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 ?


2 Answers 2


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.

  • 4
    $\begingroup$ 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
    Mar 28, 2023 at 15:10

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.
  • 1
    $\begingroup$ 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 ? $\endgroup$ Mar 28, 2023 at 7:21
  • 1
    $\begingroup$ I would be able to answer better if you can provide examples the system that you are studying. $\endgroup$
    – SakSath
    Mar 28, 2023 at 7:23
  • 1
    $\begingroup$ 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 $\endgroup$ Mar 28, 2023 at 8:18
  • 2
    $\begingroup$ -1 "Such systems are practically impossible and exist only in theory" incomplete statement $\endgroup$ Mar 28, 2023 at 11:20

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.