Lecture 17

E12 Linear Physical Systems Analysis

Prof. Emad Masroor

March 24, 2026

Free and Forced Response of linear systems

Recall that the free and forced response can be determined with the help of the Laplace Transform.

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

\[\mathcal{L}[m \ddot{x} + b \dot{x} + k x ]= \mathcal{L}[f(t)]\]

\[\mathcal{L}[m \ddot{x}] + \mathcal{L}[b \dot{x}] + \mathcal{L}[k x ]= F(s)\]

\[ \begin{aligned} \mathcal{L}[m \ddot{x}(t)] &= m \left( s^2 X(s) - s x(0) - \dot{x}(0) \right) \\ \mathcal{L}[b \dot{x}(t)] &= b \left( s X(s) - x(0) \right) \\ \mathcal{L}[k x(t)] &= k X(s) \end{aligned} \]

Rearrange to get \(X(s)\) in terms of the sum of two terms:

  1. A term dependent on \(F(s)\), and
  2. A term dependent on \(x(0)\) and \(\dot{x}(0)\).

Free and Forced Response of 2nd-order systems

\[ X(s) = \frac{1}{m s^2 + b s + k} F(s) + \frac{x(0) (ms + b) + m\dot{x}(0)}{m s^2 + b s + k} \]

\[ \begin{aligned} X(s) &= \frac{1}{m s^2 + b s + k} F(s) \\ &+ \frac{(ms + b)}{m s^2 + b s + k} x(0) \\ &+ \frac{m}{m s^2 + b s + k} \dot{x}(0) \end{aligned} \]

  1. A term that depends on \(\mathcal{L}[f(t)]\)
  2. A term that depends on \(x(0)\)
  3. A term that depends on \(\dot{x}(0)\)

The Transfer Function and the Forced Response

The Transfer Function encodes the linear relationship between the input (a.k.a forcing function) and output (a.k.a. forced response) of a linear system.

\[ \begin{aligned} m \ddot{x} + b \dot{x} + k x &= f(t) \\ X(s) &= \frac{1}{m s^2 + b s + k} F(s) \\ X(s) &= T(s) F(s) \end{aligned} \]

Free, Forced, Transient, Steady-State

  1. Free response: The part of the response that depends on the initial conditions
  2. Forced response: The part of the response that is due to the forcing function
  3. Transient Response: The part of the response that disappears at long times
  4. Steady-State Response: The part of the response that remains at long times.

  • The full solution is this combination: \[\text{Total Response } = \text{Free } + \text{ Forced}\]

  • or this combination: \[\text{Total Response } = \text{Transient } + \text{ Steady-State}\]

Procedure for identifying free, forced, transient & steady-state response

Illustrated using a first-order system with step input

\[\dot{x} + ax = f(t)\]

\[sX- x(0) + a X = F\]

\[X(s) = \frac{x(0)}{s+a} + \frac{1}{s+a} F(s)\]

\[x(t) = \mathcal{L}^{-1} \left[ \frac{x(0)}{s+a} \right] + \mathcal{L}^{-1} \left[ \frac{1}{s+a} F(s) \right] \]

\[x(t) = \text{Free Response } + \text{Forced Response} \]

  • We would like to consider a step input: \(f(t) = b \cdot u_s(t)\) \(\implies F(s) = \displaystyle \frac{b}{s}\)

\[x(t) = \underbrace{\mathcal{L}^{-1} \left[ \frac{x(0)}{s+a} \right]}_{\text{Free}} + \underbrace{\mathcal{L}^{-1} \left[ \frac{b}{s} \frac{1}{s+a} \right]}_{\text{Forced}} \]

Free Response

Use the Laplace tables to find \[\mathcal{L}^{-1} \left[ \frac{x(0)}{s+a} \right]\]

\[ = x(0) e^{-at}\]

Forced Response

Use the Laplace tables and partial fractions to find \[\mathcal{L}^{-1} \left[ \frac{b}{s} \frac{1}{s+a} \right]\]

\[ = \frac{b}{a} \left( 1 - e^{-at} \right)\]

\[x(t) = \underbrace{x(0)e^{-at}}_{\text{Free}} + \underbrace{\frac{b}{a} - \frac{b}{a} e^{-at}}_{\text{Forced}}\]

\[ = x(0) e^{-at} + \frac{b}{a} - \frac{b}{a} e^{-at}\]

  • Which parts die out when \(t \rightarrow \infty\) and which parts remain?

\[x(t) = \underbrace{\frac{b}{a}}_{\text{Steady State}} + \underbrace{\left( x(0) - \frac{b}{a} \right)e^{-at }}_{\text{Transient}}\]

Calculating Free, Forced, Transient, Steady

Show that the transient and steady-state parts of the response of a system \[\dot{x} + ax = f(t)\] to a ramp input \(f(t) = ct\) is \[\underbrace{x(0) e^{-at} + \frac{c}{a^2} e^{-at}}_{\text{transient}} + \underbrace{\frac{c}{a} t - \frac{c}{a^2}}_{\text{steady state}}\]

Stability

  • Unstable: A system is unstable if its free response approaches \(\pm \infty\) as \(t \rightarrow \infty\)
  • Stable: A system is stable if its free response approaches \(0\) as \(t \rightarrow \infty\)
  • Neutrally Stable: A system is neutrally stable if its free response neither approaches \(\pm \infty\) nor approachs \(0\) as \(t \rightarrow \infty\).
  • Stability is a property of a system, and does not depend on what input is paplied to the system.

Stability and the roots of the characteristic polynomial

  1. Unstable behavior occurs when the root lies to the right of the imaginary axis.
  2. Neutrally stable behavior occurs when the root lies on the imaginary axis.
  3. The response oscillates only when the root has a nonzero imaginary part.
  4. The greater the imaginary part, the higher the frequency of the oscillation.
  5. The farther to the left the root lies, the faster the response decays.

The Transfer Function can represent a system’s equations

A linear system can be represented either as a set of differential equations or as a transfer function.

Differential Equation

\[\ddot{x} + 5 \dot{x} + 7 x = f(t)\]

Transfer Function

\[X(s) = \frac{1}{s^2 + 5s + 7} F(s)\]

  • Sometimes, systems are specified using just their transfer function.
  • Given a transfer fucntion \(T(s)\), it is assumed that we are referring to an equation of the form \(X(s) = T(s) F(s)\).

The Transfer Function is also the impulse response

For any linear system, its transfer function is also its impulse response.

  • What does this mean?
  • Why is this true?

Equivalent Representations

The following information is equivalent, and any one of hte following fully specifies a linear system.

Differential Equation \(\ddot{x} + 5 \dot{x} + 7 x = f(t)\)
Transfer Function \(\displaystyle \frac{1}{s^2 + 5s + 7}\)
Characteristic Polynomial \(s^2 + 5s + 7\)
Characteristic Equation \(s^2 + 5s + 7 = 0\)
Roots (of the characteristic polynomial) \(s = \displaystyle \frac{1}{2} \left( -5 \pm i \sqrt{3} \right)\)
Impulse Response \(\displaystyle \frac{2}{\sqrt{3}} e^{-5 t/2} \sin \left(\frac{\sqrt{3} t}{2}\right)\)

Transfer Functions as rational functions of \(s\)

Recall that a rational function of \(s\) is a ratio \[\frac{P(s)}{Q(s)}\]

where \(P(s)\) is a polynomial of degree \(m\) and \(Q(s)\) is a polynomial of degree \(n\).

In E12, we will always have \(m < n\)

  • Recall that any polynomial of degree \(n\) is fully defined by a set of \(n+1\) coefficients
  • In E12, our polynomials will always have real coefficients.
  • If \(P(s)\) has degree \(>0\), then the system is said to have numerator dynamics
    • Typically, this happens when the output depends on derivatives of the input

MATLAB Control System Toolbox

In MATLAB, it’s possible to specify a system from its transfer function using the tf function.

>> system1 = tf([2,1],[1,5,7])

system1 =
 
     2 s + 1
  -------------
  s^2 + 5 s + 7
 
Continuous-time transfer function.
Model Properties
  1. The first argument of tf is the coefficients of \(s\) in the numerator of the Transfer Function
  2. The second argument of tf is the coefficients of \(s\) in the denominator of the Transfer Function
  • Polynomial coefficients are given in order of decreasing powers of \(s\).

The tf function

sys = tf(num,den)

Creates a ‘transfer function’ object in MATLAB using the polynomial representation of a transfer function

The step function

sys1 = tf([1],[1,3])
step(sys1)

Imposes a unit step function on the system and returns the result as a plot and as an array.

The impulse function

sys1 = tf([1],[1,3])
impulse(sys1)

Imposes a unit step function on the system and returns the result as a plot and as an array.

The lsim function

Imposes an arbitrary function of time on the system and returns the result as a plot and as an array.

sys1 = tf([1],[1,3])
tvals = 0:0.1:3
f1 = @(t) 3 .* t    % ramp function
fvals = f1(tvals)
lsim(sys1,fvals,tvals)

Create some tf objects in MATLAB

  1. Create the system
  2. Examine its impulse response
  3. Examine its step response

For the following systems:

  • \(T(s) = \frac{1}{s+3}\)
  • \(T(s) = \frac{1}{2s+3}\)
  • \(T(s) = \frac{1}{s-3}\)
  • \(T(s) = \frac{1}{s^2+5s+7}\)