Designing a Lag Compensator with Root Locus
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- čas přidán 9. 06. 2013
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This video walks through a phase lag compensator example using the Root Locus method.
Errata:
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Get back to studying, there will time for distractions after your exam! :) Actually, now that I've posted this video I've been distracted on CZcams myself. Good luck with the exam.
Even those this video series is quite old now, it's still by far the best explanation of control engineering I've ever seen. I thought I hated system control until I found these videos, and now I actually enjoy it and that's all down to you mate. Absolute top class.
I am a biomedical engineering undergrad, we don't have control systems in our syllabus and this playlist was completely new to me! i usually take one hour to finish a video in this playlist! (Taking notes, thinking rewinding) but so far it is worth it!! i learned control systems just from you!! than you so much brian!!
am watching this video just before my control final exam and believe me or not this man just demonstrate and well explained in 10 mins what my teacher can't did in several lecture ...... u the best bro
wow , thank you so much . u did in 10 min what my teachers couldn't do in a whole semester
Love the clarity in your way of teaching! Thanks.
It's 2021 and you're still helping me from Canada!
This is, once again, an amazing video; I just can't thank you enough
I came onto youtube to get onto your channel for some refreshers when I got distracted because, well, it's youtube. I feel like you knew I was suppose to be studying for my controls exam in 18 hours when you posted this. Now that I'm here let the studying commence :D Thanks for all the videos!
Thank you so much! These compensator design videos helped me in understanding my lecture notes.
You're a hero, you saved my project in 11 minutes
thank you for your job, you explained this material perfectly, but could you make one more lecture about design of lead-lag compensator?
i'm watching these videos 2 hours before my control system exam! hahaha! you are awesome sir!
Amazingly helpful! Thank you.
more simplified than Ogata ,, Thanks bro.. ^_^
Awesome video! Thank you so much!
You are a really good teacher !You should teach in college !!!
Thanks a lot for these videos !!
Your videos are much more helpful than Ogata!
And Nise!
Jacob Hiller oh yes definitely!
hello brian, how do design lead/lag compensator for min. settling time of plant greater than second order?
Could you place the required zero and pole very far from the imaginary axis instead of very close? If you put them far enough, their angles will tend to cancel each other also. Of course, maintaining the rule of "pole closer" to keep it a lag compensator. Could it be done? Or why didn't you? Thanks for your videos! They're very clarifying!
Great video and I do have few questions too? If I have a situation where I do not want a PID controller, all I want to use is a lead lag compensator and I want a faster response but then my Ess is also too large and I want it near to zero, then what should I do?
Should I use both lead and lag as lead compensator will help me gain a faster response and lag compensator will help me reduce the Ess??
My other question is that besides PID and lead lag, is there any other way to attain faster response and reduce Ess??
Thank You Sir.
You have that error equation for a step input, whereas in your system design you choose impulse input. So shouldnt the error equation change while designing the compensator, since you are using an impulse input and not step? Thanks.
If we are designing in z-space, can we still use the fact that the zero location should be 1/50 of the dominant pole?
thanks, and you're videos are the best!
Woww, Best channel ever!!!
Your videos are awesome, I think you really have "the thing" to teach others.
Why couldn't you use the same method in the Lag Compensator for Bode Plot? Instead of doing all that, just adding a 50 gain in the controller? Wouldn't this produce the same consequences?
what is the physical meaning for zeros and poles and how does it help in circuits?
As alwayys!! The Legend
Hey Brian. I have watched all of your videos and would sincerely like to thank you for your excellent explanations and intuitions. It has helped me tremendously review alot of concepts from college and wish I had studied control engineering more deeply than I did. One thing I am curious about is the relationship between time, frequency and s-plane domains and how they interact. You touch on this in this video (ie how steady state error relates with the s-plane) indirectly through your expertise in the subject but I'm having a hard time building overall intuition on this. Just thought it'd be a great video to your collection of already great videos!
Brian Douglas mentioned some things about how the location of the poles affect the system's response to an input. I forget where it was, but try among these videos: czcams.com/users/results?search_query=brian+douglas+drawing+root+locus
As for how frequency, time and s-planes are related, look up the Laplace transform. If memory serves, that is used to go from the time domain to the s-domain (since it has a factor of exp(st) in it or something). The relationship between the s-plane and frequency is by setting s = jw where j is the imaginary unit and w is omega, the frequency in radians per second. Not a perfect explanation but hopefully it can be a start.
When you place the pole for the lag compensator, doesn't the dominant pole gets shifted towards right?
Hi Brian, Great video!
I have a small doubt though. We initially said that the Lag compensator doesn't affect the location of closed loop poles a lot, and hence transient performance remains almost the same. But while showing the impulse response at 10:38, the settling time has been affected badly after using Lag compensator. But it shouldn't have been, right?
Yesss, I have noticed the same problem..
The step response is the convolution of the unit step with the impulse response.. If impulse response remains the same, how come the step response be so different?
@@sahhaf1234 notice the magnitude of the time axis. the long time period magnifys the small difference in the impulse response.
Brain, at the last few seconds, why for impulse response there is not much changes? I thought the dominant poles are regardless to input.
Damn you’re good at teaching
Hi Brian, thanks for this very helpful video !
But in this video one thing that made me confuse that in Lag Compensator we are adding pole and zero closer to the imaginary axis. And distance of these compensated pole and zero from imaginary axis is lesser than the distance of dominant poles from imaginary axis. So won't these compensated pole and zero change our required dominant poles location & system requirement ?
As you have said this things in your "Designing a Lead Compensator with Root Locus" video.
And sorry if I am in wrong direction but please do elaborate ...
Yeah I have the exact same doubt. does'nt the dominant poles change because of adding the lag compensator.
good video :)
As per your last video,
we can't place a pole to the right of dominant pole, so why are we doing it here?
If that is just because angle due to pole and zero remains nearly same, then that pole(of pole zero combine) becomes the dominant pole
Good question and I think the response doesn't change because of the fact that theta(p) - theta(z) = 0
Secondly, the pole zero combine do not become a dominant pole, instead they help you to maintain your root locus that is why the impulse response didn't change because the R-locus remains the same
take a count that lag compensator practically doesn't has impact over the open loop gain, because theta(p) - theta(z) = 0 and its gain on the compensator its approximately 1, thats why doesn't change the root locus, thus they are not dominant.
Late response, but I didn't find the 2 replies to your comment very satisfying. Because after all, we *are* adding a new closed loop pole that is closest to the imaginary axis (hence, a dominant pole). In fact, this is why the settling time for the step response is so much longer now, we've got a new closed loop pole with a low magnitude real part which is causing the system to take its time! But then why is the impulse response settling time so close to the lead only controller? The contribution of the lag compensator actually does have a long settling time in the impulse response, but the magnitude contribution is so small that it doesn't even look like it's there! Notice how the lag/lead impulse response is slightly larger than the lead compensator. This discrepancy gets smaller over time, ie it's that long time constant we've added slowly dying off.
For a more concrete view of this, try plotting the impulse response of a few versions of (s+z)/(s+p) while maintaining the z/p ratio. For small values of z&p, you'll find very low magnitude impulse responses with large settling times. For larger values of z&p, you'll find higher magnitude responses with shorter settling times! So, by adding a small z&p pair as in our lag/lead compensator, we'd expect the contribution to the impulse response to be small, but long lasting (as long as we haven't moved around the other closed loop poles).
@@thevirtualsphere Also a late thanks to you!
What about if you have 𝐺𝑐(𝑠)𝐺(𝑠)= (𝑠+𝛼)/( 𝑠3+(1+𝛼)𝑠2+(𝛼−1)𝑠+1−𝛼 ) and the requirement is to design a compensator that has the steady-state position error for a step input to be less than or equal to 10% of the magnitude of the input. I don't believe we can follow the same method.
Hi Brian. Thank you for the good series of videos first. Then a question: if you place lag compensator as close to imaginary axis possible, does it mean the pole from lag compensator becomes dominant pole? Regards, Peng
I think while finding the dominant poles, the criterion should not be just that whichever pole is most near to the origin, dominates. One should also check whether there are zeros very near to it, as they can cancel out the effect of that pole.
Ah...new zero and pole have canceled out each other mostly with some effects left which slow down the dynamic response. Is that what you meant? It kind of makes sense. Thank you!
Peng Bai Ya right.
A video on lead lag compensation please.
you are gold
how to find dominant pole? is is the PI same with lag? and PD same with lead? I'm really stuck in this chapter. help me plis.
Best ever
This feels incredibly sketchy, because it's only the DC gain of the phase lag compensator that addresses the steady-state error, and this could literally just be achieved by adding a gain stage. Is it just that the addition of the pole and the zero help to reshape the root locus to its original shape to compensate for this change in gain?
Based on your next video, it looks like that's exactly what's going on. The lag compensator reduces the gain at higher frequencies which offsets the effects of including your additional DC gain that addresses steady-state error.
at 9.51 u put -3 dominant pole,why dont u take -1. please help??
-1 is uncompensated close loop pole
-3 is designed dominant pole location
we want to make locus like P(D)=-3 not -1
I hope that this is some help for u
Hi Brian!
I fell in love with control systems engineering after watching your explanations. But I do have a question about the lag compensator
If you're placing the poles of the lag compensator as close to the imaginary axis as you can, wouldn't that mean that the roots of the compensator become the new dominant poles?
If that happens, would that mean that your system doesn't act in the right way now?
Sorry, I feel like I'm just missing a really obvious point haha but I can't figure out the answer
Hi Jason, since you're placing your lag pole close to the lag zero they will 'cancel out' and you will still maintain the dominant poles as they were before introducing the lag compensator. Hope that helps
I have the same quesiton.. I think they dont cancel out because theyre not exactly on top of each other.. i cant fully understand this
Does the added ....16. Change your reference from 125. to 124.44444
9:52 wait why is the dominant pole -3 instead of -1?
1:49 is scary! :)
you said that dominant poles doesn't move much when you bring new pair closer to origin. you are adding a new pole closer to origin than the previous dominant poles. doesn't it mean you are changing the dominant pole's position to new point?
+Uday Santosh Raju V
I also have this question !
If you find the answer through other resources in future than please don't forget to post it here, too.
thanks in advance !
+Milan Shah mmm... Yes, technically by putting the lag pole closer to the imaginary axis than the original dominant poles, you have made of it the dominant pole. When he says that the dominant pole doesn't move too much, he is talking about the original dominant poles. Using Matlab you can find that the position of the poles for the closed loop transfer function for the uncompensated system(for the Open Loop TF in the minute 6 of the video) are in p1 = -0.6111+ 0.42673i
and p2 = -0.6111 - 0.42673i. The position of the the poles for the compensated are in p1 = -0.6120 + 0.4278i , p2 = -0.6120 - 0.4278i and p3=-0.00549. As you can see, the position of p1 and p2 have changed very little. Plotting also the root locus in matlab you will see that there is practically no difference. The only one that you have is that now in the compensated system you have one additional locus branch in the real axis that goes from the lag pole to the lag zero.
Hope it helps!
+Uday Santhosh Raju V I think there is also this little trick in control theory where if you have a pole zero pair which is very close (called a dipole pair I think), they basically "cancel out" like they would if there were in the same position (so they negligible effect on transient response) so we can approximate our system to be a second order system. The exact math though, would be kind of complex.
+Prabhpreet Dua Well, in this case, if you take the step response of the compensated and the uncompensated closed loop in Matlab(for the Open Loop TF in the minute 6 of the video) , you will see that the transient response of both systems is different. The settling time is way different(around 10sec for the uncompensated, 800s for the compensated(!!)), the rise time is also very different. You can see that also in the graph that he shows in the minute 10:25, where both the rise and the settling time are way higher in the compensated system. As he says, the problem didn't have any time settling time requirement so that increases didn't matter that much. The objective was to decrease the steady state error and the Lag Compensator did its job.
The Lag compensator moves the root locus of the uncompensated system to the right(in the opposite direction to the Lead Compensator). The purpose for wanting the original root locus from not moving very much from its place is in order to not lose stability. I really don't think that you approximate the system to a second order equation!
Hey Awesome video
But how to implement compensators as algorithms?
I started from CS GO lag compensation and somehow ended up here
mhah lol
Thank you for the videos, they are great. in the video how did you determine that the zero should be at -3/50 = -0.06 ? (how did you come up with the number 50)
He has mentioned to place it near the imaginary axis but you cannot place it very near as the practical values for components that make system become impossible to achieve. So it can be taken as thumb rule to place the zero at 1/50 of the dominant pole real part i.e. real of (-3+2i)/50
i thought you said the pole that is closer to the imarginary axis is the dominant one. why are we selecting 3 as the dominant pole
+Don-Roberts Emenonye as you are thinking -1 and -3 are no more our poles. In the last part of this video we already designed lead compensator to set poles to -3+2i and -3-2i
Is there a tutorial for the root locus?
czcams.com/video/S4DhKn1fRAI/video.html
or
czcams.com/play/PLUMWjy5jgHK3-ca6GP6PL0AgcNGHqn33f.html
All heroes don't wear capes.
something is definitely wrong with the plots at @11:00
step response is the convolution of the step input with the impulse response. if the lag compensation does not change the impulse response, the step response cannot change also.. Hence I think the right plot must be wrong. In other words, lag compensation must change the impulse response...
Another way of approaching the same result: impulse response is the derivative of the step response.. But the derivative of the blue and green step responses in the left hand plot will not give the same impulse response...
CONCLUSION: When we add lag compensation to a system, the impulse response cannot remain the same. Hence the left hand side plot must be wrong..
Actually, when we add lag compensation to a system, the pole of the lag compensator becomes the dominant pole of the system..
Get this man a college
Way to fast on this one brian. You make a video about lag compensators but then go off on steady state error with out making sure people understand the error part
good video but you go way to fast lol
Your videos are hard to understand