Meaning of convolution?

Question

I am currently learning about the concept of convolution between two functions in my university course. The course notes are vague about what convolution is, so I was wondering if anyone could give me a good explanation. I can't seem to grasp other than the fact that it is just a particular integral of two functions. What is the physical meaning of convolution and why is it useful? Thanks a lot.

Answer 1

Have a look here:
http://answers.yahoo.com/question/index?qid=20070125163821AA5hyRX

...and lots of good answers here:
https://mathoverflow.net/questions/5892/what-is-convolution-intuitively

Answer 2

(This is from Lecture 8 of Brad Osgood’s lectures here, already mentioned above)

Consider signals ${ f(t), g(t) }.$ We have their Fourier transforms ${ (\mathcal{F}f)(s) = \int _{-\infty} ^{\infty} e ^{-2\pi i s t} f(t) dt }$ and ${ (\mathcal{F}g)(s) = \int _{-\infty} ^{\infty} e ^{-2 \pi i s t} g(t) dt }.$

Is there a single signal whose Fourier transform is sum of Fourier transforms ${ \mathcal{F}f + \mathcal{F}g }$ ?
By ${ \mathcal{F}(f+g) = \mathcal{F}f + \mathcal{F}g }$ we see ${ f + g }$ is such a signal.

Is there a single signal whose Fourier transform is product of Fourier transforms ${ (\mathcal{F}f)(\mathcal{F}g) }$ ?

To look for such a signal, we simplify

${ \begin{align*} (\mathcal{F}f)(s) \,(\mathcal{F}g)(s) &= \left( \int _{-\infty} ^{\infty} e ^{- 2 \pi i s t} f(t) dt \right) \left( \int _{-\infty} ^{\infty} e ^{-2 \pi i s x} g(x) dx \right) \\ &= \int _{-\infty} ^{\infty} \int _{-\infty} ^{\infty} e ^{-2 \pi i s (t+x)} f(t) \, dt \, g(x) \, dx \\ &\overset{u=t+x}{=} \int _{-\infty} ^{\infty} \int _{-\infty} ^{\infty} e ^{-2 \pi i s u} f(u-x) \, du \, g(x) \, dx \\ &\overset{\text{interchange}}{=} \int _{-\infty} ^{\infty} \int _{-\infty} ^{\infty} e ^{-2 \pi i s u} f(u-x) g(x) \, dx \, du \\ &= \int _{-\infty} ^{\infty} e ^{-2 \pi i s u} \underbrace{\left( \int _{-\infty} ^{\infty} f(u-x) g(x) \, dx \right)} _{ =: \, h(u)} \, du \\ &= (\mathcal{F}h)(s). \end{align*} }$

This gives ${ (\mathcal{F}f) \, (\mathcal{F}g) = (\mathcal{F}h) }$ where $${ h(u) = \int _{-\infty} ^{\infty} f(u-x) g(x) \, dx }.$$

Answer 3

The Wikipedia has some nice graphical explanations.