On 03/11/2013 10:00 AM, George Herold wrote:
> On Mar 9, 10:06 pm, Phil Hobbs
> <pcdhSpamMeSensel...@electrooptical.net> wrote:
>> On 3/9/2013 8:05 PM, George Herold wrote:
>>
>>
>>
>>
>>
>>> On Mar 9, 7:46 pm, Phil Hobbs <pcdhSpamMeSensel...@electrooptical.net>
>>> wrote:
>>>> On 3/9/2013 6:51 PM, bloggs.fredbloggs.f...@gmail.com wrote:
>>
>>>>> On Friday, March 8, 2013 3:59:36 PM UTC-5, Phil Hobbs wrote:
>>
>>>>>> It's actually more general than that--there's a Fourier transform
>>>>>> theorem that variances add under convolution, ...
>>>>> There is? It seems that is a result only for normal distributions.
>>
>>>> Yup. From the derivative theorem, the variance is proportional to the
>>>> second derivative of the transform, evaluated at zero frequency. To
>>>> within a constant factor that depends on your Fourier transform definition,
>>
>>>> int(-inf to inf) [t**2 * h(t)]
>>>> var(h) = ------------------------------- = H''(0)/H(0)
>>>> int(-inf to inf) [h(t)]
>>
>>>> Because the function is real, its transform is Hermitian, i.e. the real
>>>> part is even and the imaginary part is odd. Even-order derivatives of
>>>> an odd function vanish at the origin, as do odd-order derivatives of an
>>>> even function.
>>
>>>> When you convolve g(t) and h(t), the transform is G(f)H(f). The
>>>> variance of this is proportional to the normalized second derivative at
>>>> the origin, as before. Ignoring a possible constant of proportionality
>>>> that we don't care about,
>>
>>>> d(GH)/df = GH' + HG' so
>>
>>>> d**2/df**2[GH]| H(0)G''(0) + G(0)H''(0) + 2G'H'
>>>> var(g*h) = --------------| = -------------------------------
>>>> GH |f=0 H(0)G(0)
>>
>>>> G''(0) H''(0) 2G'(0)H'(0)
>>>> = ------- + -------- + ------------
>>>> G(0) H(0) H(0)G(0)
>>
>>>> The first term is the variance of G, the second is the variance of H,
>>>> and the third is zero if either G or H is an even function.
>>
>>>> There is a slightly more subtle condition that will make the third term
>>>> zero for any choice of g and h: that the first moment of either h(t) or
>>>> g(t) is zero, i.e. that at least one of them has its centroid at t=0,
>>>> which makes its first derivative zero at f=0.
>>
>>>> It is always possible to satisfy this condition by an appropriate choice
>>>> of the time origin, so if you'll allow me to slide over that rather
>>>> trivial issue(*), variances add under convolution.
>>
>>>> Cheers
>>
>>>> Phil Hobbs
>>
>>>> (*) The reason for this is the effect of a shift of origin on the
>>>> variance. If you convolve a function centred at t=5 with one centred at
>>>> t=3, the convolution's centroid is at t=8. If you're computing the
>>>> variance as the second moment about t=0, this will make a big
>>>> difference, but it doesn't change the shape of the resulting convolution
>>>> function. Forcing one of them to have its centroid at 0 gets rid of
>>>> this shift of time origin.
>>
>>> Awesome... (Be careful what you ask for.)
>>
>>> Phil, If I mix two lasers together on a photodiode and look at the
>>> bandwidth of the beat note to measure the laser bandwidth.. Is the
>>> measured bandwidth the quadrature sum of the individual lasers (as
>>> above for rise times, noise..) or does the power law nature of the PD
>>> put a twist on it?
>>
>> Hmm, no good deed goes unpunished. ;)
>>
>> The PD is square law, but it isn't E**2, it's |E**2|, so to be
>> completely safe you have to stay in real-valued functions (sin and cos)
>> and not complex exponentials. (At least at first.)
>>
>> Assuming that the relative phase of the two beams depends only on time,
>> i.e. that they're both really really single transverse mode, then
>> shining E1(x, y, t) and E2(x, y, t) on a sufficiently large photodiode
>> will get you
>>
>> i_photo(t) = int(-inf < x < inf) int(-inf < y < inf)[ R|(E1(x,y,t) +
>> E2(x,y,t)|**2]dxdy
>>
>> = R int int[ |E1|**2 + |E2|**2 + 2 Re{E1 E2*}]dxdy
>>
>> where R is some constant (R is the responsivity if E**2 has units of
>> watts per square metre, but it doesn't matter here).
>>
>> With a large enough beat frequency, we can regard E1**2 and E2**2 as DC
>> and filter them out. (Since the negative frequencies are twice as far
>> away as that, we can use the complex exponential notation with no
>> worries.) So at AC, all we get is
>>
>> i_AC = 2R Re {int int[ E1(x,y,t)E2*(x,y,t)]dxdy }
>>
>> If both lasers are single transverse mode, this integral is proportional
>> to Re{E1(0,0,t) E2*(0,0,t)}. Since the beat frequency is large, we
>> sample all relative phases of E1 and E2 much more rapidly than their
>> fluctuations, so we pretty much get an honest multiplication.
>> Multiplying them in the time domain is convolving them in the frequency
>> domain, so yes, the frequency variances add, *provided they're computed
>> around the nominal beat frequency*. Of course nobody in his right mind
>> would do anything else, but a computer might. ;)
>>
>> The only thing that modifies this significantly is the frequency
>> response of the photodiode and the electrical measurement system.
>>
>> Cheers
>>
>> Phil Hobbs
>>
>> --
>> Dr Philip C D Hobbs
>> Principal Consultant
>> ElectroOptical Innovations LLC
>> Optics, Electro-optics, Photonics, Analog Electronics
>>
>> 160 North State Road #203
>> Briarcliff Manor NY 10510 USA
>> +1 845 480 2058
>>
>> hobbs at electrooptical dot nethttp://electrooptical.net- Hide quoted text -
>>
>> - Show quoted text -
>
> Cool, thanks Phil, (In practice I just quote the laser bandwidth as
> the beat note bandwidth and ignore the above convolution.)
> Ahh, what's the line about the PD size "on a sufficiently large
> photodiode" refer too?
> (I used this tiny EOT photodiode ET-2030.)
>
> George H.
>
If the PD is too small, spatial fringes won't average out to zero, so
orthogonality doesn't quite work.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net