# What does convolution has the meaning? [duplicate]

As I know, if we want to know the LTI system output, then we do convolution between input x[n] and impulse response h[n]. but actually,in this question, I want to know what does convolution has the behind meaning.

Why do we do convolution(sum of products) not just using adder or multiplier for calculation between input signal and impulse response?

## marked as duplicate by Matt L., jojek♦, Peter K.♦Mar 21 '16 at 11:48

• have a look to this definition (it is for continuous time, but it's equivalent in discrete time domain): en.wikipedia.org/wiki/Convolution#Definition there you can see graphically the differences between convolution and normal addition or multiplication – Behind The Sciences Mar 21 '16 at 8:02

Red is a reference delta impulse (of height 1), whereas green is a typical response to that impulse, denoted by function h(t). In LTI, output is proportional to the input, that is if we have a delta impulses at the input at time 0, output at time t will be a*h(t). Now, instead of single impulse at the origin, you apply a series of input impulses at various times. What will be the output? Say, there was input impulse of height $a_1$ at $T_1$. Since current time is $t$, impulse occurred $t-T_1$ seconds ago and its contribution to the current output y(t) is $a_1 h(t - T_1)$. There is another contribution from another impulse, occurred at $T_2$. Its contribution is $a_2 h(t-T_2)$. So, $y(t) = a_1 h(t-T_1) + a_2 h(t-T_2)$. You simply add up the contributions beacuse of LTI linearity.
In general, you have $y_t = \sum_0^t {a_i h(t-i)}$. That is a convolution formula.
It also appears when you multiply to polynomials $(a_0 + a_1 z + a_2 z^2 + ...)(b_0 + b_1 z + b_2 z^2 + ...) =\sum_0^\infty {c_n z^n}$ where $c_n = a_i b_{n-i}$. That is why you tend to represent series as z-transforms. In this case you can simply multiply them and have convolution on the background.