Calculating transfer function of a linear time varying system?

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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– Peter K.
Commented May 29, 2023 at 19:52
• 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. Commented May 30, 2023 at 4:16
• 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 $t_0$. Commented May 30, 2023 at 4:50
• 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 $$h(t, t_0) = g(t_0) \, h(t-t_0)$$ Commented May 30, 2023 at 5:12
• In the example I give t is completely seperable from $t_{0}$.Does this mean the derivation is correct? Commented May 31, 2023 at 14:40

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.

• Does $\Delta(t_0)$ mean that it's an arbitrary real and non-negative function of $t_0$? Commented May 30, 2023 at 4:23
• Yes. I can add a few examples. Commented May 30, 2023 at 6:49
• 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. Commented May 31, 2023 at 0:13
• @Hilmar well in the example I give I can seperate t from $t_{0}$.Does this means what I have calculated is correct? Commented May 31, 2023 at 14:28

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 $$h(2,2)h(3,4) = 4\cdot 7 = 28 \neq 30 = 5 \cdot 6 = h(3,2)h(2,4)$$

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