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If a system is linear, then the operator $S$ mapping input signals into output signals - i.e. $y(t)=S\{u(t)\}(t)$ - is the integral of the input weighted by the impulse response: $$ y(t) = \int_{-\infty}^{+\infty} h(t,\tau) u(\tau) \mathrm{d}\tau $$ where $h(t,\tau):=S\{\delta(t-\tau)\}$, response of the system to an impulse in $\tau$, is a function of two variables: the second one $\tau$ refers to the time at which the input is applied while the first one $t$ refers to the time at which the output is observed.

If this system is also time-invariant, the system operator commutes with the time shift operator: $$S\{u(t-T)\}(t) = S\{u(t)\}(t-T) $$ that is, time shifting the input, a time shifted version of the output is produced. Linear time invariant systems have an impulse response function which depends upon a single variable, the difference between $t$ and $\tau$: $$h(t,\tau)=\tilde h(t-\tau)$$ where the $\sim$ symbol is used just to underline that $h$ and $\tilde h$ cannot be the same, because they have a different number of arguments. How can I show that the impulse response can be expressed as a function depending only on $t-\tau$ starting from the definition of time invariance?

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  • $\begingroup$ this is in the textbooks. what is the definition of $h(t,\tau)$ ? it is the linear system response, $y(t)$ to a unit impulse applied at time $\tau$ (that is $x(t) = \delta(t-\tau)$ ). then ask what is the definition of "time-invariance"? and apply that to $x(t)$ . $\endgroup$ – robert bristow-johnson Dec 1 '16 at 18:24
  • $\begingroup$ @robert bristow-johnson I have already seen this path, but I wanted to show this in terms of any input $u(t)$, not to infer it starting from $\delta(t-\tau)$ as input, since we know which is the linear operator between the input and the output, and the definition of time-invariance could be applied to it. $\endgroup$ – Vexx23 Dec 1 '16 at 18:32
  • $\begingroup$ but the definition of $h(t,\tau)$ is not the output due to any input. it is the output due to a specific input. $\endgroup$ – robert bristow-johnson Dec 1 '16 at 18:34
  • $\begingroup$ @robertbristow-johnson maybe I've misunderstood you, can you formally express what you have said before? $\endgroup$ – Vexx23 Dec 1 '16 at 18:59
  • $\begingroup$ okay, define $$\tilde{h}(t) \triangleq h(t,0)$$ if the system is both L and TI, what is the difference between $h(t,\tau)$ and $h(t-\tau,0)$? $\endgroup$ – robert bristow-johnson Dec 1 '16 at 19:03
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If the response to $x(t)$ is given by

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

then the response to $x(t-T)$ is

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

If the system is time-invariant we require

$$\tilde{y}_T(t)\stackrel{!}{=}y(t-T)=\int_{-\infty}^{\infty}h(t-T,\tau)x(\tau)d\tau\tag{3}$$

For $(2)$ and $(3)$ to be equal for any $x(t)$ we must have

$$h(t,\tau+T)\stackrel{!}{=}h(t-T,\tau)\tag{4}$$

which is equivalent to

$$h(t+T,\tau+T)\stackrel{!}{=}h(t,\tau)\tag{5}$$

From $(5)$ it is clear that for a linear and time-invariant system the value of $h(t,\tau)$ only depends on the difference of its arguments, and not on their individual values. Consequently, the impulse response can be rewritten as a function of $t-\tau$:

$$h(t,\tau)=\tilde{h}(t-\tau)\tag{5}$$

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  • $\begingroup$ Thank you, this is exactly the same path I was trying to follow! The only passage that does not convince me is the latest: when you say that, in order to satisfy $(4)$ the two variables cannot be independent, this is rather clear, I just don't get the reason why the resulting function has to be precisely that one $(5)$ $\endgroup$ – Vexx23 Dec 1 '16 at 22:32
  • $\begingroup$ @Vexx23: Try to evaluate both sides of $(4)$ using $(5)$. I think then you should see that no other option exists: the function only depends on the difference of its arguments. $\endgroup$ – Matt L. Dec 1 '16 at 22:42
  • $\begingroup$ Intuitively I see this property, considering for instance the $\mathbb{R}^2$ plane of $(t,\tau)$ couples, but something is missing when I try to prove it analyticaly. $\endgroup$ – Vexx23 Dec 2 '16 at 10:18
  • $\begingroup$ @Vexx23: Eq. (4) is equivalent to $h(t+T,\tau+T)=h(t,\tau)$, which explicitly shows that it is only the difference $t-\tau$ that is relevant for the function value. $\endgroup$ – Matt L. Dec 2 '16 at 20:46

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