I'm new to signal processing and working my way through a textbook. There is an exercise where a causal LTI system is given that responds to a rectangular pulse.

I have an exercise where a causal LTI system responds for a rectangular pulse

$$ s(t) = \Pi(t) = \operatorname{rect}(t) $$

with a triangular pulse.

$$ g(t) = \Lambda(2t) = \operatorname{tri}(2t)$$

One task is to find out what the impulse response for this system is. The solution is

$$h(t) = \frac{d}{dt} h_{\epsilon}(t) $$

because $$\delta(t) = \frac{d}{dt} \epsilon(t)$$

where $\epsilon(t)$ is the unit step function, $\delta(t)$ the unit impulse function.

and looks like this

enter image description here

I don't really understand the solution and how it turned to an endlessly oscillating rect pulse. Unfortunately, the textbook doesn't explain its approach. So, why does the output signal looks like this?


enter image description here

Translation: A causal LTI system responds to a rectangular pulse $s(t) = \operatorname{rect}(t)$ with $g(t) = \Lambda(2t)$

a) What is the response for $s_1(t) = \operatorname{rect}(\frac{t-1}{2})$?

b) What is the unit step response?

c) What is the impulse response?

And the solutions for a) - c) from the textbook

enter image description here

  • $\begingroup$ Please define all symbols you are using. $\endgroup$
    – Hilmar
    Apr 18, 2021 at 13:46
  • $\begingroup$ Done. Sorry, I thought the symbols were uniformly defined in signal processing. $\endgroup$ Apr 18, 2021 at 14:00
  • $\begingroup$ did i hit the definitions well enough, @Hilmar? $\endgroup$ Apr 18, 2021 at 17:08
  • $\begingroup$ Well, they are in my experience -- but you're not using them. Either your book or your professor is a bit idiopathic in their choice of symbols, or you're coming at this from a completely different starting point (theoretical math?). $\endgroup$
    – TimWescott
    Apr 18, 2021 at 17:10
  • $\begingroup$ At any rate -- things still aren't clear, and Robert and I both think we're experienced hands. Could you take a picture of the problem statement from the book, and post it here? Or do your best to quote it word-by-word? $\endgroup$
    – TimWescott
    Apr 18, 2021 at 17:11

1 Answer 1


EDIT : Based on RBJ's comment that the indicated the system cannot possibly be an LTI, due to the fact that the output spectrum contains frequencies that do not exist at the input spectrum..., I've looked at the input-output spectrums, and came to the following conclusion :

Given the input:

$$ s(t) = \begin{cases} { 1 ~~~,~~~-0.5 < t < 0.5 \\ 0~~~,~~~\text{otherwise} }\end{cases} \tag{1} $$

and the output: $$ g(t) = \begin{cases} { 2t+1 ~~~,~~~ -0.5 < t < 0 \\ 1-2t ~~~,~~~ ~~~~~~~~0 < t < 0.5 }\end{cases} \tag{2} $$

The corresponding spectrums (apart from linear scale factors !) are :

$$ S(\omega) = \frac{ \sin(\omega /2) }{ \omega} \tag{3}$$


$$ G(\omega) = \left( \frac{ \sin(\omega/4) }{ \omega} \right)^2 \tag{4}$$

Now, the input spectrum contains all frequencies except the zeros of the sine function which occur at $\omega = 2\pi k$, whereas the output spectrum contains all frequencies except those $\omega = 4\pi k$. Therefore, RBJ is correct in his claim that this particular input-output pair violates the LTI principle of output cannot contain frequencies that do not exist at the input.

However, in my opinion, redefining the output triangular-pulse duration as $$ \hat{g}(t) = \begin{cases} { 2t+1 ~~~,~~~ -1 < t < 0 \\ 1-2t ~~~,~~~ ~~~~~0 < t < 1 }\end{cases} \tag{5} $$

eliminates the associated problem, as the output spectrum becomes:

$$ \hat{G}(\omega) = \left( \frac{ \sin(\omega/2) }{ \omega} \right)^2 \tag{6}$$

which includes all frequencies except at those $\omega = 2\pi k$. Thus the missing frequencies at the input are also missing at the output.

Hence, I would like to assume that there was a typo at the duration of the triangular-pulse output $g(t)$, and that otherwise the analysis based on LTI remains valid... Thanks to RBJ for the reminder... EDIT-1-ENDS

EDIT-2 : However, now happens another problem. If the system is to remain LTI (so that subsequent analysis remains valid), then $\hat{g}(t)$ (the redefined $g(t)$) should be used in the remaning progress. But then the observed waveforms in the figures will be different from what's being drawn. Most notably, the step-response obtained in part-b, becomes a flat output (a unit-step with finite length triangular tails), and finally the impulse response found as step-response's derivative will be a single rectangular-pulse at the origin (and one more at positive infinity) and not a periodic waveform.

So I think the problem should be completely redefined to make any sense...


Your system is stated to be causal and LTI. And an example response for a single rectangular-pulse input $s(t)$ is provided to be a single triangular-pulse output $g(t)$:

$$g(t) = \mathcal{T}\{s(t)\} = s(t) \star h(t) $$

where $h(t)$ is the impulse response of the LTI system, and $\star$ denotes convolution.

The followings are easily deduced from the properties of LTI systems:

a-) Observe that $s_1(t)$ is given by a sum of two shifted $s(t)$ pulses side by side as:

$$ s_1(t) = s(t-0.5) + s(t-1.5)$$

then from the LTI property, conclude the output as:

$$g_1(t) = \mathcal{T}\{s_1(t)\} = \mathcal{T}\{s(t-0.5) + s(t-1.5)\} = g(t-0.5) + g(t-1.5) $$

the output for $s_1(t)$ consists of two triangular pulses repeated (and the origin is shifted).

b-) Now, instead of two pulses, consider an infinity of such rectangular pulses shifted to locate side by side. Then the output will also be infinity of triangular pulses located side by side, as given by the figures in the case b:

$$ \epsilon(t) = \sum s(t -0.5-n) \implies h_{\epsilon}(t) = \mathcal{T}\{ \epsilon(t) \}$$ $$ \begin{align} h_{\epsilon}(t) &= \mathcal{T}\{ \sum s(t -0.5-n) \} \\ &= \sum \mathcal{T}\{s(t -0.5-n) \} \\ &= \sum g(t-0.5-n) \end{align} $$

incidentally, the signal $\epsilon(t)$ is a unit-step function, and the corresponding response $h_{\epsilon}(t)$ is known as step-response of an LTI system.

c-) then the impulse response of the LTI system can be found based on the relation of step-reponse to impulse-response. Namely, if $h_{\epsilon}(t)$ is the step-response of an LTI system, then the impulse response $h(t)$ is :

$$ h(t) = \frac{ d h_{\epsilon}(t)} {dt} $$

you can see from the figure in c-) that the impulse response $h(t)$ is indeed the differentiation of the step-response; i.e, the triangular pulses are converted into bipolar rectangular pulses, and they are repeated indefinetely at the rate of triangular pulses in the step-response. Hence, the figure in case-c.

or it could as well be directly verified as the differentitation property of convolution. Actually this property is a consequence of the differentiation property of the convolution operator...

  • $\begingroup$ Fat, once you look at Aufgabe 1.1 a), you can tell that the particular system is not linear at all. it is a multiplier with $g(t)$ as the output, the periodic triangle wave is one input, and $s(t)$ is the other input. $\endgroup$ Apr 18, 2021 at 20:12
  • 1
    $\begingroup$ okay, i'm wrong. i see how it could be LTI where a $s(t)=\Pi(t)$ is the input and a single $g(t)=\Lambda(2t)$ is the output. there might be a zero divided by zero issue where the $S(f)=\operatorname{sinc}(f)$ spectrum of the input becomes a $G(f)=\frac12 \operatorname{sinc}^2(\frac{f}2)$ spectrum in the output. gonna be a very strange frequency response. I cannot see how the length of the output (one unit of time) can be the same as the length of the input (also one unit of time). how does the LTI system not ring out a little longer? $\endgroup$ Apr 18, 2021 at 20:36
  • $\begingroup$ Actually, there are frequencies of the input spectrum with zero amplitude that somehow get multiplied by the frequency response and become non-zero. so this is not a possible LTI system because the output contains frequency components that do not exist in the input. $\endgroup$ Apr 18, 2021 at 20:41
  • $\begingroup$ i am refraining from a -1. but the answer is wrong. $\endgroup$ Apr 18, 2021 at 20:44
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    $\begingroup$ @robertbristow-johnson see the edit please $\endgroup$
    – Fat32
    Apr 18, 2021 at 21:25

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