For time-invariant systems, if the input is periodic, the we can argue that the output is periodic:

If we shift the input signal one period, we come up with the same signal as before, and since the output must be shifted the same amount as a result of the system being TI, it should be periodic too.

For causal systems we know that the output signal at time $t$ depends on the input for times up to $t$. My question is, can we conclude that for causal TI systems if the input is periodic before $t_0$, the output will be periodic up to $t_0$ too?


2 Answers 2


Your line of reasoning is correct. For a (not necessarily linear) time-invariant system the following always holds: if $y(t)$ is the response to an input signal $x(t)$, then $y(t+T)$ must be the response to the shifted input signal $x(t+T)$. If the input is periodic with period $T$, this implies that $x(t)=x(t+T)$, and, consequently, $y(t)=y(t+T)$, i.e. the output signal is periodic with the same period as the input signal.

Furthermore, if the system is causal and the input up to time $t_0$ is periodic, the output up to time $t_0$ must be periodic too, because a causal system cannot know the (possibly non-periodic) input beyond time $t_0$.

This is (part of the) theory. But then comes @MBaz with his "counter-example" of an ideal integrator. The problem here is indeed, as already noted in a comment, the integrator's instability. Showing the time-invariance of a linear system involves manipulating the convolution integral. However, if for the given input signal that integral is divergent, the proof becomes invalid. And this is the case with systems having poles on the $j\omega$-axis (or, for discrete-time systems, on the unit circle). If an LTI system has a pole on the $j\omega$-axis and if you excite it with the corresponding pole frequency, you get an output signal with increasing amplitude, which is obviously non-periodic. You can also see it in the following way: the response of an LTI system to a complex exponential $e^{j\omega_0 t}$ is

$$y(t)=H(\omega_0)e^{j\omega_0 t}\tag{1}$$

Eq. (1) of course requires that $H(\omega_0)$ can be evaluated. For an ideal integrator this is not the case for $\omega_0=0$. Note that if the input signal to the integrator had no DC component, then the response to a periodic input signal would indeed be periodic.

  • $\begingroup$ I'm not convinced. The integrator's linearity doesn't play a role in my example; I think the inegrator's time invariance can be proven without using the convolution integral. Furthermore, the question does not require the system to be stable. In my counter-example, the integrator is a time-invariant system with a monotonically increasing output, which (despite's @Dilip's handwaving) cannot be considered periodic, IMO. $\endgroup$
    – MBaz
    Dec 31, 2014 at 18:22
  • $\begingroup$ @MBaz: The output of the integrator is infinite for a periodic input signal with a DC component. So there's not much sense in discussing its periodicity. The integrator is of course linear and time-invariant, and its time-invariance can be easily shown as long as its output doesn't become infinite. And it will certainly produce a periodic output for any periodic input with a zero DC component. In sum, for a periodic input signal, its response is either periodic or infinite. By the way, I didn't mean to criticize your example, I actually liked it because it's an interesting special case. $\endgroup$
    – Matt L.
    Dec 31, 2014 at 18:37
  • $\begingroup$ That summary I can agree with. I didn't feel criticized; thanks for your comment. I was just worried about being misunderstood. $\endgroup$
    – MBaz
    Dec 31, 2014 at 18:57

I believe the answer is no: a periodic input does not imply a periodic output.

An example: let the input be a square wave with amplitudes 1 and 0, and let the time-invariant system be an integrator. The system's output is monotonic, and thus not periodic.

  • $\begingroup$ Thanks for the answer. But the example system you provided is not BIBO stable, and I think can not be a legitimate example. $\endgroup$
    – user215721
    Dec 31, 2014 at 6:38
  • $\begingroup$ To echo the OP's comment, hand-wavingly, since the periodic square wave input has been there since $-\infty$, the output for all time is also infinite, and technically does satisfy the condition $y(t+T) = y(t)$ for all $t$. $\endgroup$ Dec 31, 2014 at 14:56
  • $\begingroup$ @user215721, you do not require the system to be BIBO stable in your question. $\endgroup$
    – MBaz
    Dec 31, 2014 at 18:25
  • $\begingroup$ @Dilip, I don't agree that this can be hand-waved so easily: the output can be considered monotonically increasing, so not periodic. See also my comment to Matt's answer. $\endgroup$
    – MBaz
    Dec 31, 2014 at 18:26
  • $\begingroup$ So what you are saying is that with $x(t)$ being your square wave with amplitudes $0$ and $1$, for $t_1 < t_2$, $$y(t_1)=\int_{-\infty}^{t_1}x(t)\,\mathrm dt \leq y(t_2) = \int_{-\infty}^{t_2}x(t)\,\mathrm dt\tag{1}$$ because $$y(t_2) = y(t_1) + \int_{t_1}^{t_2}x(t)\,\mathrm dt\tag{2}$$ and the integral on the right side of $2$ is $\geq 0$? Isn't this the same as the schoolboy assertion that "my infinity is bigger than your infinity because I added 1 to your infinity to get mine"? $\endgroup$ Jan 1, 2015 at 15:59

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