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
@@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?
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....
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.
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.
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.
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.. :)
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.
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...
Longshot on a response, but I'm working on a problem to design a lead compensator D(s) = K(s+z)/(s+p) with an open loop plant of G(s) = 1/s^2. As I understand, there are two poles both located at the origin, and the prompt gives a desired pole location at -2 +- 2j. The locus of the open loop plant would be two poles at the origin that extend out at 90 and 270 degrees. Given that the desired pole location is at -2, 2 and the angle contribution of both poles are 135 degrees, the resulting theta 1 and 2 add together to be 270 degrees. Obviously this is 270 does not equal 180, so the point isn't on the locus and the compensator will have to add zeros and poles. If I place a zero at -2, below the desired point, the angle contribution to the angle criteria is -90 degrees. 270 - 90 = 180, so it now satisfies the angle criteria, however there is an unaccounted for theta p for the angle contribution of the arbitrary pole. The problem is then that the contribution of the added zero made the locus pass through the desired point, meaning that the added pole angle contribution should be 0, which is not possible unless I'm missing something. I did try moving the zero one unit to the left, placing it at -3 rather than -2 directly beneath the desired point. This led to an angle contribution of 63.4 degrees, meaning that the added pole contribution would be 26.6 degrees. The problem then is that, when solving for the distance that the pole must be placed for the 26.6 degree angle contribution, the "x" value is equal to 2/(tan(26.6) which comes out to be 4, meaning the pole should be placed 4 units away from the desired point, -2. I took this to mean that the pole should be located at -6. When comparing to a solution manual, all of the above is incorrect. I've ran through the problem a few different times to make sure I've done it correctly according to the procedure shown in the video but still haven't come up with the correct result when plotting in matlab. The solution manual states that the D(s) should be K(s+2)/(s+20). When plotting, this gets close to the value but not exactly crossing, so I'm unsure how accurate the solution is. Regardless any tips would be helpful as I'm not sure where I'm going wrong here.
+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.
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 ?
+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.
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.
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?
you have videos that are connected to each other, it would be nice if you would link them in the description. for example how to find K. I dont know which video to go get that information from
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
Ohimai Musa For the zero located at -4, he just placed it there randomly (it just need to be left of the pole at -3 so that it can move the root locus towards the desired pole location). After placing the zero at -4, he can then find a suitable pole location of the compensator to meet the angle criteria. Now to find the angle from having a zero at this is location is just geometry: 63.4 degrees = tan-1(desired dominant pole imaginary location /(zero location on Real axis - desired location on real axis )) = tan-1(2/(4-3)). Remember than the tangent of an angle is the ratio between the opposite side of a triangle over the adjacent side. Note tan-1 is inverse tangent.
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.
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?
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
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.
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.
Some people were born to teach, you are one. Unfortunately most professors aren't
can I get an AMEN
no its about effort,not talent especially for teaching
I took an entire semester of controls, and in your 11min video I finally understand lead compensator. Jeez. You're great.
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
LOL
Seriously. We went through about 3 lectures on this stuff and I didn't fully understand it until I watched this video.
@@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?
@@spm04 We went through about 3 lectures and i i didn't understand anything lol.
This 14min video gave me better intuition for lead compensators than the 100-page chapter in my textbook. Thank you so much!
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....
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!
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.
Weeks of pain and lack of understanding, and this video helped me get past it. Thank you.
Thank you! I'm working on the lag design right now ... should be up by the weekend. Standby :)
you know what Brian, the entire class of my control course were watching your video for exam study last night.. you are awesome!!
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.
This is the best explanation that i have ever seen on the subject. Thanks a lot.
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.
Man, you're saving me on this quarantine period on university. Many thanks, you do a great job!
This 13 min video is equivalent to the whole semester dedicated to lead lag compensator
Dude my prof couldn't make me understand this simple thing in a month. Thanks a lot !!!!
You are so good that I feel like I am doing a bad thing watching you for free
Incredible it's been 10 years! Good explanation, thank you
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😂😂👏👏🤔🤔
this is actually a new form of tic-tac-toe. trust me, it gets pretty intense
This is a subject of Control Engeneering.
ATGplays OP !!
lol i am feeling betrayed caz someone watched my fav subjects legendary video and wont comprehend its greatness🥲
Thank you! Your lectures brought me a great help in my understanding control system.
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.. :)
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.
Got me through Electrical engineering, great content also easy to understand. Thanks !
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.
Brian legit best teacher eva.
Please, just never stop with this awesome content. Much appreciated!
Most underrated channel
This lecture is equivalent to drinking a Sprite in a hot summer day. So much better than a university lecture...
Super helpful Brian !
Nice and lucid... this is just how Control Systems need to be tackled.
Thank you so much.
You made it simple. Tomorrow is my exam! Thanks!
Hail Brian.
All hail the controls god.
Get this man an Oscar
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...
Walter Evans would be proud!
Longshot on a response, but I'm working on a problem to design a lead compensator D(s) = K(s+z)/(s+p) with an open loop plant of G(s) = 1/s^2. As I understand, there are two poles both located at the origin, and the prompt gives a desired pole location at -2 +- 2j. The locus of the open loop plant would be two poles at the origin that extend out at 90 and 270 degrees. Given that the desired pole location is at -2, 2 and the angle contribution of both poles are 135 degrees, the resulting theta 1 and 2 add together to be 270 degrees. Obviously this is 270 does not equal 180, so the point isn't on the locus and the compensator will have to add zeros and poles. If I place a zero at -2, below the desired point, the angle contribution to the angle criteria is -90 degrees. 270 - 90 = 180, so it now satisfies the angle criteria, however there is an unaccounted for theta p for the angle contribution of the arbitrary pole. The problem is then that the contribution of the added zero made the locus pass through the desired point, meaning that the added pole angle contribution should be 0, which is not possible unless I'm missing something.
I did try moving the zero one unit to the left, placing it at -3 rather than -2 directly beneath the desired point. This led to an angle contribution of 63.4 degrees, meaning that the added pole contribution would be 26.6 degrees. The problem then is that, when solving for the distance that the pole must be placed for the 26.6 degree angle contribution, the "x" value is equal to 2/(tan(26.6) which comes out to be 4, meaning the pole should be placed 4 units away from the desired point, -2. I took this to mean that the pole should be located at -6.
When comparing to a solution manual, all of the above is incorrect. I've ran through the problem a few different times to make sure I've done it correctly according to the procedure shown in the video but still haven't come up with the correct result when plotting in matlab. The solution manual states that the D(s) should be K(s+2)/(s+20). When plotting, this gets close to the value but not exactly crossing, so I'm unsure how accurate the solution is. Regardless any tips would be helpful as I'm not sure where I'm going wrong here.
Many thanks from NL. These are a great help!
So keep it up:)
A beer in Paris sound fantastic!
This is so awesome! Thank you Brian!
dude!! i got A in my control class!
Thanks !!!
Awesome videos!! You actually get me excited to learn about controls.
I honestly wish I had just watched all these videos instead of going to systems and controls lectures
Finally I could make sense out of compensators
U are the best teacher ever! 😉 from Turkey... 🤗
very useful before exams .thanks sir.......
It was one of the best lectures of you. Thanks a lot.
I can't thank you enough 🌟 , you are a life savior
dude, I love you!
Thank you Brian for your advices. I am going to apply them.
If you come in Paris one day i will be glad to paid you a beer, thanks for sharing your knowledge.
That's awesome! Very good job. Is that your last class for the summer?
Good stuff, but you headed off at an exponential rate towards the end (the bit I needed).
But, THANKS. Great stuff.
You are a life saver, Brian. :)
Really helpful and easy to follow! I am ready for my control final exam now :D
Great lecture Brian!
you are awesome ...teaching methodology loved it ....great brian !!!!!!
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.
Brian brilliant video ,but you didn't really explain how to get the pole and the zero for the lead compensator.
+Boitumelo Dikoko True... I wish he had a full formal example on the subject.
Same here because control is very rare to find. And all textbooks have different methods and explanations.
+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.
+Jonathan Tefera Endale mmm... part of my previous comment appears crossed out... I did not intend for it to be in that way...
I teach too but the way you do is quite impressive. good going brian :)
Do you have a video on converting requirements to desired pole locations?
Would love your notes
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 ?
+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.
Brilliant video, thanks very much.
Thank you very much for the great videos!
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.
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.
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
If u come to Brasil, we can pay a beer to u too. rs. Hope u continue posting this videos!
Great video and helps me lot....thank you!
This was great ! Thanks !
so how do we determine the K after finding a good z and p?
THAT
WAS
*AWESOME*
I loved it! You are amazing. Thank you so much :D
Thanks Brian , you are awesome . :)
You are the man.
super helpful! Thanks so much!
THANK YOU SO MUCH !!!
Could you please share the way that you used to get the gain K?? Thanks!
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?
Best explanations out there !! concise and fluid. where can one get your textbook?
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? 😭
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.
from magnitude ceiteria
1 + K . G(s) = 0
i.e
K = (length of poles) / (length of zeros)
great explanation, really like your videos.
When is the lag design going to be up online?
you have videos that are connected to each other, it would be nice if you would link them in the description. for example how to find K. I dont know which video to go get that information from
how to find K from Root Locus method if you don't have Zeroes --> TF = K / s(s+4)(s+3)
What an explanation
Fantastic lecture. I found treasue
tq for ur explanation it helped me
Is there a video where you found k using the magnitude criterion?
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
could someone please explain how he got the 63.4 degrees, and how he got the zero at 4 exactly? Thanks
Ohimai Musa For the zero located at -4, he just placed it there randomly (it just need to be left of the pole at -3 so that it can move the root locus towards the desired pole location). After placing the zero at -4, he can then find a suitable pole location of the compensator to meet the angle criteria. Now to find the angle from having a zero at this is location is just geometry: 63.4 degrees = tan-1(desired dominant pole imaginary location /(zero location on Real axis - desired location on real axis )) = tan-1(2/(4-3)). Remember than the tangent of an angle is the ratio between the opposite side of a triangle over the adjacent side.
Note tan-1 is inverse tangent.
AlexB Thank you!
Thank You Sir.
So how would you go about solving this if the open closed-loop transfer function was 1/2^s, making the current poles in the imaginary space?
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.
For finding position of final pole: 2/tan(18.43) =~6, not 9. Did I mess up?
Thank you so much sir
Now it's 2024...😃Brain
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?
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
how we can find the gain (K) with calculation ... I have an exam tomorrow about this need quick answer please !
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.
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.
At 8:50, shouldn't it be sum of zero angles - sum of pole angles = 180 degrees?
+Aaron Wilson Would that not be -180 degrees which is the same thing as 180 degrees?
you are ri8
Pls... which software you are using to make videos and the hardware???
I am enlightened :D