FPGA Central

bruce varley <[email protected]> wrote in message news:...
>
> Bas <[email protected]> wrote in message
> news:[email protected] ups.com...
> > Hi Groups,
> >
> > I am trying to stabilize a laser which has one slow input with a huge
> > control range and a second input which is really fast, but has a
> > limited range only. I have an error signal which I feed with an analog
> > PID-type controller directly to the fast input. The problem is that the
> > integrator of the PID saturates after some time.
> >
> > To desaturate the integrator (and to keep the fast input as close to
> > zero for another reason) I can use the slow input. This is controlled
> > by a DSP, so I do have some flexibility in how I handle that. Right now
> > I use the output of the fast analog PID as the input to the DSP, and
> > configured the DSP to be a pure integrator with a certain time
> > constant.
> >
> > /---\ /----------\
> > /-->|DSP|-->|slow |
> > /---\ | \---/ | Laser|-----*----> optical frequency
> > /-->|PID|---*---------->|fast | |
> > | \---/ \----------/ |
> > | /-----\ |
> > \---------------------|error|<-----------/
> > \-----/
> >
> > My question is if this is allowed by control theory. If I integrate the
> > output of the PID with the DSP I have roughtly a double integrator in
> > the slow branch. According to my (limited) understanding this gives a
> > delay of 180 degrees, which would mean trouble. Would this lead to
> > instabilities at low frequencies? And are there better solutions to
> > desaturate the PID?
> >
> > Thanks in advance,
> > Bas
> >
> > Experimental details:
> > The process is an external cavity laser diode, which has both a fast
> > (~1MHz) input to the laser diode current and a slow (~2kHz) input to a
> > piezo that drives a grating. I obtain an error signal from a resonance
> > of a Fabry-Perot cavity by using a fast modulation scheme (Pound
> > Drever). The error signal has a zero crossing with a really steep slope
> > at the point where we want to stabilize. The DSP is a DSpace 1104 which
> > runs at a sample rate of 40 kHz. It can be programmed via Simulink.
> >
> > I do have a background in physics with only one introductory course in
> > control theory and some practical experience, so please be gentle...
>
> Bas, A bit of clarification would help us control types to understand
your
> system better and maybe help you more effectively. For someone that
> understands optics, these questions may seem dumb.
>
> Are you only attempting to control one variable, and is the error
(roughly)
> linearly proportional to the weighted sum of the two handles, ie. error
> responds as (A * delta fast input + B * delta slow input)? I guess that
the
> thing you're trying to control is wavelength.
>
> The frequencies that you quote, are they roughly representative of the
> response times of the device to changes in that input? Are sinewaves
> actually involved, or are you just using frequency to indicate typical
rates
> of change?
>
> Based on the limited understanding I've gleaned from the OP, yes, the
double
> integration could well cause problems, but the limited range of one of
your
> inputs could also be troublesome, causing limit cycling. It looks like at
> least a Proportional + integral action in the slow path could give you a
> better response.
>
> There are standard techniques for handling this sort of thing, the
> industrial analog would be the big valve/small valve situation for
trimming
> reagent addition. If you can post a bit more info, happy to take a closer
> look, and I suspect a few others will as well. Regards
>
Put your diagram into the correct font, most questions now answered. Still
like to know about the sinewaves..... Later

Sorry, forgot to tell that you have to view the diagram in a monospaced
font.

Yes, I am indeed trying to control the wavelength, which is directly
related to the optical frequency by wavelength = (speed of
light)/(optical frequency), so these two are equivalent.

I also assume that the output is proportional to the weighted sum of
the two inputs, so the optical frequency nu = k_fast*V_fast+
k_slow*V_slow+nu_0.

I did measure the transfer function of both inputs, and both have
roughly a flat response up till the given frequencies (~2kHz and ~1
MHz), after which the response drops of as some higher order filter.
There are no sinewaves involved, I just indicated the bandwidth of the
two inputs.

The error signal is linear to the deviation from the optical frequency
of the resonance we want to lock to: e = K*(nu-nu_resonance), and has a
first order low-pass character of some 100 kHz.

Cheers,
Bas

[email protected] wrote:
> Sorry, forgot to tell that you have to view the diagram in a monospaced
> font.
>
> Yes, I am indeed trying to control the wavelength, which is directly
> related to the optical frequency by wavelength = (speed of
> light)/(optical frequency), so these two are equivalent.
>
> I also assume that the output is proportional to the weighted sum of
> the two inputs, so the optical frequency nu = k_fast*V_fast+
> k_slow*V_slow+nu_0.
>
> I did measure the transfer function of both inputs, and both have
> roughly a flat response up till the given frequencies (~2kHz and ~1
> MHz), after which the response drops of as some higher order filter.
> There are no sinewaves involved, I just indicated the bandwidth of the
> two inputs.
>
> The error signal is linear to the deviation from the optical frequency
> of the resonance we want to lock to: e = K*(nu-nu_resonance), and has a
> first order low-pass character of some 100 kHz.
>
> Cheers,
> Bas

What kind of accuracy are you after?
And what do you mean by the integral is saturating? Do you mean the PID
has an integral limit which is being hit, or do you mean that the
integral term is saturating the output excessively?

It might be best to totally remove the analog PID and replace it with a
DSP implementation. That way you will have full control over it for the
task you wish to use it for. PID loops are very easy to code, it's just
the tuning that is the difficult part. And it sounds like you'll have
just as many problems trying to tune the double loop situation.

Why not filter the high frequency control signal from the low frequency by
putting a high pass/low pass filter pair with a cutoff frequency of say 50 Hz.
You would need to be careful about the phase response of the the filters
which would make the digital processor mentioned in an earlier post a great
idea.

In article <xCQbf.1546$[email protected]>, Bevan Weiss
<[email protected]> wrote:
>[email protected] wrote:
>> Sorry, forgot to tell that you have to view the diagram in a monospaced
>> font.
>>
>> Yes, I am indeed trying to control the wavelength, which is directly
>> related to the optical frequency by wavelength = (speed of
>> light)/(optical frequency), so these two are equivalent.
>>
>> I also assume that the output is proportional to the weighted sum of
>> the two inputs, so the optical frequency nu = k_fast*V_fast+
>> k_slow*V_slow+nu_0.
>>
>> I did measure the transfer function of both inputs, and both have
>> roughly a flat response up till the given frequencies (~2kHz and ~1
>> MHz), after which the response drops of as some higher order filter.
>> There are no sinewaves involved, I just indicated the bandwidth of the
>> two inputs.
>>
>> The error signal is linear to the deviation from the optical frequency
>> of the resonance we want to lock to: e = K*(nu-nu_resonance), and has a
>> first order low-pass character of some 100 kHz.
>>
>> Cheers,
>> Bas
>
>What kind of accuracy are you after?
>And what do you mean by the integral is saturating? Do you mean the PID
>has an integral limit which is being hit, or do you mean that the
>integral term is saturating the output excessively?
>
>It might be best to totally remove the analog PID and replace it with a
>DSP implementation. That way you will have full control over it for the
>task you wish to use it for. PID loops are very easy to code, it's just
>the tuning that is the difficult part. And it sounds like you'll have
>just as many problems trying to tune the double loop situation.