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I am studying signal processing and currently doing signals & systems. While going through convolution and especially the impulse response , there are problems where LTI systems wherein the input is an unit impulse at $ t=0$ ,but the output pulses are generated at $t\ge 0 $ as well as $t\lt 0 $ as well. Consider the following example:

Impulse Response of a non-causal system

How can an input at $ t=0$ can generate an output at $t\lt 0 $ , how much ever I think about this , cant wrap my mind around this , is there an intuitive way this can be explained ?

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As mentioned in SakSath's answer a system with $h[n]\neq 0$ for $n<0$ is non-causal. Such a system cannot be implemented in real-time. However, you could use such a system for offline processing. Also, the index $n$ doesn't always need to refer to time. It might as well be space, such as in image processing, or anything else. In these cases, a non-zero impulse response for $n<0$ doesn't have anything to do with causality.

The specific impulse response you refer to corresponds to a zero-phase (non-causal) low pass filter. Such an impulse response could be obtained from an ideal filter (lowpass, in this case) by applying a symmetric window. Since the ideal filter is non-causal, the windowed finite length impulse response is also non-causal. The next step for implementing such a system in real time is to simply shift it (i.e., add delay) in order to make it causal.

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    $\begingroup$ And if you're doing offline processing, and you'd prefer not to add a delay to all of your output timestamps, it's okay to leave it non-causal. $\endgroup$
    – hobbs
    Commented Mar 28, 2023 at 15:10
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Output starting at point before the start of input is possible:

  1. In case of "non-causal systems". Non causal systems present output depends on future inputs. Such systems are practically impossible and exist only in theory. Let $y(t)$ represent the output of a system for input $x(t)$ then the following equation represents a simple continuous time non-causal system

$$y(t) = x(t) + \alpha x(t + T)$$

where $\alpha \in R $ and $T > 0$

  1. This can also happen if the system's output is constant irrespective of input $$y(t) = C $$ where C is non-zero value.
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    $\begingroup$ Thanks for your answer. Ok , so no such system is possible in reality , then why are such systems extensively studied in DSP and Signals & Systems subjects ? what is the point ? ,any reason ? $\endgroup$
    – Madavan Viswanathan
    Commented Mar 28, 2023 at 7:21
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    $\begingroup$ I would be able to answer better if you can provide examples the system that you are studying. $\endgroup$
    – SakSath
    Commented Mar 28, 2023 at 7:23
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    $\begingroup$ Please consider an example system below for which the impulse response $h(n)$ is given below , it can be noted that the input impulse is given at $t=0$ but there is an output pulse extending in to $t \lt 0$. qph.cf2.quoracdn.net/main-qimg-6c722c6f63d530f9cbba9ecf052e65f2 $\endgroup$
    – Madavan Viswanathan
    Commented Mar 28, 2023 at 8:18
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    $\begingroup$ -1 "Such systems are practically impossible and exist only in theory" incomplete statement $\endgroup$
    – OverLordGoldDragon
    Commented Mar 28, 2023 at 11:20

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