Designing a Lead Compensator with Root Locus

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This video walks through a phase lead compensator example using the Root Locus method.
Errata:
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[@jeremyviromek]

Some people were born to teach, you are one. Unfortunately most professors aren't

[@runnerkidal]

can I get an AMEN

[@DarkAvegger]

no its about effort,not talent especially for teaching

[@cWeeks1992]

I took an entire semester of controls, and in your 11min video I finally understand lead compensator. Jeez. You're great.

[@jeremyviromek]

cWeeks1992 yea im an electrical engineer and i come back to these videos when i forget something. Universities need to work harder on hiring good teachers rather than that standard professor that has no idea how to make it sink in

[@pankajkumar-ry5tp]

LOL

[@spm04]

Seriously. We went through about 3 lectures on this stuff and I didn't fully understand it until I watched this video.

[@sanjaytumati]

@@jeremyviromek And why may I ask should they do that? What's in it for them? They are already getting your dollars. Will they get much more of those dollars if they do it your way?

[@umutyldrm9696]

@@spm04 We went through about 3 lectures and i i didn't understand anything lol.

[@rahulshiv2945]

This 14min video gave me better intuition for lead compensators than the 100-page chapter in my textbook. Thank you so much!

[@BrianBDouglas]

Great! It's impossible to cover all of the topics of classical controls in a single semester (and make sense of it all) so professors have to pick and choose what they think is important. Congrats on the A. It's sounds like you have a knack for control theory.

[@32tonybh]

Thank you, seriously thank you! Now it's all about finding the proper way of inputting z and p values to not disturb the dominant poles and still have access to a proper response to the closed loop system....

[@blizzconquer1109]

I typed locus to watch black ops 3 locus sniper gameplay but i decided to watch this and... ..... I have no idea what this is😂😂👏👏🤔🤔

[@indygon]

this is actually a new form of tic-tac-toe. trust me, it gets pretty intense

[@lechbrtube]

This is a subject of Control Engeneering.

[@subinnair3835]

ATGplays OP !!

[@rahulkulk745]

lol i am feeling betrayed caz someone watched my fav subjects legendary video and wont comprehend its greatness🥲

[@BrianBDouglas]

Thank you! I'm working on the lag design right now ... should be up by the weekend. Standby :)

[@Logantan90]

you know what Brian, the entire class of my control course were watching your video for exam study last night.. you are awesome!!

[@jayc3861]

Weeks of pain and lack of understanding, and this video helped me get past it. Thank you.

[@theboy12344]

In just 3 mins of watching this video I have learned 3 months of work from university. It's really not difficult when you explain it. Thank you Sir!

[@chinmaychhatre1170]

Sir can you include more examples in your videos? i.e start with some transfer fuction.. then design lead compensator and then drawing toot locus of the same.. AS ALWAYS love your video.. :)

[@shalaiot7420]

Thank you! Your lectures brought me a great help in my understanding control system.

[@chescondo6460]

man. big ups on your lecturing for this subject. it is the most concise overview i have come across. thanks

[@dobkhnistheking]

no spring semester at the University of North Dakota and here is the awesome thing, the prof. did not cover PID controllers in class at all so i did a presentation covering almost everything in control systems in 30 mins at the last day of class last week, and most of the material was yours, i introduced PD controls in class from your root locus lecture.

[@BrianBDouglas]

Hi LouAnn, A lead compensator can only add up to 90 degrees of phase (and that's only if the pole is way off to the left towards -inf). But realistically a lead compensator can really only add about 55 deg of phase due to restrictions in the size of the components needed to make it. Therefore, in order to get 90 deg of phase you'll need to add two lead compensators. Try designing one compensator to add 45 degrees of phase and then create a double pole and zero at that point.

[@adityakamath6624]

Got me through Electrical engineering, great content also easy to understand. Thanks !

[@prateek6502-y4p]

This 13 min video is equivalent to the whole semester dedicated to lead lag compensator

[@geofreymirongamoruri911]

Incredible it's been 10 years! Good explanation, thank you

[@pranshugarg4627]

Dude my prof couldn't make me understand this simple thing in a month. Thanks a lot !!!!

[@hesamembedded710]

i have this lab course(control system) in this semester i really hated this course but finally i really enjoy it by using your videos..tnx a lot.

[@pushpakrajss108]

You made it simple. Tomorrow is my exam! Thanks!

[@yousifalhajery]

just turned in a project involving the design of a compensator for a robotic arm. Although I used a PID controller. Your video are actually really good. I'm guessing it's just because of the type of content that isn't so popular.

[@qdunker]

Hail Brian.

[@jakelevsen887]

All hail the controls god.

[@GAment_11]

Please, just never stop with this awesome content. Much appreciated!

[@gufo__4922]

You are so good that I feel like I am doing a bad thing watching you for free

[@siegfriedsas]

Brian legit best teacher eva.

[@kairu48]

Super helpful Brian ! Nice and lucid... this is just how Control Systems need to be tackled. Thank you so much.

[@vacagonzalo]

dude, I love you!

[@coenbeeksma3475]

Many thanks from NL. These are a great help! So keep it up:)

[@LouAnn333]

Thank you Brian for your advices. I am going to apply them.

[@danslanker8880]

Wow these are really good videos. Big thanks Brian for all the work on these! I never really understood how to draw a Root Locus. I have a Masters in Engineering. Odd that I see quite a few people saying they had 3 or 4 lectures on RL and I seem to remember the same. Very good job of explaining. I have also been review Dr. Lums lectures, helpful also. I learned a lot a lot from these lectures and I like your style, it very effective. Truely thanks! Dan...

[@danslanker8880]

Walter Evans would be proud!

[@BrianBDouglas]

That's awesome! Very good job. Is that your last class for the summer?

[@Aryan91191]

It was one of the best lectures of you. Thanks a lot.

[@hrdv2]

This is so awesome! Thank you Brian!

[@srinivaschowdaryappasani2248]

very useful before exams .thanks sir.......

[@blakeyoungblood7814]

Awesome videos!! You actually get me excited to learn about controls.

[@SandeepChaudhary-vx9zy]

Most underrated channel

[@nicholaslee6098]

This lecture is equivalent to drinking a Sprite in a hot summer day. So much better than a university lecture...

[@dobkhnistheking]

dude!! i got A in my control class! Thanks !!!

[@jyothishkv]

Finally I could make sense out of compensators

[@andyhype2546]

Get this man an Oscar

[@BrianBDouglas]

A beer in Paris sound fantastic!

[@luisgonzaloandrade]

At 9:50 when he is adding all the angles of the poles for the angle criterion, why doesn't he add the angle made by the complex conjugate pole. That would add an extra 90 deg making the total 315 deg instead of 225 deg. Please and thank you for the clarification.

[@linkmaster959]

for anyone else wondering, those are not complex conjugate poles. They are the points where he would like to move the poles on the real axis.

[@eraykosan6747]

U are the best teacher ever! 😉 from Turkey... 🤗

[@iremcantas6602]

Thank you for the video, I really appreciate it. My question is: Why do we have to determine K for what? thank you in advance

[@afoose]

I honestly wish I had just watched all these videos instead of going to systems and controls lectures

[@serath1542]

I can't thank you enough 🌟 , you are a life savior

[@michaelbills7401]

Great lecture Brian!

[@DontLookLikeAFool]

Really helpful and easy to follow! I am ready for my control final exam now :D

[@vishalmeena9245]

you are awesome ...teaching methodology loved it ....great brian !!!!!!

[@milan_shah]

Very helpful video. Thanks Brian !! But I have a few doubts... 01. How did you determine to put dominant poles on -3+2j and -3-2j ? 02. How to determine dominant pole location(s) ? 03. How it can be proved that "In order for a point to exist on the root locus, the sum of the angles of all of the poles - the sum of the angles of all of the zeros should be 180 degree ?

[@kirar2004]

+Milan Shah Dominant poles are those poles which have a greater or a longer effect on the transient response, i.e., the poles which are nearer to the zero on s-plane are dominant poles. The general rule followed for differentiating between dominant poles and insignificant poles is a factor of 5.

[@zhuokaizhao5926]

Could you please share the way that you used to get the gain K?? Thanks!

[@bbq436]

so how do we determine the K after finding a good z and p?

[@careerladderupsc]

I teach too but the way you do is quite impressive. good going brian :)

[@georgederleres8489]

Brian, why are the angles drawn between poles of the open loop system and the closed loop system ? I think that part of "chosing the dominant poles" is not mathematically well-established. Very helpful video though ! Thank you.

Refer to the video about the introduction to the root locus method. You plot the open loop poles and zeroes and then start from there.

[@BoitumeloDikoko]

Brian brilliant video ,but you didn't really explain how to get the pole and the zero for the lead compensator.

[@32tonybh]

+Boitumelo Dikoko True... I wish he had a full formal example on the subject.

[@BoitumeloDikoko]

Same here because control is very rare to find. And all textbooks have different methods and explanations.

[@jonahum92]

+Boitumelo Dikoko Somehow yes he does say how to get the pole and the zero. From 10:14 on, he explains that in order for the desired poles(the green crosses) to be inside the locus branches, the angle that they make with the poles of the loop have to fulfill the 180º requirement. For the uncompensated system with poles in -1 and -3, the sum of the angles for the desired pole was 225º so he needed to place the lead pole and zero so he could subtract 45º. The position of the zero was chosen arbitrary(the only requirement that he gave was for it to be at the left of the original poles but not very far from them), in this case in -4. Once you have chosen the position of the zero, there is only one posible position where we can place the lead pole if we want the desired pole to be in the root locus(again, it must fulfill the 180º angle requirement). First you have to obtain the angle that the desired pole makes with the zero. In this case tan(AngZero) = 2/(-3-(-4)) = 2 -> angZero = atan(2) = 63.4. Knowing that 225 + angPole-angZero = 180 -> angPole = 180-225+63.4 = 18.4. angPole is the angle that the desired pole has to have with respect the lead pole. tan(angPole) = 2/(distance between the poles in the real axis), -> distanceBetweenPolesRealAxis = 2/0.333 = 6. The lead pole will be then in -3-6=-9 . The Lead compensator will be then (s+4)/(s+9) Sorry for my English! Hope it helps.

[@jonahum92]

+Jonathan Tefera Endale mmm... part of my previous comment appears crossed out... I did not intend for it to be in that way...

[@Ancientlaws]

Would love your notes

[@treyvives3207]

Is there a video where you found k using the magnitude criterion?

[@chillingmeko]

You are a life saver, Brian. :)

[@BigAL-x7f]

Best explanations out there !! concise and fluid. where can one get your textbook?

[@tomdoerschler7065]

Thank you very much for the great videos!

[@94rainbowx33]

If you come in Paris one day i will be glad to paid you a beer, thanks for sharing your knowledge.

[@piyughosh4542]

THAT WAS *AWESOME*

[@leonoberholzer2805]

THANK YOU SO MUCH !!!

[@sandeepsrikonda7352]

phase lead drags the root locus to the left only if the added pole and zero are farther than the existing poles right? If the added ploe and zero are closer to the origin then it might drag to the right.....4:20

[@trotion]

Finding K? Could you please refer me to the video where you calculated it from the root locus? I know it has something to do with 1+KL=0.

[@KIMIRAIKKONE198N4]

from magnitude ceiteria 1 + K . G(s) = 0 i.e K = (length of poles) / (length of zeros)

[@vishnunarayanv526]

Hey Brian! Small clarification. Sum of angles of open loop poles - sum of angles of open loop zeroes should be 180 degrees, because, for 1+kG(s) to be 0, D(s)/N(s) = -k. So, to account for the minus part, the net angle should be 180 and the actual gain k is decided by the frequency at which it is measured, And so to reduce the angle from 225 to 180, we add more angle in the N(s) and less in D(s) through a lead compensator. Is that correct?

[@pradipvairal4183]

Great video and helps me lot....thank you!

[@GaViNa352]

at 7:20 , why would you want to move the open loop poles to the right ever? If you're trying to inject your own dominant poles, and poles are more dominant if they are closer to the imaginary axis, wouldn't you want to direct all the other poles in the system away from the imaginary axis?

[@NolanDodson]

You are the man.

[@salehalr]

great explanation, really like your videos. When is the lag design going to be up online?

[@mnada72]

Great video. It's all under 30 seconds from 7:25 to 7:55 why I didn't these 30 seconds in my past life? 😭

[@javadmaleki9483]

This was great ! Thanks !

[@College19]

Now it's 2024...😃Brain

[@messedmushroom]

For finding position of final pole: 2/tan(18.43) =~6, not 9. Did I mess up?

[@hamzahkhan4357]

super helpful! Thanks so much!

[@ShreyFadia]

I am still not clear about first zero. We can place it on the negative real axis left to the open poles, right?

[@Krb68691]

Why does my system completely lose its integrator when I try to apply a lead compensator? When I compute the new closed loop transfer function, the denominator has a new K * (s + z) term, which eliminates the integrator. Now my type 1 system goes down to type 0. What the hell

[@vishalmishra7568]

when we add zero at -4 and pole at -9 then centroid shifts to -4.5 (so does the assymotote) but you passed it through -3 !! why??

[@jamaloubasslam7914]

yes! it can be, just your are ensuring that those poles are in the root locus. If you compute the breking point, you will find it between -1 and -3. So, when the asymptote branches start to deflect, they pass trough the new dominant poles point. Grate question!! its just an error of scale.

[@MrUsername7376]

At 8:50, shouldn't it be sum of zero angles - sum of pole angles = 180 degrees?

[@GenericPurpleTurtle]

+Aaron Wilson Would that not be -180 degrees which is the same thing as 180 degrees?

[@tajammalnawaz5552]

you are ri8

[@kirtikansal6946]

plss explain whether the poles and zeroes of a phase lead compensator must be near dominant poles or zeroes or not...but if not then how can they introduce a significant angle to design the system..

[@kungwankaipiphuannok8872]

how to find K from Root Locus method if you don't have Zeroes --> TF = K / s(s+4)(s+3)

[@arjunb5941]

tq for ur explanation it helped me

[@salim786dua]

Thanks Brian , you are awesome . :)

[@saminhas8495]

Kindly any one elaborate me how does a pid controller effects systems poles and zeros and thus stability? In my opinion PID controllers only add gains to remove errors.

[@AvinashMauryaShripal]

at 9:58, the angle is 90 + 90 + theta2, wouldn't that be (90+90+90+theta2), what about the angle by its conjugate pole?

[@luisgonzaloandrade]

Avinash Maurya Shripal I had the same question. Did you ever figure it out?!

[@Rahul-ge3ml]

Luis Andrade theta2 is nothing but 90 - theta1. Theta1 was split into 90 degrees and theta2 for the sake of simplicity.

[@aminashamil3210]

Can we place zero and pole of compensator in another location, let's say -5 for zero and smth futher for pole and still get the same result ?

[@mzakirbaig]

Hello Sir, Huge fan of yours. Thumbs up for your hard work

[@AbhilashIngale]

Probably late but no, looks correct to me. -3 and -1 both are poles and their sum evaluates to -4 which divided by 2 is -2.

[@aldobaits6984]

Thank you so much sir

[@basics7930]

Pls... which software you are using to make videos and the hardware???

[@akshaykolap5331]

If the angle contributed is more than 60 degree then we consider it is cascade of two lead compensator But why 60 degree? Why not other angle?

[@pallavbakshi612]

I loved it! You are amazing. Thank you so much :D

[@augustomunoz6138]

Hello Brian! Thanks for your amazing videos! I have a question that might sound dumb, but when you determined the angles of your desired poles and saw thet they were 225º (and therefore you had to substract 45º to make it 180º), why not simply add a zero at 45º? Why add a zero AND a pole?

[@agentl3r]

I don't suppose you found an answer to this?

[@luizpadovani]

But if your angles sum 225º shouldn't you be using a lag compensator in order to subtract 45º and attend the angle condition? I guess the confusion here is due to the fact that you subtract the zero angle instead of adding it (10:18). As far as I know, the zero angles should be added and the pole angles should be subtracted because of their positions in the numerator and denominator, respectively, in the feedfoward transfer function.

[@LouAnn333]

Hi, how about the following system: H= 1/(s+1)^3 I to design want a compensator such that overshoot and settling time to a step response are respectively 15% and Tr95% = 3s. So the dominant pole will be -1+1.6536i. When I try to design a lead compensator such that transient performance are met, I find uncompensated angle of the system is -270°, that means lead compensator must contributes for 90°. How to determine the compensator zero and pole in this particular case. Thanks for your answer.

[@anuhyajyothirmaiannamraju8054]

How can we calculate the angle betwwn open loop poles and the desired pole if one of the open loop poles is a complex conjugate?

[@thegoatzander]

please where did you get S+4 and S+5 from??? what if you have S+Zc and S+Pc instead how do u solve for Zc and Pc. please can kindly reply or someone please explain

[@tarunghosh5101]

Thank You Sir.

[@w8liftr615]

When you place the first zero, is it possible to place it on top of, or at least extremely close to, the pole already in place at -3? That way you don't need to worry about the angle from that zero or the pole it is on top of as they should cancel out.

[@Jnglfvr]

I realize that this is five years late but...yes you can!. Placing the compensator zero precisely at -3 and a pole at -5 with a gain of 8 will give you roots at -3+/- 2j.