Lecture 10

E12 Linear Physical Systems Analysis

Author

Prof. Emad Masroor

Published

February 19, 2026

The Natural Frequency

Second-order systems have a natural tendency to oscillate. The frequency at which this occurs is called the natural frequency.

Symbol Quantity Units Definition
\(\omega\) or \(\omega_n\) Natural (angular) frequency \(\mathrm{rad} s^{-1}\) \(\omega = \sqrt{k/m}\)
\(f\) or \(f_n\) Natural frequency \(s^{-1}\) \(\omega = 2 \pi f\)
\(T\) Period \(s\) \(2\pi/\omega\)

Energy Considerations in the Free Response

Where else do second-order systems arise?

  • Applying Kirchhoff’s Laws:

\(\displaystyle \boxed{LC\frac{d^2 i}{dt^2} + i = 0}\)

\(\displaystyle \boxed{LC\frac{d^2 v}{dt^2} + v = 0}\)

Derivation of equations for LC Circuits

Second-order systems in the frequency domain

\[\ddot{x} + \omega^2 x = f(t) \tag{1}\]

  • Taking the Laplace Transform, we get \[s^2 X + \omega^2 X = F\]

  • We can therefore write the transfer function as \[\frac{X(s)}{F(s)} = \frac{1}{s^2 + \omega^2}\]

  • Recall: The transfer function of a system is also its ‘impulse response’

  • Impulse Response of system in Equation 1 is: \[\frac{1}{s^2+\omega^2} = \frac{1}{\omega} \frac{\omega}{s^2+\omega^2}\]

The Impulse Response of \(\ddot{x}+\omega^2 x = f(t)\) is a sine function

\[\frac{X(s)}{F(s)} = \frac{1}{s^2+\omega^2} = \frac{1}{\omega} \frac{\omega}{s^2+\omega^2}\]

What is the amplitude and period?

The impulse response of a 2nd order system

\[m \ddot{x} + k x = f(t)\] \[f(t) = \delta(t)\]

In-class task: Find the impulse response in (1) the time domain and (2) the frequency domain.

Impulse response: “An expression in terms of (1) \(t\) and (2) \(s\) that describes how this system behaves when \(f(t) = \delta(t)\), the unit impulse.”

Recall: The unit impulse response is the transfer function

Answer: \[ \frac{X(s)}{F(s)} = \frac{1}{ms^2 + k} = \frac{1/m}{s^2 + k/m} = \frac{1}{\sqrt{mk}} \frac{\omega}{s^2+\omega^2} \]

In the time domain: \(\displaystyle \quad \frac{1}{\sqrt{mk}} \sin \omega t\)

Putting some numbers on it

\[m \ddot{x} + k x = f(t)\] \[f(t) \approx 2 \delta(t-t_1)\]

\(m =3.5\mathrm{kg}, k = 1.2 \mathrm{N/m}, t_1 = 2 \mathrm{s}\)

The mass is struck with a force described by the graph above.

In-class task: Sketch a graph of position vs. time (with numbers!)

Free and Forced Response

\[\ddot{x} + \omega^2 x = f(t)\]

  • Applying the Laplace Transform with \(x(0) \neq 0, \dot{x}(0) \neq 0\) \[ s^2 X - s x(0) - \dot{x}(0) + \omega^2 X = F(s) \\ \implies (s^2+\omega^2)X = F + \dot{x}(0) + s x(0) \]
  • Dividing through by \(s^2 + \omega^2\), \[X(s) = \underbrace{\underbrace{\frac{1}{s^2 + \omega^2}}_{\text{T.F.}} F}_{\text{Forced Response}} + \underbrace{\frac{\dot{x}(0)}{s^2 + \omega^2} + \frac{s x(0)}{s^2 + \omega^2}}_{\text{Free Response}}\]

In-class task: Express the Transfer Function as a sum of fractions.

Step Response of a Second-order system

\[ \ddot{x} + \omega^2 x = \begin{cases} a & t > 0 \\ 0 & t < 0 \end{cases} \] \(x(0) = \dot{x} = 0\)

In-class task: If \(a\) is some positive number constant in time, describe the motion of this object.

Step Response of \(\ddot{x} + \omega^2 x = f(t)\)

Calculating the step response of \(\ddot{x} + \omega^2 x = a u_s(t)\)

Step function with magnitude \(a\) applied to r.h.s.

\(\ddot{x} + \omega^2 x = a\)

Procedure in time domain

  • Guess a solution: \(x(t) = ba\)
  • Substitute: \(\frac{d^2(ba)}{dt^2} + \omega^2 ba = a\)
  • \(\implies b = 1/\omega^2\)
  • \(x(t) = \frac{a}{\omega^2}\) is a solution to \(\ddot{x} + \omega^2 x = a u_s(t)\)
  • Add the solution to \((\ddot{x} + \omega^2 x =0)\) to above
  • \(x(t) = \left( x_0 - \frac{a}{\omega^2} \right) \cos \omega t + \frac{v_0}{\omega} \sin \omega t + \frac{a}{\omega^2}\)

\(s^2 X(s) + \omega^2 X(s) = F(s)\)

Procedure in frequency domain

  • Take Laplace Transform of \(u_s(t)\)
  • Determine forced response as a function of \(s\)
  • Use partial fractions
  • Convert into time domain using Inverse Laplace Transform.