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?

  • $\begingroup$ 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 $x(t)$ . $\endgroup$ Dec 1, 2016 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, 2016 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. $\endgroup$ Dec 1, 2016 at 18:34
  • $\begingroup$ @robertbristow-johnson maybe I've misunderstood you, can you formally express what you have said before? $\endgroup$
    – Vexx23
    Dec 1, 2016 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 $h(t-\tau,0)$? $\endgroup$ Dec 1, 2016 at 19:03

1 Answer 1


If the response to $x(t)$ is given by


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


If the system is time-invariant we require


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


which is equivalent to


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 $t-\tau$:


  • $\begingroup$ 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, 2016 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, 2016 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, 2016 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, 2016 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.