Designing a Lead Compensator with Bode Plot
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- Опубліковано 19 жов 2024
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This video walks through a phase lead compensator example using the Bode Plot method.
A great reference to read if you're interested in learning more on this topic: teal.gmu.edu/~g...
Errata:
Around 7:00 I added an additional '+1' in the denominator of the Ess equation. The denominator should be 0.2s^2 + s + k.
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If you have any questions on it leave them in the comment section below or on Twitter and I'll try my best to answer them.
I will be loading a new video each week and welcome suggestions for new topics. Please leave a comment or question below and I will do my best to address it. Thanks for watching!
Thanks for the comment! My background is dynamics and controls in aerospace engineering from university. But the majority of my working experience is with satellite guidance, navigation, and control.
You're an amazing teacher. And video editor. The fact that I don't have to wait to see you write something down, and you're writing it down as you're saying it, along with one fluid document and nice colors, makes this series amazing. Thank you so much
I just realised I'm never impatient and skipping forward because he is always going at a great pace. The editing really is spot on
Brian, I am a ME grad student with one of my concentrations focused on controls. I just want to say without your videos I would not fully understand the concepts that are being fully covered in class. Thank you very much you are awesome!
wow 8 years down the line and you're still as amazing as ever. You have my respect
Thank you so much for posting these videos! On behalf of a large portion of the UBC engineering students in the control system class we are soooo grateful!! You are awesome!!
Hello again! Yes you are correct it is obvious from the graph that the DC gain is +6dB :-) Thanks for pointing that out. I'll add an annotation to correct it. And you are right that you wouldn't have a gain crossover frequency in your case. And that means that your system (assuming this was a Bode plot for an open loop system that you were planning on closing) would have infinite phase margin! Or no matter how much delay the system had it would still be stable.
120k views and I'm responsible for at least a half of them
Brian , if there is an equivalent of the oscars for control engineering you should get an oscar for this a series of videos . These videos are a masterclass in control engineering !
You can use the command margin(sys) in the command window to bring up a Bode plot with the gain and phase margins in the title. Another way to show margins is to type bode(sys) and right click on the graph, -> Characteristics, -> All Stability Margins.
Thank you so much, you have told me everything I need to solve my projects at Faculty. Keep making videos like this one. Best regards. From Mexico
Thank you.You saved me.I had a project about this but the lecturer told us to find out by ourselves how to find the compensator.
I just want to say, thank you for all of your videos. I'm taking controls this summer and it has been a very fast pace class to be in. Your videos and step by step instructions have made studying 10x easier. Thank you!
Your videos are fantastic, I am currently doing my maters in Controls Engineering and all we do is complex math to solve these. Which takes away from the application and process. Thank you !
Thanks for the example. I always need an example to understand anything I am trying to learn.
This is more of a trip than a video. Really solid knowledge you've been putting out man!
You are the best. I got this so fluently. I wish we had prof like you.
Not only have you saved my exam, you got me like the subject too. Big thanks
Also, the -3dB gain, which can be called the bandwidth of a system, is the frequency at which the output signal is sqrt(2)/2 times smaller than the input. This is traditionally thought of the as the frequency above which the system effectively attenuates the signal so much that it doesn't pass it through. Of course there is a little attenuation at lower frequencies and some passing at higher frequencies but it gives a good measure. So if your signal cut all gains in half at all frequencies ...
thankyou very much for this good explaination. i have to translate a little here and there to the way my teacher thought me but basically what he explains in 6 hours., you explain more clearly in 15 minutes! so thanks for that!
all of the videos are very helpful for my exam
super big thanks Brian
Excuse me. I have a comment.
Brian Douglas.
You're an American hero.
Wonderful video! I have been struggling with compensator design for a while now and this is a tremendous resource.
Oh my god could you please teach my university professor how to teach the course? You are just amazing
Once again you saved my bacon on a control systems assignment. Awesome video. Awesome.
i think you meant to say +6dB:) Awesome videos BTW.
@ 3:33 for other's reference :)
Man, you really deserve that money on patreon, i will start to danate, you're so good in this!
Really excellent Brian ! You’re videos are awesome!
Very strong and very dense video. I find my self not able to absorb the amount of info in one video in single session.
Nice i'm about to start electrical soon and watching is helping get a grasp on whats ahead so thanks!!!!
man your are awsome, thank you very much. you explain better that a lot of professor I had
How did you get wm to equal 22.2?? Thank you for all you help! I love your videos.
Based on the color he used (yellow) to write the phase margin he chose, I assume he also chose wm at that value. If I'm not wrong, usually there are constraints, like the sensitive of the closed-loop system to variations, that will guide us to chose the right value for the wm frequency. Please someone correct me if this is not correct.
Your videos really are incredible. Extremely informative, clear and easy to understand.
Just makes it so blatantly obvious how backwards our higher education system is. We are taught by people who don't know how to teach. Most of my professors may be extremely knowledgeable in their field, but their method of conveying information and teaching is stale. And that is because, unfortunately,, our system favours those with more publications rather than ability to teach. The primary interest and focus of my professors lies in their research - lecturing to them is a thing they do on the side. They strive to improve their research and produce more publications, but few of them strive to improve their teaching methods.
Brian, thank you so much for this video, really concise and informative.
Best video i ever seen for control system 🔥🔥🔥🔥
TOP tier videos Brian!
You're an amazing teacher.....Really thank you for sharing the videos....I have a doubt....K > 50 right?
Dude do more of these! so helpful
First of All Thanks a lot for the videos. I really admire your effort and I'm very grateful for these lectures. At 6:54 Step 2, Where's that extra 1 (adding with K) in the denominator is coming from ( right most equation in white colour), when trying to find the value K.
Best teacher ever, thought only indians are good in teaching, u r creating competition 😂✨
seriously good work, recommending you to all my friends.
Brian's mentioned a couple of times that component values become too big or too small to be practical, can someone give me a resource that shows how you would select component values and actually implement a continuous controller?
Used both Lead compensator to further improve 10% of current phase margin of the response.
(You can used Scilab to draw the bode plot).
hey you referenced a professor at my school! awesome!
Thank you for your elaboration!
Thank you Brian
You kinda sound like Ben Wyatt from Parks and Rec, which is awesome btw
Thanks for making these videos they are superb!
at 6:57 you have 0.2s^2 + s + K + 1 in the denominator. When I do the math the 1 is not there. Am I doing something wrong?
Im 6 years late but yes, he made mistake. He mentions it in the description box.
@@aki_gunting4644 Better late than never. I had the same question.
You could invert colors and print in pdf your classes! Would be great for study have all that you write in paper =D
By the way, congratulations for your classes! It saved my semester in control.
Greatings of Brazil!
Hey Brian. At 6:59, shouldn't the equation be (0.2s^2 + 1)/(0.2s^2+s+k)? how comes there is an additional "+1" in the denominator?
Im 6 years late but yes, he made mistake. He mentions it in the description box.
@@aki_gunting4644 well I am 8 years late but thanks for your comment. I thought that I made a mistake somewhere when I try to calculate myself.
@@aki_gunting4644 Thank you, I was wondering the same.
Thank you! Was about to comment, been wracking my brain with finding out where that '+1' came from. Glad to see i'm not alone.
you should add to title "PM GM" soo it will be possible to find with search.
this is an exelent lecture!
thanks.
This makes so much more sense now then only in the time domain. xD
Dear Brian, tnx for the video, as a question:
Is this problem (we cannot add more than 55 deg of Phase Margin using lead compensator, otherwise the components become unrealistic and big) also includes digital implementation? Because in digital control, we do not use capacitors and opamps anymore, and therefore the implementation of phase margins larger than 55 might be easy.... what is your opinion about this?
Very good lecture man
Hey Brian. Gain Margin and Phase margins are defined for the "closed-loop" using the "open-loop transfer function's" Bode plot. And for the finding the bandwidth we need to see the bode plot of Closed-loop transfer function. But in the video (at 3:36) you mention the Gain margin, Phase margin, and Bandwidth on the same bode plot. I think Bandwidth should be found from the Bode plot of the Closed-loop transfer function. Just a thought. Could you please help me with this question?
Awesome video man! Thank you!
Brian Hi , I am an Electrical student , I really appreciate your videos and work your are doing , BUT can you please sort and number all your lectures in order , so when you recall in your particular video(some others , when you say for e.g. "in the previous video") we could easier find it .
Thanks again,
Ron from Boston , (Ma)
.
at 12:10 the value of wm is selected as 22.2 rad/sec how it is selected and second thing how is this table over a2 id built
Too late but might help some other people - you can find that value by calculating the extra gain:
extra gain = 20*log(1/sqrt(alpha))
and then finding its corresponding frequency in the bode plot of the uncompensated system :)
@@666blablablabla666 Is alpha a2?
THIS IS GOLD!!!
Hey Brian,
I think you have an error in the phi max relation w/ Alpha equation at 10:50. it should be:
phimax = sin^-1(1-a/1+a)
no because he uses a > 1
There's a subtle difference between the two ways I presented the data and so I can see now that it is very confusing. In this video I am talking about the steady state ERROR to a ramp input. If you have a type 0 system with a ramp input the final ERROR in the system will be inf, which means your system does a poor job of tracking that input. In the other video I was talking about the steady state value of the OUTPUT, not the error. In a type 0 stable system the final output is 0 to an impulse.
then the concept of bandwidth doesn't make much sense. Also, once thing that drives me crazy is that the term bandwidth is used in so many applications that it's hard to keep track of the meaning one person is using versus another. Therefore, I should have just said the -3dB gain frequency and left it at that. And if there is no -3dB gain frequency then it doesn't exist.
very true
i love this content. thank you so much!
Hello and thanks for the video, at 9:43, you write the upper cutoff freq. eq. and marked it on the lower side on the phase plot, and you marked the upper side for the lower cutoff, shouldn't it be the reverse ? :)
Hello Ersin, you are quite right! I accidentally called the upper cutoff frequency the lower one and vice versa. I placed an annotation in the video there so others aren't confused. Thanks for catching that and pointing it out.
Dear Brian, Thanks for sharing such a nice and easy way presentation.
1. Can you correct me that in the first 2 steps you actually introduce an Integral I controller by placing the pole at origin and adjusting the required Gain "Ki" ?
2. Can we say a lead compensator is actually a PD controller ? if so then you first introduced Integral controller and then PD so would that mean it is indeed some how a PID controller design video ?
3. Can we say PD = Lead compensator, and PI = Lag compensator so they are just two different names of the same controllers if not than what are the difference actually ?
Thanks
Regards
Bandwidth is mostly defined in the closed loop frequency response plot. Since you mentioned along with PM and GM, it may create some confusion..just a suggestion
I think K should be greater than or equal to 50 using the steady state error. Anyway you still got the correct K value.
Did you plot the bode plot for Plant G(s), or did you include K/s as well while plotting?
Super helpful! Thank you so much
Thank you Brian! The reference link in the description seems to be dead though.
That's so cool! Could you tell me what subjects you feel would be the most important for me to grasp during uni, in order to be successful at space engineering?/Do you have any advice for becoming a succesful aerospace engineer :)
Hi Brian, Thanks for the great lecture! You mentioned some disadvantages of adding a double integrator to the controller design to get zero Ess. Could you please justify those with an example if possible?
Thanks for doing these videos Brian they are awesome and very helpful! I was wondering though how did you derive K > 49?
6:30
Thank you SOOO much !! This video helped me a lot :)
Is it possible to go with step input instead of ramp input, and leave the system as type ( 0 ) ?? which means without adding an integrator ?
What is your background if you dont mind me asking? Very interesting videos!!
Can you make some vides on filters area? kalman, particle filter.. Thanks! Keep working on this beautiful classes!
Great instructor....Slute You Brian
Hi Brian, thanks for the video !
In this video at 4:47, one thing is very contradict with what you said in your "Final Value Theorem and Steady State Error" video. In that video you have said for type 0 system -> FV = 0, for type 1 system -> FV = real value & for type 2 & later systems -> FV tends to infinite.
And in fact for a ramp input don't FV will tends to infinite for any type of systems as laplace transform of ramp signal(input) is 1/s^2 so for any type of system there will be at least one 's' in the denominator !!
Sorry if I am wrong but please do elaborate...
+Milan Shah The reason the properties seem to be opotise is that you evaluate FV when t->inf, but evalueate ess(t) when s->0.
The Final Value Theorem formula expresses that.
Hope it helps.
Hi Milan,
I have the same confusion as you, do you get the answers or do you figure it out? Thanks.
Hi Milan,
I found the answer, maybe you've already known since it's your 2 years ago question...
The reason is we use the wrong formula of ess, ess=lim s*{U(s)/(1+G(s))}. Here U(s)=1/s^2 for example. For type 1, G(s) has a denominator of 1/s, then ess= lim s*{(1/s^2)/(1+1/s)}=lim 1/(1+s)=1 or finite. For type 2, ess = lim s*{(1/s^2)/(1+1/s^2)}=lim 1/(s+1/s)=0.
Hope this help others who have same question.
Brian, the lead compensator has anything to do with type 2 compensator used in Power Electronics Control, for example?
At 6:55, you have an edit that corrects the denominator of the S.S.E. equation. If s goes to 0, wouldn't you be left with (1/K) < 0.02? Which would mean that K > 1/0.02 which means K would have to be greater than 50. So is using 50 correct? 49 would've been the right gain value had you not made the correction to the denominator. I'm not sure how much this changes the entire solution.
where does the 1 come from in the denominator at 7 minutes? isnt it (0.2s+1)/(0.2s^2 + s + k)50
The 1 comes from the blue equation just before it, where the steady state error is lim as s goes to inf. times R(s)/(1+G(s)). Now the question is where did THAT 1 come from? :) It came from re-arranging the block diagram in terms of the error, E(s), as the output instead of Y(s). If you start from a standard unity feedback block diagram with G(s) the transfer function in the forward direction (like I have drawn around 6:08) we can re-arrange it like this. E(s) = R(s) - Y(s), where Y(s) is the signal just to the right of G(s). Which means Y(s) = E(s) * G(s). Substitute and get E(s) = R(s) - (E(s) * G(s)). You can solve for E(s) and get E(s) = R(s)/(1+G(s)). This is the function that we solved the steady state error on and that is where the 1 came from. I hope this cleared it up for you and didn't just make it more confusing!!
Brian Douglas I think Peter meant the +1 (the one written in white color) in the denominator. Could be an error? For me too the denominator is (0.2s^2+s+k) and not (0.2s^2+s+k+1) as it's written in the video. Anyway I want to say thank you so much for your videos it really helps.
Brian Douglas
I concur with the comment above.
Ess = lim[ (1 / s) * ((0.2s^2 + s)/( 0.2s^2 + s + k)) ]
s->0
instead of:
Ess = lim[ (1 / s) * ((0.2s^2 + s)/( 0.2s^2 + s + k + 1)) ]
s->0
By the way, thanks so much for your videos, they have been really helpful.
yorchais I also noticed this and agree with OP. Again, thanks so much for your videos! They are currently saving me in my university controls course!
Trevor Fritz You all are correct! I completely messed that up and didn't realize what was being asked earlier. Thanks for catching it again and reminding me to review it. I'll add an annotation right now and add it to the errata list for this video. Cheers!
Are you using your mouse to write all this???? If yes, I am in complete awe.
Great video but in your first example you state a second order system but it is clearly 3rd order as the phase shift is 270 degrees.
Brian , say you have a particular piece of hardware , how do you derive the transfer function for the plant . You assume the transfer function for the plant but how can this be derived experimentally ?
Brian, At 3:35 in the video, you say DC gain is about -6dB, did you mean to say positive +6dB?
Hi Brian, I'm a teacher, and really fascinated me the tool you use to make your presentation, I would like to know what program or tools do you use to produce them??
amazing lectture!
Amazing job👍👍 .. thanks alot
Thank you. I've posted a video on how I make a control system video just for that purpose. Just search for that video title with my name and it should come up. I hope your own presentations turn out well!
Sir i think there is a mistake at 6:52. There should't be +1 at the denumerator side
hi, Brian, one more question: how do we determine phase margin requirements if we only have some settling time specifications?
How in the denominator you got 0.2s^2+s+k+1? Shouldn't it be 0.2s^2+s+k.
Hi! Thank you so much for this video. I still don't get how you were able to calculate wm, did you just chose it? Does anybody know ho he did it? Thanks in advance.
You don't compute wm, you set it equal to the zero crossover frequency. Cause we need to increase the phase at that particular frequency i.e increasing the phase margin !
@@omaraissani5910 Thank you su much for your answer!
awesome, amazing ...
Could you add videos for compensator design using matlab or could you provide the codes in the description below. It would be great!!!
Thanks for your video!! Although I'm a bit unsure why having more phase is a good thing?
Answered: you want your phase to be > -180 degrees at your crossover frequency, otherwise the closed loop system is on the verge of being unstable. This lead filter is designed to bump up the phase at the frequency so that the closed loop will be stable and robust.
I was confused with the "steady state error". The ramp response characterizes the s.s.velocity error, which is given by ve(t) = | (1-G(0))*t - G'(0) | , where G(s=0) is your transfer function when s=0 and G'(s) it's derivative.
So what you mean is velocity error,that is different from steady state position error, which is obtained from the step response as pe = | 1 - G(0)|. And by making ve=0 you also make pe=0, which makes sens.
But I'm still troubled, because the ve(t) equation I give above doesn't yield the results you mention relative to system type. The ve(t) of the system you mention at 5:01 yilded ve = 0.2, which is constant and not infinite as you say.
Now I get it... I think you weren't clear in distinguishing the open loop Gain (GL(s)) from the unit feedback gain (Go(s)).
So, if the GL system is of type 0, like the function at 5:01,with GL(s)=1/(.2s+1) then the corresponding Go(s)=.2/(.2s+2) wount yield a finite ve. In this case the velocity error is ve(t)=.5*t+0.05 which tends to infinity.
I checked in the book Chen - "Analog and Digital Control System Design", on section "6.3.2 System Types-Unity-Feedback Configuration". There it is explained why and how the type influences the velocity error.
I dont understand why you say the plant is k/s.. that's the controller the plant would be (1/(.2s+1)) right? Secondary question: if I have a requirement on both phase margin and crossover frequency which should I design for first? I would assume the phase margin?
hi brian douglas,here my question is that why we are adding only 15 degrees to pi(max).Why don't you add any other....?
@Brian Douglas Is the Gain @Wm in dB or is this the K-Factor? And is this Formula really correct?
thanks you so much. I got a weird question, what if the reference signal is 0? For example we want hold a rob on a moving car always straight up by a motor on the bottom of the rob and thus the reference is 0 degree