Your teaching method is much better than all other channels ever exist in UA-cam who make Enginneering Videos.Please continue uploading relevant lecture.
thanks you actually covered an example with complex poles that nobody else seems to want to do, this is the only sort of problem that matches my homework lol
I have seen so many vedios on this topic but I feel all difficult but finally I gone through your explanation is very very easy and you made the questions as very easy thank you sir
very very useful video, helped a lot, u r so greatt, so much efforts u have taken to bring such wonderful output, may god bless you , u will reach great heights
while calculating centroid, shouldn't we only consider the real parts of poles and zeros ? in this particular problem the imaginary parts are getting cancelled out but in other cases it will be a problem...won't it ?
doubt: to check root locus point are we supposed to consider only real poles or for example there is a conjugate in 2+4j 2 is on real line so 2 also shall i consider?
I think that you are asking about finding the root locus on the real line. If so the complex roots don't matter because they always appear in conjugate pairs anyways. So if you add them in you will always get an even number (2) added to what ever poles and zeros are to the right of the point you are evaluating. Since adding 2 to an odd number leaves it odd and adding 2 to an even number leaves it even you can ignore the complex roots for this purpose.
You should consider only the on real axis.(i.e. only real numbers) not the complex poles that lies on the s- plane. 2+4j is a complete pole. You should not consider only 2.
clear all; clc; s = tf('s'); disp('The given transfer function is,'); Gs = (s+9)/(s*(s^2+4*s+11)) rlocus(Gs,'k'); axis([-6 2 -6 6]); sgrid; I think it is like this... Just try this in Matlab
Zeros are two know because u considered S= 0 in poles Z1=0 Z2=-9 Replay bro I want clarity ????????????????????????? And zero is even num But u considering it's an odd number? Replay 😊
You explain everything without confusing the student. It's very rare and very valuble skill.
Your teaching method is much better than all other channels ever exist in UA-cam who make Enginneering Videos.Please continue uploading relevant lecture.
ua-cam.com/video/r3TlKbs7TKQ/v-deo.html
Yes sirrr
@@srivallipadhamavathi😂😂🎉
thanks you actually covered an example with complex poles that nobody else seems to want to do, this is the only sort of problem that matches my homework lol
I have seen so many vedios on this topic but I feel all difficult but finally I gone through your explanation is very very easy and you made the questions as very easy thank you sir
Welcome 😊
Sir you are a Smart Engineer indeed. Thank you for saving my life, all the way from South Africa ✌
U are always a magical teacher...thanks a lot
Thank you so much
Seriously he is
Ara bata thanku ki koi jarurat nahi
Even explained those terms which were not asked in this question but we may need in some other questions👏👏, Thank you!!
This was extremely well explained! Love from Canada.
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gone through lot of videos but this one is the right one. nice explanation
Do share to your class group
@@smartEngineerShinton Ok bro!
small doubt if zero is there in b/w poles then how to calculate break away or in point?
@@abhiramambati3346 break away point exists btw two adjacent poles ,break in point exists btw two adjacent zeros ,btw pole and zero they don't exist
very very useful video, helped a lot, u r so greatt, so much efforts u have taken to bring such wonderful output, may god bless you , u will reach great heights
Made my day 😇
Thanks Biradar
✌🤟❤
U make it simpler
Now I can complete my assignment. 😜
Really thanks
ua-cam.com/video/r3TlKbs7TKQ/v-deo.html
at 19:38 ans= -16
17 is crt
This is the best video I have ever seen on this. Also much better than anything my professor explains, she confuses me lol
Thankyou sir, for explaining every detail, without confusing
I'm blessed.......as i found this video.
Thanku you so muchhhhhh sir.
sir number of asymtotes is not the number of poles .It is poles-zeros(p-z)
You r write
Sir ur channel didnt show up for some days I was worried what had happened. Thank God finally it came back.
Got hacked😭
@@smartEngineerShintonthen how did you get back bro because i too last my account how did you recover
The best video I saw on this topic 👏
thanks a lot brother
helped me a lot before the exam
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Прекрасный урок! Спасибо огромное!
Such a detailed explanation 👏👏👏🏻👏🏻
Explanation is tooo Good👏👏👏
Thank you so much!! I understand so much better now!
Control System Needs So Much Attention. Active Mind Whole Book.
Like on of the best explainations i have watched ❤️
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Sir, your explanation was clear and satisfyng thank you
sir why we take +-90 and +-270....It should we 90 and 270 ?
Sir please reply
ur... our garden angle 🔥
Do share to your class group
Why u have take plus or minus 180 in asymptodes , as in first problem it's not there
Sir angle of departure value is -16⁰
The best teacher 👍❤️ i😭 iam madly searching for the best video then I found u tqq uu best video
Welcome 🤗 and do share to your class group
Why +- only in this question...in asymptote
Explained well thank you sir..
Is the angle of arrival/departure measured from the centroid?
God level explaination
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Super Bhayya..Well Explained
Deserves more subscribers
Ur are the 🔥 fire sir no words to tell about ur teaching sir
how you got angle of departure 17 degree, it is 58 degree according to that formula you have calculated
Osm lecture sir continue uploading...
Brilliant man
It helped a lot.
GREAT VIDEO, keep doing this
while calculating centroid, shouldn't we only consider the real parts of poles and zeros ? in this particular problem the imaginary parts are getting cancelled out but in other cases it will be a problem...won't it ?
Extraordinary bhaiya
Excellent lecture sir. Is this from nagoor kani text book sir
sir why you consider 90* and 270* as same angle with same line
Yaa I'm also confuse
you are good teacher thank you......
Welcome 🤗
@@smartEngineerShinton never
No. Of asymptotes =P-Z
=>3-1= 2
q=0, 1
Thank you sir🙏
Thank you 🤗
Ma Man thank u sm
Your lecture is exllent sir👏🙏 but the marker pen your are using is not good please do explain with good marker🙂sir
Awesome video
Thank you for really helpful this video
Aiml medha job's gurinchi oka video chy Anna skills ame narchukovalo chappu anna
Super explanation 👌
ua-cam.com/video/r3TlKbs7TKQ/v-deo.html
Thank you 💖
ua-cam.com/video/r3TlKbs7TKQ/v-deo.html
Thank you so much sir 🙏
Thank you so much sir❤
How to find angle of arrival 😊
sir if we got the zero at rightside to the pole then what is the root locus on real axis
Plz do a qustion with zero exist and non comblex
great explain bro
Thank you so much sir 💥🥰
I think angle of departure is -57.79 degree
ua-cam.com/video/r3TlKbs7TKQ/v-deo.html
Yes you are right 👍
Brother this video very good, but slove difficult problem to solve
Angle of departure is 58 or 59 degree ,your calculated ans is wrong
sir what about lack of zeros? how does it affect my test point?
When is break away point?
Only if two poles are in adjacent
nicely explained ☺️
what if a pole present on right side of real axis can we find gain stability
doubt: to check root locus point are we supposed to consider only real poles or for example there is a conjugate in 2+4j 2 is on real line so 2 also shall i consider?
I think that you are asking about finding the root locus on the real line. If so the complex roots don't matter because they always appear in conjugate pairs anyways. So if you add them in you will always get an even number (2) added to what ever poles and zeros are to the right of the point you are evaluating. Since adding 2 to an odd number leaves it odd and adding 2 to an even number leaves it even you can ignore the complex roots for this purpose.
You should consider only the on real axis.(i.e. only real numbers) not the complex poles that lies on the s- plane. 2+4j is a complete pole. You should not consider only 2.
@@saineethureddy6195 thanks but it's been 2 months since I've completed my xam
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@@mallamgurudeep1341 😂
sir if we got k value is negative then what we need to do sir?
When zero is ended with in real axis
Then there is no need to draw root locus from o to infinity?
ua-cam.com/video/r3TlKbs7TKQ/v-deo.html
In finding centroid only real part of the poles and zeroes is to be taken
So angle of departure and break away point pass through same point as per cross imaginary axis sir?
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Amazing video
ua-cam.com/video/r3TlKbs7TKQ/v-deo.html
Sir, what if the k value is negative is that okay?
best video 😇
While finding q why you are doing p-z-1
ua-cam.com/video/r3TlKbs7TKQ/v-deo.html
how to know poles are complex or not ?
Thank you so much 😊
Sir dsp videos (teaching) continue karoo
Sir how can we do code of this in the matlab ???
clear all;
clc;
s = tf('s');
disp('The given transfer function is,');
Gs = (s+9)/(s*(s^2+4*s+11))
rlocus(Gs,'k');
axis([-6 2 -6 6]);
sgrid;
I think it is like this... Just try this in Matlab
Sir thee thanne 🎉
Athe athe... Mattanal alle mwuthe 😌
@@akaparz4109 mattannal endha machaa
Sir how imaginary part solve in caslo fx-991EX plz tell them...
Love you sir
Sir i have options with root locus part and time response steady state analysis which shall i choose
Don't we write asymptotes as P-Z???
Thats the number of asymptots
Zeros are two know because u considered S= 0 in poles
Z1=0
Z2=-9
Replay bro I want clarity ?????????????????????????
And zero is even num
But u considering it's an odd number?
Replay 😊
ua-cam.com/video/r3TlKbs7TKQ/v-deo.html
.
Bro why you consider j value is -1
he considered j^2 value as -1 and j^3 value as -j
Thanks for clean explanation.
thankyou sir
VERRRRRY GOOOD
14:00 need link
Superrrrrrrrrrrrrrrrrrrrrr explanation
Thank you so much sir
Centroid will be -2.5 not 2.5
Thank you
Thank u souch sir
Thanks 🙏
thank you sir