In an example, an equation describing a causal LTI-system is

$$ (D^2 + 5D + 6) y(t) = (D+1) x(t) $$

where $y(t) = y_{zs}(t) + y_{zi}(t)$ and the initial conditions are $y(0^-) = 2, \dot{y}(0^-) = 1$.
$x(t) = e^{-4t}u(t)$ and we want to calculate $y(t)$.

My teacher said in an example of a solution that because $x(t) = 0, \quad t < 0$ we can take the one-sided/unilateral laplace transform of the RHS (since then the unilateral and bilateral laplace transform are the same, I guess ). Further, the explanation continued with that because the system is causal, the impulse response $h(t) = 0$ for $t < 0$ and therefore $y_{zs}(t) = (x*h)(t) = 0$ for $t < 0$. Therefore, supposedly, $y(t<0)=0$ and we can take the one-sided laplace transform of the LHS too.

I am questioning the part in boldface, because what about the zero-input response $y_{zi}(t)$, how do we know its value for $t<0$? I would like to belive it is not equal to $0$ for $t<0$ due to the initial conditions given at $0^-$ not being zero and $0^- < 0$. If it is not equal to $0$ for $t<0$ how can we know we can take the one-sided laplace transform of the LHS?

Note: I also read this question in which it is stated that non-zero initial conditions make the system non-linear and time-varying, which also makes me think the equation at hand is inconsistent with the fact that it describes an LTI-system?

Edit 1 in response to the comment about Lathi's linear systems and signals:
I did not remember that this example was indeed from the book, but I have read the example and the section "Comments on initial conditions at $0^-$ and at $0^+$ " before. The section explains that we cannot expect the total response $y(t)$ to satisfy the initial conditions given at $0^-$ at $0$, which makes perfect sense I think since $0^- \neq 0$. It goes on to say that there is another version of the laplace transform, $L_+$ that is not as convenient to work with. The author's discussion might have gone a bit over my head, because unfortunately I can't understand how it answers my questions, that is how we can assume $y(t)$ to be causal (in order to use the unilateral laplace transform on both sides of the equation) and the note about the non-zero initial conditions implying the system to not be LTI.

The reason I have not accepted the answer given so far is that it is a bit to advanced for me to judge its correctness and therefore I wanted to wait a bit in case there would be more input for my question. But eventually I will just assume it is correct and accept it.

  • $\begingroup$ That example can be found in Lathi's book Linear Systems and Signals. Have you read the section "Comments on Initial Conditions at $0^−$ and at $0^+$"? There your question is actually answered, at least as far as I can see. If not, then please explain any remaining doubts by editing your current question. $\endgroup$ – Matt L. Dec 5 '20 at 11:43
  • $\begingroup$ @MattL. Thanks for your comment, I have edited my question. $\endgroup$ – DancingIceCream Dec 5 '20 at 12:44
  • $\begingroup$ Do you have doubts about the interval $t\in (-\infty,0^-]$ or about the infinitesimal interval $t\in [0^-,0]$? $\endgroup$ – Matt L. Dec 5 '20 at 13:07
  • $\begingroup$ @MattL. I have doubts about the interval $t \in (-\infty, 0^-]$, I think the other interval would not matter since it is infinitesimal. $\endgroup$ – DancingIceCream Dec 5 '20 at 13:11

If it is not equal to 0 for t<0 how can we know we can take the one-sided laplace transform of the LHS?

Unfortunately, the way the problem is framed it is very difficult to do in a rigorous manner. You can, by remembering that $0^-$ is shorthand for $-\epsilon$ as $\epsilon \to 0$, and taking limits everywhere, and having expressions where you're looking at intervals of $t$ from $-\infty$ to $-2\epsilon$ and other maddening things.

The easy way to do it is to separate the problem: solve the problem for the $x(t)$ that's given. Then augment the problem by finding an $x(t)$ that is zero everywhere except in $0^- < t < 0^+$ that will result in $y(0^+) = 2$ (note the $0^+$ instead of $0^-$) and $\dot y(0^+) = 1$. This involves setting $x(t)$ to a linear combination of $\delta(t)$ and $\delta^2(t)$ (as if $\delta(t)$ weren't wacky enough, $\delta^2(t) = d/dt\ \delta(t)$).

Then add the two solutions together. It may not be what your prof had in mind, but it takes the least amount of hand-waving, and if you really want rigor you can break $\epsilon$ and put it to work.

By doing the above trick of expressing $x(t)$ as a linear combination of the specified $x(t)$ plus whatever sum of the Dirac impulse and it's derivatives, you cast the problem into one where the system is linear -- you're just using physically impossible values for $x(t)$.


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