Can I use root locus when the input is the negative feedback?

[curiousPep]

Homework Statement:

I have the ODE $$m*y''+c_v*y'+c_p*u = c_p*u$$, where u = -k*y. I need to identify the range of static feedback gains u = −ky that guarantee stability of the closed loop system.

Relevant Equations:

$$T_{u \to y}(s) = \frac{1}{s^{2}m+c_{v}s+c_{p}}$$

I have used root locus before but my confusion now is that the input is the negative feedback. Usually when I have negative feedback I consider the the error between the input (ideal) signal and the observed signal.
Also, in this case what is the tranfer function since u = -k*y, and what does the transfer function represent since the only variable is the equation is y?

For the case where we use the equation m*y''+c_v*y'+c_p*u = c_p*u, the transfer fucntion is $$T_{u \to y}(s) = \frac{1}{s^{2}m+c_{v}s+c_{p}}$$. I am no sure what to do when I consider u = -k*y.

Can someone provide me some hints please?

 

[DaveE]

Homework Statement:: I have the ODE $$m*y''+c_v*y'+c_p*u = c_p*u$$, where u = -k*y. I need to identify the range of static feedback gains u = −ky that guarantee stability of the closed loop system.
Relevant Equations:: $$T_{u \to y}(s) = \frac{1}{s^{2}m+c_{v}s+c_{p}}$$

I have used root locus before but my confusion now is that the input is the negative feedback. Usually when I have negative feedback I consider the the error between the input (ideal) signal and the observed signal.
Also, in this case what is the tranfer function since u = -k*y, and what does the transfer function represent since the only variable is the equation is y?

For the case where we use the equation m*y''+c_v*y'+c_p*u = c_p*u, the transfer fucntion is $$T_{u \to y}(s) = \frac{1}{s^{2}m+c_{v}s+c_{p}}$$. I am no sure what to do when I consider u = -k*y.

Can someone provide me some hints please?

Sorry,I don't really have the time right now to respond well. But, a few comments to clarify your question:

$m*y''+c_v*y'+c_p*u = c_p*u$ is the same as $m*y''+c_v*y'= 0$ which has no input and thus no feedback. So, can you fix that typo?

Also $m*y''$ looks like a convolution operator, not multiplication, to controls people like me. It's not a huge deal, I know you meant multiplication like $my''$ or maybe $m⋅y''$. But it is a distraction to readers to use that symbol.

The short answer is that you substitute your feedback for the $u$ input to get a characteristic equation in terms of only y. Then you evaluate the stability and it's dependence on the constant parameters.

Yes, you can use Nyquist plots. I never like it myself, and I've never actually met a real EE that uses it except in HW problems. I would look at the poles of the characteristic equation myself. If you have a RHP pole, it's unstable. IDK, I guess it's the same thing, except all you care about is the sign of the real part of the poles.

 

[curiousPep]

Sorry,I don't really have the time right now to respond well. But, a few comments to clarify your question:

$m*y''+c_v*y'+c_p*u = c_p*u$ is the same as $m*y''+c_v*y'= 0$ which has no input and thus no feedback. So, can you fix that typo?

Also $m*y''$ looks like a convolution operator, not multiplication, to controls people like me. It's not a huge deal, I know you meant multiplication like $my''$ or maybe $m⋅y''$. But it is a distraction to readers to use that symbol.

The short answer is that you substitute your feedback for the $u$ input to get a characteristic equation in terms of only y. Then you evaluate the stability and it's dependence on the constant parameters.

Yes, you can use Nyquist plots. I never like it myself, and I've never actually met a real EE that uses it except in HW problems. I would look at the poles of the characteristic equation myself. If you have a RHP pole, it's unstable. IDK, I guess it's the same thing, except all you care about is the sign of the real part of the poles.

I apologise it was meant to be $my''+c_vy'+c_py=c_pu$.

I understand what you say, but the goal is to find the range of k wt which the system reamains stable. Other than considering the point (-1,0) in the Nyquist the number of encirclements help to determine the stability, thus the Root locus comes in handy.

However, in the case of Root locus we consider usually the 1+kG(s) which corresponds to the poles of the close loop poles. In this case where the eqaution is only defined by y's what is the transfer function since the input (u) is 'converted' to y through u = -ky.


Thank you

 

[DaveE]

You do not need an input for a system to be classified as stable or unstable. In fact in a linear system with feedback, the input is essentially irrelevant since it's not dependent on anything within the system. Look at the characteristic equation of the system, that is what determines stability.

There are lots of web pages that deal with this subject, here are the first three from my google search:
https://web.vscht.cz/~cermanj/file/TR/Kapitola 9-Stability.pdf
https://www.tutorialspoint.com/control_systems/control_systems_stability.htm
https://www.javatpoint.com/stability-conditions

 

[curiousPep]

You do not need an input for a system to be classified as stable or unstable. In fact in a linear system with feedback, the input is essentially irrelevant since it's not dependent on anything within the system. Look at the characteristic equation of the system, that is what determines stability.

There are lots of web pages that deal with this subject, here are the first three from my google search:
https://web.vscht.cz/~cermanj/file/TR/Kapitola 9-Stability.pdf
https://www.tutorialspoint.com/control_systems/control_systems_stability.htm
https://www.javatpoint.com/stability-conditions

I see what you mean, but I try to understand it using root locus because I find it very intuitive and easy visually once the plot is made to determine the renge of k where the system is stable. However, in this case I am not use which the transfer function is to use the formula 1+kG(s) to perform the analysis.
Can you please explain me what I am missing to use the root locus methods? In the cases I am familiar with the input is not the feedback, usually the error is calculated using the input and the feedback.

 

[DaveE]

I see what you mean, but I try to understand it using root locus because I find it very intuitive and easy visually once the plot is made to determine the renge of k where the system is stable. However, in this case I am not use which the transfer function is to use the formula 1+kG(s) to perform the analysis.
Can you please explain me what I am missing to use the root locus methods? In the cases I am familiar with the input is not the feedback, usually the error is calculated using the input and the feedback.

I don't think I can explain root locus better than what you will find on the web. Anyway, I don't feel like writing a textbook.

You will find an expression for the loop gain of the system T(s) = 1 + kG(s). This is the gain "seen" by an electron that travels completely around the feedback loop(s) of the system. Then you will evaluate T(s) to find its roots, these will depend on system parameters like k. The root locus is a plot of where those roots are in the complex plane as a parameter, like k, is varied. Stability is analysed based on the position of those roots; i.e. for which values of the parameter you have roots leave the left-half plane.

 

[curiousPep]

I don't think I can explain root locus better than what you will find on the web. Anyway, I don't feel like writing a textbook.

You will find an expression for the loop gain of the system T(s) = 1 + kG(s). This is the gain "seen" by an electron that travels completely around the feedback loop(s) of the system. Then you will evaluate T(s) to find its roots, these will depend on system parameters like k. The root locus is a plot of where those roots are in the complex plane as a parameter, like k, is varied. Stability is analysed based on the position of those roots; i.e. for which values of the parameter you have roots leave the left-half plane.

Thank you that is what I have been trying the only part that I am missing is the actual transfer function. Since u=-ky once I apply the Laplace transform the only variables are Y(s). Is this wrong? Do I apply first the Laplace transform, calculate the transfer function and the substitute u=-ky?

 

[DaveE]

It's not about the transfer function. It's the loop gain that matters. You don't need an input or an output to be unstable.

 

[curiousPep]

It's not about the transfer function. It's the loop gain that matters. You don't need an input or an output to be unstable.

So in the case of Nyquist diagram I would use $$\frac{k}{s^{m}+sc_{v}+c_{p}}$$ to do the analysis?