In the context of signal processing, consider the following system: $$ x\left( t \right) \to System\, A \to z\left( t \right) \to System \, B \to y\left( t \right) .$$

Let the global system be the "union" of system A and system B. The input and output of System A are represented by the equations:

$$ \frac{d^2 \, z(t)}{d \, t^2} + 6 \frac{d\, z(t)}{d\, t} + 8 z\left( t \right) = \frac{d\, x(t)}{d\, t} + 2x\left( t \right) $$

and the unit impulse response of System B is defined as follows: $$ h_B \left( t \right) = t e^{-3t} u\left( t \right) .$$

Q1: What is the unit impulse response of the global system?
Q2: For a periodic input of the form $$ x\left( t \right) =e^{-t} \quad ; \quad -1<t<0 .$$ with a period of $T=1$, what will the output signal $y\left( t \right) $ be?
Q3: What is the frequency spectrum of that last signal?

Note: In this problem $u(t)$ denotes the unit step function. And I found these three articles that I think are related.

Edit: With the help of @KwinvanderVeen , I tried to solve it by finding the transfer function for both systems. $$ H_A(s)= \frac{Z\left( s \right) }{X\left( s \right) } = \frac{\mathcal{L}\{z\} \left( s \right) }{\mathcal{L}\{ x\}\left( s \right) }= \frac{\int^{\infty}_{0} z\left( t \right) e^{-st} dt}{\int^{\infty}_{0} x\left( t \right) e^{-st} dt }$$

And for the system B: $$ H_B(s)= \frac{Y\left( s \right) }{Z\left( s \right) } = \frac{\mathcal{L}\{y\} \left( s \right) }{\mathcal{L}\{ z\}\left( s \right) }= \frac{\int^{\infty}_{0} y\left( t \right) e^{-st} dt}{\int^{\infty}_{0} z\left( t \right) e^{-st} dt }$$

Applying the properties of Laplace transforms:

$$ \mathcal{L}\{ \frac{d^2 \, z(t)}{d \, t^2} + 6 \frac{d\, z(t)}{d\, t} + 8 z\left( t \right) \} = \mathcal{L} \{ \frac{d\, x(t)}{d\, t} + 2x\left( t \right) \} $$ $$ \Rightarrow s^2 \mathcal{L}\{ z\} +6s \mathcal{L}\{ z\} +8 \mathcal{L}\{ z\} = s\mathcal{L} \{ x\} +2\mathcal{L} \{ x\} $$ $$ \Rightarrow H_A(s)= \frac{\mathcal{L} \{ z(t) \}}{\mathcal{L} \{ x(t) \}} = \frac{s+2}{(s+2)(s+4)}=\frac{1}{s+4} $$ $$ \Rightarrow h_A=e^{-4t} $$

(in this last line, the original expression was wrong. It was fixed thanks to @KwinvanderVeen) for $s\neq -2$. We also have that:

$$ H_G(s)=\frac{Y(s)}{X(s)} = \frac{Y(s)}{Z(s)} \frac{Z(s)}{X(s)} = H_B H_A = \mathcal{L} \{ (h_B*h_A )(t) \} $$ $$ \Rightarrow \mathcal{L}\{ h_G\}=H_G(s)= \mathcal{L}\{ (h_B*h_A)(t) \} $$ $$ \Rightarrow h_G=h_B*h_A = \int_{0}^{t} h_B(\tau ) h_A(t-\tau )d\tau = e^{-4t}(t e^t-e^t+1) $$

(this last line was also fixed thanks to @KwinvanderVeen). And I think that last line would the the unit impulse response, and the answer to question 1.

Now, I tried to solve question 2 with the suggestions of the same commenter. If the input is periodic, then I think it follows that:

$$ x(t)= e^{1-t} \quad ; \quad 0<t<1 \quad ; \quad T=1 $$ $$ x(t)=e^{2-t} \quad ; \quad 1<t<2 \quad ; \quad T=1 $$

And in general, for $n \in \mathbb{N} $: $$ x(t)=e^{n-t} \quad ; \quad n-1<t<n \quad ; \quad T=1 $$

Then using the formula for the convolution integral: $$ y(t)= \left( h_G*x \right) (t) =e^{n-t} \int_{n-1}^{n} e^{-3\tau } (\tau e^{\tau} -e^{\tau} +1 ) d\tau $$

for $n-1<t<n $. Is that the answer? Can i leave the expression like that?

  • $\begingroup$ Your notation is confusing. What is $u(t)$ is it the input to the second system? Hence, I would assume it to be equal to $z(t)$. $\endgroup$ Jul 8, 2019 at 4:32
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    $\begingroup$ You might want to read about convolution integral and Laplace transform. $\endgroup$
    – obareey
    Jul 8, 2019 at 6:10
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    $\begingroup$ There are multiple ways of going about this, but I would first work out the transfer function for both systems. $\endgroup$ Jul 8, 2019 at 17:07
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    $\begingroup$ @evaristegd but what are the transfer functions, in terms of $s$ explicitly when using what is given about system $A$ and $B$? $\endgroup$ Jul 8, 2019 at 18:41
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    $\begingroup$ you are welcome to submit this question to the Signal Processing Stack Exchange if you want to. it's what it's for. $\endgroup$ Jul 9, 2019 at 3:38

1 Answer 1


The transfer function and impulse response for system A seem to be correct; good that you also noticed the pole zero cancellation. You could manually calculate the impulse response from the transfer function using the inverse Laplace transform, but in practice it is often easier/quicker to use a Laplace transform table such as this one. From such a table you should also be able to derive what the (minimal) transfer function should be for system B.

In order to answer question 1 you could indeed do the convolution of the two impulse responses, however I believe that one (me) can easily make mistakes when solving integrals. Instead I would recommend to decompose the total transfer function into a sum of simple terms that appear in a Laplace transform table. For example if the total transfer function is given by

$$ G(s) = \frac{a_2\,s^2 + a_1\,s + a_0}{(s + b_1)^2\,(s + b_2)}, $$

then you would want to solve for $A$, $B$ and $C$ such that

$$ G(s) = \frac{A}{s+b_1} + \frac{B}{(s+b_1)^2} + \frac{C}{s+b_2}. $$

The total impulse response can then be found by summing the impulse responses of each of the three terms, where impulse response of each term should be able to be found on a a Laplace transform table.

Question 2 does not have a completely well defined answer, because no initial conditions are given for system A and B. One way you could interpret this is with all initial conditions set to zero, which can be obtained by evaluating the convolution integral between the input and the impulse response of the total system. Another way is to assume that the output has settled into a period signal as well, which can be obtained by first evaluating the Fourier series of the input

$$ x(t) = \sum_{k=0}^\infty a_k \cos(2\,\pi\,k\,t) + b_k \sin(2\,\pi\,k\,t) = \sum_{k=0}^\infty c_k \cos(2\,\pi\,k\,t + \phi_k), $$

the output can then be obtained using

$$ y(t) = \sum_{k=0}^\infty c_k\,|G(j\,2\,\pi\,k)| \cos(2\,\pi\,k\,t + \phi_k + \angle G(j\,2\,\pi\,k)), $$

so the total transfer function is evaluated at $s=j\,2\,\pi\,k$, with $j$ the imaginary unit.

This second approach would basically already give the answer to question 3, so I guess the first approach, with the convolution integral, is probably the intended answer.

  • $\begingroup$ Thank you so much for all your help! One question, when you say $\angle G(j2\pi k)$, you mean the arg function for complex numbers ? $\endgroup$
    – evaristegd
    Jul 9, 2019 at 3:14
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    $\begingroup$ @evaristegd Yes, I also added a reference to hopefully make it more clear. $\endgroup$ Jul 9, 2019 at 3:19
  • $\begingroup$ Thank you again! Regarding your first line, I did get some terms like $-sz(0^{-})-z' (0^{-}) $, but I just assumed they were zero without a specific reason. Why is it that that pole zero cancellation occurs? $\endgroup$
    – evaristegd
    Jul 9, 2019 at 4:02
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    $\begingroup$ @evaristegd For transfer functions it is indeed assumed that all initial conditions are zero (other wise you can't define a ratio between the Laplace transform of in the input and output). $\endgroup$ Jul 9, 2019 at 13:10
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    $\begingroup$ @evaristegd Note that $x(t)$ is a piecewise function so $y(t)$ will be probably as well. One can find an analytical expression for $y(t)$ when $t$ is an integer and between those times you can evaluate the remaining convolution integral. $\endgroup$ Jul 11, 2019 at 0:48

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