Assume that you have a system in a feedback loop:

The transfer function, G(s), is given by:

It is required that the closed loop system have poles at s = -1 + 2j and s = -1 - 2j.

• Is it possible to get poles at that position using a proportional gain?  Why not?  (Reasoning based on the root locus probably gets this answer fastest.)
• Using proportional control, the root locus moves verticaly up/down from the poles at s = -0.1.  Thos root locus lines are straight and do not pass through the desired pole location.
• If a compensator is used with a zero at s = -1 (an arbitray choice), can a pole be placed to give the required closed loop poles?
• We need to examine whether it is possible.  The angle from -0.1 to the desired pole is 114.2^o.  Let's assume the zero is at s = -1, then the angle from -1 to -1 + 2j is 90^o.  Then, if the angle from the pole is f, we have:
• -f + 90^o - 114.2^o  - 114.2^o = -180^o.
• Solve for f and we have f = 270^o - 228.4^o =  41.54^o.
• This picture sums it up.  Clearly, we can find a place where the angle will be correct.  It's not as far to the left as it is pictured.

For a pole with an angle of 41.54^o, we have to have tan(41.54^o) = 2/x = 0..886.  That means x = 2.257, and the pole is at s = -3.257.

• What is the gain, Kc, for the compensator, assuming the compensator is of the form Gc(s) = Kc(s-z)/(s-p)?  (And, we have chosen z = -1, right?)
• We need the product of pole distances divided by the product of zero distances to get the root locus gain.
• Root Locus Gain = 4.8x5.8/2 = 7.25
• Since there is a gain of 10 in the system (and the system is in "root locus form".) the compensator gain is .725.
• Sketch the root locus for the system + compensator.