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<w

# 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 .

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