Luckily i know how to use the integration feature on my calculator but assuming i didn't a problem on my homework requires finding the entropy of a continuous time signal.

fx(a)= 0.7979e^(-2a^2)

where

the entropy H(x)= -integral(fx(a)*log2(a) da)

so the equation is H(x) = -integral(.7979*e^(-2a^2)* log2(0.7979*e^(-2a^2)) da)

is there an easy way to solve this by hand? been awhile since MA 241

4/9/2008 11:12:20 AM

I've got some ideas to suggest, but first I wanted to ask: How did you get from here:

H(x)= -integral(fx(a)*log2(a) da)

to here:

H(x) = -integral(.7979*e^(-2a^2)* log2(0.7979*e^(-2a^2)) da)

Is there a typo? They don't seem to be the same. The first has a factor of log2(a) in the integrand, while in the second the factor becomes log2(fx(a)). Which one should it be?

4/9/2008 1:57:17 PM

Actually, though, whichever factor it should be, I tend to think it's impossible to find a nice antiderivative for it. What are the bounds of integration? There may be some tricks, given certain "nice" specific bounds. Otherwise, numerical approximation may be the only way to go.

[Edited on April 9, 2008 at 2:10 PM. Reason : Added last sentence]

4/9/2008 2:09:34 PM

isn't that a Gaussian integral?

Int(exp(-a^2)da) = sqrt(pi)

something like that, the integral is over all z (-infinity to infinity)

[Edited on April 9, 2008 at 2:14 PM. Reason : .]

4/9/2008 2:13:33 PM

Quote :

"H(x)= -integral(fx(a)*log2(a) da)"

yeah i made a mistake should have been

H(x)= -integral(fx(a)*log2(fx(a)) da) da

Mathman was right it does have something to do with a gaussian integral not sure how though

4/10/2008 11:39:38 AM

I don't know if this will be helpful or not, but what I see is this:

Most of the constants can be dealt with by straightforward methods, except for possibly the .7979 inside the logarithm, so I'll mostly ignore them. Any two logarithms of different bases differ only by a constant factor, so to simplify things I'll treat log2 as the natural logarithm ln. Considering that, the integral simplifies to:

integral(e^(-a^2)*ln(C*e^(-a^2))da

(where C=.7979)

ln of a product is the sum of two logarithms, so this integral becomes:

integral(e^(-a^2)*ln(C))da

plus

integral(e^(-a^2)*(-a^2))da

The first integral is basically Gaussian; the second integral becomes Gaussian after a basic integration by parts step (u = a, dv = -a*e^(-a^2)da).

4/10/2008 7:21:43 PM