Routh-Hurwitz Criterion, Beyond Stability

[@yvonnesun3070]

you deserve my tuition, not the prof at school who can't explain anything

[@sreeraj1980]

cant agree more

[@vaguebrownfox]

Exactly!! These people (all yt teachers) are the ones who are truly educating us with concepts in a very sensible way. I don't understand why my parents pay so much money, I just sleep and bored to death in classes.

[@Douglas.Kennedy]

My professor in control theory was actually pretty good, but I still find these videos helpful and interesting :D

[@ryuuzSd]

don't know if you are still reading comments , but you really helped me ! Thank you , from Tunisia.

[@BrianBDouglas]

You can specify your requirements either way. Sometimes it makes more sense to specify time to half or double, especially when there are humans in the loop. For example, if you let go of your steering wheel your car would eventually veer off the road even if the road was straight. In this case the plant (car) is slightly unstable. As long as the time to double is small enough for the driver to compensate for the veering then the closed loop system with the driver will be stable and safe.

[@TabletopWargamer]

I haven't seen this in a textbook. This is the kind of info I'm here for. Thanks!

[@BrianBDouglas]

I don't know what I was thinking!! Of course it would simplify to that since the s^1 line has a zero in the second column. Thanks for finding that and letting me know. :-)

[@vanickkam1255]

You'll help me pass my exam. It was postponed due to the covid 19 crisis and I'm thankful this made me come across your videos. Thank you Brian.

[@bl4ckm4mb4f0rlif5]

You sir have a unique gift in your ability to choose a minimal yet appropriate combination of words that usually present a more than adequate level of insight into a concept that makes one feel as though one might have to get zapped by a Laplace Transformer from Cybertron to be able figure out what the pole is going on..

[@jeevanraajan3238]

Brian..U r the legend !! That rockstar thing was the most awesome example ever !!!

[@k4piii]

2016 and still helpful, thx Brian :D

[@bibekshah3701]

Didn't notice that it was made in 2012 until I saw this comment. I'm watching this in 2017. Thanks Brian for such a wonderful work.

[@tonee899]

2018

[@VenkatChidambaram]

it's 2018 bitch

[@AJ-et3vf]

2021 🙋🙋🙋

[@oguzserin]

2024

[@ilyassessasnaoui7478]

2019 7 years and it's still useful , sir Brian Douglas i might not ever meet you but i consider you as my teacher and i'm sure many of the future generations will benefit from this ! your name will be remembered .

[@PatrikDezseri]

update 2023 it is still usefull and bonus point, it was my teacher who told us to watch this youtube channel to help our study

[@Cynxstry]

I was in middle school when you posted this video, 8 years later I'm using this to study for senior midterms for my undergrad. crazy

[@firestormv91]

Thank you making these videos they are really helpful for everyone who wants to learn....

[@JordanEdmundsEECS]

Wow the last 2 minutes were mind-blowing. You could use this to find the gain of your system when any poles cross any line along the imaginary axis. That's epic.

[@jyo3337]

Sir your job save millions of engineers. Huge Thanks to you

[@j.f.4859]

Awesome videos. You are able to very efficiently communicate information in an understandable manner. There is no way I could learn at this rate with my textbook and there is no way I could learn at all with my professors.

[@BrianBDouglas]

Hello Scott, I didn't do a good job explaining that. The impulse response of a pole is e^st (where s is the pole location in the s-plane). So if your pole is at s = -1, then the impulse response would be e^-t. And in this case the time to half would be approximately 0.7 seconds. That's because at t=0 the response would start at 1, at t=0.7 the response would be 0.5, and at t=1.4 the response would be at 0.25 and so on. "Time to half" is equivalent to stating "poles must be to the left of".

[@Ming_Qiu]

Thank you so much for the explication. However, i still have some question. To ensure the time to half, as you said that “time to half” is equivalent to poles must be to the left, then, when t=0, the response should be 0.5, not 1. isn’t it?

[@max_4310]

You are the best. My course take 1.30 hours for 12 minutes from your clips but you can describe better than its.

[@houssemjebari7153]

2020 and Brian is still a Hero

[@elliscld1386]

I I've all three of your routh there and just wanted to say thank for making it understandable and enjoyable

[@longalexislong]

hey great lectures, but you made that last entry in the table way more complex than it needed to be. It's ((42-k/3)(24+k)-0)/(42-k/3), so just cancel 42-k/3 rather than expanding the numerator and you're left with a nice neat 24+k

[@harshpatel5280]

I find ur lectures more practical. i appreciate it! You are saving time of us for good cause with short super videos

[@wonduek92]

Thank you for making this videos. I can finally understand something haha

[@bharthdk1139]

James Lee gv fun n fun v the f got FB f fun for hello

[@adem5484]

2018 and till now helpful, You really help an engineer ! Thanx From Turkey :)

[@deneyimadam2926]

It is 2022 and i am so glad that you have made that videos !

[@himanshukaushik5761]

Thank you so much Brian!! You have done commendable work. Hats off to your teaching methods.

[@awstyno6019]

Thank you for doing what my tenured professor with a PhD could NEVER.

[@vaguebrownfox]

Seriously man... why on earth am I going to uni? We need more people like you.

[@KadirErturk]

This is awesome work. Thank you for making this video. That would be great if you mention how to find the Transfer function with an real life example.

[@tstamir5981]

even though at the begining - you had mistakes with fourier trans and then corrections and it was very annoying, i have to say - those lectures are gold! and you are a great teacher. thank you very much

[@richardkroesen]

Thank you so much sir, you are really good at making concepts visually which is how I perceive the world. And you are clear and concrete, thank you again :)

[@BD-Research-Den]

2019 and still helpful (Y)

[@ChamathEdi]

2020 it still is

[@AJ-et3vf]

2021 and learning it in advance for my Control Systems I subject.

[@koushikghosh8770]

2019 and its really great.. thank you man..

[@sullivan3503]

I'm impressed that you can do algebra while lecturing so smoothly.

[@ohayuhanna]

"If the roots have both real and imaginary parts, lets say the real part is negative the time response will be a sinusoid that decays according to the envelope determined by the real part" This is so well explained! I have been studying these topics for some time now, so much so that I've arrived at the Routh Hurwitz criteria, but this explanation made it clear for me how this works

[@BrianBDouglas]

Yes it is! Try it out and you'll see.

[@muazmasood1664]

Mr Brian you have done an amazing job at explaining these control system concepts that seemed pretty complex. Hats off! stay awesome ! absolutely love these videos! keep up the great work :)

[@tungdinh4114]

2016 and these lectures are still helpful :D thank you very much! Keep doing amazing videos :D

[@imacoder3160]

2024 here))

[@GeorgeVidal]

That was incredible. I love music and everything sound related which is why im studying electrical engineering. And that example about having the gain just right just caught my attention so much. I wish my professor would use examples like this. He is terrible.

[@aniketnitj]

It’s a privilege listening to you. Let me know if i can ever be any help to you… Thank you so much 👍

[@mourgelas4510]

I gratefully concur with Vicente here-below. Thank you Brian!

[@BaselogicsBlogspotIn]

Very good Control lectures.... very basic and very well taught..

[@aallingsation]

Thanks for the fabulous lectures, Brian. From South Korea :)

[@jahansaid6382]

Thank you very much for all of your effort to make a video like this !

[@Manchi_]

Amazing videos, tomorrow is Automatic Control final and I'm learning this now lol

[@trcnsz]

Thank you for these videos , you are much more better than my professor.

[@jessstuart7495]

The Routh-Hurwitz Method tests if there are any polynomial roots with Real Part > 0. This means the test will detect the presence of roots where Re(z) > 0. Because we want to know if roots are to the right of the line Re(s) = -a, this gets mapped to the z=0 axis. s=-a is mapped to z=0 s+a=0=z, z=s+a. To apply this transformation, you must replace s with z-a. Now you need to calculate how far you shift the vertical axis for the time to half calculation. exp(-a*t) = 1/2 -a*t = ln(1/2) a*t = ln(2) a = ln(2)/t Change to variable z (s=z-a) and run the Routh-Hurwitz test on the polynomial in the variable z. If you find RHP roots in z, that means you have roots to the right of the Re(s)=-a line in the s-plane.

[@jasonleng2490]

helped me a lot on my control theory class! thank you so much Brian!

[@satyanshshukla9535]

its really helpful,please increase the number of videos on basic topics

[@harshgoel6860]

I will make a control system that detects the position of such people and sends them nobel prize

[@oskarlukasik9278]

thanks you. i have a signal and control system exam soon, your videos are helping alot. I wonder if you have more signal videos also

[@bhanuvardhanch7324]

100 10-minute videos sound and feel much more fun than 10 100-minute videos tbh.

[@Vanshkharbanda1]

Amazing lectures. Thank you so much.

[@scotttn1990]

How did you determine that all the poles left of -1 on the real line meets the "time to half requirement"? Can you elaborate on the relationship between "time to half" and the value of the line?

[@VictorMilare]

Thank you Brian awesome lessons, still helpful in 2017! And to a viewer from Brazil

[@andrewmartens6776]

There is a little trick my professor shared where when you are performing the figure 8 math talked about in the video and have a L shape pattern (with a zero in the bottom right hand corner) then the answer to the figure 8 math is the value of the top right hand corner.

[@SnapperTwentySix]

Thats ok happy to help, especially since your videos have hhelped me with University quite a bit :)

[@dan20pesos]

Thank you so much!!! I don't know why are you not a teacher of control systems at harvard! or maybe you are

[@kasvi0484]

Great explanation 😍hope to see more videos of more subjects

[@adityachaudhari_0156]

Best it's so easy to understand as practically

[@kuan926]

Brian, you are a god amongst men

[@nirajkelkar6708]

Its really really helpful. Thank you Brian

[@ruochensu]

Thank you so much, these videos are so useful!

[@danieaucamp]

Brilliant explanation! Thank you very much!!

[@manmohankrishna2407]

very nice explanation and thorough content.

[@amrabdelghanyy]

2017 and still saving us

[@TimPicarTV]

Thank you so much for the lecture video. It really helped me.

[@SnapperTwentySix]

6:06 the s^0 equation in the 1st column should simplify to 24 + K

[@victorbarroscoch]

The last coefficient of even rows is always the same.

[@Imrishabh168]

very precise and up to the mark.

[@anishjha4499]

thanks...for...your..Lectures.....Sir...

[@bakasa3429]

Just when I understand Routh-Hurwitz very well he said we will almost never use it :)

[@alilabbene8166]

Wonderful videos with an extraordinary and really intuitive approach. Thank you Brian! Thanks a lot! Just one request if I may. Would you please avoid using the dark purple? It is not easily readable over a black background.

[@kylejsears]

These videos are Fantastic! When doing the Routh Array shouldn't the final value at the bottom of the table should always be equal to the s^0 coefficient?

[@vanshgoyal-uo5cf]

Thanks and love from India

[@appubanerjee2727]

superb..explanation.....sir i beg to say you...that if u give few more ...conceptual lectures.... punching .....Circuit theory and Control system...then it would be of great help to me.....because i am still now searching a connecting line between Ckt theory and Control systm....

[@garychen6367]

thank you very much, Brian

[@ankursingh2630]

you are the best i've seen till date oh my god just simply awesomeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

[@abdulmajeedmohameed949]

2018 and still helpful, thx Brian :D , wish u luck

[@deepaksree6]

please do a video on steady state error calculation of non unity feedback

[@vukrosic]

Ty for carrying my degree

[@ify6ify]

You, Sir, are AWESOME! Thank You and God bless

[@mgmichael85]

Hey, Brian Thank you for the lecture: you are really helping a lot of guys around the world., B-) (Hope that helps you sleep better at night). haha.., So just one question about the time to half requirement. If you are given a numerical value: say 5 seconds. i.e the output signal amplitude should decay to 50% of its peak amplitude in 5 seconds. How does one calculate the appropriate Z value from this information so that one is able to translate the S-line to the new Z-line.

[@suneeldatta]

The impulse response of a pole is e^st (where s is the pole location in the s-plane). So if your pole is at s = -1, then the impulse response would be e^-t. And in this case the time to half would be approximately 0.7 seconds. That's because at t=0 the response would start at 1, at t=0.7 the response would be 0.5, and at t=1.4 the response would be at 0.25 and so on. "Time to half" is equivalent to stating "poles must be to the left of -1" ...... This was the reply given by author in below comments.

[@madhavkrishna2287]

@@suneeldatta so for quicker response times we can increase s

[@madhavkrishna2287]

sorry decrease s

[@medotedo8410]

6:22 how did you obtain the minus sign in last row -(-k/^2/3 ........)/k/3-42 , I mean the outer negative sign from where ?!!!

[@PasanJayaweeraYashoda]

thank you for this lecture!!

[@aswath2481]

you are a life saviour man! thank u!!!

[@rakeshreddy8733]

Beautiful! Just beautiful!

[@kkpncl912]

Thank you for covered most important topic in "Robust Control". could you please explain " Kharitonov's Theorem"

[@bhavanapanchal3807]

Good

[@TheChikabum]

Thank you very much, there is a video where he shows the whole process :)

[@bad8brains8]

Many thanks, dear sir.

[@TheRealLukeOlsen]

What about auxiliary equations for finding the marginally stable poles?

[@AvinashMauryaShripal]

Brian, Please can you explain why you preferred -1, not any other number ? Please answer!

[@tharindumalawaraarachchi1132]

you are awesome. thanks Brian

[@hamzahasan1141]

Salute to you, Sir!

[@1991Aski]

Hi Brian, I'm studying AC at the moment and have an exam on Wednesday for it. I don't really understand the concept of a "gain", every definition I've read about it keeps saying it's a "signal amplifer". But that still doesn't clear my understanding, please could you explain what a gain is? Much appreciated! Cheers, Keep up the great work!

[@TheJq32]

Bump. Would love to hear someone explain gain as well.

[@AbrarShaikh2741]

+Akshay Malde I don't know if you still need the definition of gain but since I read your comment I would like to make my humble attempt to clear the definition of Gain. Gain, as the name implies, is increasing the amplitude of the signal. Why we do that? Let's say I whisper to you standing 10 m away. Case 1 (open loop): If you don't give me feedback of mine being inaudible then you will never hear what I said, that is unless you have super sonic ears (hehe). Case 2 (closed loop): If you give me feedback that I am inaudible and ask me to raise my voice (In other words, to increase the amplitude of my voice signal OR in other words, to increase the positive gain of my voice signal) then I will use my vocal controller to increase the amplitude until my voice is audible to you. This was a small and very weak example for gain, but the underlying concept remains the same. Your signal strength (amplitude) must be sufficient enough to meet the current task. A 10mA signal cannot run a 100mA motor. So you need to add a K=10 multiplier gain block to meet the requirements. With the help of Routh Array you can check if the this new block of 'K' would not ruin your systen stability. I your system remains stables then "Green signal" - go ahead and add an amplifier (a device which increases gain) of Gain K=10. Hope I helped atleast 1%... :)

[@asheshshrestha]

I have understood it as: say you have a robot and you want to move its leg. You can move your leg very slowly by taking 10 mins to get to the point. But you could also get there in 1 sec. You get different times by gaining velocity. But if you move your robots leg very fast, your robot will fall down. But you don't want to wait 10 mins to get leg to specific position. So suitable gain will get the robot leg as fast as possible at specific position without falling down.

[@lionellye1337]

Solve for K and choose the smallest value, I got -24

[@pramodsrinivas1822]

I also got the same values . I guess since we are trying to build an amplifier whose purpose is to multiply the amplitude , I would choose a value closer to 126 . A value of -24 would still result in a stable system but would attenuate the input signal .

[@jeevanraajan3238]

THanks from India !!

[@valentinfinozzi9358]

I didnt really understand how the time to half requirement corresponds to that specific z line. Other than that, great video, thank you so much!

[@viharigandrakota1783]

hello brian, shouldn't that arbitrary 'Z' axis be equal to "S-1"? It was mentioned 's=z-1' in the lecture. And you also told "a pole on Z line means its -1 from the S line " . So it should be S=Z+1, shouldn't that?

[@yongjianrong8587]

Does k > -24 matter for this question when solving the conditions to be stable?

[@moshe778950101]

Everything is awesome! Thank you!

[@sarahbarendswaard1741]

When you found the gain K for which the system attains marginal stability, you named it gain margin, is this the same gain margin as is found in the Bode Plot?

[@AyushKamal-h9m]

at 5:05 How Brian wrote the combined transfer function?

[@AyushKamal-h9m]

Well I was thinking before multiplying directly the transfer function of gain block we require it to be converted to S- domain by Laplace Transform then K/s would be the transfer function. Here Brian considered K to be in S-Plane only so no need for the further transformation