is problem is a modification of Problem 1 of Chapter 9 Given

is problem is a modification of Problem 1 of Chapter 9. Given the plant transfer function G(s) = 1/(s + 1)^2 (s + 10) Given this plant is in a unity feedback system so that the block diagram is as follows. where D(s) = k, plot the squareroot locus of the system with respivt to the parameter k, Do this first by yourself and then check using Mat lab. Don\'t cheat and do it with Matlab first! Determine that value of k such that the damping ratio is zeta = 0.6. You should be able to see from your squareroot locus plot that a ray extending from (0,0) with the proper damping angle will, in fact, cross the squareroot locus. The rav itself will be proportional to s­_0 = - zeta + j Squareroot 1 - zeta^2. In order to calculate the required value of k, you will have to determine the exact point at which the ray crosses the squareroot locus. The following procedure should wok for you: Write a subprogram in Mullah to evaluate the angle of the transfer function in squareroot locus form, that Is. L(s). To do this you can just pass the routine an array of zerus, an array of poles, tin- value of zeta and a multiplying factor, say 2. You can then evaluate the angle of the trunsfor function at two points, one at s x multiplier and one at s/multiplier. Then you ran look at tlw angle of the transfer function modulo 2Pi. If one of the angles is more than Pi and the other angle is lew than Pi. you linve bounded the range where von have to  look for the the correct value of s. That is. if you supplied the multiplier value of 2. then you will know that the proper value for the pole is between 2 times s_0 and 0.5 times s_0. If the two phases you found do not span Pi, then just double the multiplying factor and try again. Very soon you will succeed. You ran write a Matlah routine to do this for you. Define the value of a for whirh the angle exceeded Pi as s_hight and the the value of s for which the angle was leas than Pi as s_low. The value you want Is Iteiween those two points on the ray. In order to find the correct point, you ran use a procedure railed divide and conquer, or if you are very fancy, via Dedekind cuts. Here is the idea. Define s_test = s_high + s_low/2. Use your angle routine to find the angle of L(s) at s_text. If the resulting value is less than Pi, put S_low = s_lost otherwise put s_high = s_text. Repeat the previous 3 steps until the angle of L(s) is sufficiently close to Pi. At that point s_low, and s_high will be almost equal. Agreement within about 16 decimal jmints is just as easy as to 6 decimal points. Note that whenever you do divide and conquer, you ave to make sure you are actually bound-ing a desired point or you are going to just make a big mess.

Solution

(1) 0.5

(2) 0.01*so

(3) a