Ideal filter time-domain impulse response is $$h(t) = 2B \operatorname{sinc}(2Bt)$$ It is not causal nor BIBO stable. Is it time invariant? And how can i prove it?


1 Answer 1


If your system is defined by its time-invariant impulse response $h(t)$, then it is by definition a linear time-invariant system (LTI). The output of the system is the convolution of the input signal with its impulse response. This holds, since convolution is translation-invariant (https://en.wikipedia.org/wiki/Convolution#Properties).

A time-invariant impulse response can be recognized, if it only depends on one time-variable, i.e. $h(t)$. For time-varying multipath channels, the impulse response can become time-variant, and is in this case expressed as $h(t,\tau)$ and the output of the system is given by

$$y(t)=\int x(t-\tau)h(t,\tau)d\tau.$$

However, these systems are more advanced and not often studied in fundamental courses.

  • $\begingroup$ Thank you for answer. Maybe i used "impulse response" term wrong, it could be "transfer function". It is an ideal LPF filter which is pulse in frequency domain, and a sinc function in time domain. Is it still time invariant? $\endgroup$
    – ylcn
    Mar 29, 2017 at 7:02
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    $\begingroup$ Yes, the transfer function is the Fourier Transform of the system's impulse response. So, they are mapped one-to-one, and the system is time-invariant and linear. $\endgroup$ Mar 29, 2017 at 7:03
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    $\begingroup$ I would suggest adding that the impulse response should also be time- invariant in order to have an LTI system. A multipath channel, for instance, is defined by a time-varying impulse response but is not LTI. $\endgroup$
    – msm
    Mar 29, 2017 at 9:41
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    $\begingroup$ @msm I agree and added some sentences on time-variant impulse responses. $\endgroup$ Mar 29, 2017 at 10:17
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    $\begingroup$ Yes, exactly like that. just search for the proof that convolution is translation-invariant for more details. $\endgroup$ Mar 29, 2017 at 13:24

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