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 ($\delta(t)$ ) and the lake  is a transfer function $h(t)$.

We don’t what exactly is $h(t)$ 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,  $\delta(t)$ convolute with $h(t)$ 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 ($h(t)$), you suppose to see two responses.

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

Condition: $a>w$

# 4

However, if your projection time interval is shorter than the $h(t)$ response time, you will the superposition of two waveforms.

Condition: $a

# 5

Now, a little bit imagination …

If there are a multiple of $\delta(t)$ function convolute with $h(t)$, 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,  $\Delta \to 0$ ….

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

Then, the convolution of this rectangular pulse and $h(t)$ 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 .