This lecture discusses several examples on sketching Root Locus in undergraduate controls course (ME 352 @ SDSMT). Instructor's Webpage: webpages.sdsmt....
thanks soo much. i was having root locus depression before watching this. now im smilling to the ear. im going to nail the exam. thanks from South Africa
Nice job!!! According to my calculations, I agree with the value you found at 37:14 (which had been contested by @Thatipally Raghava). I had found the value for K equal to 7.71, however with a different approach, by using 2nd order expression: K/(s^2 + 2ζωo.s + ωo^2), Which related to the closed loop general expression (G(s)/1+G(s)) applied to the open-loop function K/s(s+5), yields the following expression for denominator: (s^2 + 5s + K) Therefore: K = ωo^2 2ζωo = 5 So, 2(0.9)ωo = 5 (1.8)ωo = 5 ωo = 2.777... √K = 2.777... K = (2.777...)^2 ============== K ≈ 7.71 ==============
good.. but some mistakes in find K value in example problem 4 it is not 7.7 something but it is root of 7.7 ,it means around 2.65 will be the K value. but it is really good ....thanks
Dear sir Ali Heydari or someone else. Can you explain the phrase "this is part of the root locus because it is to the left of an odd number of poles and zeros" more precisely? Because I don't completely understand it. Thanks in advance and thanks for the great video, it already gave me a lot of insight while rapidly watching it :)
it is simply a rule in drawing root locus, we consider the real axis part on the left of an odd number of zeros or poles as part of the root locus, if you are still not sure watch the "sketching root locus part 2" by Brian Douglas in youtube
Hi Thomas, As Abdul says it's a rule , if you have 3 poles located at -1, -2 and -3. the area between -1 and -2 is in locus becouse it has only one pole to the left that is -1. now -2 to -3 is not becouse the line you will draw has two poles to the left_ -1 and -2. -3 to inf is a part of the locus it has -3,-2,-1 to the left.
Safa Okşaş In example 5 there are 3 poles and a zero, and can be arranged as follows -10,-5,-4,0. As per the rootlocus rule, root locus exit to the left of an odd number of poles or zeros, (Remember when counting you must consider poles and zeros). if we start counting from the left by giving 1 to zero and 2 to -4 and 3 to -5 ...., the odd numbers are given to 0 and -5 hence root locus exist to the left of zero and to the left of -5 ans is marked as shown in the video.
Sir,what will happen when two open-loop poles meet just before they got separated from each other from the real axis? For example,in the very first problem,for small value of K,the poles will be at 0 & -4 whereas for larger values of K the poles gradually move close to each other & then separated. My question is that,for a particular value of K,the two poles meet just before the separation.What will happened in that case?Is this the case of repetitive poles for a single value of K?
In exp-3 , asymptotes angle , suppose we take i=-1 , as we took in first two examples. so we get pi and -pi ..!! but in video only pi ?? could you please tell me how it is ?
In the first example what is the best value for K? Isn't that what this is all about? Yet this point is not made. On top of that the real part of the closed loop poles determine how the error will decay. Notice that no matter what value of K you choose the error will decay at the rate of exp(-Re(poles)). In other words the open loop transfer function determines the best response but the video doesn't say that. My point is that if you are the control guy you are screwed by the plant designed by the mechanical or chemical engineers. If the break away point is a -2 the best decay rate you can achieve is exp(-2*t). There is NOTHING the control guy can do without using some tricks that go beyond what you learn from root locus. In the first case a PD or PID controller is required, NOT a simple K or proportional controller., if you want to move ALL the closed loop poles to the left of the break away point with the result of a faster response. My point is that the root locus is useless except to know where the breakaway point is because to get the best response you need to use more than a simple K gain or proportional gain. This is another example of instructors teaching something that really isn't relevant and wasting students time and money.
You explain really well, this is the best tutorial of Root Locus I found. Thank you.
thanks buddy, 10 years and you still the best one who explains
Bless you Sir, this is the best tutorial about Root locus
thanks soo much. i was having root locus depression before watching this. now im smilling to the ear. im going to nail the exam.
thanks from South Africa
Watching this whole video because of your comment :)
last year it was your turn, this year it's my turn to nail it too. lol, I'm also from S.A by the way
All my friends must know this! Need Nyquist and Bode classes! Thank you a lot!
Best examples Best explanation in TH-cam.
You are wonderful. Thank you very much sir!
This video has 44 dislikes because 44 other root locus explainers got jealous of how well you did. Nice job. Subscribed
You explained it really well. I was looking for the exact tutorial I found best on youtube. Thank you very much for such tutorials!
This is actually a pretty darn good explanation. Appreciate it Ali.
You are a scholar and a gentleman. Many thanks.
Ma sha Allah.....really great job...Barak Allahu fik
Special thanks from Kuwait, I have better understanding for Root locus now thanks to you sir.
Great lecture Professor! Really appreciate the examples you give!
Ammazing, informative! the language is clear, video is clear, examples are clear. excellent work really. watching in 2019
This video easily got me an extra letter grade on my controls theory final exam. Thank you!
Nice job!!!
According to my calculations, I agree with the value you found at 37:14 (which had been contested by @Thatipally Raghava). I had found the value for K equal to 7.71, however with a different approach, by using 2nd order expression:
K/(s^2 + 2ζωo.s + ωo^2),
Which related to the closed loop general expression (G(s)/1+G(s)) applied to the open-loop function K/s(s+5), yields the following expression for denominator:
(s^2 + 5s + K)
Therefore:
K = ωo^2
2ζωo = 5
So,
2(0.9)ωo = 5
(1.8)ωo = 5
ωo = 2.777...
√K = 2.777...
K = (2.777...)^2
==============
K ≈ 7.71
==============
Respected SIR your lecture is very helpful and i learned alot from it. thank you very much and may ALLAH pak bless you. live long. From PAKISTAN
ัyou're awesome man i've an exam today. you're my savior
One of the best tutorial I have ever seen... :) Thanks a ton
you are so good in what you are doing Sir. Thank you
I've looked for so many videos on root locus, this is the best!! You can really explain. Do you have a video on root locus design?
this is some real stuff on root locus. #rootlocus_demystified!
good.. but some mistakes in find K value in example problem 4 it is not 7.7 something but it is root of 7.7 ,it means around 2.65 will be the K value. but it is really good ....thanks
Good man Thatipally yup ow dat
Thank you so much for the tutorials, this is the best video for this topic!
Well Articulated Ali - Thanks alot man! U just made me ready for the Exams!!
very good compilation of problem. greatly helpful
Greetings from Brazil!, God bless you extremely helpful lesson!
Very helpful !!! big thanks Mr. Ali :D
best explanation on root locus
Dear sir Ali Heydari or someone else. Can you explain the phrase "this is part of the root locus because it is to the left of an odd number of poles and zeros" more precisely? Because I don't completely understand it. Thanks in advance and thanks for the great video, it already gave me a lot of insight while rapidly watching it :)
it is simply a rule in drawing root locus, we consider the real axis part on the left of an odd number of zeros or poles as part of the root locus, if you are still not sure watch the "sketching root locus part 2" by Brian Douglas in youtube
Hi Thomas, As Abdul says it's a rule , if you have 3 poles located at -1, -2 and -3. the area between -1 and -2 is in locus becouse it has only one pole to the left that is -1. now -2 to -3 is not becouse the line you will draw has two poles to the left_ -1 and -2. -3 to inf is a part of the locus it has -3,-2,-1 to the left.
Abdul Hardy but in example 5 , pole -5 is not an odd number but he put the root locus left of it
Safa Okşaş In example 5 there are 3 poles and a zero, and can be arranged as follows -10,-5,-4,0. As per the rootlocus rule, root locus exit to the left of an odd number of poles or zeros, (Remember when counting you must consider poles and zeros). if we start counting from the left by giving 1 to zero and 2 to -4 and 3 to -5 ...., the odd numbers are given to 0 and -5 hence root locus exist to the left of zero and to the left of -5 ans is marked as shown in the video.
Abdul Hardy Ah I counted them separately.Now I get it.Thanks man :)
I understood very well. Thank you very much.
thankyou !! one of the best lectures i found
Best video for the topic...Thanks :D
Awesome video Ali. Thank You so much.
Excellent video and explanation! Thank you so much!
الشكر الجزيل أوفره ....
Watching in 2022. Amazing !!!!!!!
you are damn good. even as a black panther and root locus
excellent teaching sirr
it is very help full
Superb Explanations!
Thanks so much! Fantastic presentation!!
Excellent explanation, thank you
thank you so much..best explanation ever
thnku sir u r wnderful teacher...u explained so gud...
ey val dash ali very helpful thanx
AWESOME Lecture!
Awesomee sir .....thank you !!
Great video, thanks
Thanks so much for your assistance!
sir you make my exam day
thank you sir ,very good explaination,
really helpful..god bless you
Sir,what will happen when two open-loop poles meet just before they got separated from each other from the real axis?
For example,in the very first problem,for small value of K,the poles will be at 0 & -4 whereas for larger values of K the poles gradually move close to each other & then separated.
My question is that,for a particular value of K,the two poles meet just before the separation.What will happened in that case?Is this the case of repetitive poles for a single value of K?
Thank you, sir!
great video i was not understanding well you clarify it thanks
Thanks Doc , Fardad from Iran Sahand university
Awesome... Thank you so much !!
This was very helpful, thank you.
Perfect . Thanks a lot !
This is amazing. Thank you so much.
very useful! thank you so much
extremely helpful!
thank you Very helpful
well explained. thanks
just what the doctor ordered!
Thanks a lot for helping
I failed the course because the root locus part is not examined...
Thank you! You're the best =)
Thanks from Pakistan
Thanks, bro! I appreciate it. Allah bless you.
its closed loop block diyagram we have not to make it open loop block ( without feedback ) befor ?
very informative, thanks! :)
Thank you, you helped me a lot :)
awesome!!!
how nice!
Thank you so much Sir :)
thank you sir
ali ur awesome....thumbs up !!
Very good
Awesome thanks
Dear Sir, what are the rules of root locus plotting? Can you please post the rules to the description part of the video? Thank you.
thanks!
thank you brother
THANK YOU SIR
thank u
you are perfect
GREAT
that was helpful thank you
Thanks
thank's sir
One 'thumb up' isn't enough.
for ex2, are you trying to say -10 and 0 are odd numbers?
The pole at 0 is the first(odd) pole and the pole -10 is the third( also odd) pole. he is referring to the position, not to the value of the poles.
thanks alot
Lau emang paling locus bos!
In exp-3 , asymptotes angle , suppose we take i=-1 , as we took in first two examples. so we get pi and -pi ..!! but in video only pi ?? could you please tell me how it is ?
Because in the end -pi and pi are the same. If you rotate 180 degrees to the left or to the right it will not matter.
Jagdpz5 thank you ☺☺ Jagdpz5
thanks a lot!!!!!!!
In the first example what is the best value for K? Isn't that what this is all about? Yet this point is not made. On top of that the real part of the closed loop poles determine how the error will decay. Notice that no matter what value of K you choose the error will decay at the rate of exp(-Re(poles)). In other words the open loop transfer function determines the best response but the video doesn't say that. My point is that if you are the control guy you are screwed by the plant designed by the mechanical or chemical engineers.
If the break away point is a -2 the best decay rate you can achieve is exp(-2*t). There is NOTHING the control guy can do without using some tricks that go beyond what you learn from root locus. In the first case a PD or PID controller is required, NOT a simple K or proportional controller., if you want to move ALL the closed loop poles to the left of the break away point with the result of a faster response.
My point is that the root locus is useless except to know where the breakaway point is because to get the best response you need to use more than a simple K gain or proportional gain.
This is another example of instructors teaching something that really isn't relevant and wasting students time and money.
thanks too much
Please tell me which software r u using for writing on the screen .. please !!
Cool