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I have a LTV (linear and time-varying) system. So, $h(\tau, t)$ is the "instantaneous impulse response" at time $t$ such that if the input signal is $x(t) = \delta(t - t_0)$ (an impulse at time $t_0$), the output signal should be $y(t) = h(t - t_0, t_0) = h(\tau, t_0)$ where $\tau = t - t_0$ is the delay/lag from the impulse.

This makes sense to me, intuitively. Obviously, if you input an "impulse at $t_0$", the output signal will be the "impulse response at time $t_0$".

This also has the nice property that if the input is some sort of comb (sum of many impulses with arbitrary complex-valued amplitudes), $$ x(t) = \sum_i A_i \delta(t - t_i) $$ then the output signal is $$ y(t) = \sum_i A_i h(\tau, t_i) $$ which also seems very reasonable to me.


However, this does not agree with the treatment given in most literature that deals with LTV systems where: $$ y(t) = \int_{-\infty}^{+\infty} h(\tau, t) x(t - \tau) d\tau $$

As above, if we have an impulse at $t_0$ such that $x(t) = \delta(t - t_0)$ then $$ \begin{align} y(t) &= \int_{-\infty}^{+\infty} h(\tau, t) \delta(t - \tau - t_0) d\tau \\ &= h(t - t_0, t) \\ &= h(\tau, t) \end{align} $$

This does not match what I expect above!


Am I misunderstanding something? I have a nagging feeling that this is somehow just a matter of definition/convention.

However, I want to make sure I am not accidentally making an implicit assumption of some sort that I don't have to make.


Edit:

Okay, if I define the integral differently, I get what I expected: $$ y(t) = \int_{-\infty}^{+\infty} h(\tau, t - \tau) x(t - \tau) d\tau $$

Is this wrong? How does this differ from the usual treatment of time-varying impulse response?

One immediate problem is that this no longer looks like a "convolution".

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  • $\begingroup$ If $x(t)=\delta(t-t_0)$, then $x(t-\tau) = \delta(t-\tau-t_0)$. $\endgroup$
    – Ash
    Oct 6, 2022 at 21:55
  • $\begingroup$ Yes, that's what I used to compute the integral. I edited my question to show this is what I am doing. $\endgroup$
    – XYZT
    Oct 6, 2022 at 21:56
  • $\begingroup$ Your last equation is correct. The response for an impulse at a delay of $\tau'$ is only valid for the impulse response defined at that time delay (i.e. $h(\tau, t-\tau')$). A response $h(\tau, t)$ would only be realized by an impulse occuring at $\delta(t)$, not $\delta(t-\tau')$. Variable overload, adding prime for input delay $\endgroup$
    – Ash
    Oct 6, 2022 at 22:11
  • $\begingroup$ Hmm, but it does not look like a convolution - which I thought would be the case if it was correct. $\endgroup$
    – XYZT
    Oct 6, 2022 at 22:24
  • $\begingroup$ Your analysis agrees with this blog post. $\endgroup$
    – Ash
    Oct 6, 2022 at 22:43

1 Answer 1

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The problem is that there are two common definitions of the impulse response of an LTV system, resulting in the following input-output relations:

$$ y(t)=\int_\tau h_1(t,\tau)x(\tau)d\tau\tag{1}$$

and

$$ y(t)=\int_\tau h_2(t,\tau)x(t-\tau)d\tau\tag{2}$$

In the first one, the impulse response (integration kernel) $h_1(t,\tau)$ is the response at time $t$ to an impulse at time $\tau$. In the second, $h_2(t,\tau)$ is the response at time $t$ to an impulse at time $t-\tau$. The impulse response $h_2(t)$ is also called the input delay-spread function.

The relationship between $h_1(t,\tau)$ and $h_2(t,\tau)$ is

$$h_1(t,\tau)=h_2(t,t-\tau),\quad h_2(t,\tau)=h_1(t,t-\tau)\tag{3}$$

The conditions for time-invariance are as follows. For the first definition we require that $h_1(t,\tau)$ only depends on the difference $t-\tau$:

$$h_1(t,\tau)=\tilde{h}(t-\tau)\tag{4}$$

The second definition results in an impulse response that is independent of $t$:

$$h_2(t,\tau)=\tilde{h}(\tau)\tag{4}$$

The causality conditions are $h_1(t,\tau)=0$ for $t<\tau$ and $h_2(t,\tau)=0$ for $\tau<0$.

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