Routh Stability criterion special case 2

[@Okfolifeful]

Correct me if i'm wrong. You made a mistake around 5:50 while calculating for the first element in s1. You divided by 8 when you should be dividing by 6.

[@techgurukula]

+Stephan Khamis You are correct! Thank you for taking time and letting me know. will be correcting in a day.

[@ashaygatade4842]

@@techgurukula when is that day coming ?

[@sowmyag2045]

Thanks for the detailed explanation...In this case, if the entire row is zero..can we conclude that the system is either marginally stable or unstable based on last given examples?

[@ahmedhisham484]

first term in s^4 should be (-2). Thanks for the explanation by the way.

[@malharjajoo7393]

Just to add on further to his explanation - 1. The routh array may have a row of zeros, 2. When you find the auxilliary polynomial , it is either an even or odd polynomial ( has only even or odd powers of "s" ). Any even or odd polynomial can have 3 cases or roots as mentioned towards the end of the video. Two of them are unstable , and only one is marginally stable. Hence we need to check the auxilliary polynomial for its zeros ( and hence decide which case it is )

[@prathamgupta338]

The auxiliary polynomial always has only even powers of 's'. You made a mistake there.

[@ηοτιη]

@@prathamgupta338 No.

[@sumit_chouhan97]

how did you form the auxiliary polynomial?

[@abuhanif3991]

Thanks a lot of you ....

[@ShivamRaj-pt6sb]

thank you so much :)

[@cutietifftiff]

thanxxxx a mill :-)