Shot noise factor of 2

The mysterious factor of 2 is due to the analytic signal convention, where we use complex exponentials but chop off negative frequencies. All test equipment uses it AFAIK.

The Fourier transform of a 1-second boxcar is sin(pi*f)/(pi*f), which is

The rules are:

1. Leave the DC alone
2. Double the amplitude at positive frequencies
3. Chop off the negative frequencies
4. Switch back to sines and cosines before analyzing anything nonlinear, e.g. computing the power.

Cheers

Phil Hobbs

the tricky part is relating the bandwidth to the time T. So here the auth or states that BW = 1/(2*T)

"Ding Ding".. (bells going off in my head) Thanks very much Phil. (I was kinda hoping you'd have the answer.)

George H.

i_n^2 = 2*e*I_avg*BW

Where e is the electron charge and BW is the bandwidth. I?ve been trying to derive the factor of 2 in the formula. But I?m not quite getting it.

A typical derivation looks at random pulses measured over some time T.. the tricky part is relating the bandwidth to the time T. So here the author states that BW = 1/(2*T) and says that?s related to the Nyquist sampling theorem.

But I don?t see it? Is that correct? Help!

TIA George H.

That kind of 'reasoning' has no basis whatsoever in reality. A derivation grounded in reality is found in Apendices C & D here:

You can read all about the Hilbert transform stuff in many different places (I discuss it in my book, for instance). I claim that the four rules above give the correct analytic signal from any not-too-pathological two-sided complex transform.

Can you give a counterexample?

Cheers

Phil Hobbs

Thanks Fred, I think Phil nailed it for me.

But other derivations are always nice to look at.

George H.