2 (s+2)(s+1)* % overshoot in the step response y(t) is less than 5% • Settling time of the step response y(t) is less than 6 seconds. • For a unit ramp

reference input, the steady state error must be less than 0.12. Write down the steps you followed in the sisotool as a numbered list. Include the following in those steps: a. Your error calculations. b. Paste the snapshots of all the root loci diagrams leading to the final one. c. Expression of your final controller K(s). d. The plot of the step response with settling time and overshoot marked in it with a cursor. e. The plot of ramp response with a cursor (data tip) marked in steady state (say around 10s). Make sure the (x, y) coordinates in the data tip i.e. (time, plant output value y) are easily visible. This is how we can estimate the steady state error. 3. Consider a plant G(s) = Design a controller K(s) for it with the following specifications. Please note that the difference between the x and y coordinates is the error which hopefully is less than 0.12. Why? The ramp reference of unit slope is a function r(t) = t. So, at time 10s, the reference value is 10. Hence, in the y(t) response's data tip, the x coordinate (time) is the same as the reference value. Hence, the difference between the x and y coordinates is the error. Extra credit: Above problem has no restrictions on the value of input u(t) that goes into the plant. Repeat the above problem with a 4th specification: the input u(t) always remains below 10 for a step reference command of value 3. Also, explain mathematically, the risks for careless use of a 1st order pre-filter in this problem, on the ramp error.