Problem no -4 ( Design a lead Compensator using Bode Plot)

[@Navyaah-mm4he]

While writing the open loop transfer function You have included the attenuation factor but in problem 3 you skipped it why?

[@kowsiteaches]

In problem no 3, the phase margin is 13°(positive) but in problem no 4 ,the phase margin is -44° ( negative).This is the reason behind this

[@aniketrenjarla1353]

why the attenuation factor has not taken in series on step 6

[@kowsiteaches]

Here the system is unstable ( refer to step 3.Phase margin is negative). Therefore we are not including the attenuation factor

[@aniketrenjarla1353]

Thank you Ma'am I cleared CSD subject because of your teaching

[@kowsiteaches]

Most welcome ☺️

[@deeslifeuncut6928]

Ma'am incase in LAG compensator the PM is negative, similarly we should not add 1/ Beta in the OLTF block diagram??@@kowsiteaches

[@kowsiteaches]

Yes,if P.M is negative,we should not include the attenuation factor

[@kaboggozakenny5614]

mam, on 19:00, isn't it supposed to be -60log (30÷15) - 23?

[@kowsiteaches]

Yes ,you are correct.it should be 30/15.extremely sorry.Thanks for the info

[@anvi_05]

Transfer function of lead compensator 15:19 formula is changing problem to problem

[@kowsiteaches]

Check step no 4.Here the angle is greater than 60 ⁰.So we are using two compensators in series.when you combine 2 blocks in series,you have to multiply as per block diagram reduction.this is what done here

[@amruthavarshanxo7080]

isnt it 59 at 13:16

[@kowsiteaches]

No, it is 69° ( 20-(-44)+5)= 20+44+5= 69°

[@ManvithaNukala]

First and secod magnitude plot is coinciding mam am i wrong or should proceed ?

[@kowsiteaches]

I think you are not correct.kindly check each and every step

[@YashikaGarg-cp8fo]

Mam why we take t.f of lead compensator (s+1/t)/s+1/alpha t But in next question take square of t.f of lead compensator

[@kowsiteaches]

Can you explain your doubt in detail.I can't get

[@YashikaGarg-cp8fo]

Mam in the lead compensator of bode plot we take problem no 3 of the t.f compensator And in problem no 4 we will square of t.f of compensator

[@aryarohanrachamalla2730]

Mam why are you specifically taking the lead angle is > 60* ? i didn't get it

[@kowsiteaches]

It is asking per the procedure

[@ManvithaNukala]

Mam What if the numerator contains number instead of k? Should we consider the given number equal to k ? Like g(s) = 4/s(s+2) Can we take k = 4 directly without the formula ?

[@kowsiteaches]

Yes .you can

[@rohithadapad7]

Maam from where do you start your omega on x axis, im getting wrong ans because of graph scale ig, can you help me maam

[@kowsiteaches]

I started omega at 0.01 on x axis

[@JA2022_]

pdf ah covert panni description la podunga

[@kowsiteaches]

Ok