Question

Natural frequency of oscillations of the following transfer function is \(\frac{C(s){R(s)} = \frac{1}{(s^2 + 0.5s + 1)}\):}

• A. \(\omega_n = 0.5\)
• B. \(\omega_n = 2\)
• C. \(\omega_n = 0.25\)
• D. \(\omega_n = 1\)

Hint:

Always look directly at the constant term in the denominator of a normalized second-order system (where the coefficient of \(s^2\) is 1). The square root of that constant term is your natural frequency \(\omega_n\). Here, \(\sqrt{1} = 1\), which takes less than two seconds to identify!

Answer 1

The Correct Option is D

Concept: A standard second-order system transfer function is represented in control systems as: \[ T(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} \] where:
• \(\omega_n\) is the natural frequency of oscillations.
• \(\zeta\) is the damping ratio. By comparing the characteristic equation (the denominator polynomial) of the given system with the standard second-order characteristic equation, we can directly solve for the natural frequency parameter.

Step 1: Extract the characteristic equation from the given transfer function. The transfer function provided is: \[ \frac{C(s)}{R(s)} = \frac{1}{s^2 + 0.5s + 1} \] The denominator polynomial equated to zero gives the characteristic equation of the system: \[ s^2 + 0.5s + 1 = 0 \]

Step 2: Compare with the standard standard form to find \(\omega_n\). The general second-order characteristic equation is: \[ s^2 + 2\zeta\omega_n s + \omega_n^2 = 0 \] By comparing the constant terms on both sides: \[ \omega_n^2 = 1 \] Taking the positive square root since frequency is a positive physical quantity: \[ \omega_n = \sqrt{1} = 1\text{ rad/s} \] Thus, the natural frequency of oscillations is exactly equal to 1. This directly matches option (D).