# Finding points on root-locus by hand

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1. Aug 13, 2016

### Jacer2

1. The problem statement, all variables and given/known data

I am trying to answer two questions:

1. Solve part (b) in the image above by hand.

2. What are the differences between transfer function (a) and transfer function (b).

2. Relevant equations
cos(Im(s) / Re(s)) = ζ

3. The attempt at a solution

1. For part one given the damp I know that cos-1(ζ) = 45. Thus the magnitude of the real and imaginary parts of the root has to be the same. However I can't seem to think of a way to do this by hand. I can plot the root locus in Matlab.

Once I find the root locus I know I can find K by using:
$$K = \frac{|denominator \quad of \quad tf|}{|numerator \quad of \quad tf|}$$

2. The transfer functions are $$K = \frac{KGH}{1+KGH}$$ and $$K = \frac{KG}{1+KGH}$$ . Besides a different output response would the root locus and bode plots of the both transfer functions be the same?

2. Aug 13, 2016

### Jacer2

The only solution I could think of is using the angle condition:

The phase should all add up to 180 degrees + a multiple of 360:
SInce the damping results in a 45 degree angle, the real and imaginary parts of s are the same. Thus:
$$s = -x + xj$$

I can then apply the angle criterion by getting the phase of each pole:
$$\sum\angle(zeroes) - \sum\angle(poles) = 180 \pm 360*n$$
$$-2*tan^{-1}(\frac{x}{-x+2})-tan^{-1}(\frac{x}{-x+4})-tan^{-1}(\frac{x}{-x+1}) = 180$$

I had to brute force and solve this in my calculator and got x = 2.203. I had to then try setting the equation equal to -180 as opposed to 180:
$$-2*tan^{-1}(\frac{x}{-x+2})-tan^{-1}(\frac{x}{-x+4})-tan^{-1}(\frac{x}{-x+1}) = -180$$
and got x = .907672 which is the right answer.

Is there a better way to do this without numerically solving it?

Last edited: Aug 13, 2016