Question

Impulse response for \(t \ge 0\) of a second order control system when damping ratio = 1 is

• A. \( \omega_n \sin(\omega_n t) \)
• B. \( \omega_n^2 t e^{-\omega_n t} \)
• C. \( \omega_n^2 \sin(\omega_d t) \)
• D. \( \omega_n e^{\omega_n t} \sin(\omega_d t) \)

Hint:

Critically damped system (\(\zeta=1\)): Denominator of \(H(s)\) has repeated real roots \(-(s+\omega_n)^2\).
Impulse response for critically damped second-order system: \(h(t) = \omega_n^2 t e^{-\omega_n t} u(t)\).
Inverse Laplace Transform pair: \(\mathcal{L}^{-1}\left\{\frac{k}{(s+a)^2}\right\} = k t e^{-at} u(t)\).

Answer 1

The Correct Option is B

The standard transfer function of a second-order control system is \( H(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} \), where \(\zeta\) is the damping ratio and \(\omega_n\) is the natural undamped frequency. The impulse response \(h(t)\) is the inverse Laplace transform of \(H(s)\). When the damping ratio \(\zeta = 1\), the system is critically damped. In this case, the denominator becomes \(s^2 + 2\omega_n s + \omega_n^2 = (s+\omega_n)^2\). So, \( H(s) = \frac{\omega_n^2}{(s+\omega_n)^2} \). To find the inverse Laplace transform, we use the standard pair: \(\mathcal{L}^{-1}\left\{\frac{1}{(s+a)^2}\right\} = t e^{-at} u(t)\). Here, \(a = \omega_n\). So, \(h(t) = \mathcal{L}^{-1}\left\{\frac{\omega_n^2}{(s+\omega_n)^2}\right\} = \omega_n^2 \mathcal{L}^{-1}\left\{\frac{1}{(s+\omega_n)^2}\right\}\). For \(t \ge 0\) (since impulse response is usually causal for physical systems, or \(u(t)\) is implied): \[ h(t) = \omega_n^2 t e^{-\omega_n t} \] This matches option (b). Other cases:
\(\zeta<1\) (underdamped): \(h(t) = \frac{\omega_n}{\sqrt{1-\zeta^2}} e^{-\zeta\omega_n t} \sin(\omega_d t)\), where \(\omega_d = \omega_n\sqrt{1-\zeta^2}\).
\(\zeta>1\) (overdamped): Response is sum of two decaying exponentials.
\(\zeta = 0\) (undamped): \(h(t) = \omega_n \sin(\omega_n t)\) (Matches option (a) if \(\zeta=0\)). \[ \boxed{\omega_n^2 t e^{-\omega_n t}} \]