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?
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$\begingroup$ What do you mean zero input? It looks more like there are 3 inputs. $\endgroup$ – Olli Niemitalo Feb 24 '16 at 6:35
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$\begingroup$ @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
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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.
