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I am confused regarding the concept of 'natural frequecny' in control systems. Is it only dependent on the denominator or is it also dependent on the numerator? For example, I have '8' in both numerator and denominator of my open loop transfer function but the values of numerator and denominator of the closed loop transfer function are not the same as I get 8 in the numerator but I get 16 in thedenominator.

I am very confused. What will be the natural frequency of my system? Will it be the square root of 8 or will it be thesquare root of 16?

My MATLAB code:

clc
clear
close all
num=[8]%numerator of open loop transfer function
den=[1 2 8]%denominator of open loop transfer function
sys_open_loop=tf(num,den)%open loop transfer function
sys_close_loop=feedback(sys_open_loop,1)%close loop transfer functio
step(sys_close_loop)%step response of close loop transfer function
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    \$\begingroup\$ Your transfer function doesn't have any \$s\$ in the numerator so, what you see is what you get: the single term in the denominator is the \$\omega_n\$ you seek. \$\endgroup\$ Nov 3, 2022 at 14:05
  • \$\begingroup\$ @aconcernedcitizen Are you aware that there is a relatively large set of changes/improvements in the new LTspice 17.1.5 version (no longer is it "XVII"?) If you want to test it, the MSI installer link is here. It includes a new "FRA" component (aka Middlebrook) to support loop gain and output impedance analysis with a single added component. The viewer is a lot faster. Various bugs fixed, etc. It's beta, though. \$\endgroup\$
    – jonk
    Nov 3, 2022 at 21:24
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    \$\begingroup\$ @jonk Yes, I've seen the announcement in the group. The FRA looks like a new toy but, after playing with it a bit, I think it's simply this or something very similar (possibly a sum of sines and different time windows; just speculating) but, the resolution vs time it takes, leaves to be desired, IMHO. I do like the way it churns, though, I tried some finicky schematics and it didn't cough at all, while the CPU looked quite relaxed. If only some of the ...let's say "wrinkles" would be ironed out, too, and it would look promising. \$\endgroup\$ Nov 3, 2022 at 22:06

4 Answers 4

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So i am very confused. What will be natural frequency of my system? Will it be square root of 8 or Will it be square root of 16?

If the Transfer Function has 8 in the numerator and 16 in the denominator you can regard it as having a gain factor of 0.5 in the numerator and hence, numerically, it looks like 8 when you should consider it as \$\frac{1}{2}\cdot\omega_n^2\$

So, \$\omega_n^2\$ is still hiding in there there but "the half" (gain factor) is masking it numerically.

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  • \$\begingroup\$ Where is this gain factor(0.5) mentioned? How can i manipulate it ? \$\endgroup\$
    – DSP_CS
    Nov 4, 2022 at 7:18
  • \$\begingroup\$ @engr in what context do you mean? Mathematically? Physically? With another control loop i.e. in the s plane? \$\endgroup\$
    – Andy aka
    Nov 4, 2022 at 9:38
  • \$\begingroup\$ i meant physically? will this gain block be apparently visible in block diagram? \$\endgroup\$
    – DSP_CS
    Nov 4, 2022 at 11:40
  • \$\begingroup\$ I've not seen the block diagram @engr \$\endgroup\$
    – Andy aka
    Nov 4, 2022 at 11:54
  • \$\begingroup\$ @engr That gain value (k) of 0.5 is the closed loop dc gain. It isn't a particular block in either the open loop or closed loop system. It is the dc gain equal to Vout/Vin of the closed system. It can be changed by either changing the forward gain (the gain, k in the open loop transfer function) which will change the value of your open loop transfer function's numerator or alternatively by changing the feedback transfer function H(s) which you currently have set to unity. Usually the feedback transfer function would be changed by varying the resistor values in the feedback network (beta). \$\endgroup\$
    – user173271
    Nov 4, 2022 at 12:24
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The first thing you need to do is find the poles. Will be the frequency at one of the poles. Another thing you could do is do a body plot and find any peaks in the plot

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wn is the square root of the last term in the denominator of the closed loop transfer function when the denominator is in standard form (coefficient of s^2 = 1). So your value for wn is sqrt(16) = 4.

The numerator of your closed loop transfer function (when in standard form) is equal to k.wn^2 where k is the value for the dc closed loop gain.

Therefore your closed loop transfer function's numerator could be written as k.wn^2 = 0.5 * 16. So again wn = 4 and the dc closed loop gain is equal to 0.5

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Mathematically, the open loop system (forward) transfer function is: $$G(s)=\frac{8}{s^{2}+2s+8}$$

Compare to the parametric standard form: $$G(s)=\frac{8}{s^{2}+2\zeta\omega_{N}+\omega^{2}_{N}}\tag{1}$$

So the open loop natural frequency \$\omega_{N}=\sqrt{8}=2\sqrt{2}\$.

Closing the loop changes the system. You have another system.

The reverse (feedback) transfer function is \$H(s)=1\$.

The closed loop transfer function is: $$M(s)=\frac{G}{1+GH}=\frac{8}{s^{2}+2s+16}$$

So the closed loop natural frequency \$\omega_{N}=\sqrt{16}=4\$.

The open or closed natural frequency is always determined from the characteristic polynomial that is the denominator when the transfer function is in the standard form as shown in equation 1.

What will be the natural frequency of my system? Will it be the square root of 8 or will it be the square root of 16?

There are two natural frequencies.

Open loop: \$\omega_{N}=\sqrt{8}=2\sqrt{2}\$.

Closed loop: \$\omega_{N}=\sqrt{16}=4\$.

Closing the loop makes the system faster.

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