Fourier transform of the auto correlation function to get energy

[thomas49th]

Homework Statement

The Fourier transform of the auto correlation function is the energy spectral density (ESD) of a signal. Here is the "apparent" proof:

$$\int e^{-jwT} [ \int g(t)g(t+T)dt] dT$$
=> $$\int g(t)[ \int g(T+t)e^{-jwT}dT] dt$$

What happened here? Why did the second integral change from t to T, why did t+T turn to T+t and the exponent change from jwt to jwT?

Thanks
Thomas

 

[Sourabh N]

Your first expression looks wrong.

$$\int e^{-jwt} [ \int g(t)g(t+T)dt] dt$$

IF the first integral (the one inside []) is over t, then the resulting function is independent of t, so second integral with the same t will be trivial. You might want to check the source again for a possible typo. Here is what I think it should look like

$$\int e^{-jwT} [ \int g(t)g(t+T)dt] dT$$

From which the second expression can be obtained.

 

[thomas49th]

you're right - i amended it in the first post. but I'm still stuck on it :\

 

[vela]

They're just changing the order of integration.

 

[thomas49th]

sorry... i don't see? How can you just flip around variables of function arguments?

 

[vela]

You can pull the exponential into the inside integral to get

$$\int e^{-jwT} \int g(t) g(t+T)\,dt\,dT = \int \int e^{-jwT} g(t) g(t+T)\,dt\,dT$$

Now switch the order of integration and then pull out the factors that don't depend on T from the inside integral.

 

[thomas49th]

what do you mean by order of integration. A quick Google seems to point me at double integration. Something which I haven't done yet. If it is double integration I question the material of my course, seeing as they haven't introduced that concept yet.

 

[vela]

Well, you have two integral signs, right? That's a double integration, one with respect to t and one with respect to T.