# Why do I need to shift time for getting a response?

Well this is a basic doubt.

Let us take an example of a discrete time impulse response of a system as h[n] which has components at n= 0,1,2.

Now my input signal is made of x[0]+x[1]

I know the summation says to shift h[n] by 1 which is h[n-1] and then get output response as

y[n]= x[0].h [n]+x[1].h [n-1]

But why do I have to shift this h[n]? Is it because I have a input at x [1] also? If so then h[n] also has a component at n=1 so why the time shift is required?

• Perhaps this answer will help. Commented May 25, 2022 at 2:41

The impulse response has components at n = 0, 1, and 2, which means for any scaled impulse at the input at any given time m, we will get an output at time m, m+1, and m +2 as the scaled impulse response:

Arbitrary inputs can be viewed as a sequence of weighted impulses. Given an assumed Linear Time-invariant system, each impulse will produce a response according to the impulse response and its own weighting. The continued response of past inputs will add to the response from the current input with a result as the summation of all the responses, each delayed by when they appeared at the input:

This result is the convolution of the input with the impulse response, and explains intuitively why we do the time reversal, delay and add operations for graphical convolution. Please also see this post which details this further.