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.
Hi Justin the calculation are correct. Please check by substituting s= -4.7 in K equation. k=-s^3-9s^2-18s k=(-(-4.7)^3) -(9*(-4.7)^2)-(18*(-4.7)) Answer is k= -10.39
−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
−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
Two ways to check the s valid or not 1. Root Locus Check: If the value of 's' is on the root locus plath it’s valid. 2. 'K ' Equation :substitute 's ' value in 'K' equation. If 'K' is positive, 's' is valid; if K is negative 's' isn’t valid.
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.
−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
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°
Go through this link. th-cam.com/video/_GnUs2wwZ6k/w-d-xo.html th-cam.com/video/Q1aNPwwbr8c/w-d-xo.html In this problem I have explained to calculate breakaway point
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Thank you sir.. from Nigeria.. having a paper in the afternoon🙏
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Same bro we have paper tmmrw in india
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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.
Yes sir
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@@as-tutorials5537sir all sums of root locus aasa ee solve hoonga method toon same rahegaa na
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Nice explanation sir, but in root locus how to get 's' values i didn't understand
Plz Go through the steps
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Hi sir great explanation, for the s=-4.7, the value of k is -230.723, I think might be calculations might be wrong..
Hi Justin the calculation are correct. Please check by substituting s= -4.7 in K equation. k=-s^3-9s^2-18s
k=(-(-4.7)^3) -(9*(-4.7)^2)-(18*(-4.7))
Answer is k= -10.39
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but sir is k value 10.39 how sir😢
Sir tqsm
Sir after dk/ds how to find value of s im stuck here... Please help me... S=1.21 how???
−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
Thankyou so much sir
Thank you sir,very helpful 🙂
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Sir your taking s=-1.26 not understand Plz exam Ripley
−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
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Understandable.Thank you sir!!
Tqsm sir 👍🏻
thank u sir
Thank you sir taking the effort n explaining the concept very clearly
How did u get the -1.26 anwser as a valid s ?
Two ways to check the s valid or not
1. Root Locus Check: If the value of 's' is on the root locus plath it’s valid.
2. 'K ' Equation :substitute 's ' value in 'K' equation. If 'K' is positive, 's' is valid; if K is negative 's' isn’t valid.
Thanks.......
Thank you
How do we find out the values for K for which the closed loop poles are real?
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.
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How to get s value
Thank you sir...
checking valid break-away and break-in point is interesting
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Sir I didn't understand how s=-1.26& s= -4.7
−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
if there is no s in the denominator how to solve
mrans s/(s+3)(s+6)
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
Go through this link. th-cam.com/video/_GnUs2wwZ6k/w-d-xo.html
th-cam.com/video/Q1aNPwwbr8c/w-d-xo.html
In this problem I have explained to calculate breakaway point
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