Could a continuous time-limited and absolute integrable function be the output of a causal continuous-time Linear and Time-Invariant system (CT-LTI)??
I am trying to understand if there exists any restrictions for the derivative of a continuous and time-limited signal due if it is described by a causal function. So looking for causality-induced properties, I was studying the Laplace transform since I remember from my student years that some characteristics of the transform could be used to figure out if the signal is causal or not.
So, studying the book "Signals and Systems, 2nd Edition" (Alan V. Oppenheim, Alan S. Willsky, with S. Hamid) [1] – I am using a Spanish translation-, following its notation the Laplace Transform is defined as $X(s) = \int\limits_{-\infty}^\infty x(t)\,e^{-st}\,dt$ for complex $s \equiv \sigma + iw$, which converges only for determined values of $s$ which are called its "Region of Convergence" or ROC, and both a required to determine the Laplace Transform of a function (same $X(s)$ but different ROC will define different functions $x(t)$).
- On chapter 9 section 9.2, on property 3 is said that (translated): "If $x(t)$ is of finite-duration and it is absolutely integrable, then the ROC is the whole complex plane defined by $s$".
But in the same chapter, on section 9.7 starts talking about Linear and Time-Invariant Systems (LTI), and when starts talking about causality in section 9.7.1, there is a box highlighting the following:
- "The ROC associated with the function of the system for a causal system is a right-side semi-plane"
And it is explained why (and it seems "right"), but right know I figure out that it looks as a contradiction for time-limited functions, or more probably, I have a big misunderstanding about what is said on the book – but it could be also a translation issue so please follow the next explanation:
Let say $x(t)$ is a continuous time-limited function, so there is a starting point $t_0$ and an ending point $t_F$ from where $x(t) = 0,\,t<t_0$ and $x(t) = 0,\,t>t_F$, so is of finite-duration, and also, is compact-supported so since is also continuous, is also bounded $\sup_t |x(t)|<\infty$. Also assume that $x(t)$ is absolutely integrable $\int\limits_{t_0}^{t_F} |x(t)|\,dt < \infty$, so the first property follows and the ROC of $X(s)$ is the whole complex $s$ plane.
Here it looks right so far, since is similar to the property that time-limited signals have Fourier Transforms with unlimited bandwidth, but caution must be taken since the ROC is determined by its real part "$\sigma$", and convergent Laplace Transforms could have ROC that don´t includes the "$iw$" axis for which the corresponding Fourier Transform is not convergent (it is a property that for a function $x(t)$ for having a convergent Fourier Transform $\mathbb{F}\{x(t)\}(w)$, its Laplace Transform $X(s)$ has to have a ROC that includes the "$iw$" axis). Maybe in this property is raising my misunderstanding, so please keep it in mind.
But in the other hand, if this time-limited signal $x(t)$ is also a LTI system, with an impulse response $h(t)$ when having as input another function $y(t)$, it will be meaning that $x(t)= h(t)\circledast y(t)$, or equivalently in the Laplace domain, $X(s) = H(s)\cdot Y(s) \rightarrow H(s) = \frac{X(s)}{Y(s)}$. Now, here, if we follow the second property, if my function $x(t)$ is also a causal function, I will have that $H(s)$ is going to have a ROC that is only defined from the right side of the vertical line defined in the $s$-plane by some specific value of $\sigma_0$. So, in the left side of $X(s) = H(s)\cdot Y(s)$ I have a function $X(s)$ defined for every value of $s$, and in the numerator of the right side I have a function that don´t converges for any value of "s" with $\sigma<\sigma_0$ so it impose a restriction on $Y(s)$ that has to make converge the fraction (I don´t understand yet what this is saying about $Y(s)$, so If you can please explain it).
Now thinking in that many general time-limited functions could be described as $x(t)=\int_{-\infty}^t x'(t)\,dt = x'(t)\circledast \theta(t)$ with $\theta(t)$ the standard unitary step function, I will have that $X(s)$ has a ROC in the whole plane, and $\Theta(s) = \frac{1}{s}$ will have a ROC defined only in $\Re\{s\}>0$ (or $\sigma>0$ in the book notation), and since here the function that is acting as the impulse response is $x'(t)$ which has Laplace transform $sX(s)$, for one property I will requiring that $sX(s)$ will have ROC in the whole $\mathbb{R}$-line, but in the other hand the causality requirement will be asking that $sX(s)$ must have a ROC defined only for a right-side interval $[\sigma_0;\,+\infty)$.
So, I have a big misunderstood here, because following this line of thought I have a contradiction:
- Or no causal (continuous and absolute integrable) CT-LTI system could be of finite-duration (time-limited)??
- Or no time-limited (continuous and absolute integrable) function could be the output of a causal CT-LTI system?
And neither of these statements seems legit to me, since I believe it is totally possible to have a continuous time-limited and absolute-integrable function as the output of a causal CT-LTI system. But all the continuous time examples of the book are related to linear ordinary differential equations, which outputs cannot be time limited (since its solutions are analytical), so maybe there is something I am missing.
So please tell me where I am making my mistake. Beforehand thanks you very much.
Best Answer
You seem to be confused by the following two statements:
The correct conclusion from these statements is that the Laplace transform of a causal time-limited function converges everywhere (because it is time-limited). So what about the right half-plane? Well, it's a right half-plane to the right of $-\infty$, which is the same as the whole complex plane.
Systems built with (ideal) lumped elements like resistors, capacitors and inductors always have an infinitely long impulse response. But it's easy to come up with a (theoretical) system that is causal and has an impulse response of finite length. The output of such a system will be time-limited if the input is time-limited. As you know, time-limited signals have infinite bandwidth, so they don't exist in a mathematically exact sense. However, they can be approximated very well.
As an example, take an ideal averaging filter with impulse response
$$h(t)=\begin{cases}\displaystyle\frac{1}{T},&0<t<T\\0,&\textrm{otherwise}\end{cases}$$
The output of this system will be time-limited for any time-limited input. Such a system can't be implemented exactly in continuous-time, but it can be implemented in discrete-time.