Optimize pow() function

Am 23.09.2016 um 08:15 schrieb Jack:

> start using integer math instead of float.

Good advice in general, but hardly practical while he's still dead-set 
on computing, of all things, a fractional power of his input value.

Am 23.09.2016 um 09:23 schrieb pozz:
> I thought about exponentiation, because it is simple to calibrate.

Simple to calibrate it may seem -- but then, so is almost every curve 
you can compute at all, with only 3 measurements and 3 parameters to 
determine from them.

But if all you're really after is a "small correction of the square 
law", why lock yourself so thoroughly on the idea that tweaking the 
exponent should be that small correction, rather than, say, a small 
shift of the zero point, or a bit of a cubic correction term?

Let's face it: of all the functions in the usual collection of standard 
math library functions (as in: the Standard C <math.h>), pow() is 
_by_far_ the least simple.  You really don't want to use that beast 
unless your machine has (for a task as hinted here) huge large amounts 
of FPU cycles you really cannot find more practical use for.

On 9/23/2016 12:23 AM, pozz wrote:
>> What is your *load*?  If purely resistive, does it have a temperature
>> coefficient?  (e.g., if you left it operating at Pmax, would you see
>> changes)
>>
>> In operation, do you expect the load to be "static"?  Or, time varying?
>> Does the load's impedance change based on how *it* is driven?  etc.
>
> I don't care the load.  I'm only working on the power transducer.  If the load
> changes (because of temperature, degradation, ...), the output signal of
> transducer will change accordingly and the power reading will change too.

(sigh)  You really don't understand the problem you're trying to solve.

I have a voltmeter, here, good to 5-1/2 digits.  It was calibrated 4 minutes
ago to an NBS traceable reference.  It's sitting in a temperature controlled
housing.  yada-yada-yada-yada.

It is presently displaying 1.92341 volts.  To what "power" does that
correspond?

[pull black curtain to hide my actions...]

Now, it's displaying 1.84421 volts.  What power level does THIS represent?
Is it MORE or LESS than the power above??  (think carefully before
answering).

How does your "professional instrument" (acting as your authoritative
reference) measure the power in the load?  Why do you trust *it* -- yet
KNOW that your measurements are incorrect?  What does *it* know that
you don't?  What is it *doing* that you aren't?

On 23.9.16 10:03, pozz wrote:
> Il 22/09/2016 23:31, rickman ha scritto:
>> On 9/22/2016 5:10 PM, pozz wrote:
>>> Il 22/09/2016 16:17, Don Y ha scritto:
>>>> On 9/22/2016 6:36 AM, pozz wrote:
>
>> How
>> are you getting the power measurement to compare to your calculations?
>
> Power meter.
>
>
>> What is the device which is the load for this calculation?
>
> The load is an RF amplifier.


A RF power amplifier usually reacts non-linearly to changes
in main supply voltage (plate/clooector/drain feed). In AM
transmitters special measures are taken to create a linear
realtionship between modulating voltage and output amplitude,
which would be the square law you're after.

For measuring the RF output, nothing beats a RF directional
wattmeter.

Please note that if the load behaves non-linearly or changes
its impedance, the relation between supply voltage and output
power will also change, often in a non-predicatable way.

-- 

-TV

On 9/23/2016 3:08 AM, pozz wrote:
> Il 22/09/2016 23:44, rickman ha scritto:
>> On 9/22/2016 5:10 PM, pozz wrote:
>>> Il 22/09/2016 16:17, Don Y ha scritto:
>>>> On 9/22/2016 6:36 AM, pozz wrote:
>>>>> I have to convert voltage measure (Volt) into power (Watt).  As you
>>>>> know, the
>>>>> physics says I need to square the voltage measure.
>>>>>
>>>>> I have to calibrate the mathematical law and I have two points:
>>>>> 0W -> 0V
>>>>> <nominal power> -> <voltage at nominal power>
>>>>>
>>>>> That could be enough, but we noticed a different relation between volt
>>>>> and watt
>>>>> at intermediate points. I, as the software engineer, have to try to
>>>>> compensate
>>>>> differences from the theory, without increasing complexity during
>>>>> calibration.
>>>>>
>>>>> My idea is to calibrate the exponent: it is normally 2.00, but I can
>>>>> calibrate
>>>>> from 1.00 to 3.00.
>>>>> In this way, the two old calibration points are the same as before.
>>>>> So the
>>>>> exponent (the third calibration measure) can be changed at last.
>>>>>
>>>>> The problem is pow() function with double arguments takes a lot of
>>>>> time to
>>>>> finish.  Do you suggest a mathematical method to reduce its
>>>>> complexity?
>>>>>
>>>>>
>>>>> uint16_t power_nominal = 1500;   // In Watt
>>>>> uint16_t volt_nominal;   // Volt measured at nominal power
>>>>> signed uint8_t exp_corr = -100 to +100;
>>>>> double exp = 2.00 + exp_corr;
>>>>>
>>>>> uint16_t volt = adc_convert(...);  // Actual volt measure
>>>>> double alpha = (double)power_nominal / pow((double)volt_nominal, exp);
>>>>> uint32_t watt = alpha * pow((double)volt, exp);
>>>>>
>>>>>
>>>>> alpha can be pre-calculated, because power_nominal, volt_nominal and
>>>>> exp are
>>>>> constant (after calibration). The last instruction is the problem.
>>>>
>>>> No, the "problem" lies in your transducer/measurement error.
>>>> The definition of a "watt" (power) doesn't change just because your
>>>> observations of that APPARENT relationship (V ~ W) "has issues".
>>>>
>>>> Perhaps your load's characteristics are changing as a function
>>>> of operating point.  Or, nonlinearities in your measurement system.
>>>> Or, ...
>>>>
>>>> Ask yourself WHY the exponent needs to be changed.  I suggest you
>>>> don't have an accurate model of your load and its excitation.
>>>> Until you do, how will you know that this is the CORRECT means of
>>>> compensating your model to agree with your observations?
>>>
>>>
>>> I know the problem is in the poor quality of the transducer, but I'm
>>> only in charge of the software and the hardware engineer can't fix it
>>> (often the software solves hardware issues).
>>>
>>> So I need a more complex law. The quadratic law (one parameter) can be
>>> calibrated with a single measure (nominal value). The generic power law
>>> (exponent not exactly equals to 2, two parameters) needs two calibration
>>> measures and has a nice feature: it's very simple to calibrate in two
>>> steps.
>>>
>>> The input section has an hardware calibration, a simple digital trimmer.
>>> So the first step is to calculate the power from the measured voltage
>>> with the following formula:
>>>   P = Pn / (Vn ^ 2) * (V ^ 2)
>>> where Pn is the nominal power and Vn is the voltage needed at nominal
>>> power.  The user increase/decrease the trimmer pot until the correct
>>> power is displayed.
>>>
>>> Second step. Change the power to another value (half of nominal power,
>>> for example). If the transducer works as in theory, the measured power
>>> is correct. Otherwise the user increase/decrease the exponent until the
>>> power displayed is correct. The power is calculated as:
>>>   P = Pn / (Vn ^ exp) * (V ^ exp)
>>>
>>> After the second step, we are sure the first calibration point is yet
>>> valid, because that formula is correct when V=Vn, even if exp isn't 2.
>>
>> I think your equation is not written in a clear way.  I think what you
>> mean is this...
>>
>> P = Pn * (V ^ exp) / (Vn ^ exp)
>
> IMHO, they are identical mathematical expressions.

It is equivalent, but the way you wrote it the order of operations has 
to be implied by the standard rules.  I typically don't write a division 
that way because often the equation really intended is

A = B / (C * D) rather than A = (B / C) * D

which are *not* equivalent.  So

A = B * D / C is more clear, to me anyway.  Or supply the parentheses.


>> Now the equation makes sense to me.  But that still doesn't mean it will
>> give close to correct values anywhere other than the calibration points.
>
> Yes, I have tried and this mathematical law could be ok for my application.
>
>
>> Have you considered...
>>
>> P = Pn * (V / Vn) ^ exp
>
> Again IMHO, this is identical to the above expressions.

Yeah, I got wrapped around the axle on this thinking of logs.  A 
division before taking the log is equivalent to adding the logs of the 
two values.

Still, again, this is a more clear way to write what you are doing.  The 
value Vn is just a scale factor.  Why complicated it by taking the 
exponent and dividing after the exponent is taken of the value measured? 
  At least this is more clear to me.


>> Then there are an infinite set of other corrections you could use.  The
>> question is which ones are more accurate?
>
> I don't need to increase accuracy.  I only need a little compensation of
> the square law.  The calibration process of the mathematical model
> should be simple.

Isn't the "compensation" to get better accuracy?


>> Have you looked at a range of data to see what the correction curve
>> needs to be?
>
> Yes, and it is somewhat a square law.  In some cases, I need to change
> the shape of the square law a little.

"A little" is a poor way to describe this.  There are many ways to 
correct your data.  You have chosen one that you find hard to calculate 
quickly enough.  So why not consider others which can be evaluated more 
quickly?  Others have given you other approaches.  I suggest you 
consider them more carefully and not wed yourself to using a variable 
exponent.  You say it is easy for the user to calibrate, but really you 
are saying it is easy for *you* to write the calibration algorithm.  If 
you can do the calibration with two measurements with your variable 
exponent, you can do a calibration with an x^3 term with two 
measurements.  The user doesn't need to know what the software is doing 
with those measurements.

-- 

Rick C

On 9/22/16 9:36 AM, pozz wrote:
> I have to convert voltage measure (Volt) into power (Watt).  As you
> know, the physics says I need to square the voltage measure.
>
Whoever told you that didn't give you the full message, as it applies in 
only very limited cases.

Electrical Power is Voltage * Current, not Voltage squared.

IF you have a resistive load, then we have that

Voltage = Current * Resistance.

 From this we can get to Power = Voltage * Voltage / Resistance.

This means that for a system with constant resistance we can say that 
Power goes as Voltage squared, with a scale factor of 1/Resistance.

If the resistance of the system isn't constant, this relationship 
absolutely doesn't hold. I have built systems where (within an operating 
band) the Power consumed goes up with dropping voltage.

If all you can measure in the field is the Voltage, and the system if 
fairly stable, then what you can do is in the lab go to a NUMBER of 
points and measure voltage and power to determine how your system 
responds and come up with some sort of curve to fit the results.

You need MORE points than unknowns to be able to sanity check your fit. 
You also should perform it under varying conditions (like ambient 
temperature) to make sure the results are stable.

On 25.9.16 13:50, Richard Damon wrote:
> On 9/22/16 9:36 AM, pozz wrote:
>> I have to convert voltage measure (Volt) into power (Watt).  As you
>> know, the physics says I need to square the voltage measure.
>>
> Whoever told you that didn't give you the full message, as it applies in
> only very limited cases.
>
> Electrical Power is Voltage * Current, not Voltage squared.
>
> IF you have a resistive load, then we have that
>
> Voltage = Current * Resistance.
>
> From this we can get to Power = Voltage * Voltage / Resistance.
>
> This means that for a system with constant resistance we can say that
> Power goes as Voltage squared, with a scale factor of 1/Resistance.
>
> If the resistance of the system isn't constant, this relationship
> absolutely doesn't hold. I have built systems where (within an operating
> band) the Power consumed goes up with dropping voltage.
>
> If all you can measure in the field is the Voltage, and the system if
> fairly stable, then what you can do is in the lab go to a NUMBER of
> points and measure voltage and power to determine how your system
> responds and come up with some sort of curve to fit the results.
>
> You need MORE points than unknowns to be able to sanity check your fit.
> You also should perform it under varying conditions (like ambient
> temperature) to make sure the results are stable.


The situation of the OP is more complex: He's attempting to measure
the output power of a RF amplifier by measuring the DC feed voltage
and relating the output power to it.

There are simply too many moving parts in the equation, it will not
work with some mathematical trickery.

-- 

-TV

On 09/25/16 13:36, Tauno Voipio wrote:

>
> The situation of the OP is more complex: He's attempting to measure
> the output power of a RF amplifier by measuring the DC feed voltage
> and relating the output power to it.
>
> There are simply too many moving parts in the equation, it will not
> work with some mathematical trickery.
>

This has been a long thread, but fundamentally, you can't determine
output power into the load without measuring the load vswr. Depending
on the output power, a simple directional coupler could be a short 
length of coax, measuring forward and reverse power simultaaneously,
which could be calibrated via a lookup table and interpolation to the
required accuracy. No need for complex math in the software...

Why does it have to be such hard work ?...

Regards,

Chris

On 26.9.16 20:30, Chris wrote:
> On 09/25/16 13:36, Tauno Voipio wrote:
>
>>
>> The situation of the OP is more complex: He's attempting to measure
>> the output power of a RF amplifier by measuring the DC feed voltage
>> and relating the output power to it.
>>
>> There are simply too many moving parts in the equation, it will not
>> work with some mathematical trickery.
>>
>
> This has been a long thread, but fundamentally, you can't determine
> output power into the load without measuring the load vswr. Depending
> on the output power, a simple directional coupler could be a short
> length of coax, measuring forward and reverse power simultaaneously,
> which could be calibrated via a lookup table and interpolation to the
> required accuracy. No need for complex math in the software...
>
> Why does it have to be such hard work ?...
>
> Regards,
>
> Chris


 From a directional coupler, you get the component amplitude, which
has to be squared for power. (It is kosher here, due to telegraph
equations). The math happens to go so that the net output is available
as forward power - reflected power.

-- 

-TV

Il 25/09/2016 15:36, Tauno Voipio ha scritto:
 > On 25.9.16 13:50, Richard Damon wrote:
 >> On 9/22/16 9:36 AM, pozz wrote:
 >>> I have to convert voltage measure (Volt) into power (Watt).  As you
 >>> know, the physics says I need to square the voltage measure.
 >>>
 >> Whoever told you that didn't give you the full message, as it applies in
 >> only very limited cases.
 >>
 >> Electrical Power is Voltage * Current, not Voltage squared.
 >>
 >> IF you have a resistive load, then we have that
 >>
 >> Voltage = Current * Resistance.
 >>
 >> From this we can get to Power = Voltage * Voltage / Resistance.
 >>
 >> This means that for a system with constant resistance we can say that
 >> Power goes as Voltage squared, with a scale factor of 1/Resistance.
 >>
 >> If the resistance of the system isn't constant, this relationship
 >> absolutely doesn't hold. I have built systems where (within an operating
 >> band) the Power consumed goes up with dropping voltage.
 >>
 >> If all you can measure in the field is the Voltage, and the system if
 >> fairly stable, then what you can do is in the lab go to a NUMBER of
 >> points and measure voltage and power to determine how your system
 >> responds and come up with some sort of curve to fit the results.
 >>
 >> You need MORE points than unknowns to be able to sanity check your fit.
 >> You also should perform it under varying conditions (like ambient
 >> temperature) to make sure the results are stable.
 >
 >
 > The situation of the OP is more complex: He's attempting to measure
 > the output power of a RF amplifier by measuring the DC feed voltage
 > and relating the output power to it.

I think many misunderstandings derive from this assumption.  I never 
wrote (or I think I never wrote) that I need to calculate the output 
power of an amplifier knowing only the DC feed voltage.

I'm trying to calibrate a measurement device that, in my case, is a 
directional coupler.  For me (the firmware engineer), it is a voltage 
signal that is related to the output power by a square mathematical law 
(at least, this said the RF engineer).

By comparing our results with a professional instrument, we noticed that 
the square law (P=kV^2) is somewhat good for our necessity. 
Unfortunately, if it is calibrated at the nominal power (k), at low 
power levels the results deviate a little from the theorical square law.

This is why I'm trying to introduce a second parameter in a mathematical 
law P=f_k,h(W) that is *the* square law when the second calibration 
parameter (h) isn't touched.
That formula is:
    P = k * (V / Vn) ^ (2 + h)
When h=0 (second calibration parameter not touched) the law *is the* 
theoric square law.

If the operator notices a big error at, i.e., Pn/2, he tries to change 
h. In this way, he is sure the measure at Pn/Vn will not change.

 >
 > There are simply too many moving parts in the equation, it will not
 > work with some mathematical trickery.
 >