Sketch (by hand) the root locus having the open-loop transfer functions given be

Question

Sketch (by hand) the root locus having the open-loop transfer functions given below. In addition to the sketch, specifically solve for (if applicable) asymptote centers, breakaway points, and values of s at any crossings of the imaginary axis.

Problem 3: Sketch by hand the root locus having the open-loop transfer functions given below. In addition to the sketch, specifically solve for (if applicable) asymptote centers, breakaway points, and values of sat any crossings of the imaginary axis K (s+1) s(s+2) s (s+10) +1 (d) Verify each root locus in a) (c) with MATLAB

Explanation / Answer

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

In the open loop transfer function, G(s)H(s), we have n=2 finite poles, and m=0 finite zeros, therefore we have q=n-m=2 zeros at infinity.

Angle of asymptotes at odd multiples of ±180°/q, (i.e., ±90°)

There exists 2 poles at s = 0, -3, ...so sum of poles=-3.
There exists 0 zeros, ...so sum of zeros=0.
(Any imaginary components of poles and zeros cancel when summed because they appear as complex conjugate pairs.)

Intersect of asymptotes is at ((sum of poles)-(sum of zeros))/q = -1.5.
Intersect is at ((-3)-(0))/2 = -3/2 = -1.5 (highlighted by five pointed star).

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