Question

Given the following unit-step response of a second-order system:

c(t)=1-1.25e⁻⁰.⁷⁵⁽ᵂ⁺²ᵗ⁾sin((W+2)t +53.1°)
W=0
Find the rise time tr, peak time tp settling time ts (5% criterion) and maximum (percent) overshoot Mp

Answer 1

The rise time is 0.847 seconds, the peak time is 0.411 seconds, the settling time is 10.676 seconds, and the maximum overshoot is 56.522%.

Step-by-step explanation:

The rise time (tr) is defined as the time it takes for the system output to rise from 10% to 90% of its final value. In this case, we can find tr by solving the equation 0.9 = 1 - 1.25e^(-0.75(W+2)tr). By substituting W = 0 and solving for tr, we find that tr = 0.847 seconds.

The peak time (tp) is the time it takes for the system output to reach its first peak. In this case, we can find tp by solving the equation (W+2)tp + 53.1° = π/2. By substituting W = 0 and solving for tp, we find that tp = (π/2 - 53.1°)/(2 radians/second) = 0.411 seconds.

The settling time (ts) is the time it takes for the system output to stay within a certain range of the final value. In this case, we need ts to satisfy the condition 0.95 = 1 - 1.25e^(-0.75(W+2)ts). By substituting W = 0 and solving for ts, we find that ts = 10.676 seconds.

The maximum percent overshoot (Mp) is the maximum percentage by which the system output exceeds the final value. In this case, we can find Mp by solving the equation 1 - 1.25e^(-0.75(W+2)t) = (1 + Mp/100) for t. By substituting W = 0 and solving for Mp, we find that Mp = 56.522%.