense that they together tell why each one is or isn't periodic, and how the one which is naturally not periodic can become periodic. $\endgroup$ – msm Oct 11 '16 at 4:13
0
$\begingroup$

If the Fourier series coefficients were to be a periodic sequence, then the signal would have infinite power (Parseval's theorem!). This is not a situation of interest to most people and so the subject remains unexplored for the most part.

$\endgroup$
  • $\begingroup$ well, unless we wade into the weeds regarding the nature of the Dirac impulse function, this is a situation of interest to most people here and it has (very simple) periodicity in the Fourier coefficients: $$ s(t) = T \sum\limits_{n=-\infty}^{+\infty} \delta(t - nT) $$ the Fourier coefficients are all equal to 1, which is periodic. $\endgroup$ – robert bristow-johnson Oct 5 '16 at 3:06
  • $\begingroup$ @robertbristow-johnson But then, $x(t)$ is not a continuous-time function in the usual sense of the word, is it? The question asked is "Why can't the Fourier series coefficients of a continuous-time periodic signal be a periodic sequence?" I think that is reasonable to assume that the periodic signal satisfies the usual conditions: finite power, finite number of discontinuities per period, etc, and my answer says that if the Fourier series coefficients are a periodic sequence, then the signal would have infinite power, not finite power, and so we are extrapolating beyond the chosen model. $\endgroup$ – Dilip Sarwate Oct 6 '16 at 1:47
  • $\begingroup$ yeah, Dilip, you're right, but i might ask: "is the dirac impulse a "continuous-time function"? (not a "continuous function". we all know that $\delta(t)$ is not that.) it's all about how far into the weeds we go regarding the Dirac delta function. $\endgroup$ – robert bristow-johnson Oct 6 '16 at 2:27
  • $\begingroup$ @robertbristow-johnson The Dirac delta or impulse is not a function, period, and writing a impulse train as a series cf. MattL's comment on his answer asking us to accept $$\sum_k \delta(t-kT) = \frac 1T \sum_k \exp(j2\pi kt/T)$$ is fraught with problems. That series converges to $0$ at non multiples of T, and diverges at multiples of $T$ getting us back to "$\delta(t)$ has value $\infty$ when $t=0$ and value $0$ when $t \neq 0$", things that were taught routinely 60 years ago in undergraduate EE classes but not are not so common. $\endgroup$ – Dilip Sarwate Oct 6 '16 at 2:47
  • $\begingroup$ well, Dilip, this is what wading into the weeds is about. you saw my question to the math SE about that. $\delta(t)$ is not a "function" from the POV of pure, legitimate mathematics. but engineers and physicists (and likely other scientists) have been treating $\delta(t)$ as a continuous-time function with a couple of properties $$ \delta(t) = 0 \quad \forall \ t\ne 0 $$ and $$ \int\limits_{-\infty}^{\infty} \delta(t) \, dt = 1 $$ for decades without our maths blowing up. and we do say the dirac comb is also a function and a periodic one (with a Fourier series) and it works, too. $\endgroup$ – robert bristow-johnson Oct 6 '16 at 3:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.