[time-nuts] New Question on HP3048A Phase Noise Test Set

Bruce Griffiths bruce.griffiths at xtra.co.nz
Thu Jan 24 00:13:25 EST 2008


BriMDavis at aol.com wrote:
> Bruce Griffiths wrote:
>   
>> After low pass filtering
>>
>> Vo(t) = (A2/2)*sin((w1-w2)t)
>>
>>
>> Thus the amplitude of the discrete spur related component in the low
>> pass filtered phase detector output is 1/2 (6dB) the amplitude of the
>> discrete spur itself.
>>
>>
>> This result is independent of any additional calculations that may be 
>> done to derive L(f) from the observed spectrum.
>>
>>     
>  
>  I agree completely that the phase detector output amplitude 
> of your ideal multiplier is half the input spur amplitude.
>  
>  But I believe that particular 0.5 drops out later along the 
> way in the computation of S_phi(f) and L(f), as converting
> from voltage to phase requires division by the phase detector
> constant Kd, which is also 1/2 for your ideal multiplier.
>   
Yes I realised that as I fell asleep and I was going to comment on it
but I thought that I would gather it all together in pdf file for easier
reference.
To generate a beat frequency signal for calibration requires one to use
an RF signal which is offset from the LO frequency, so naturally their
is no difference between this and Martyn's case.
>  
>  Here's my attempt at continuing on to calculate S_phi(f) and L(f)
> for your example phase detector ( somewhat simpler than a real 
> world calculation requiring gain and calibration corrections ) :
>  
>
> S_phi is defined as the square of the RMS phase
> :
> :  S_phi = phase_rms ^ 2
> :
>  
> RMS phase is the RMS voltage over the phase detector constant:
> :
> :  phase_rms =  Vrms / Kd 
> :
>  
> Substituting to get S_phi in terms of Vrms and Kd:
> :
> :  S_phi = [ Vrms / Kd ] ^ 2
> :
>  
> Given your earlier calculation:
> :
> :  Vphase_detector_out = 0.5 * Vspur
> :
>  
> The rms voltage at your ideal phase detector output is therefore:
> :
> :  Vrms = 0.707 * Vphase_detector_out
> :       = 0.707 * 0.5 * Vspur
> :
>  
> The phase detector constant, Kd, is 0.5 for your ideal multiplier:
> :
> :  Kd = 0.5
> :
>  
> Plugging in, we get:
> :
> :  S_phi = [ Vrms / Kd ] ^ 2
> :        = [ 0.707 * 0.5 * Vspur / 0.5 ] ^ 2
> :        = [ 0.707 * Vspur ] ^ 2
> :        = 0.5  * Vspur^2
> :
> :      L = 0.5  * S_phi         ( small angle approximation L ~= S_phi/2 )
> :        = 0.5  * 0.5 * Vspur^2
> :        = 0.25 * Vspur^2
> :
>  
>  As these are powers, S_phi(f) is -3 dB, and L(f) is -6 dB
>  
>  Which is exactly what Martyn's 3048A has been telling him !
>  
>
> Brian
>  
>
> p.s.
>  
>  One interesting detail of the calculation of S_phi 
> is that the phase detector constant Kd depends upon
> the _peak_ value of the phase detector output.
>  
>   
This is the Achilles heel of that system, real phase detectors may not
have a sinusoidal response to the phase difference.
This is why NIST developed a phase angle modulator to calibrate phase
measurement systems without switching amplifier gains or disabling PLLs etc.
>  In an actual system with the phase detector constant 
> calibrated by a beat note measurement, if the system
> instrumentation measuring the beat note amplitude 
> returns an RMS value, that tosses another 1.414 into 
> the calculation of Kd to convert from RMS to peak.
>  
>  This is the source of the 'mystery' 3dB correction 
> John Miles mentioned elsewhere in the recent discussions:
>   
>> According to the notes written by the guy whose office 
>> door the 3048A authors would have knocked on for advice 
>> (see _www.ke5fx.com/Scherer_Art_of_PN_measurement.pdf_ 
>>     
> (http://www.ke5fx.com/Scherer_Art_of_PN_measurement.pdf)  page 12),
>   
>> you need to subtract 6 dB from the noise trace for two reasons.  
>> 3 dB of it comes from a mysterious "Accounts for RMS value of 
>> beat signal (3 dB)" clause in Scherer's app note.
>>
>>     
>  
> The top of page 35 ( original page numbering ) of the following 
> HP app note describes the beatnote correction :
>  
>  _http://www.home.agilent.com/upload/cmc_upload/All/5952-8286E.pdf_ 
> (http://www.home.agilent.com/upload/cmc_upload/All/5952-8286E.pdf) 
>  
The above analysis accurately describes the origin of the 6dB reduction
in the relative amplitude of the spur as seen by the phase measurement.

Bruce



More information about the time-nuts mailing list