There are may ways to explain convolution. But I prefer this example.

# 1

Imagine you are throwing a stone into a lake.

As the stone hit to the lake, you will see some ripples on the surface of the lake.

That is the *response* of the lake towards your projection of stone.

# 2

Now, let say we want to model this.

Your stone is an **unit impulse** ( ) and the lake is a transfer function .

We don’t what exactly is yet.

However, when your stone hits into the lake , we see a ripple or response . That ripple is the response of the lake towards your stone.

Or, mathematically, convolute with and we see an output something like the figure below.

*(Of cause, the response of the lake is not a pure rectangular wave form. This is just for our ease of illustration.)*

# 3

Now, you throw two stones (two impulse) one after another into the lake (), you suppose to see two responses.

Say the interval of your projections (interval of consecutive ) is longer than response time, you will two distinct rectangular waveforms.

Condition: …

# 4

However, if your projection time interval is shorter than the response time, you will the superposition of two waveforms.

Condition: …

# 5

Now, a little bit imagination …

If there are a multiple of function convolute with , you will see something like a triangle below as below.

# 6

Further imagine …

Let say the gap between any two consecutive impulses become very small, ….

Or in another words, we say the these impulses become a rectangular waveform …

Then, the convolution of this rectangular pulse and will become a very smooth triangle …

# —

I will try to derive more proofs for Section #6 in future, especially the part where the sequence of delta functions become a rectangular pulse .