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If we excite a LTI system with the Dirac delta $\delta(t)$, the system outputs the impulse response $h(t)$. For a LTI system, it doesn't matter when we excite the system with the Dirac delta, we will always get the same impulse response.

However, that is not the case for a linear time-varying (LTV) system. If, say, we excite a system at $t_{0}$ then its impulse response will be different than if we excite the same system at $t_{1}$ so the impulse response $h(t,t_{0})$ is a function of $t_{0}$. However, since the $t$ part is completely separable from the $t_{0}$ part, we can express the impulse response of a system as the multiplication of $2$ signals $g(t_{0})$, which shows the dependence of the impulse response on the moment of excitation and $h(t)$ which is the impulse response if the system was linear and time invariant so $h(t, t_{0}) = g(t_{0})h(t)$

Suppose now we have a system for which it is true that if $t_{2} = t_{1} + t_{0}$ then $$ h(t,t_{2}) = h(t,t_{1}) + h(t,t_{0}) $$ By substitution, we get $$ g(t_{2})h(t) = g(t_{1}) h(t) + g(t_{0}) h(t) \to g(t_{2}) = g(t_{1}) + g(t_{0}) $$ which is interesting because it means that $g(t)$ is a linear signal. Because it is a linear signal, we can choose to express in any form of a linear signal but lets say for simplicity $g(t) = At$.Because we have the values for $g(t_{1}),g(t_{0}),t_{1},t_{0}$ we can determine $A$.


Example: Suppose we have the linear time varying system with a impulse response of $h(t,t_{0}) = \frac{t_{0}}{t}$. We can write $h(t,t_{0}) = g(t_{0})h(t)$ where $h(t) = \frac{1}{t}$ and $g(t_{0}) = t_{0}$. We can easily prove that this system satisfies the condition of the article.So now let's find $g(t)$. Well $A = \frac{g(t_{1})-g(t_{0})}{t_{1}-t_{0}} = 1$ so $g(t) = t$.Now $g(t)\cdot h(t) = t\cdot t^{-1} = 1$.

Since the system is linear we can safely head to the Laplace domain in which $H(s)$ or the transfer function will be $\frac{1}{s}$. So now have I calculated the transfer function of a time varying system? Is my process correct?

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Signal Processing Meta, or in Signal Processing Chat. Comments continuing discussion may be removed. $\endgroup$
    – Peter K.
    Commented May 29, 2023 at 19:52
  • $\begingroup$ Peter, I am not wanting to disregard your instruction, but I wanna see the math rendered. - - - - @Jazzmaniac , maybe the simplest way to put it is that, if the LTV system varies sufficiently slowly, the frequency response of the system during some time close to $t_0$ would be: $$ H(f, t_0) = \int\limits_{-\infty}^{\infty} h(t, t_0) \, e^{-j 2 \pi f t} \ \mathrm{d}t \ \cdot \ e^{j 2 \pi f t_0}$$ The $e^{j 2 \pi f t_0}$ factor at the end it compensate for the delay $t_0$ exactly as we would if $h(t, t_0) = h(t-t_0)$ and it became LTI. Just for consistency. $\endgroup$ Commented May 30, 2023 at 4:16
  • $\begingroup$ I guess, by extension, that means that, if the LTV system varies sufficiently slowly, the Laplace-like transfer function would be: $$ H(s, t_0) = \mathscr{L}\Big\{ h(t, t0) \Big\} \ \cdot \ e^{s \, t_0} = \int\limits_{-\infty}^{\infty} h(t, t_0) \, e^{-st} \, \mathrm{d}t \ \cdot \ e^{s \, t_0}$$ That way if the system degenerates to an LTI system (because it varies extremely slowly), then the transfer function $H(s, t_0)$ is not a function of $t_0$. $\endgroup$ Commented May 30, 2023 at 4:50
  • $\begingroup$ We need to be clear that, in LTV system theory, $h(t, t_0)$ means the impulse response of the system, measured at time $t$, of a unit impulse that was applied at time $t_0$. If the LTV system degenerates to one that doesn't vary much and then is considered LTI, then $$h(t, t_0) = h(t - t_0) \ .$$ So, I think that if $h(t,t_0)$ is separable, then we need to say that $$ h(t, t_0) = g(t_0) \, h(t-t_0) $$ $\endgroup$ Commented May 30, 2023 at 5:12
  • $\begingroup$ In the example I give t is completely seperable from $t_{0}$.Does this mean the derivation is correct? $\endgroup$
    – Volpina
    Commented May 31, 2023 at 14:40

2 Answers 2

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However, since the $t$ part is completely separable from the $t_0$

Where does that come from? In most cases the time variance is not separable from the a "time-invariant" impulse response.

A simple (and very common) example of a time variant impulse response is a variable delay, i.e.

$$h(t,t_0) = \delta(t-\Delta(t_0))$$

where $\Delta(t_0)$ is the delay as a function of time.

Physically this happen when either source and receiver are moving or if the propagation medium changes temperature/pressure/impedance, etc.

For example if you have microphone on a stand and the stand wiggles a bit, you get something like

$$h(t,t_0) = \delta \left[ t-\frac{R+r_v\cos(\omega_v\cdot t_0)}{c_0} \right]$$

where $R$ is the bulk distance between the microphone and the loudspeaker, $c_0$ the speed of sound, $r_v$ the amplitude of the microphone movement and $\omega_v$ the frequency of the microphone movement.

You can certainly construct examples where the time invariance is separable, but it's not the general case.

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  • $\begingroup$ Does $\Delta(t_0)$ mean that it's an arbitrary real and non-negative function of $t_0$? $\endgroup$ Commented May 30, 2023 at 4:23
  • $\begingroup$ Yes. I can add a few examples. $\endgroup$
    – Hilmar
    Commented May 30, 2023 at 6:49
  • $\begingroup$ Hil, remember that when $t=t_0$, that's the beginning of the impulse. So, if the impulse is delayed by $t_0$, so also is the impulse response. If it's LTV, then it is delayed and changed. So a wire (which is also LTI) would have $h(t,t_0)=\delta(t-t_0)$. I think your microphone on a wiggling stand might be: $$h(t,t_0) = \delta \left( t-t_0 - \frac{R+r_v\cos(\omega_v\cdot t_0)}{c_0} \right)$$ and I think that $R$ should be the bulk distance between the vocalist's lips and the loudspeaker when the microphone is in the middle of its wiggle. $\endgroup$ Commented May 31, 2023 at 0:13
  • $\begingroup$ @Hilmar well in the example I give I can seperate t from $t_{0}$.Does this means what I have calculated is correct? $\endgroup$
    – Volpina
    Commented May 31, 2023 at 14:28
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However, since the t part is completely separable from the t0 part

To add to @Hilmar's answer, and give you a way to devise your own counter-examples, as you requested in the chat, if that were true then for values $t_0, t_1, t_2, t_3$, and assuming $h()$ is defined for all 4 combinations, you'd have: $$h(t_0,t_1)h(t_2,t_3) = g(t_1)h(t_0)g(t_3)h(t_2) = h(t_2,t_1)h(t_0,t_3)$$ Now consider (out of thousand others) $h(t,t_0) = t+t_0$. Then $$h(2,2)h(3,4) = 4\cdot 7 = 28 \neq 30 = 5 \cdot 6 = h(3,2)h(2,4)$$

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  • $\begingroup$ in the example I give t is completely seperable from $t_{0}$.Does this mean my derivation is correct? $\endgroup$
    – Volpina
    Commented May 31, 2023 at 14:41
  • $\begingroup$ @volpina Do you want to know if it’s correct for the specific example you gave? $\endgroup$
    – Jdip
    Commented May 31, 2023 at 15:49
  • $\begingroup$ yes I would like to know. $\endgroup$
    – Volpina
    Commented May 31, 2023 at 15:54

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