Answer to Question #233465 in Chemical Engineering for Lok

Question

A step change of magnitude 10 is introduced into the transfer function

Y(S)/X(S)= 4/(2S²+0.3S+0.5),

find the following

a. Percent overshoot.

b. Rise time

c.Maximum value of y(t)

d. Ultimate value of Y(t)

e. Period of oscillations.

Answer 1

The magnitude transfer function of 10 is given by:

Y(S)/X(S)= 4/(2S²+0.3S+0.5),

First, it is worth noting that an overshoot simply means that an output exceeds its final, steady-state value. In the case of the presented step input change, the percentage overshoot (PO) therefore:

·        Is the maximum value minus the step value divided by the step value. In the case of the unit step, the overshoot is just the maximum value of the step response minus one.

 

Therefore:

 (a) Percent overshoot = 25.4%

 

(b) Rise time = 1.0814 s

 

(d) Ultimate value of Y(t) = Y(∞) = 10

 

(c) Maximum value of Y(t) = 12.54

 

(e) Period of oscillation = 3.4278 s