# How can an impulse generate an output in the past time frame?

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:

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 ?

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.

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