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?
Best Answer
There is no problem with the computations above, but there is definitely a misunderstanding in the way they are interpreted.
$$y(t)=Ce^{-2t}+\frac{e^{2jt}}{2j+2}\equiv y_0(t)+y_p(t)~~,~C\in\mathbb{R}$$
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 $\propto e^{-2t}$.
$$\Psi(s)=H(s)X(s)\iff y(t)=\int_0^{t}dt' h(t')x(t-t')\iff y'(t)+2y(t)=x(t)~,y(0)=0$$
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.