Final Value Theorem (Solved Problem)

[@mikeybolts4278]

Hands down best channel for engineering students ❤

[@HimmatSingh-ju8jk]

1. Function have complex poles with real part so I think the output should have be defined for final value theorem. i.e is 4 option (d) 2 (d)

[@raghunandanmutalikdesai299]

Ans is (d) for both homework problems

[@darshanrs399]

correct !

[@inderjitsaini9303]

How though? Like I got -1 +/- j4

[@kulasekarans5428]

@@inderjitsaini9303 I think as it its mentioned that an unit step input is applied, multiply the given G(s) with 1/s and as per the Final value theorem multiply with s in the numerator. s and 1/s will get cancelled and now put s=0 as per the theorem. You will get 20/5 = 4.

[@kulasekarans5428]

@@inderjitsaini9303 for factorization you should have obtained -1 +/- 2j I guess.

[@PARi0070]

​@@kulasekarans5428 Final value theorem applicable when system given in OLTF here in Q it is in CLTF

[@iamblue8272]

your chanel is a gold mine

[@makisxatzimixas2372]

My friends I would like to thank you for your teaching style, you are the only teacher in my life who has shown the WRONG answers, and it is important to avoid them. Respect!

[@PowerShellWizard]

For the second problem you could "shortcut" by looking at the characteristic equation for R-H necessary criteria. No vanishing powers of s, check. Alternating sign for the final term (-2) therefore system is NOT stable and hence output is expected to blow up, and as such be unbounded. :)

[@makelearningeasy111]

Answer1: 4 Answer2: unbounded

[@evanbyrne7020]

This channel has saved me from having to redo a very poorly organised and taught Control Systems unit. Thank you so much!

[@wadoudazer6906]

&² thank you for this beautiful course , i found the first question 4 , because lim s*G*(1/s) (when s=>0) = 4 and there are two poles in the left hand , the second question we have a pole in origin and another one in the right , so its unbounded due to the right one , because one pole in the origin will have no effect , if we have more poles in origin this is a problem , i hope i am right and keep going !

[@darshanrs399]

please do upload answers for homework questions so that we can verify with ours!

[@komalchaudhari1753]

Got us in the first half ngl 😂

[@ujjwalartsiit]

Explained very well ! Thank you 😊 sir!

[@globalglimpses786]

Ans D) for both question

[@RajithaWijesooriya]

Answer (d) for both

[@mnada72]

This is an excellent course I hope you complete. Thank you

[@PARi0070]

First Q CLTF given Break into OLTF =20/S²+2S+5-20 =20/S²+2S-15

[@pramudithathilakarathne2301]

I did so and worked for the OLTF which you have indicated above. When that happens the steady state output with final value thearom can not be applied since one out of two poles becomes positive in 20/(S+5)(S-3). am i not correct?

[@Abvht1656]

Homework Q1- 4 Q2- unbounded

[@eruditecoder5215]

Q1: 4 Q2: unbounded

[@عايشبصمت-ح1ي]

hw junction; q1: b) 4. q2: unbonded

[@s.m.senses7199]

thanks Sir🙏❤

[@aaqibrashidle6655]

Uh r really great sir Ur lectures helps us a lot

[@vineetsingh6941]

In 1st Q it is given as CLTF so why aren't we considering the original factor as told earlier for calculating transfer fuction?

[@ahmedalzaabi5848]

Answer d for both

[@liamkneeson8866]

Great video thank you

[@samiranchanda2459]

Good

[@zahidhasankhoka1649]

1.D 2.D

[@PunmasterSTP]

Right-handed pole? More like "The virtues of this channel I extol!" Thanks for making so many high-quality videos, and sharing them for free!

[@hadisyuaib8613]

for problem 1 I got 4, but can anyone tell me if the CLTF had a factor in the calculation?

[@mm_31]

No factor would be there

[@satyamyadavvlogs3185]

and d for both

[@b4ms142]

First Homework answer is :4 Second one is : -1 Am I right?

[@piyushsoni4100]

In 2nd example, the system T.F is having one pole in R.H.P, it is unbounded.

[@b4ms142]

@@piyushsoni4100 should we make Lim s.Y(s) s->0

[@piyushsoni4100]

@@b4ms142 we can't apply final value theorem for unbounded functions, that's what we discussed in the previous lecture.

[@b4ms142]

@@piyushsoni4100 Thank you

[@Earthwrmio]

@@b4ms142 how did you solved the first one can you please please explain it please!

[@nandyalanaveenreddy7598]

Sir for first home work problem it can be unbounded know sir as we get poles on jw axis

[@debdutmajumder82]

how to calculate poles in the imaginary axis??

[@barish.pakoray]

Option a is correct... And fir 2nd it is - 1

[@radhas.1171]

1.d

[@shriram5494]

Damn I fell for the trick both the times. First I thought it was B and then C. Goddamnit

[@bugracetin3422]

Why we said input 1/s I couldn’t understand.

[@afaqahmad4518]

since we are given our system in frequency domain ...and input is unit step u(t) ..so we need to take input also in frequency domain before multiplying which is laplace of u(t)=1/s

[@anandharsh1773]

Ans(d) for both problems

[@AmanGupta-wb8mj]

Mine was 500th like... 😊

[@devilking1673]

4

[@triposat]

Please upload the notes of DATA STRUCTURE as soon as possible

[@dhruvkanthaliya9052]

👌👌

[@pawankpx]

Both are unbound output so final value is not definite .

[@devilking1673]

D

[@karunasonawane6682]

1st = 4 2ns = -1

[@javeriashaikh5274]

how ? can you explain to me, please

[@anti-tankartur677]

2nd ans is undefined. it has a right sided pole

[@EhsanKeshavarz-wi8pb]

the second system is unbounded since it has a pole in RHP

[@108ahah]

👍👍

[@aaqibrashidle6655]

Bouth bouth bouth bouth bouth bouth bouth zabardas

[@3bdo3id]

DD

[@gaudok3281]

mf had us in first half and second half

[@mohamedelaminenehar333]

😊👏😍

[@azizouldkherroubi4178]

a7laf w nagaz?

[@justpaulo]

HW: (d) (d) (d)(d) double D 😁

[@amangoheer8824]

terrible video. atleast explain properly