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I am confused about linear time-varying system. For a time varying system, the output is given by \begin{align} y(t)=\int x(\tau) h_{\tau}(t) d\tau, \end{align} where $ h_{\tau}(t)$ is the output of the system given the input $\delta(t-\tau)$. By manipulating the integration above, we have \begin{align} y(t)=\int x(t-\tau) h_{t-\tau}(t) d\tau \end{align} I found that in some refeneces, we can write above as $y(t)=\int x(t-\tau) h(t,\tau) d\tau$, where $h(t,\tau)=h_{t-\tau}(t)$. To avoid confusion, here I let $g(t,\tau)=h_{t-\tau}(t)$. Thus, \begin{align} y(t)=\int x(t-\tau) g(t,\tau) d\tau \end{align}

From my understanding, the $g(t,\tau)$ is just obtained by substituting the variables in $ h_{t-\tau}(t)$. However, I am not sure whether we can treat $g(t,\tau)$ as the system output of some input signal.

Based on the definition of $h_{\tau}(t)$, though $ h_{\tau}(t)$ is the output of the system given the input $\delta(t-\tau)$, $h_{\tau}(t)$ in fact defines a function over $(t,\tau), \forall t,\tau$. Therefore, the value of $h_{t-\tau}(t)$ can be obtained by substituting the pair $(t,t-\tau)$ in to $h_{\tau}(t)$.

However, can we treat $h_{t-\tau}(t)$ as the output of $\delta(t-(t-\tau))$, i.e., $\delta(\tau)$?

Here, how can we understand the input $\delta(\tau)$? In my view, I think the input of a system should be a function over $t$, and $\tau$ is just a parameter. Does this mean that the input is a constant?

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  • $\begingroup$ Did you not forget to mention linearity? $\endgroup$ May 7, 2023 at 10:02
  • $\begingroup$ t is the time reference for input and output. For time variant systems, same input at t1 or t2 causes different outputs. To avoid hanging around with t1 t2 .. the relevant times needed, as you correctly mention, as parameter TAU, are also time, but it's the second time reference needed to characterize time variant systems. $\endgroup$ May 7, 2023 at 15:55

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As you've correctly noted, there are two common ways to describe the relation between input and output of a linear time-varying system. The first is

$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t,\tau)d\tau\tag{1}$$

In this case the kernel $h(t,\tau)$ is the response at time $t$ to an impulse at time $\tau$. I've always found it instructive to look at a few special cases:

  • LTI: the system is time-invariant if $h(t,\tau)$ only depends on the difference $t-\tau$, i.e., $h(t,\tau)=h'(t-\tau)$.
  • causality: the system is causal if $h(t,\tau)=0$ for $t<\tau$.
  • modulation: according to definition $(1)$, a modulator with modulation function $m(t)$ is described by $h(t,\tau)=m(t)\cdot\delta(t-\tau)$.

The second common description of the input-output relation of a linear time-varying system is

$$y(t)=\int_{-\infty}^{\infty}x(t-\tau)h(t,\tau)d\tau\tag{2}$$

In this case, the response to an input $\delta(t-t_0)$ is given by $h(t,t-t_0)$. This means that $h(t,\tau)$ is the response at time $t$ to an impulse at time $t-\tau$. Let's take a look at the same special cases as above:

  • LTI: the system is time-invariant if $h(t,\tau)$ is independent of $t$, i.e., $h(t,\tau)=h'(\tau)$.
  • causality: the system is causal if $h(t,\tau)=0$ for $\tau<0$.
  • modulation: according to definition $(2)$, a modulator with modulation function $m(t)$ is described by $h(t,\tau)=m(t)\cdot\delta(\tau)$.
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