At 4:41, it is mentioned that asymptotically stable in the large. From my understanding, for one to say a linear system is "asymptotically stable in the large" or an equilibrium point is "asymptotically stable in the large" for a nonlinear system, it is necessary that V(X) should tend to infinity as norm of X tends to infinity. If we don't mention that point, we can say only asymptotic stability. Kindly reply if I am true.
Please note the 3rd point. It says Vdot(X) does not vanish along state trajectory. It means that V(x) will always be present, even at infinity. This automatically implies V(x) tends to infinity as |x| tends to infinity. :) PS: The theorem is defined as such by Lyapunov.
@@Topperly Vdot(X) refers to the derivative of V(X), once it starts. When we mention about asymptotic stability in the large, we need to consider an infinite radius circle in the two dimensional state space which in turn represents the entire state plane. So that initial point of the state trajectory can start even at infinity. "It says Vdot(X) does not vanish along state trajectory. " This "DOES NOT" mean that V(X) is always present. It "just means that" once the state trajectory starts Vdot(X) doesnot vanish along the trajectory. The starting point of the state trajectory is not mentioned. So it is necessary to mention about infinity. You can also mention as V(X) is radially unbounded. Also for your kind information, the theorem as mentioned by Lyapunov is as follows for Asymptotic stability in the large: Assume that there exists a scalar function V of the state X, with continuous first order derivatives such that 1. V(X) is positive definite 2.Vdot(X) is negative definite 3.V(X) tends to infinity as norm of X tends to infinity then the equilibrium point at the origin is globally asymptotoically stable. This is referred from Applied Nonlinear Control by Slottine and Li Many authors have modified the statements according to their understanding.
Wish to add one more good reference Chapter 7 Stability of Nonlinear Systems of "Control of Nonlinear Dynamic Systems : Theory and Applications by J.K.Hedrick and A.Girard
@parvathy r When I added the third point, "Vdot(x) does not vanish identically at t>t0 for any t0", I thought it was enough as it mentions(or atleast I believed) a starting point. Thank you for correcting me. As they say, we always learn from our mistakes :) PS : Thanks for the reference, I'll surely look through.
@@Topperly Again, any "t0" does not point to the fact that the initial point of the state trajectory can start at infinity. We should keep in mind that asymptotic stability in the large is a way of representing "Global Stability". Hence it is essential that the unboundedness of V(X) should be highlighted in its definition.
7:50 x2 = 0 => x2(dot) = 0; How is this possible? Because the state x2 is constantly varying and finally die down to 0 at origin. x2 = 0 can be a point in state-space where the phase trajectory cuts the x1 axis. If x2(dot) = 0 at this point, then the value of x2 would not change further. How can this be possible?
Thank you so much sir for your dedication towards teaching.. Actually I am having Control Systems exam on day after tomorrow. As I dont understand anything from my class, your class means a lot to me.. May god bless you (If its there)... Thank you sir.. Now I have enough confidents that I will pass... Please make videos for other topics also sir..
The lectures are very helpful and the concepts are well explained and understood. Please do make more videos! which one is your previous video? can you provide the link for the previous video
dV/dt = 2×( [x1]^2 + [x2]^2 )^2, right? So even if x1 = -x2, we have dV/dt = 2×( [-x2]^2 + [x2]^2 )^2 = 8×[x2]^4 and this is a positive function everywhere except at origin. So it is positive definite
Please Sir, can you help to get this book, A. A. Matrynyuk: “A theorem on polystability,” it look like is not in the system, when i use pdfdrive i dont get any results. Please your response is essential to me
i dont think i understand this. you mean that if V' is positive then maybe there is also a lyapunov function that its negative? but if we find that V'>0 this means that the lyapunov function has a sign and in this case its >0. so we can say that the equilibrium point is unstable
Sorry, that's actually a mistake. x = 0 doesn't necessarily mean that the gradient of x will also be zero. Please give me some time to figure out the actual reason. It's been a while since I refreshed my Non linear topics :)
@@Topperly sure. I understand that it has been almost two years since this video was posted. I'll eagerly wait for your response. Thanks for the response!
V(x, t) is an generalised energy function of system(having all properties of energy functions) defined in terms of state variables. So negative rate of change of V(x, t) means system energy is decreasing and therefore will eventually settle at any equilibrium point where V= 0.
What do u mean by just positive? If dV/dt is positive, the change in energy of system is positive. This means that energy of system is increasing and such a system is unstable. However, Inorder to apply Lyapunov Theorems, it's best to stick to theorem conditions :)
to be specific , in the example that i am studying , the dv/dt is equal to zero only on the circle x^2+y^2=1 , and positive in every other point of the plane (the equilibrium point i am studying is (0,0) ). It is not positive definite. Even though i cannot strictly apply the theorems, i can say that the point is unstable right?
@@Topperly yes exactly. How do i check the stability of the circle? I thought lyapunov stability was only for equilibrium points Also, what is the prerequisite that the V function has to follow, so that the dv/dt shows the change in the energy of the system? Is it just that the V function must be positive does it have to be positive definite in a region? Btw i am a total beginner in this subject and i want to thank you , both for your video and for your answer!
To analyse a system, we have many methods - Phase Plane Analysis, Describing Function, Lyapunov Stability etc are few of them. The analysing technique we use depend on the situation and nonlinearity. With Lyapunov, you have to choose a Lyapunov function V such that it satisfies the conditions of system. I assume that you have done the Lyapunov Analyisis in this case (please refer video 19 and 20, they are about how to find V) and identified that there's the possibility of a limit cycle( because the circle has zero dv/dt and rest has positive dv/dt meaning the phase trajectories are going to circle around there). Now, confirm the existence of limit cycle using describing function and you have to check the stability of that limit cycle. I have explained that in lecture 12 of this series. Please watch that :)
Hello Sir, I need to present my deepest appreciation and respect for this matter as this particular video and the rest of the videos are exceeding the level of perfectness. However, if there was a rating, I would give a 4.5/5 and that is for two main very important reasons. the first is that the sound quality is not of the best so I was sometimes missing some of the words; yet, I understood the lecture perfectly. The second point is that I realised that you used to refer to your previous videos but without precisely mentioning to which video do we need to go, that could be in the form of a link may be and would be good to mention what duration of a particular video to save time. If these videos are for money, then I would definitely pay as it saves time and it informs a lot! all the best in your future endeavours my friend :))
We can't express how much your words mean to us. Thank you for taking time to give us a detailed feedback. We are aware of the problem with sound quality, but currently we have no income sources for the upgradation of our microphone. As soon as youtube start paying, we'll surely switch to a better quality mic. I'm sorry if I missed some, but usually when I refer to previous videos, I also add a card at the top right corner of the video which refers viewers to the mentioned video. However, if u didn't see one, please drop a comment (with the timestamp of instant where I'm referring to another video) and we are happy to help you. We, team Topperly, is two students in final year of B.Tech. UA-cam videos has always helped us in understanding our courses better. This is our way of helping others in a similar way. We really appreciate that you are ready to help but as of now this is not a commercial project. Thanks again :)
At 4:41, it is mentioned that asymptotically stable in the large. From my understanding, for one to say a linear system is "asymptotically stable in the large" or an equilibrium point is "asymptotically stable in the large" for a nonlinear system, it is necessary that V(X) should tend to infinity as norm of X tends to infinity. If we don't mention that point, we can say only asymptotic stability. Kindly reply if I am true.
Please note the 3rd point. It says Vdot(X) does not vanish along state trajectory. It means that V(x) will always be present, even at infinity. This automatically implies V(x) tends to infinity as |x| tends to infinity. :)
PS: The theorem is defined as such by Lyapunov.
@@Topperly Vdot(X) refers to the derivative of V(X), once it starts. When we mention about asymptotic stability in the large, we need to consider an infinite radius circle in the two dimensional state space which in turn represents the entire state plane. So that initial point of the state trajectory can start even at infinity.
"It says Vdot(X) does not vanish along state trajectory. " This "DOES NOT" mean that V(X) is always present. It "just means that" once the state trajectory starts Vdot(X) doesnot vanish along the trajectory. The starting point of the state trajectory is not mentioned. So it is necessary to mention about infinity. You can also mention as V(X) is radially unbounded.
Also for your kind information, the theorem as mentioned by Lyapunov is as follows for Asymptotic stability in the large:
Assume that there exists a scalar function V of the state X, with continuous first order derivatives such that
1. V(X) is positive definite
2.Vdot(X) is negative definite
3.V(X) tends to infinity as norm of X tends to infinity
then the equilibrium point at the origin is globally asymptotoically stable.
This is referred from Applied Nonlinear Control by Slottine and Li
Many authors have modified the statements according to their understanding.
Wish to add one more good reference
Chapter 7 Stability of Nonlinear Systems of "Control of Nonlinear Dynamic Systems : Theory and Applications by J.K.Hedrick and A.Girard
@parvathy r When I added the third point, "Vdot(x) does not vanish identically at t>t0 for any t0", I thought it was enough as it mentions(or atleast I believed) a starting point. Thank you for correcting me. As they say, we always learn from our mistakes :)
PS : Thanks for the reference, I'll surely look through.
@@Topperly Again, any "t0" does not point to the fact that the initial point of the state trajectory can start at infinity. We should keep in mind that asymptotic stability in the large is a way of representing "Global Stability". Hence it is essential that the unboundedness of V(X) should be highlighted in its definition.
Straight to the point with great explanation ....
Thank you! Means a lot :)
7:50 x2 = 0 => x2(dot) = 0; How is this possible? Because the state x2 is constantly varying and finally die down to 0 at origin. x2 = 0 can be a point in state-space where the phase trajectory cuts the x1 axis. If x2(dot) = 0 at this point, then the value of x2 would not change further. How can this be possible?
EDIT : @04:25 I accidentally said "negative definite" for "negative semi-definite". Please note that it is negative semi-definite only. :)
The lectures are very helpful and the concepts are well explained and understood.Please do make more videos! Thanks again!:)
Thank you! Means a lot :)
Good to know
Thank you soooooo much for the videos! They are very crisp and precise!! A great starting point for non linear controls
Thank you! Your words mean a lot to us :)
this was really helpful . please do more
Thank you! Means a lot :)
Thank you so much sir for your dedication towards teaching.. Actually I
am having Control Systems exam on day after tomorrow. As I dont
understand anything from my class, your class means a lot to me.. May
god bless you (If its there)... Thank you sir.. Now I have enough confidents that I will pass... Please make videos for other topics also sir..
Thank you! Means a lot :)
Yeahhhh
@kiran g krishnan
Thanks 4 the video. at 10:00, could U please consider closing )'s ? Best Wishes.
The lectures are very helpful and the concepts are well explained and understood. Please do make more videos!
which one is your previous video? can you provide the link for the previous video
Thank you! :) This is the link for previous video on sign definiteness ua-cam.com/video/AqkVIEaT48A/v-deo.html
at 11:44 when x1=-x2 system becomes positive semidefinite...?so how the theorem is valid?
I'm sorry but the video length is only 11:44 minutes. Can you please specify the timestamp again?
@@Topperly sorry sir!i mean at 10:04
When x1=-x2..the system can have positive semidefinite ...?
dV/dt = 2×( [x1]^2 + [x2]^2 )^2, right? So even if x1 = -x2, we have dV/dt = 2×( [-x2]^2 + [x2]^2 )^2 = 8×[x2]^4 and this is a positive function everywhere except at origin. So it is positive definite
@@Topperly yeah bro thanku so much..didnt see the square...I appreciate it.:)
Please Sir, can you help to get this book, A. A. Matrynyuk: “A theorem on polystability,” it look like is not in the system, when i use pdfdrive i dont get any results. Please your response is essential to me
great presentation.
Thank you :)
Nice explanation.
Glad you liked it :)
You can't say equilibrium point is unstable rather it's possible you have chosen your lyapunov candidate isn't suitable for that case.
Yes, what you said is right. That's why I specifically said 'Unstable in Lyapunov Sense' :)
i dont think i understand this. you mean that if V' is positive then maybe there is also a lyapunov function that its negative? but if we find that V'>0 this means that the lyapunov function has a sign and in this case its >0. so we can say that the equilibrium point is unstable
what is date for next lecture?
I'm afraid only after a month. I'm currently making videos on Describing Functions.
At 7:54, it is mentioned that as x2 =0, x2_dot also equal to zero. Why? How did we deduce this?
Sorry, that's actually a mistake. x = 0 doesn't necessarily mean that the gradient of x will also be zero.
Please give me some time to figure out the actual reason. It's been a while since I refreshed my Non linear topics :)
@@Topperly sure. I understand that it has been almost two years since this video was posted.
I'll eagerly wait for your response. Thanks for the response!
The video is very nice.I believe you could add subtitles to make it better.Anyways you are awesome!!!!!!! Waiting for more videos!❤
Yeah. Will do that soon ;)
what is the physical significance between V(x,t) and the state equations ?
V(x, t) is an generalised energy function of system(having all properties of energy functions) defined in terms of state variables. So negative rate of change of V(x, t) means system energy is decreasing and therefore will eventually settle at any equilibrium point where V= 0.
@@Topperly thanks again:)
what if V(x) is positive definite in the region , but dv/dt is just positive. Can we then make a conclussion about the stability ?
What do u mean by just positive? If dV/dt is positive, the change in energy of system is positive. This means that energy of system is increasing and such a system is unstable. However, Inorder to apply Lyapunov Theorems, it's best to stick to theorem conditions :)
to be specific , in the example that i am studying , the dv/dt is equal to zero only on the circle x^2+y^2=1 , and positive in every other point of the plane (the equilibrium point i am studying is (0,0) ). It is not positive definite. Even though i cannot strictly apply the theorems, i can say that the point is unstable right?
It seems like the system is falling into a limit cycle. Check for the stability of that limit cycle.
@@Topperly yes exactly. How do i check the stability of the circle? I thought lyapunov stability was only for equilibrium points
Also, what is the prerequisite that the V function has to follow, so that the dv/dt shows the change in the energy of the system? Is it just that the V function must be positive does it have to be positive definite in a region?
Btw i am a total beginner in this subject and i want to thank you , both for your video and for your answer!
To analyse a system, we have many methods - Phase Plane Analysis, Describing Function, Lyapunov Stability etc are few of them. The analysing technique we use depend on the situation and nonlinearity. With Lyapunov, you have to choose a Lyapunov function V such that it satisfies the conditions of system. I assume that you have done the Lyapunov Analyisis in this case (please refer video 19 and 20, they are about how to find V) and identified that there's the possibility of a limit cycle( because the circle has zero dv/dt and rest has positive dv/dt meaning the phase trajectories are going to circle around there). Now, confirm the existence of limit cycle using describing function and you have to check the stability of that limit cycle. I have explained that in lecture 12 of this series. Please watch that :)
Helpful video sir
Glad to hear that :)
what if the derivative is semi positive? (V'>=0). is the equilibrium point unstable?
In that case, equilibrium point is unstable as the rate of change of energy is positive meaning the energy of system is increasing.
Very helpful ......
Thank you! Means a lot :)
Hello Sir, I need to present my deepest appreciation and respect for this matter as this particular video and the rest of the videos are exceeding the level of perfectness. However, if there was a rating, I would give a 4.5/5 and that is for two main very important reasons. the first is that the sound quality is not of the best so I was sometimes missing some of the words; yet, I understood the lecture perfectly. The second point is that I realised that you used to refer to your previous videos but without precisely mentioning to which video do we need to go, that could be in the form of a link may be and would be good to mention what duration of a particular video to save time. If these videos are for money, then I would definitely pay as it saves time and it informs a lot!
all the best in your future endeavours my friend :))
We can't express how much your words mean to us. Thank you for taking time to give us a detailed feedback.
We are aware of the problem with sound quality, but currently we have no income sources for the upgradation of our microphone. As soon as youtube start paying, we'll surely switch to a better quality mic.
I'm sorry if I missed some, but usually when I refer to previous videos, I also add a card at the top right corner of the video which refers viewers to the mentioned video. However, if u didn't see one, please drop a comment (with the timestamp of instant where I'm referring to another video) and we are happy to help you.
We, team Topperly, is two students in final year of B.Tech. UA-cam videos has always helped us in understanding our courses better. This is our way of helping others in a similar way. We really appreciate that you are ready to help but as of now this is not a commercial project. Thanks again :)
🙏🙏🙏
:)
bhai title change krde, abhi paper m lypapunov likh deta
Changed it.Thanks a lot man :)