Question

Given an open-loop transfer function \(GH = \frac{100}{s}(s+100)\) for a unity feedback system with a unit step input \(r(t)=u(t)\), determine the rise time \(t_r\).

Hint:

For second-order systems, always calculate the damping ratio \(\zeta\) first. If \(\zeta>1\) (overdamped) and one pole is significantly closer to the origin than the other (a rule of thumb is by a factor of 5 or more), you can approximate the system as a first-order system using the dominant pole to quickly estimate parameters like rise time and settling time.

Answer 1

Step 1: Formulating the Characteristic Equation
The closed-loop transfer function (CLTF) for a unity feedback system is given by \(T(s) = \frac{G(s)}{1 + G(s)}\). \[ T(s) = \frac{\frac{100}{s(s+100)}}{1 + \frac{100}{s(s+100)}} = \frac{100}{s^2 + 100s + 100} \] The characteristic equation is \(s^2 + 100s + 100 = 0\).
Step 2: Identifying System Parameters
Compare with the standard second-order form: \(s^2 + 2\zeta\omega_n s + \omega_n^2 = 0\).

• \(\omega_n^2 = 100 \implies \omega_n = 10\) rad/sec.
• \(2\zeta\omega_n = 100 \implies 2\zeta(10) = 100 \implies \zeta = 5\).

Since \(\zeta = 5>1\), the system is overdamped.
Step 3: Calculating Poles and Time Constants
Using the quadratic formula for \(s^2 + 100s + 100 = 0\): \[ s = \frac{-100 \pm \sqrt{10000 - 400}}{2} = \frac{-100 \pm 97.98}{2} \] The poles are \(p_1 = -1.01\) and \(p_2 = -98.99\). The system response is dominated by the pole closest to the origin (\(p_1 = -1.01\)). The time constant for the dominant pole is \(\tau = 1/|p_1| = 1/1.01 \approx 0.99\) s.
Step 4: Deriving Rise Time (\(t_r\))
For a heavily overdamped system, the rise time (\(t_r\)) is defined as the time taken for the response to go from 10% to 90% of its final value. This is dominated by the slow pole: \[ t_r \approx 2.2 \tau = 2.2 \times 0.99 \approx 2.178 \text{ s.} \]