Question about difference between White Noise and Gaussian White Noise

[JohnLai]

I am confused by the power sense of the White Noise and Gaussian White Noise. Just look at the average powers of this two types of signals:

1) For White Noise: S_nn(f)=N/2 and the total power P_average = infinity

2) But for Guassian White Noise, the average power can be expressed as

P_average = E[|n(t)|²] = Var[n(t)], which is a finite number, because it is just the variance of a Gaussian random variable .

Can anybody please explain? Thank you.

 

[rhearty1]

I don't think P_average = infinity for white noise.

 

[FvM]

I don't think P_average = infinity for white noise

It is - by definition. Finite power density multiply infinite bandwidth equals infinite power. But it's a purely theoretical construct.

 

[rhearty1]

It is - by definition. Finite power density multiply infinite bandwidth equals infinite power. But it's a purely theoretical construct.

Thanks, I just read about that. Now I'm also interested in an answer to his question...

 

[iVenky]

I will tell you about Gaussian White noise first- Take a microphone (if you have one) and don't speak anything. Just record it using MATLAB and plot it. What you would see is some kind of noise. Now if you plot the histogram(pdf) of it what you would get is a Gaussian distribution. Similarly if plot the frequency spectrum of that, you would get same value in all the frequencies. That's the reason we call this as Gaussian White Noise. Now coming to your question- I don't think Pavg is infinite for white noise. Whatever operation you do, you always limit the bandwidth due to some filtering so that Pavg will never be infinite. I have read somewhere that at some higher frequencies there is no white noise due some kind of quantum effect that I don't remember and it was theorized by Max Planck (It's somewhat related to Planck's constant). So Pavg will never be infinite even though you have an infinite bandwidth.

 

[FvM]

Now coming to your question- I don't think Pavg is infinite for white noise. Whatever operation you do, you always limit the bandwidth due to some filtering so that Pavg will never be infinite. I have read somewhere that at some higher frequencies there is no white noise due some kind of quantum effect that I don't remember and it was theorized by Max Planck (It's somewhat related to Planck's constant).

This is a reasonable consideration about physical reality, but hasn't to do with the theoretical concept of white noise. If your mathematics teacher is asking about infinite numbers, he hardly won't accept the answer, that infinity doesn't exist in reality...

So Pavg will never be infinite even though you have an infinite bandwidth.

That's wrong in any case. Pavg of a real noise signal is finite due to finite bandwidth.

 

[dave9000]

the average power can be expressed as P_average = E[|n(t)|²] = Var[n(t)], which is a finite number

Yes, but the power/variance itself is a function of the bandwidth - a typical example is the thermal noise.

 

[JohnLai]

Thanks Folks, even though the physics of Gaussian White noise is band-limited, but the question is how you calculate the average power. Let's review the definition of average power for random noise:

Pave = 1/T * \[\int\] ( |n(t)|^2 ) dt = E (|n(t)|^2) = Var ( |n(t)| ), where T is the calculation time period.
= Rnn(0) = \[\int\] ( Snn(f) ) df , where Snn(f) is the power spectrum of random noise.

If the noise is strictly white, the Snn(f) = N/2 for all frequencies, then Pave = Rnn(0) = N/2*\[\delta\](0) is infinity.

But for a Gaussian White noise, Pave = Var (|n(t)| ) = 1 for standard Gaussian noise.

So the puzzle is that
the bandwidth of the Gaussian white noise is wide but bandlimited, Pave should be a huge number and can not be that small as 1 in the standard case.