R-H Criteria (Solved Problem 2)

[@LIVE_LIVELY]

H.W. problem also by shorcut method. Ans = 3/4 = 0.75 12:14

[@amnaiqbal8016]

This video really helped me in my finals.Similar question was in exam. Thanks alot. Your teaching style is tremendous.

[@ravindrasahu8738]

your contents are far better than other so-called gate preparation classes

[@PunmasterSTP]

I'm just curious; what gate preparation classes have you used?

[@karthikkj3421]

So true, this is direct and to the point honestly enough to do the gate questions

[@zahidkatarya]

Solution to the h/w problem is 0.75

[@UEI_HimadriShekharChakraborty]

H/W problem ans : K = 3/4

[@radhakrishna....1271]

How? Can you explain the steps

[@ano777g]

God bless you ❤✨✨ Really thankful to you. You explain far better than my tutor. And the way you explain background information properly makes things easy for even a beginner or a person who has forgotten basic concepts or a person who don't know how to use basic concepts. ❤ Really thank you sir.

[@mhz5749]

i think there is a shortcut method, when you get the characteristic equation, and you already know that +-2j is two of the solutions, then (s^2+4) must be a factor of the characteristic equation, you can sub in +-2j and get the coefficients of the equation, but what i did is you can deduce that the characteristic equation = (s^2+4)(s+a) by long division or otherwise, then you get K+1=4a and K+2=4 as the simultaneous equations comes from the fact that long division equals the characteristic equation is a identity.

[@PunmasterSTP]

Solved problem 2? More like "This stuff I never knew"...before I started watching Neso Academy! Thanks again for making and sharing so many amazing videos.

[@dakshsingh5891]

cringe comment.

[@PunmasterSTP]

@@dakshsingh5891 Yes, I’m aware.

[@verissakissiedu1302]

I had K=-3/4. When I use s^2 but when you solve for s^1, K=3/4.please what is the right answer? Please help me soon

[@nostalgic7752]

i didnt understand your doubt, you wont get k in negative at all, when you solve Auxillary eqn, you get k = 3/4, when you put the s^1 element of routh array you still get k = 3/4 and when you use shortcut method, you still get k=3/4

[@SatyamMishra-zq4mn]

sir upload more videos of control system as soon as possible.

[@kyawgyikyawgyi2603]

qooj(o()pkiuyyy%🙂🙂🙂🙂🙂🙂

[@ankool9794]

Hey Do you know if this Control system playlist is worth watching to prep for Gate ECE?

[@tarupratikshapandurang1536]

k= 0.75 Is this correct?

[@sanskarkumar6484]

thanks Sir🙏❤

[@ee19gauravshrivas18]

K = 0.75

[@sachinkotagond]

If the solved example is followed in this video, then the solution to Home Work problem can be obtained in 2 steps !!! Given, CE = 1+G(s) = s^3+ Ks^2 +4s+3=0 We already know that frequency of oscillation is the magnitude of roots of Auxillary Equation. Given frequency is 2 rad/s. So, roots of AE, s = (+/-) 2j s^2 = -4 Substituting s^2 = -4 in CE, we get, -4s -4K + 4s + 3 = 0, K=3/4 = 0.75 !!!!

[@adarshsoni8226]

HW problem ans : k = 3/4

[@abdurrahmanabdulaziz5900]

ans = 3/4

[@komalsah6010]

K=3/4

[@etilasamwel8862]

solution 0.75

[@LIVE_LIVELY]

11:46 Shortcut Method used for this 😂

[@mashapoguajay3322]

3/4

[@21khalid8]

can k be negative

[@aProudHinduatani]

Why should k can not be equal to -1......?

[@mhz5749]

your question is very inspiring, i looked into it, i had some answers for that. when k=-1, the characteristic equation will be come a Type 1 transfer function's denominator which equals s(ss+as+1) and in this case, if a >=0, it WILL be a marginally stable system, however, omega, the imaginary part of the poles on imaginary axis, will never equal to +-2j, only possible condition except zero is when a = 0, this makes +-1j the j omega. that's why you dont k is not -1. furthermore, row of zeros should only appear at odd power rows of s. thats why we dont have go through all the above to prove this case is invalid.

[@sanjaykryadav6157]

H.w ans 3/4

[@real4788]

k = 0.75

[@053_waseemahmadwani_ee3]

0.5

[@suneeshpanicker536]

3/4

[@subodhsingh9287]

2

[@saurabhchaudhari6818]

k=3/2