I am taking a Signals & Systems course in my Telecommunications Engineering degree, and now I am trying to put together everything I have seen. I came across this basic problem of determining the output of a system whose transfer function is $H(s)=\frac{1}{s+2}$ when it is fed with a signal $x(t)=e^{2jt}$ (where $j^2=-1$). As every other student of S&S I know that exponentials are eigenfunctions to LTI systems, hence I tried this $$y(t)=H(s_0)e^{s_0t}=H(2j)e^{2jt}=\frac{1}{2j+2}e^{2jt}=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}e^{2jt}$$ "Nice!" I said. But then I wanted to test the convolution theorem, and so I happily did $$Y(s) = H(s)X(s) = \frac{1}{s+2}\frac{1}{s-2j}=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}\left(\frac{1}{s-2j}-\frac{1}{s+2}\right)$$ Seeing this, and that its time-domain counterpart would be $y(t)=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}\left(e^{2jt}-e^{-2t}\right)$ I thought I made a mistake, but Wolfram Alpha spat the same. Where does the $e^{-2t}$ term come from?. My first idea was that maybe I was messing it up with the ROC's or something, but all the present signals are right-sided.

Then I tried the convolution with the impulse response, which is precisely $h(t)=\mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\}=e^{-2t}$. $$(x\ast h)(t)=\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau=\int_{0}^{\infty}e^{-2\tau}e^{2j(t-\tau)}d\tau=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}e^{2jt}$$ I calmed down a bit when I saw that something matched what was predicted by the eigenfunction property. After this I attempted the same by applying the commutativity of the convolution: $$(h\ast x)(t)=\int_{0}^{\infty}e^{2j\tau}e^{-2(t-\tau)}d\tau$$ And it diverges, but I found out that $$(h\ast x)(t)=\int_{0}^{t}e^{2j\tau}e^{-2(t-\tau)}d\tau=\int_{0}^{t}e^{-2\tau}e^{2j(t-\tau)}d\tau=\frac{\sqrt{2}}{4}e^{-j\frac{\pi}{4}}\left(e^{2jt}-e^{-2t}\right)$$

I must admit that I do not quite understand the difference between integrating from $0$ to $\infty$ and from $0$ to $t$ in the convolution, and therefore I am unable to explain why with the integration interval $[0,\infty)$ the convolution gives the same as the inverse Laplace transform. And I definitely cannot understand why all this methods do not give the same answer.

On the other hand, I tried feeding the system with $e^{-2t}$. The convolution with the interval of integration $[0,\infty)$ is divergent, but the following is certainly true: $$\int_{0}^{t}e^{-2\tau}e^{-2(t-\tau)}d\tau = te^{-2t} = \mathcal{L}^{-1}\left\{\frac{1}{(s+2)^2}\right\}$$ Beside, a (very simple) "tool" I created in geogebra which shows graphically the convolution between two signals showed the same function (and it has worked very well for every pair of real signals I have tried).

But I don't understand how to apply the property of eigenfunction here since $H(-2)=\frac{1}{0}$.

May someone help me understand this?

  • $\begingroup$ As far as I know is the transfer Function $H(s)$ $2\pi$ periodic. Clearly $\tfrac 1 {s+2}$ is not $2\pi$ periodic, so I would assume that $H(s)$ is the $2\pi$ periodic extension of $\tfrac 1 {s+2}\cdot \mathbb 1_{[0,2\pi)}$. So now we would have $H(-2)=\tfrac 1 {2 \pi}$. Does this makes sense? $\endgroup$
    – Salfalur
    Feb 19 '21 at 18:39
  • $\begingroup$ Furthermore I am a bit confused, because shouldn't the input Signal $x(t)=e^{2jt}$ satisfy this condition. $\sum_{t\in \mathbb Z} |x(t)|<\infty$? But we have $\sum_{t\in \mathbb Z} |e^{2jt}|=\sum_{t\in \mathbb Z} 1=\infty$. So maybe $x(t)$ is no valid input Signal. Or am I totally wrong? $\endgroup$
    – Salfalur
    Feb 19 '21 at 18:48
  • $\begingroup$ @Salfalur I think you are mixing here the discrete-time version of the Laplace transform, which is the Z-transform and is, indeed, periodic; but the Laplace transform isn't. Here I am refering to continuous-time systems/signals. $\endgroup$
    – Arget
    Feb 19 '21 at 18:51
  • $\begingroup$ Ohh, yeah I think you am right, then nevermind. $\endgroup$
    – Salfalur
    Feb 19 '21 at 19:33

There is no problem with the computations above, but there is definitely a misunderstanding in the way they are interpreted.

  • When one considers the response $\Psi(s)=H(s)X(s)$ of the system , you are essentially saying that the system is satisfying a differential equation with the unperturbed initial condition $y(0)=0$, in your case the equation is $y'(t)+2y(t)=x(t)~,~ y(0)=0$. Laplace transforming this equation gives the exact expression above. The solution you have found is correct and satisfies this problem, and you can check that explicitly.
  • On the other hand saying that the solution of this problem is $x(t)/\lambda$ where $\lambda$ is the eigenvalue of the function $x(t)$ with respect to the linear operator $T=\frac{d}{dx}+2$ is NOT correct. What this function represents is a particular solution to the inhomogeneous ODE, and in no way a complete solution to the problem. In your case it can be shown that the full solution to the ODE mentioned above is

$$y(t)=Ce^{-2t}+\frac{e^{2jt}}{2j+2}\equiv y_0(t)+y_p(t)~~,~C\in\mathbb{R}$$

where $C$ is an arbitrary constant which is determined by the initial conditions. The only way $y_p(t)$ can be a solution by itself is if there is an initial condition that sets $C=0$. In this case no real number can achieve this, so the inhomogeneous solution will always be accompanied by some sort of term $\propto e^{-2t}$.

  • As far as the convolution integrals are concerned, since all signals are right-sided one can show that the following three expressions are equivalent:

$$\Psi(s)=H(s)X(s)\iff y(t)=\int_0^{t}dt' h(t')x(t-t')\iff y'(t)+2y(t)=x(t)~,y(0)=0$$

which you have shown to be true. Exchanging the order of $h,x$ does not change anything.

However, the interpretation of the same convolution integral from $(0,\infty)$ is completely different. To obtain that integral you secretly assume that $x(t)$ is not right-sided, and instead has preexisted for all times $t\in \mathbb{R}$. This means that the system started getting kicked in the infinite past, which means that it must have reached the equilibrium solution at any finite time (all transients have died since the system is causal) and that is the reason why you obtain the equilibrium solution when you perform the integral in $(0, \infty)$ instead. Indeed you can demonstrate this using the ODE representation:

$$y'(t)+2y(t)=x(t)~,~ y(t_0)=y_0 \Rightarrow y(t)=e^{2t_0}\left(y_0-\frac{e^{2jt_0}}{2j+2}\right)e^{-2t}+\frac{e^{2jt}}{2j+2}$$ Taking the limit $t_0\to -\infty$ requires the transients to vanish $$\lim_{t_0\to-\infty}y(t)=\frac{e^{2jt}}{2j+2}$$

and only the particular solution remains. It is not fair to ask both convolution integrals to give you the same answer, because they are designed to answer different problems. The method of particular solutions-eigenfunctions yields the long-time behavior of a causal system, the Laplace transform is useful for studying the full response of a system.

  • Finally, in the last example the method of eigenfunctions simply does not apply, because $e^{-2t}$ is a solution to the homogeneous ODE $y'+2y=0$, which doesn't allow the application of the method. Regardless of that subtlety, if we follow the same procedure we used above, we see that in this problem the initial condition $y(0)=0$ sets the arbitrary constant to zero, and thus renders the two solutions equal, but this is a coincidence owing to the fact that the particular solution $y_p(t)=t e^{-2t}$ has $y_p(0)=0$.
  • $\begingroup$ Ok... I have used this couple of days to review differential equations. Now I can see where the initial conditions apply and how. But I am very surprised to see that therefore this must be wrong, and that confuses me a lot. $\endgroup$
    – Arget
    Feb 23 '21 at 14:00
  • 1
    $\begingroup$ It's not wrong... as I mentioned above plugging eigenfunctions yields the correct response when the signal has been present for all time, and the professor mentions that in the first slide after the timemark you provided. If the signal is right sided, not having transients is a peculiarity, not the rule- the trick is only going to provide the behavior as $t\to\infty$. $\endgroup$ Feb 23 '21 at 18:21
  • $\begingroup$ Oooh okay, I didn't pay enough attention to notice the "(for all time)" annotation. Then all my doubts are solved, thanks! $\endgroup$
    – Arget
    Feb 23 '21 at 18:56

Your computations are correct. Your assumption that the "Laplace transform of the convolution is the product of the Laplace transforms is incorrect" for the definition of convolution you are considering. The convolution you are considering has integral limit infinity whereas the theorem applies to the convolution with integral limit $t$. There is an important difference between these two formulas: Note that restricting the integral limit to $t$ is equivalent to having the integral limit to infinity but with $x(t)$ defined only for $t>0$. That's why it is the Fourier (and not the Laplace) transform that dominates signal & systems analysis (including telecommunications)

EDIT (on the convolution integral limits):

In linear time invariant (LTI) systems theory you have a system with an impulse response $h(t)$ and an input signal $x(t)$. In general, these two functions (signals) are defined for all $t\in \mathbb{R}$ and the output of the system is given by the convolution integral

$$ y(t) = \int_{-\infty}^{\infty} h(\tau) x(t-\tau)d\tau, t\in \mathbb{R}. $$ Note that the output $y(t)$ is in general non-zero for both positive and negative values of $t$.

The above is the most general formula for the output of an LTI system and applies to arbitrary $h(t)$ and $x(t)$ (as long as the integral converges).

Now, if the support of $h(t)$ and/or $x(t)$ is finite, you can change the integral limits to reflect that.

Let's start with the impulse response defined only for $t\geq 0$, i.e., $h(t)=0,t<0$, corresponding to a, so called, causal system. Then, it is easy to see that the integrad in the above expression is zero for all $\tau<0$ and thus we can write

$$ y(t) = \int_{0}^{\infty} h(\tau) x(t-\tau)d\tau, t\in \mathbb{R}. $$

Let, in addition, the input $x(t)$ only defined for positive $t$, i.e., $x(t)=0,t<0$. Notice that the term $x(t-\tau)$ appearing in the integrad will be non-zero only whenever it holds $t-\tau \geq 0$ or, equivalently, $\tau \leq t$. But this condition effectively imposes a restriction to the upper integral limit leading to $$ y(t) = \int_{0}^{t} h(\tau) x(t-\tau)d\tau, t\in \mathbb{R}. $$ Finally, note that the above integral will only be non-zero for $t\geq 0$, ultimately leading to $$ y(t) = \int_{0}^{t} h(\tau) x(t-\tau)d\tau, t\geq 0, $$ with the understanding that $y(t)=0,t<0$.

The last convolution integral is effectively the one assumed when working the Laplace transform, i.e., all systems and input/output signal are causal.

  • $\begingroup$ Ok, thank you so much! But taking this into account, which one is the correct answer? I would suppose it should be the one given by the convolution between 0 and t, but the eigenfunction way should give also the correct answer. $\endgroup$
    – Arget
    Feb 20 '21 at 9:11
  • $\begingroup$ On the other hand, the convolutions I did had as lower limit 0 too, so for them the signal is also defined only for $t>0$. Can you explain a little further about the difference between convolving in $[0,\infty)$ and $[0,t)$? $\endgroup$
    – Arget
    Feb 20 '21 at 9:14
  • $\begingroup$ @Arget Please see my edit $\endgroup$
    – Stelios
    Feb 20 '21 at 22:54
  • $\begingroup$ Ok, didn't know that integrating in $[0,t]$ was the same as to assume $x(t)=0,\forall t<0$. Of course I was getting two different answers, I was integrating two different signals. $\endgroup$
    – Arget
    Feb 21 '21 at 8:53

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