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.
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.
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.
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
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 !!!!
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.
H.W. problem also by shorcut method.
Ans = 3/4 = 0.75 12:14
This video really helped me in my finals.Similar question was in exam. Thanks alot. Your teaching style is tremendous.
your contents are far better than other so-called gate preparation classes
I'm just curious; what gate preparation classes have you used?
So true, this is direct and to the point honestly enough to do the gate questions
H/W problem ans : K = 3/4
How? Can you explain the steps
Solution to the h/w problem is 0.75
Sir , you are expert , I appreciate you and really thank you so much ,Allah razı olsun :)
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.
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.
cringe comment.
@@dakshsingh5891 Yes, I’m aware.
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.
sir upload more videos of control system as soon as possible.
qooj(o()pkiuyyy%🙂🙂🙂🙂🙂🙂
Hey Do you know if this Control system playlist is worth watching to prep for Gate ECE?
thanks Sir🙏❤
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
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
11:46 Shortcut Method used for this 😂
K = 0.75
k= 0.75
Is this correct?
ans = 3/4
K=3/4
HW problem ans : k = 3/4
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 !!!!
can k be negative
solution 0.75
3/4
H.w ans 3/4
Why should k can not be equal to -1......?
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.
k = 0.75
0.5
3/4
k=3/2
0.75
2
K = 0.75