Time Response of Second-Order System

The closed-loop transfer function of the second-order system is given by

\(\boxed{\displaystyle \mathrm{CLTF} = G(s) = \frac{C(s)}{R(s)} = \frac{\omega_{n}^2}{s^2 + 2\zeta\omega_{n}s + \omega_{n}^2}}\)

where \(\zeta\) is the damping ratio; and \(\omega_{n}\) is the undamped natural frequency, expressed in radians per second (rad/s).

The damping case is determined by the damping ratio \(\zeta\) from the poles.

The closed-loop poles of the second-order transfer function \(G(s)\) are

\(\boxed{\displaystyle s = -\zeta\omega_{n} \pm \omega_{n}\sqrt{1 - \zeta^2} = -\sigma_{d} \pm j\,\omega_{d}}\)

where \((\alpha = \sigma_{d})\) is the damping attenutation, expressed in nepers per second (Np/s); and \(\omega_{d}\) is the damped natural frequency, expressed in radians per second (rad/s).

See: Second-Order Circuits

Step Response

Using the unit-step signal \(u(t)\) as an input \(r(t)\) to the second-order system,

For a unit-step signal, \(r(t) = u(t)\), the Laplace transform of \(r(t)\) is

\(\boxed{\displaystyle R(s) = \frac{1}{s}}\)

Then, the unit step response can be found using \(C(s) = R(s)\,G(s)\), followed by the inverse Laplace transform.

\(\boxed{\displaystyle C(s) = \frac{\omega_{n}^2}{s\left(s^2 + 2\zeta\omega_{n}s + \omega_{n}^2\right)}}\)

Undamped Case \((\zeta = 0)\)

The two poles of \(G(s)\) are imaginary, \((s = \pm j\omega_{n})\).

Substituting \((\zeta = 0)\) to the transfer function \(C(s)\),

\(\displaystyle C(s) = \frac{\omega_{n}^2}{s\left(s^2 + \omega_{n}^2\right)} = \frac{1}{s} - \frac{s}{s^2 + \omega_{n}^2}\)

Apply the inverse Laplace transform,

\(\boxed{\displaystyle c(t) = \left[1 - \cos(\omega_{n}t)\right]\,u(t)}\)

So, for undamped case \((\zeta = 0)\), the transient response does not die out, and the unit step response will be a continuous time signal with constant amplitude and frequency.