# [time-nuts] Re: Phase noise with a lock-in amplifier.

Javier javier at nebulosa.org
Sun Apr 17 14:30:14 EDT 2005

```Hello,

I think that there is a problem with the mixing of the oscillator with a
delayed version of itself. Suppose that the noise affecting the
oscillator were exclusively a square wave lets say at 1KHz rate, lets
say modulating the oscillator at +/- 1Hz. At any given instant, and
suppossing that the cable length is far less that to produce a 1ms delay
(1ms delay would be around 200Km of RG-58/U, if I remember welll), the
frequencies entering the mixer would be exactly the same, producing a DC
- you would only have glitches at a 1KHz rate, corresponding to the
instants when the two frequencies differ.

This can be translated to that for measuring the phase noise mixing the
signal with a delayed version of itself, you must have a delay at least
comparable to the lowest noise frequency you want to determine, in order
to have truly uncorrelated phase noise at the two mixer inputs.

Regards,

Javier, EA1CRB

David Kirkby wrote:

> Bill Hawkins wrote:
>
>> Dr. David Kirkby wrote,
>>
>> "I was not thinking of impedance matching at all. If the mixer is
>> 50Ohm input (as most are) and the oscillator has a 50 Ohm output, the
>> cable length would have no effect on this at all."
>>
>> The cable length has no effect on frequency (if the output device is
>> buffered with an amplifier).
>
>
> But what if a 10MHz (for example) oscillator is not perfect?  The
> output is no longer a pure sine wave at 10MHz, but could be considered
> as an infinite number of oscillators, all of different frequencies,
> with amplitudes and phases that are essentially random. Those
> oscillators closer to 10MHz would on average have higher amplitudes,
> but all oscillators will have some amplitude.
>
> So if the two inputs are fed to a mixer have different path lengths,
> the inputs to the mixers will *not* be identical.
>
>> As I understand it, a phase detector that has both inputs fed by the
>> same frequency is only capable of measuring systematic errors, such as
>> may be caused by differences in the FET response times. Of course, it
>> will also measure differences in cable length.
>
>
> But the inputs will not be of the same frequency normally, due to the
> phase noise on the oscillator.
>
>
>> There was a time when obscure discoveries could be rediscovered when
>> the right opportunity arose. These days, with information at the speed
>> of the Internet, the process of natural selection works much faster.
>> The phase noise question has been around longer than the Internet, so
>> it seems to be true that the best solutions have been found.
>
>
> Well, it never hurts to consider new ideas. That might be silly,
> fatally flawed in some way.
>
>
> I'm 99.9% sure the mixer, with two different cable lengths from a
> 10MHz oscillator will produce two ranges of frequencies. One will be
> close to DC, but will extend up a few hundred kHz or so. The other
> will be close to 20MHz, but extend a few hundred kHz either side.
>
> I'm pretty sure if you low-pass filter the output of that mixer, you
> will see the oscillator phase noise shifted from 10MHz down to near DC.
>
> I think if you put an audio spectrum analyser on the output of the
> mixer, after the low-pass filter, you would see the phase noise of the
> oscillator.
>
> In fact, would you not hear the oscillator phase noise on a speaker if
> you amplified the low-pass filtered signal from the mixer? (I'm
> ignoring the fact the noise on the audio system might be lower than
> the oscillator phase noise).
>
> I'm less sure, but it it possible a lock-in might be able to see this
> phase noise, by setting the reference to the offset you want.
>
> Question, what would you see on the lock-in if you used its internal
> oscillator as as reference and added to this (op-amp configured as an
> adder) the output from the mixer? The op-amp would measure its own
> reference amplitude, but you would have introduced noise on it from
> the 10MHz reference, so the lock-in should see that noise too.
>
> Perhaps I'll sketch out what I mean and put a circuit diagram.
>
> The Standford dual-phase lock-in has functions to measure the noise on
> the signal.
>

```