The Concept of Dominant Pole

[@bbugraisik]

Guys, as I understood there is only one requirement to make a eliminate non dominant pole. And this requirement is first one. Then we shape the approximated function in order to obtain the same dc gain. So, after eliminating non dominant pole, we assign 0.5 value to numerator, so that we can get dc gain same as before. I hope I could explain what I mean.

[@lokendrasinghlodhi718]

I also understood the same thing bro! Thank You for clearification.😊

[@makisxatzimixas2372]

yes, it looks like he skipped a step there.

[@makisxatzimixas2372]

the way he eliminated the pole is that he divided s+10 by ten, and then he multiplied the whole parenthesis with 10. If you do this the numerator becomes 0.5

[@PunmasterSTP]

Dominant pole? More like "Your videos are on a roll!" Thanks again for making and sharing such high-quality educational content.

[@qozia1370]

pathetic reply

[@PunmasterSTP]

@@qozia1370 Why do you think that?

[@vinceangeloupancho8728]

Happy Teacher's Day!

[@nesoacademy]

Thank you! 😃

[@bharathgopalakrishnan3739]

how did the last step figure out. You just calculated the lim tending to 0 for the transfer function and got the answer as 0.5. how was that equal to the same equation upon eliminating the insignificant pole ?

[@aritrabiswas1523]

from what i understood, he substituted the approximated system as B/(s+1)... now if we calculate B it comes out to be 0.5 hence the gain is unchanged

[@deepakmaurya-sk7fh]

Awesome concept in very good way👌👌

[@sankadharmapala4064]

really helpfull channel for students..... thank you....

[@sumiasameer5238]

After eliminating how come T(s)=0.5/(s+1). It should be 5/(s+1),right?

[@lokendrasinghlodhi718]

no !

[@aura3167]

How did you approximated the transfer function just give a reply how did the 0.5 came on top in approximated transfer function?

[@YashTarwani]

i have same doubt

[@s.m.senses7199]

thanks Sir🙏❤

[@sammutyala2251]

Plz create video for block diagram reduction techniques

[@zahidansari9745]

Hello sir can you tell me how we got DC gain 0.5 after removal of (S+10).

[@uttarkarpujitha102]

actually by doing partial fractions we get 5/9((1/s+1)-(1/s+10)) so by removing 1/s+10 we get (5/9)(1/s+1) and 5/9 = 0.555

[@bhyllw]

Transfer function 5/[(s + 10)(s + 1)] turns to 5/[(0 + 10)(0 + 1)] because, in order to get the DC gain, or k, you have to evaluate it when s tends to 0, so: 5/10 = 0.5

[@naveenchandra960]

@@uttarkarpujitha102 why we are doing partial fraction, we are talking about removal of s+10 insignificant term

[@vicentehernandez3936]

just get rid of 's' not 's+5' and then symplify the expression 5/10(s^2+s+1) (5/10)/((10/10)*(s^2+s+1)) 0.5/(s^2+s+1)

[@ayishaasghar5186]

After eliminating the insignificant pole: 5/(0=(insignificant pole)+10)*(s+1))=(5/10)*(1/(s(dominant pole)+1)

[@harsteinmani]

Nice sir👌👌👌

[@zaksahu2652]

Sir...please post all videos of the topic so hum woh topic ek hi din me kahtam krke uspe questions kr le🙏🏻

[@avirashridhere5708]

How does gain remain unchange 0.5?

[@PunmasterSTP]

Keeping the gain the same is one of the two criteria necessary for making a valid approximation (the ratio of the insignificant to the significant pole is the other). So what you need to do is figure out the DC gain of the original system, and then make sure that is the same gain as the first-order approximation. Then since the first-order approximation only has one linear factor in the denominator, it will just be K/(s + p), where K is the DC gain and p is the (negative of the) dominant pole.

[@bretthaddin8071]

very very thank you sir , increase more and more lectures on Control System

[@muhammadmehran1202]

Great resource of knowledge....blessings

[@tylane6192]

Instablaster.

[@DSCS_AkshataChavan]

Your vedios are so helpful for me

[@dastran2731]

Closer to Orgin or jw axis, which is it? Are both effectively same?

[@just4coments355]

Can a dominant pole be at the origin? If i want to dimension a controler should i use the pole at the origin or the one next to it?

[@zakyvids6566]

Assalamualaykum Sir I really like your way of teaching it’s very clear and helpful. I would appreciate if you can make a video in python 3 course You are an awesome teacher 👍❤️

[@DSCS_AkshataChavan]

Plzzzz upload vedios on Algorithm

[@hasibulhasan7716]

how did the 0.5 came on top in approximated transfer function?

[@ignatiusdio7264]

In order to have the same value of K, the numerator should be 0.5

[@appa1018]

great work! you saved my day ahahhaah

[@gowriv944]

Sir it's nice.upload more and more vedios about control system.thank you sir.

[@samuel-wb4jp]

Hello. is it apply to 2nd order system too??

[@बाहुबली2.0]

Hlo..sir.., I have a doubt about on dominant pole topic.., as we know in D.p..if the ratio of insignificant pole to significant pole is equal or greater than 4..then we neglected insignificant pole ..with the same value of d.c gain of the system.. right ..my question is .. if insignificant play a important role in stability of system because of less time constant and high bandwidth..and speed of the response is also incresed by insignificant pole.. then why we neglected insignificant pole...or why significant pole is a dominant pole instead of insignificant pole..?

[@karasanimonika5498]

Thank you sir! Helped me a lot!

[@osamahebala6923]

thanks for this nice explanation

[@palli6458]

In second condition What if my TF= 1/(s+100)(s+4)

[@PunmasterSTP]

Based on your TF, what poles would you have, and what would their ratio be?

[@022_firdousnabi5]

Thank u soooo much sir

[@hanaa.r_]

1/(s*(s+1)*(s+5)*(s+6)) i get 1/30(s+1) it's that true? but it's not 2nd order system, i can't get the damping ratio

[@humusekz]

how it work with zeros?

[@sayedirfanzain7702]

Can we study it for Semester exam ?

[@nesoacademy]

Yes :)

[@Frost_Gaming_23]

How the D.C gain =0.5,Please explain!!!!!

[@LaplacianFourier]

You make s go to zero so you end up with 5/[(0+10)(0+1)] which is 5/10 = 0.5

[@raghunandanmutalikdesai299]

@@LaplacianFourier but after eliminating the insignificant pole, how come the DC gain is still 0.5? With limit s-->0 & removal of (S+10), will the DC gain not be 5?

[@bpavankumar9597]

@@raghunandanmutalikdesai299 I also have the same doubt.

[@zahidansari9745]

@@bpavankumar9597 yes...bro i also have same doubt....if anybody know the solution of this let us know

[@golinagasandesh4464]

Assume s=jw and substituting in place of s in the given transfer function, except for dominant poles. Now finding the limit w->0 in the magnitude of the substituted transfer function, we get the DC gain :) Watch this video for more clarity: kzbin.info/www/bejne/r2KlqX-Cia19has

[@vicentehernandez3936]

Just get rid of 's' not 's+5' and then symplify the expression 5/10(s^2+s+1) (5/10)/((10/10)*(s^2+s+1)) 0.5/(s^2+s+1)

[@glob.entertainment]

What if the pole is zero? Does zero automatically become a dominant pole