The teacher is so excellent with his ausum tutorial vedio so helpful❤

Am greatly humbled by d lecture

Easily understandable 👍

Very much helpful content sir......

Very helpful , thanks sir!

Interesting and easy to understand ❤

Good content and explanation sir 👍

Loving the video

tq sir its really helpful 👍

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Tqew Sir. For urs wonderful explanation ♥

Thankyou sir because detailed explanation

Thank u sir for your wonderful explain

Thank you sir Nice explanation you made it so easy to understand step by step 👍

Very helpful. Tq sir👍

Thank you sir.. from Nigeria.. having a paper in the afternoon🙏

@rakshithv2370

9 ай бұрын

Hope you did well, Nigga

@Rohit2403-ek4sg

6 ай бұрын

Same bro we have paper tmmrw in india

@PiyushSharma-zz1ut

6 ай бұрын

​@@Rohit2403-ek4sg mine too

Explanation is very nice

Thank you so much sir 👍

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Lot of help Sir 🎉

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great Explanation

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Thank You Sir•

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You make the concepts simple,clear and easy to understand."Thank you so much sir!!!"👍🏻

@anshulbajpei935

7 ай бұрын

Aapki engineering complete ho gai

@gayathriar942

7 ай бұрын

@@anshulbajpei935 yes

@anshulbajpei935

7 ай бұрын

​@@gayathriar942very nice but aap konse state se ho me Ahmedabad se hu

@gayathriar942

7 ай бұрын

@@anshulbajpei935 Thank you , I'm from Karnataka

@anshulbajpei935

6 ай бұрын

​@@gayathriar942 So you must not know Hindi to speak.?

Superb ❤️

Understood sir 👍

Thank u sir ☺️

wonderfull

Thank You sir

Tqsm sir 👍🏻

EASY METHOD For intersection with imaginary axis Just substitute s=jw in the characteristic equation and equate the real and imaginary part to zero. And find the value of w, which is the intersection of root locus with both +ve and - ve imaginary axis.

@daravishnu4728

2 жыл бұрын

Yes sir

Thank you

Thanku sir ❤

Understandable.Thank you sir!!

Thank you sirr

Thanks.......

Thank u sir

nice explanation

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Thanku sir

Sir tqsm

Thanks 🌝 sir

Thnks sir

Thank you so much sir🙏so calmly and nicely you have taught

@as-tutorials5537

Жыл бұрын

Thanks and welcome

@jethalalfunTV

7 ай бұрын

​@@as-tutorials5537sir all sums of root locus aasa ee solve hoonga method toon same rahegaa na

@as-tutorials5537

7 ай бұрын

Yes

Thank you so much sir 🙂 it's very helpful sir

@anshulbajpei935

7 ай бұрын

Aapki engineering complete ho gai ?

👌

👍👍👍👍👍

☺☺👍

Nice

Thank you sir Tomorrow is my exam

@as-tutorials5537

Ай бұрын

All the best👍

Nice explanation sir, but in root locus how to get 's' values i didn't understand

@as-tutorials5537

Ай бұрын

Plz Go through the steps

Sir after dk/ds how to find value of s im stuck here... Please help me... S=1.21 how???

@as-tutorials5537

2 жыл бұрын

−3s^2−18s−18=0 For this equation: a=-3, b=-18, c=-18 −3s^2−18S−18=0 Step 1: Use quadratic formula with a=-3, b=-18, c=-18. S=(−b ± (√b^2−4.a.c))/2.a Substitute a,b and c value in the above formula S=(−(−18)±√(−18)^2−4(−3)(−18))/2(−3) S=(18±√108)/−6 S=−3−√3 and s=−3+√3 S=-4.7 and s=-1.26

@TGnong

2 жыл бұрын

Thankyou so much sir

How do we find out the values for K for which the closed loop poles are real?

@carultch

4 ай бұрын

Those are called break-in and break-away points. Given a transfer function of N(s)/D(s), the breakpoints will be found when the following equation holds true: N(s)*D'(s) - N'(s)*D(s)=0 We can do this for a system with 1 zero and 2 poles as an example. (s+6)/[(s + 5)*(s + 2)]. It is very difficult to do this analytically for higher order systems, as it means solving higher order polynomials, that may be impossible to solve by hand. N(s) = s + 6 N'(s) = 1 D(s) = (s + 5)*(s + 2) D'(s) = 2*s + 7 Construct equation to equate to zero: (s + 6)*(2*s + 7) - 1*(s + 5)*(s + 2) = 0 Solve for s: s = -8, the break-in point s = -4, the break-away point Given a closed loop made from K and G(s), the closed loop transfer function will be K*G(s)/(1 + K*G(s)). The denominator of this will equal zero at the two closed-loop poles. 1 + K*G(s) = 0 When G(s) = N(s)/D(s): 1 + K*N(s)/D(s) = 0 Solve for K: K = -D(s)/N(s) Thus, for our example: K = -(s + 5)*(s + 2)/(s + 6) Plug in breakpoints for s: at s = -4: K = -(-4 + 5)*(-4 + 2)/(-4 + 6) = 1 at s = -8: K = -(-8 + 5)*(-8 + 2)/(-8 + 6) K = 9 So at 1 and 9, we have poles that transition between being real, and being a complex conjugate pair. When K=9, the poles are both real.

checking valid break-away and break-in point is interesting

Thank you sir taking the effort n explaining the concept very clearly

Sir your taking s=-1.26 not understand Plz exam Ripley

@as-tutorials5537

2 жыл бұрын

−3s^2−18s−18=0 For this equation: a=-3, b=-18, c=-18 −3s^2−18S−18=0 Step 1: Use quadratic formula with a=-3, b=-18, c=-18. S=(−b ± (√b^2−4.a.c))/2.a Substitute a,b and c value in the above formula S=(−(−18)±√(−18)^2−4(−3)(−18))/2(−3) S=(18±√108)/−6 S=−3−√3 and s=−3+√3 S=-4.7 and s=-1.26

but sir is k value 10.39 how sir😢

if there is no s in the denominator how to solve

@siddheshwagh2300

Жыл бұрын

mrans s/(s+3)(s+6)

@as-tutorials5537

Жыл бұрын

No 's' in the denominator means no Poles at the origin. For s/(s+3)(s+6) in this numerator one zero that is s=0 and denominator 2 Poles that is s=-3 and s=-6. Therefore one pole s=-3 reaches to zero and another pole s=-6 move towards infinite path at an angle 180°

How to calucat breaking point

@as-tutorials5537

Жыл бұрын

Go through this link. kzread.info/dash/bejne/kXuit9Vrp9rAZs4.html kzread.info/dash/bejne/g2WVsLKwp8XYaMY.html In this problem I have explained to calculate breakaway point

Sir I didn't understand how s=-1.26& s= -4.7

@as-tutorials5537

Жыл бұрын

−3s^2−18s−18=0 For this equation: a=-3, b=-18, c=-18 −3s^2−18S−18=0 Step 1: Use quadratic formula with a=-3, b=-18, c=-18. S=(−b ± (√b^2−4.a.c))/2.a Substitute a,b and c value in the above formula S=(−(−18)±√(−18)^2−4(−3)(−18))/2(−3) S=(18±√108)/−6 S=−3−√3 and s=−3+√3 S=-4.7 and s=-1.26

Love from Pakistan 🎉🎉🇵🇰🇵🇰🇵🇰

T che valu chukle

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@anshulbajpei935

7 ай бұрын

Aapki engineering complete ho gaya kya?