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

Rise Time Calculation

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

Step 1: Determine the Closed-Loop Transfer Function

For a unity feedback system, \( H(s) = 1 \). The closed-loop transfer function \( T(s) \) is given by:

$$ T(s) = \frac{G(s)}{1 + G(s)H(s)} $$

Substituting the given \( G(s) \):

$$ T(s) = \frac{\frac{100}{s(s+100)}}{1 + \frac{100}{s(s+100)}} = \frac{100}{s(s+100) + 100} $$

Simplifying the denominator:

$$ T(s) = \frac{100}{s^2 + 100s + 100} $$

Step 2: Analyze System Parameters

Compare the denominator with the standard second-order characteristic equation:

$$ s^2 + 2\zeta\omega_n s + \omega_n^2 = s^2 + 100s + 100 $$

By equating coefficients, we find the natural frequency (\( \omega_n \)) and damping ratio (\( \zeta \)):

• \( \omega_n^2 = 100 \Rightarrow \mathbf{\omega_n = 10 \text{ rad/s}} \)
• \( 2\zeta\omega_n = 100 \Rightarrow 2\zeta(10) = 100 \Rightarrow 20\zeta = 100 \Rightarrow \mathbf{\zeta = 5} \)

Observation: Since \( \zeta = 5 \) (which is \( > 1 \)), the system is Overdamped.

Step 3: Calculate Rise Time

Unlike underdamped systems, overdamped systems do not have a single standard analytical formula for rise time (0-100%). For overdamped systems, rise time is conventionally defined as the time taken for the response to go from 10% to 90% of the final value.

First, we find the poles of the system by solving \( s^2 + 100s + 100 = 0 \):

$$ s = \frac{-100 \pm \sqrt{100^2 - 4(1)(100)}}{2} = \frac{-100 \pm \sqrt{10000 - 400}}{2} $$ $$ s = \frac{-100 \pm \sqrt{9600}}{2} \approx \frac{-100 \pm 97.98}{2} $$

The two poles are:

• \( s_1 \approx -1.01 \) (Dominant Pole)
• \( s_2 \approx -98.99 \) (Insignificant Pole)

Dominant Pole Approximation

Since the pole at \( s_1 = -1.01 \) is much closer to the imaginary axis (by a factor of nearly 100) than \( s_2 \), the system behaves almost like a first-order system with a time constant determined by the dominant pole.

The equivalent time constant \( \tau \) is:

$$ \tau \approx \frac{1}{|s_1|} = \frac{1}{1.01} \approx 0.99 \text{ s} $$

For a first-order system, the rise time (10% to 90%) is approximated as:

$$ t_r \approx 2.2 \tau $$

Substituting \( \tau \):

$$ t_r \approx 2.2 \times 0.99 \approx 2.178 \text{ s} $$

Final Answer

The rise time of the system is approximately:

$$ \mathbf{t_r \approx 2.2 \text{ seconds}} $$