# Phase shift issue in Dominant Pole Compensation strategy....

1. Feb 16, 2016

### brainbaby

The next hurdle in my understanding of frequency compensation comes as following…
As the text says…

My problem is that as we move from point 1 to 2, the frequency increases so the phase shift should also increase..(as phase shift depends on frequency)… but the text however says that the phase shift will be constant at 90 deg with a 6db roll off..so what is this paradox..??

Either the text should say that "open loop gain falls at 6db/ octave with a varying phases shift of 90 deg..”
Isn't...!!

2. Feb 16, 2016

### LvW

No - why do you think that the phase shift is ALWAYS increasing with frequency?
Two examples:
* For poore resistive circuits the phase hift is always zero
* For an ideal integrator (or a RC network for very large frequencies) the phase shift is constant at -90deg.

Example second order:
H(s)=A/(1+as +bs²)

For very high frequencies we have H(s)=A/s²= - A/w² .
And the "-" sign is identical to a constant phase shift of -180deg.
(This is what the text says).

However, the text is somewhat simplifying. The term "constant" refers to the asymptotic lines of the real curve phase=f(frequency). As the above function H(s) shows: The phase assumes 180 deg for infinite frequencies only.

3. Feb 16, 2016

### brainbaby

Due to increasing frequency the stray capacitance both inside and outside the amplifier comes in action hence cause the phase shift or phase delay to occur....by why it increases..I certainly don't have any idea of it..

4. Feb 16, 2016

### LvW

OK - correct. Parasitic effects are always existent and - as a result - there will be always a negative phase shift for rising frequencies, even for an ohmic voltage divider.
I did not know that you were speaking about such effects.
And, of course, the text as cited by you did not take into account such effects.
So - what is your problem?
As a - more or less - phliosophical aspect: In electronics there is NO formula or rule that is correct by 100% (and this is true even for the "resistive" voltage divider).
Everything contains some simplifications and neglects unwanted effects which - hopefully -become effective for frequencies only which are beyond the working range.

5. Feb 16, 2016

### brainbaby

Oh I am really sorry..I forgot to attach an image...which will present my point of query..

What I mean to say is that if we move from point 1. to point 2. frequency is increasing...correct..so the phase shift should also become more prominent..
But in the text it is written that or means that for a 6 db roll off the phase shift would be a constant 90 degree....(as we progress down the slope the frequency on the horizontal axis increases)

So why they said that the phase shift would be constant...either they should say "open loop gain falls at 6db/ octave with a varying phases shift of 90 deg..”

6. Feb 16, 2016

### Jony130

In short. The dominant pole capacitor (single capacitor) can provide only 90 deg phase shift max.
And we select dominant pole capacitor in such a way that the loop gain drops to 1 with slope 6dB per octave at a frequency where the poles of uncompensated amp contributes very small to the total phase shift. This ensures that the phase shift is greater than -180 deg and we have a stable amplifier.
See the plot (single pole)

And notice that for F > 10*Fc phase shift is constant and equal to -90 deg.
Where Fc is a cutoff/corner frequency Fc = 1/(2 * pi * R*C).
Also do you know why roll-off is equal to 6dB per octave (20dB per decade) ? If not you should really back to basics.

And here you have a example of a second order (two poles) frequency response

Last edited: Feb 16, 2016
7. Feb 16, 2016

### jim hardy

@brainbaby
Remember asymptotes ?

Here's where technicians get an advantage
my high school teacher had us boys calculate, tabulate and plot, calculating by slide rule , on log paper
Vout/Vin for this simple circuit
from a couple decades below to a couple decades above break frequency

Vout/Vin = JXc/(R+jXc)

now - when you struggle through that exercise with all its slide rule polar to rectangular conversions
you'll see the pattern emerging
At low frequency Xc >> R so that fraction's denominator is very nearly jXc
and fraction jXc/(nearly jXc) is nearly 1, with almost no phase shift because the j's almost cancel (pardon my math liberty?)

At high frequency Xc is becoming vanishingly small
numerator is that small number multiplied by j
denominator is R plus the comparatively small numerator
so the fraction becomes j X (a vanishingly small numerator)/(an almost constant denominator)

jXc/(R+j(almost_nothing)) is very nearly: jXc/R ,
only one operator j is left for all practical purposes.
while you can't by math get rid of the j in denominator you can (by observation of the calculated numbers in your exercise) see that it becomes insignificant ,
for all practical purposes it vanishes
so phase shift is asymptotic to 90 degrees not equal , but slide rule accuracy is accepted as 3 figures.
Then it's intuitive that a single pole gives 90 degrees
and another pole would give 90 more

Once you believe that it gives you confidence in the math that edit Jony is Jony and Lvw are presenting so well

But i wouldn't have ever learnt it without that high school technician's exercise.
Maybe you should try it - do about twenty frequencies so it sinks in.
Plodders like me need to be taught in sequence What→How→Why , not Why→What→(How left to you to figure out)...

sophiecentaur always says "Work the maths!" . This is a case where doing it with real numbers will be more instructive than alphabet juggling.
Once it's soaked in you can make an elegant derivation, I expect that would feel great.

hope it helps

old jim

Last edited: Feb 17, 2016
8. Feb 17, 2016

### brainbaby

Oh It seems I get it......
Actually the phase shift depends directly upon the occurrence of poles and indirectly on increase in frequency …because if a pole occurs that means another capacitor comes into action …however the role of frequency in causing phase shift depends on the time when that frequency creates a pole…..
so rather we should say that
by increasing the frequency the number of poles increases, and if pole increases then only then phase shift increases…
so its the frequency which initiates a capacitor to cause the phase shift..

So now my inference seems to agree with what LvW stated in post 2 first lines....

Hence the word "constant" in the text is right...
Isn't..??

9. Feb 17, 2016

### LvW

...indirectly? What does that mean?
In frequency-dependent circuits, the phase shift between input and output directly depends on frequency (is a function of frequency).
However, how this function looks like (first order, second order, poles only or poles and zeros) depends on the circuit and its transfer function - expressed using the pole and zero location. OK?
..on the time?

No - the number of poles is a property of the circuit (a property of the transfer function) and has nothing to do with the applied frequency.
Again: The function of phase shift vs. frequency depends on the circuit and its frequency-dependence only.
For finding the approximate form of the phase function (in form of asymptotic lines) we are using the pole and zero location because we know what happens at these specific frequencies: At a pole (zero) the slope becomes more negativ (positive) by 20dB/dec.

10. Feb 17, 2016

### brainbaby

As we all know that a capacitor introduces a 90 deg phase shift...right..yes ....I agree that phase shift depends upon frequency...
Being indirectly I mean that... see the figure in post 5..here for uncompensated curve at point 2 ..a 90 deg phase shift is introduced...and in between point 2 and point 3 the frequency is increasing but the phase shift is same i.e 90 deg...,now at point 3 again another phase shift of 90 deg is introduced...and the accumulated total phase shift is 180 deg..but again in between point 3 and point 4 frequency is increasing but phase shift is yet the same 90 deg....
So this behaviour felt to me like that the points (2,3,4) where there is a roll off(pole) the phase shift depends on the frequency and for the rest intermediate positions frequency is independent of phase shift...

However by saying that frequency is independent of phase shift I never meant that it does not depends on phase shift....yes it depends ..but it was just a way of me inferring the situation....as simple as that....

Last edited: Feb 17, 2016
11. Feb 17, 2016

### LvW

To be exact: A capacitor introduces a 90 deg phase shift between voltage and current!
For an RC element we have a frequenc-dependent shift which reaches 90 deg for infinite frequencies only!

No - the phase cannot abruptly change its value (...a 90 deg phase shift is introduced).
Study the phase response in Fig. 6 - it tells you everything you need to understand what happens.
Such "phase jumps" are only introduced in the drawing for the asymptotic lines as a help for constructing the real and smooth phase function.

Frequency is always independent on phase shift. I suppose you mean the inverse: Phase shift independent on frequency.
But this is NEVER the case. Phase shift ALWAYS depends on frequency.
Don`t mix the real phase response with asymptotic lines which serve only one single purpose: A help for roughly constructing the real curve. See the phase diagram in Fig. 6 and realize that the final value at -90 deg is reached for infinite frequencies only.

12. Feb 17, 2016

### brainbaby

My inference from figure 6....
1. Phase shift non linear function of frequency..

2. Phase shift is independent of poles because it started changing from point x before the first pole have occurred.

3. Between point 1 and 2 the phase shift is 90 deg which is said in the text to be constant because they are talking in terms of accumulated phase shift which is 90 deg..(135-45=90) < as phase cannot arbitrarily change its value>..but with each change in frequency the value of phase shift is different as seen from the curve (for f2 and f3 phase shift is p1 and p2)..…which further signifies that phase shift totally depends on frequency…however that dependence is non linear..

Am I right ?

13. Feb 18, 2016

### LvW

1.) The term "constant" in the text (your post#1) is not correct. We have a constant phase shift of 90 deg for an IDEAL integrator only (pole at the origin). However, such a circuit does not exist.
2.) Your point (2) is false. It is the pole which causes the phase to deviate from its starting value (0 deg) already for very low frequencies.
3.) For my opinion, the diagram shows everything you should know to understand what happens. If there would be no second pole, the red curve (phase) would approach the -90 deg line (which acts as an asymptotic line in this case).

14. Feb 19, 2016

### brainbaby

After rigorous analysis I came to the following conclusion...

Phase shift depends on the circuit and frequency. Phase shift shows dependence on poles which is a circuit parameter from the fact that the phase shift begins to change one decade before the pole and stops changing one decade after the pole and ends at 90 degree and this happens prior and after to each pole.
The phase shift at the pole frequency is -45 degrees or
A single filter pole adds a maximum of 90 degree phase shift for the frequencies far away from its turnover frequency(3db frequency), but the shift is only 45 degrees at the pole (-3db point).

This behaviour can be further illustrated and explained as.....

But Why does this happens...??
how phase shift knows well in advance that pole is about to come.......and its time to change....and after the zig zag happens its time to be parallel to the asymptote ...and change if another pole comes otherwise show indifference.....??...

15. Feb 19, 2016

### jim hardy

Bravo !

let's take a plausible example
R = 1000 ohms
time constant = 1 millisecond
pole then is at 1000 radians per sec = 159.15 hz
and Xc at 1000 radians/sec is 1000 ohms

a decade away, at 15.915 hz what is transfer function's magnitude and phase ?
Vout/Vin = JXc/(R+jXc)
Xc = 1/(2pifC) = 1/(2 ⋅ π ⋅ 15.915 ⋅ 1X10-6) = ⋅1.0X104
jXc/(R+jXc) = j104/(103+j104)
which = 104∠-90°/1.005X104∠-84.3° = 0.995∠-5.7°

Phase shift doesn't know anything in advance. Mother nature just built math that way.
Circuit guys struggling to do above math with slide rules figured it out and observed it's always 5.7 degrees a decade out
so they devised those predictive rules for drawing a frequency & phase plot. No magic, it's just a graphical approach that saves wear and tear on slide rules.
One Mr Hendrik Bode made the approach popular in 1930's. I don/t know if he was the absolute first to think of it but he sure advanced the field of control systems

and that's why it's called a Bode Plot.
https://en.wikipedia.org/wiki/Bode_plot

Any help ?

16. Feb 19, 2016

### LvW

No - remember the simple RC lowpass with a pole at wp=1/RC. The decrease of amplitude with a corresponding phase shift will start already for f=1E-12 Hz (and even below). However, it will be hard to measure it. But that is not the question.
No - this is true for a first order lowpass only.

No - all the poles influence the amplitude and phase response also for very low frequencies, but this influence sometimes can be neglected.
Have a look on a 3rd-order or 4th-order transfer function. Why do you think that terms like w³T³ have no influence for very low frequencies?
Perhaps the influence is small - OK. But for your understanding it is important to know that there is an influence.

17. Feb 19, 2016

### LvW

Brainbaby - here are some additional information:
(1) There is a formula (called "BODE" integral), which has some relations to the "Hilbert transformation".
This formula exactly describes the relationship between the amplitude and phase response.
However, this formula is too complicated for writing it down at this place.
(2) For real (non-idealized) lowpass systems there is only one single frequency where the SLOPE is exactly -20 dB/Dec (-40 dB/Dec).
At this frequency the phase exactly assumes the value of -90 deg (-180 deg).
However - normally, this frequency is not known, but this knowlege is exploited for constructing a rough phase function curve based on the magnitude response (which also is known only as a approximation based on the pole location and the slope information about asymptotic lines).

Last edited: Feb 19, 2016
18. Feb 19, 2016

### brainbaby

In the above paragraph I was not talking about variation of amplitude with phase shift...rather I talked about variation of phase shift with frequency..

So you mean to say that the above conclusion is just an approximation ...however...it can be accepted on general terms....but may be rejected on precise basis....

19. Feb 19, 2016

### brainbaby

The reason why I said the above was because as the frequency grows along the graph phase shift accumulates ...and further ahead in frequency a pole exist...ok..and there an increase in phase shift is observed....and so on...
so they framed a theory based on experimental observation.....however in this thread what we have discussed so far is the "what"
and most probably Ive got what happened...now its time to know the "why".....and you have brought mother nature into consideration..

I appreciated that you worked out the maths....thanks for it...but that maths is just a representation of a phenomenon... what I believe is that
“When you understand something, then you can find the math to express that understanding. The math doesn't provide the understanding.”

and that understanding "why" is paramount for me.....

20. Feb 20, 2016

### LvW

As I have mentioned (BODE integral) gain (magnitude) and phase responses are correlated to each other

21. Feb 20, 2016

### brainbaby

Yes I agree ..but going in detail for Bode integral requires another vast separate thread ....anyways

finally ......any singular theory which describes behaviour of phase response of a two pole system for now..??

22. Feb 20, 2016

### LvW

I think, it is not necessary to have a "theory".
We just neeed the definition of the phase function which is arctan(phi)=Im[H(jw)]/R[(H(jw)]

23. Feb 20, 2016

### brainbaby

Now would you be humble enough to define this.......or again I have to scratch my head from bottom....

24. Feb 20, 2016

### LvW

Sorry - but I suppose that you know: Im:imaginary part and R:real part of a complex expression.

25. Feb 20, 2016

### brainbaby

Actually from starting I wanted to shape the discussion into this order like suppose a phenomenon happens then he understands how it happened.. then after understanding he uses maths to formulate what has happened.....
But your approach is exactly reverse...first you talk about an expression and then you give a meaning to it and in the end you tell what will happen through it...
so....most of the time I struggled synchronising both the approaches....

and talking about complex numbers....I knew what the are meant for ...but how they will behave in the following formulae to give meaning to the phase function..thats quite obscure to me....