Finding the cutoff frequency of regulator

Thread Starter

salhi

Joined Nov 29, 2023
79
Let the following transfert function : \( C(p) = 0.5926 \frac{1+1.205p}{1+0.4233p} = K \frac{1+a \cdot \tau p}{1 +p \cdot \tau p}\) with \( a = 2.848, \tau = 0.4233, K = 0.5926 \) the marge of phase of this corrector is 28.7° i want to find the cutoff frequency of this correct, the answer was already given to be 0.648 rad/sec, but i want to know how did they get it, infinite thanks!
 

MrAl

Joined Jun 17, 2014
11,687
Let the following transfert function : \( C(p) = 0.5926 \frac{1+1.205p}{1+0.4233p} = K \frac{1+a \cdot \tau p}{1 +p \cdot \tau p}\) with \( a = 2.848, \tau = 0.4233, K = 0.5926 \) the marge of phase of this corrector is 28.7° i want to find the cutoff frequency of this correct, the answer was already given to be 0.648 rad/sec, but i want to know how did they get it, infinite thanks!
Hello,

What is the variable 'p' in those expressions. Is that the complex frequency variable?
 

MrAl

Joined Jun 17, 2014
11,687
I thought that was obvious enough no? regardless yes it is
Well more often we see the variable 's' being used, but yes 'p' is also used sometimes.
I thought it was that, but the reason I asked is because the transfer function you provided acts more like an oscillator, and although that would have a specific frequency, it would not be called the cutoff frequency it would be called the center frequency or something like that. That's why I wanted to be absolutely sure it was the complex frequency variable.

However, that is when interpreting 'p' as the complex frequency variable in your second expression on the first line, and that seemed reasonable because you defined 'tau' and tau is multiplied by 'p' also. That puts one p in the numerator and p^2 in the denominator. If you look at the function then, we see that there will be a cosine term (the p in the numerator) and a sine term (the '1' in the numerator), and that constitutes an oscillator.

But then the first expression only has one 'p' in the denominator, not p^2.
So which is it?
If we only have p to the first power in the denominator, then we have something like a filter which would have a transfer function and a cutoff, and would probably look like a high pass filter.
 

Thread Starter

salhi

Joined Nov 29, 2023
79
Well more often we see the variable 's' being used, but yes 'p' is also used sometimes.
I thought it was that, but the reason I asked is because the transfer function you provided acts more like an oscillator, and although that would have a specific frequency, it would not be called the cutoff frequency it would be called the center frequency or something like that. That's why I wanted to be absolutely sure it was the complex frequency variable.

However, that is when interpreting 'p' as the complex frequency variable in your second expression on the first line, and that seemed reasonable because you defined 'tau' and tau is multiplied by 'p' also. That puts one p in the numerator and p^2 in the denominator. If you look at the function then, we see that there will be a cosine term (the p in the numerator) and a sine term (the '1' in the numerator), and that constitutes an oscillator.

But then the first expression only has one 'p' in the denominator, not p^2.
So which is it?
If we only have p to the first power in the denominator, then we have something like a filter which would have a transfer function and a cutoff, and would probably look like a high pass filter.
Hey , well the transfer function i expressed was for a PD regulator, i think that counts as a filter i guess so, but yeah how did we get its cutoff frequency? thanks again!
 

MrAl

Joined Jun 17, 2014
11,687
Hey , well the transfer function i expressed was for a PD regulator, i think that counts as a filter i guess so, but yeah how did we get its cutoff frequency? thanks again!
Hi,

Here is what I think is the most general way to calculate the cutoff frequency.

1. Replace 's' (or 'p' in your case) with j*w, 'j' being the imaginary operator.
2. Find the amplitude A(w) by taking the norm of that (square root of the sum of the squares of the real part and imaginary parts: sqrt(r^2+i^2)). A is now a function of 'w'.
3. Check to see what kind of response it is by plotting the amplitude function A(w) vs 'w' (LP, HP, BP, etc).
4. Determine what amplitude A you regard as in the passband, call that A1, which will be a constant.
5. Set the amplitude function A(w) equal to A1/sqrt(2).
6. Solve that for 'w'. There may be more than one 'w' for a filter.
7. If you want the frequency in Hertz, divide 'w' by 2*pi.

Try that with your first expression see what you get.

If it is a high pass filter you will likely set A1 equal to A(w) with 'w' approaching infinity A(inf).
If it is a low pass filter you will likely set A1 equal to A(w) with 'w' equal to zero.
After that, the value 'w' you solve for in #6 above will be the -3db cutoff frequency, divide by 2*pi to get the frequency 'f' in Hertz.

This would not be valid with your second expression because it will be a sinusoidal function which means it oscillates up and down repeatedly so there is no real 'cutoff' point. That's the expression with p^2 in the denominator.

I guess I will assume that your first function is the right one, with just p^1 in the denominator.

Just to note, in #2 above the real part 'r' and imaginary part 'i' are really both functions of 'w':
A(w)=sqrt(r(w)^2+i(w)^2)

If you need the phase shift then use the two variable inverse tangent:
Ph(w)=atan2(i(w),r(w))
 
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