Why is the impulse response flipped in the convolution definition?

[spaghetti3451]

I am trying to understand wikipedia's definition of convolution: http://en.wikipedia.org/wiki/Convolution#Definition .

I'm wondering why g(tau) is flipped in the definition.

 

[mathman]

I am trying to understand wikipedia's definition of convolution: http://en.wikipedia.org/wiki/Convolution#Definition .

I'm wondering why g(tau) is flipped in the definition.

Try to be more precise in your question. The wikipedia article looks fine.

 

[spaghetti3451]

That's for sure! I just don't understand why g(tau) has to be flipped. (why has convolutio been defined in this way?)

 

[micromass]

Because everything works that way. A better question might be "what are the uses for convolution". On example I know is from probability theory. Say you have a random variable X and a random variable Y, then you might want to figure out the distribution of the random variable X+Y. That is, you might want to ask questions like, what is

$$P\{X+Y\leq 0\}$$

It appears that the distribution of X+Y is exactly the convolution of the distributions of X and Y. If tau wasn't "flipped" in the definition, then this wouldn't work anymore.

Also, for the convolution to have the many nice properties it has now, we must have defined the convolution this way and not another If "tau were flipped", then I don't think the convolution would have been commutative for example. (but you should check this).

 

[MisterX]

If the -tau in g(t - tau) is were positive instead, it would be something that is called http://en.wikipedia.org/wiki/Cross_correlation" .

One thing you may know about convolution is the output of an LTI system is the convolution of the input signal with the impulse response of system.

Let's say f is the input and g is the impulse response of the system. So, if f(tau) was an impulse at tau = 0, the output of the system should just be g(t), and the convolution integral is also equal to g(t). But what if the input f(tau) was instead an impulse at tau = 1? Then, I would expect the same response, only delayed by 1, so the response should be g(t - 1). The convolution integral in this case is equal to g(t - 1), just like I would expect.

If instead we used the cross correlation of the impulse at 1 and g, the integral would be equal to g(t + 1).

 

[paulfr]

In Electrical Engineering we learn that
Convolution in the Time Domain (TD) = Multiplication in the Frequency Domain (FD)

As said above
"One thing you may know about convolution is the output of an LTI system is the convolution of the input signal with the impulse response of system."
This is the Time Domain.

If you take the Laplace Transform of the input and then the Impulse Response and multiply them, you get the Output in FD. Take the inverse Laplace Xform of this Output and you get the result of the convolution in the TD.