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Is there any possible periodic signal can exist(even mathematically) whose period gets change after differentiation?

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  • $\begingroup$ Including points of discontinuity or excluding them? $\endgroup$ – hops Jul 15 '20 at 6:03
  • $\begingroup$ general mathematical case possible $\endgroup$ – user215805 Jul 15 '20 at 6:17
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The point about discontinuity was meant to hint at the following example. Consider the (unbounded) signal that is defined by $$ x(t) = \lceil{t}\rceil. $$

It's derivative is $$ x'(t) = \sum_{k=-\infty}^{\infty} \delta(t-k).$$

This is periodic with period $T=1$ whereas $x(t)$ is aperiodic. Admittedly, this example is similarly impractical for most real-world applications in the same vein as the example given by V.V.T.

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  • $\begingroup$ Except that both your $\lceil t \rceil$ can be taken as a ramp plus a sawtooth, and the sawtooth is periodic. Both this an VVT's example are then of "things that can be easily decomposed into functions with periodic elements". So while, strictly speaking it's not periodic -- it's got a strong periodic character to it. $\endgroup$ – TimWescott Jul 15 '20 at 19:30
  • $\begingroup$ It is true (and interesting) that this can be decomposed into those signals and one of them is periodic with the correct period, but that doesn't change the fact that the signal $x(t)$ is not periodic (and doesn't even have a well-defined Fourier transform). So, I think this example and the example proposed by V.V.T. still meet the question criteria. I tried to offer the disclaimer that in practice unbounded signals like this aren't that interesting since they don't occur in practical systems. $\endgroup$ – hops Jul 15 '20 at 20:35
  • $\begingroup$ But then the more general answer is to take any periodic signal and add a ramp. Then poof! you have an aperiodic signal that differentiates into a periodic one. $\endgroup$ – TimWescott Jul 16 '20 at 0:21
  • $\begingroup$ I see your point. $\endgroup$ – hops Jul 16 '20 at 0:31
  • $\begingroup$ @TimWescott, hops, That's VVT's observation. An interesting fact, but the OP question starts "Is there any possible periodic signal ..." so technically, this is a aside, but I am not recommending the OP change this accepted answer. $\endgroup$ – Cedron Dawg Jul 16 '20 at 10:51
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No, in a conventional sense of a "periodic signal" phrase, but, if you permit me to delve into a math subtlety, differentiation can turn an aperiodic waveform to a periodic one: $$ \frac{\mathrm{d}}{\mathrm{d}t}(a\cdot t + b\cdot \cos(\omega\cdot t)) = a-b\cdot\omega\cdot\sin(\omega\cdot t) $$ Notwithstanding a dubious usefulness of this excursion.

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  • $\begingroup$ Is there any example of other way around i.e period changed into a periodic on differentiation ? $\endgroup$ – user215805 Jul 15 '20 at 6:26
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    $\begingroup$ Not dubious at all. Repeated differentiation (different ways, some smooth as well) is "detrending" on steroids, i.e. will ultimately flatten any polynomials while preserving sinusoidals. @user215805 the reverse, repeated accumulations will introduce polynomials to fit initial conditions. $\endgroup$ – Cedron Dawg Jul 15 '20 at 15:29
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No, it's not possible.

Proof: let $f : \mathbb{R} \to \mathbb{R}$ periodic, i.e. there exists $T\in\mathbb{R}$ such that $$f(t+T) = f(t) \quad\forall t\in\mathbb{R}.$$ Then it follows that the derivative $f'$ has the same periodicity, because $$ f'(t+T) = \lim_{h\to0}\frac{f(t+T+h)-f(t+T)}h = \lim_{h\to0}\frac{f(t+h)-f(t)}h = f'(t). $$

The other direction is a bit more subtle, because as was shown by earlier answers there are functions which aren't periodic at all yet have a periodic derivative. However, this only works for unbounded functions, which can't be signal functions. (And if $f$ is continuous and periodic with any period, then it is also bounded.)

We can make this statement: if $f$ is bounded, i.e. $f(t)\in[f_\text{min},f_\text{max}]$, and its derivative $f'$ is periodic with period $T$, then $f$ is also periodic with $T$. By the fundamental theorem of calculus: $$ f(t+T) = C + \int\limits_0^{t+T}\!\mathrm{d}\tau\: f'(\tau) = C + \int\limits_0^t\!\mathrm{d}\tau\: f'(\tau) + \int\limits_t^{t+T}\!\mathrm{d}\tau\: f'(\tau) = f(t) + \underbrace{\int\limits_t^{t+T}\!\mathrm{d}\tau\: f'(\tau)}_{=: e(t)}. $$ Lemma: $e(t) = 0$. Note that $e$ is constant, because $$\begin{align} e(t) =& \int\limits_t^{t+T}\!\mathrm{d}\tau\: f'(\tau) \\ =& \int\limits_t^{T\cdot\lceil{t/T}\rceil}\!\mathrm{d}\tau\: f'(\tau) + \int\limits_{T\cdot\lceil{t/T}\rceil}^{t+T}\!\mathrm{d}\tau\: f'(\tau) \\ =& \int\limits_{t+T}^{T\cdot\lceil{t/T+1}\rceil}\!\mathrm{d}\tau\: f'(\tau-T) + \int\limits_{T\cdot\lceil{t/T}\rceil}^{t+T}\!\mathrm{d}\tau\: f'(\tau) \\ =& \int\limits_{t+T}^{T\cdot\lceil{t/T+1}\rceil}\!\mathrm{d}\tau\: f'(\tau) + \int\limits_{T\cdot\lceil{t/T}\rceil}^{t+T}\!\mathrm{d}\tau\: f'(\tau) \\ =& \int\limits_{T\cdot\lceil{t/T}\rceil}^{T\cdot\lceil{t/T+1}\rceil}\!\mathrm{d}\tau\: f'(\tau) \\ =& \int\limits_0^T\!\mathrm{d}\tau\: f'(\tau + T\cdot\lceil{t/T}\rceil) \\ =& \int\limits_0^T\!\mathrm{d}\tau\: f'(\tau) =: e_0. \end{align}$$ Now assume $e_0$ is nonzero, without loss of generality, positive. Then it follows that $$\begin{align} f\left(T\cdot\left\lceil\frac{f_\text{max}-f_\text{min}}{e_0}+1\right\rceil\right) =& f(0) + \left\lceil\frac{f_\text{max}-f_\text{min}}{e_0}+1\right\rceil\cdot e_0 \\\geq& f_\text{min} + \left(\frac{f_\text{max}-f_\text{min}}{e_0}+1\right)\cdot e_0 \\=& f_\text{max} + e_0 \\>& f_\text{max}, \end{align}$$ which is a contradiction to the assumption that $f(t)\leq f_\text{max}\quad\forall t\in\mathbb{R}$.

Consequently, we must have $e(t) = e_0 = 0$ and therefore $f(t+T) = f(t)$.

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  • $\begingroup$ I don't care for your convention of putting the differential adjacent to the integral instead of on the outside which is the more common convention. It places a slight parsing burden on both a human and any algorithmic approach. Me, cuz I'm not used to it, algorithmically because you have to evaluate each operator and assess its operator precedence level. So I am wondering if this was an individually developed style or you picked it up somewhere, perhaps how you were taught. I'm not saying it is wrong. $\endgroup$ – Cedron Dawg Jul 17 '20 at 15:37
  • $\begingroup$ @CedronDawg it's pretty standard in theoretical physics, and i like it because when you have multiple nested integrals it's immediately clear which domain belongs to which integration variable. Also the integration variable is introduced before being used, resembling lambda calculus style, i.e. something like integral (λτ ↦ f(τ)). $\endgroup$ – leftaroundabout Jul 17 '20 at 15:44
  • $\begingroup$ I appreciate you sharing your rationale. T.P. isn't my stomping grounds except for maybe some popular sites, and usually the math doesn't surface there. So, I presume you learned it on the outside then adopted adjacent. It's just another endian issue. $\endgroup$ – Cedron Dawg Jul 17 '20 at 17:42
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Conceptually, I think this can be seen as the defining question for understanding Fourier series.

Every well enough behaved periodic signal can be expressed as a Fourier series. Which means summing up a fundamental tone and all its harmonics, frequencies of whole number multiples of the fundamental. Each harmonic, including the first, the fundamental tone, could have zero amplitude in the mix.

The derivative of each harmonic preserves its frequency and introduces none. Since the derivative of a sum is the sum of the derivatives, there is no way to introduce new frequencies.

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No.

Differentiation is a linear operator. Linear operations can't add new frequencies to a signal.

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  • $\begingroup$ You mean linear time invariant. It's perfectly possible to define a linear operator that does change frequencies, for instance $f \mapsto (t\mapsto f(2\cdot t))$ is easily shown to be a linear operator, but it doubles every frequency. It's just not a time-invariant operator. $\endgroup$ – leftaroundabout Jul 15 '20 at 14:18

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