For the zero input + zero state response in continuous time linearly time-invariant systems, why is the $y(t)$ equation not "technically" considered an LTI? I read this in a journal and there was no justification. How would you change the equation to make it a true LTI?

  • $\begingroup$ What do you mean zero input? It looks more like there are 3 inputs. $\endgroup$ – Olli Niemitalo Feb 24 '16 at 6:35
  • $\begingroup$ @OlliNiemitalo: You are right that the question is not clearly formulated, but I suppose that $y_0(t)$ corresponds to a non-zero initial condition, which makes the system non-linear. $\endgroup$ – Matt L. Feb 24 '16 at 6:51

The response of the system is given by

$$y(t)=h(t)\star x(t)+y_0(t)\tag{1}$$

where $x(t)$ is the input signal, $h(t)$ is some well-behaved function, $y_0(t)$ is a function which is independent of $x(t)$, and $\star$ denotes convolution. The model $(1)$ can be used for describing systems with non-zero initial conditions that would be LTI were it not for the non-zero initial conditions (such as an initially charged capacitor).

For $y_0(t)=0$, $h(t)$ would be the impulse response of the corresponding LTI system. For $y_0(t)\neq 0$, $h(t)$ is not the impulse response, neither can the system be characterized by an impulse response.

For $y_0(t)\neq 0$, the system described by $(1)$ cannot be linear, which is easily shown by considering a response $y_1(t)$ to an input signal $x_1(t)$. For a linear system, the response to a scaled version of $x_1(t)$ must be a scaled version of $y_1(t)$, with the same scaling factor. So the response to $ax_1(t)$ must be $ay_1(t)$ if the system is linear. But from $(1)$ we have

$$y_1(t)=h(t)\star x_1(t)+y_0(t)$$

and for the scaled input we get

$$y(t)=h(t)\star ax_1(t)+y_0(t)=ay_1(t)+(1-a)y_0(t)\neq ay_1(t)$$

for $y_0(t)\neq 0$ and $a\neq 1$. Consequently, the system described by $(1)$ is not linear.

In a similar way it can be shown that the system is also not time-invariant. The reason is again the function $y_0(t)$, which is independent of the input signal $x(t)$. If $x(t)$ is shifted, the output doesn't shift accordingly, because $y_0(t)$ doesn't shift.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.