Question

For the system described in problem AP5.1, do the following. First, do part a), the steady-state error to a unit step. For part b) remove both the zero and the third pole, normalize the system, and use Matlab to plot and print the step response. Then both graphically and analytically determine the percent overshoot, the peak time and the settling time. For part c), include the pole and the zero, and use Matlab to plot and print the step response. Then graphically determine the percent overshoot, the peak time and the settling time. Next, to see the effect of the zero alone, remove the pole, normalize, and plot the step response. Finally, to see the effect of the pole alone, remove the zero, normalize, and plot the step response.

AP5.1

A closed-loop transfer function is
T(s)= Y(s)/R(s) = 108(s+3)/(s+9)(s² +8s+36)


Determine the steady-state error for a unit step input R(s)=1/s.

Answer 1

The steady-state error to a unit step can be determined using the final value theorem, which involves taking the limit of the input divided by the transfer function as s approaches 0. In this case, the steady-state error is found to be 2/3.

Step-by-step explanation:

In order to determine the steady-state error for a unit step input, we can use the final value theorem. The final value theorem states that the steady-state value of the output of a system can be found by taking the limit as s approaches 0 of the Laplace transform of the input divided by s times the transfer function of the system. In this case, the input is R(s) = 1/s and the transfer function is T(s) = 108(s+3)/(s+9)(s²+8s+36). Plugging these values into the final value theorem, we get:

Steady-state error = lims⇒0 [1/s * 108(s+3)/(s+9)(s²+8s+36)]

Taking the limit and simplifying, we find that the steady-state error is 6/9 = 2/3.