# "Instantaneous impulse response" in a linear time-varying system

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".

• If $x(t)=\delta(t-t_0)$, then $x(t-\tau) = \delta(t-\tau-t_0)$.
– Ash
Oct 6, 2022 at 21:55
• Yes, that's what I used to compute the integral. I edited my question to show this is what I am doing.
– XYZT
Oct 6, 2022 at 21:56
• 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
– Ash
Oct 6, 2022 at 22:11
• Hmm, but it does not look like a convolution - which I thought would be the case if it was correct.
– XYZT
Oct 6, 2022 at 22:24
• Your analysis agrees with this blog post.
– Ash
Oct 6, 2022 at 22:43

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 $$\tau<0$$.