angle of arrival and angle of departure concept in root locus of control system
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For the open loop transfer function, G(s)H(s):
We have n=2 poles at s = 0, -3. We have m=0 finite zeros. So there exists q=2 zeros as s goes to infinity (q = n-m = 2-0 = 2).
We can rewrite the open loop transfer function as G(s)H(s)=N(s)/D(s) where N(s) is the numerator polynomial, and D(s) is the denominator polynomial.
N(s)= 1, and D(s)= s2 + 3 s.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) = s2 + 3 s+ K( 1 ) = 0
Completed Root Locus

Root Locus Symmetry
As you can see, the locus is symmetric about the real axis
Number of Branches
The open loop transfer function, G(s)H(s), has 2 poles, therefore the locus has 2 branches. Each branch is displayed in a different color.
Start/End Points
Root locus starts (K=0) at poles of open loop transfer function, G(s)H(s). These are shown by an "x" on the diagram above
As K→∞ the location of closed loop poles move to the zeros of the open loop transfer function, G(s)H(s). Don't forget we have we also have q=n-m=2 zeros at infinity. (We have n=2 finite poles, and m=0 finite zeros).
We have n=2 poles at s = 0, -3. We have m=0 finite zeros. So there exists q=2 zeros as s goes to infinity (q = n-m = 2-0 = 2).
We can rewrite the open loop transfer function as G(s)H(s)=N(s)/D(s) where N(s) is the numerator polynomial, and D(s) is the denominator polynomial.
N(s)= 1, and D(s)= s2 + 3 s.
Characteristic Equation is 1+KG(s)H(s)=0, or 1+KN(s)/D(s)=0,
or D(s)+KN(s) = s2 + 3 s+ K( 1 ) = 0
Completed Root Locus

Root Locus Symmetry
As you can see, the locus is symmetric about the real axis
Number of Branches
The open loop transfer function, G(s)H(s), has 2 poles, therefore the locus has 2 branches. Each branch is displayed in a different color.
Start/End Points
Root locus starts (K=0) at poles of open loop transfer function, G(s)H(s). These are shown by an "x" on the diagram above
As K→∞ the location of closed loop poles move to the zeros of the open loop transfer function, G(s)H(s). Don't forget we have we also have q=n-m=2 zeros at infinity. (We have n=2 finite poles, and m=0 finite zeros).
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