Lecture 5

E12 Linear Physical Systems Analysis

Prof. Emad Masroor

2026-02-03

What do we do when \(f(t)\) is more complicated?

  • For first-order systems, \[\dot{x} + a x = f(t).\]

  • If integrating \(f(t)\) is too difficult, we can …

    • give up, and integrate using ode45 or solve_ivp
    • or … ?

  • Numerical solutions mask deeper insight and do not give a full picture of the solution.

  • Need a new way of looking at problems …

  • The Frequency Domain

Laplace Transform

  • The Laplace Transform of a function transforms the function from the time domain to the frequency domain.
    • What is the domain of a function?
    • “The set of inputs accepted by the function”

\[\boxed{f(t) \rightarrow F(s)}\]

  • Lowercase letters: time domain functions
  • Uppercase letters: frequency domain functions
  • \(s\) is complex, actually (but don’t worry about it for now!)

Time domain and Frequency Domain

Figure 1: Schematic for Laplace Transform

Defining the Laplace Transform

  • The Laplace Transform of a function \(x(t)\) is defined as the following function of \(s\): \[\lim_{T \rightarrow \infty} \int_0^{T} x(t)e^{-st} dt \tag{1}\]
  • This is a function of \(s\), not a function of \(t\).
  • We give the expression in Equation 1 the name \(X(s)\) \[X(s) = \boxed{\lim_{T \rightarrow \infty} \int_0^{T} x(t)e^{-st} dt} = \int_0^{\infty} x(t)e^{-st} dt\] \[x(t) \rightarrow \text{Laplace Transform} \rightarrow X(s) \] \[x(t) \rightarrow \mathcal{L}[\cdot] \rightarrow X(s)\] \[\boxed{\mathcal{L}[x(t)] = X(s)}\]

Derive the Laplace Transform for a simple function \(x(t) = 3\)

If \(x(t) = 3\), a constant, what is \(\mathcal{L}[x(t)]\), i.e., what is \(X(s)\) ?

\[ \boxed{ \begin{aligned} \mathcal{L}[3] &= \int_0^{\infty} 3 e^{-st} dt \\ &= 3 \times \frac{1}{-s} \left. e^{-st} \right|_0^{\infty} \\ &= \frac{3}{-s} \left( e^{-\infty} - e^0 \right) \\ &= \frac{3}{s} \end{aligned} } \]

Derive the Laplace Transform for a simple function \(x(t) = e^{-at}\)

If \(x(t) = e^{-at}\), what is \(\mathcal{L}[x(t)]\), i.e., what is \(X(s)\) ?

\[ \boxed{ \begin{aligned} \mathcal{L}[e^{-at}] &= \lim_{T \rightarrow \infty}\int_0^T e^{-at} e^{-st} dt \\ &= \lim_{T \rightarrow \infty} \int_0^T e^{-(s+a)t} dt \\ &= \lim_{T \rightarrow \infty} \left[ \frac{-1}{s+a} e^{-(s+a)t} \right]_{t=0}^{t=T} \\ &= \frac{1}{s+a} \end{aligned} } \]

Table of Laplace Transforms

Properties of Laplace Transforms

The Inverse Laplace Transform

  • Time domain → Frequency domain: Laplace Transform
  • Time domain ← Frequency domain: ??
  • Inverse Laplace Transform

\[\boxed{F(s)\rightarrow f(t)}\]

  • We will not learn how to calculate the inverse transform directly. Instead:
  • Pattern matching!

Identifying the inverse Laplace transform

  • We have already learned that

\[ \begin{aligned} \mathcal{L}[3] &= \frac{3}{s} \\ \mathcal{L}[e^{-at}] &= \frac{1}{s+a} \end{aligned} \]

  • So the inverse Laplace transforms are

\[ \begin{aligned} \mathcal{L^{-1}}\left[\frac{3}{s}\right] &= 3 \\ \mathcal{L^{-1}}\left[\frac{1}{s+a}\right] &= e^{-at} \end{aligned} \]

Linearity Property of the Laplace Transform

\[ \boxed{ \begin{aligned} \mathcal{L}\left[ a f(t) + b g(t) \right] &= a \mathcal{L}\left[ f(t) \right] + b \mathcal{L} \left[ g(t) \right] \\ &= a F(s) + b G(s) \end{aligned} } \]

  • The Laplace Transform of the sum of two functions equals the sum of the Laplace Transform of those two functions.
  • The Laplace Transform of \(k \times\) a function equals \(k \times\) the Laplace Transform of that function.
  • The inverse Laplace Transform is similarly linear: \[ \boxed{ \begin{aligned} \mathcal{L^{-1}}\left[ a F(s) + b G(s) \right] &= a \mathcal{L^{-1}}\left[ F(s) \right] + b \mathcal{L^{-1}} \left[ G(s) \right] \\ &= a f(t) + b g(t) \end{aligned} } \]

Using the Linearity Property to find \(\mathcal{L}[x(t)]\)

Given \[x(t) = 6 + 4 e^{-3t},\] find \(\mathcal{L}[ x(t)]\)

a.k.a. \(X(s)\)

a.k.a. “Laplace transform of \(x(t)\)”.

\[ \boxed{ \begin{aligned} \mathcal{L}\left[ a f(t) + b g(t) \right] &= a \mathcal{L}\left[ f(t) \right] + b \mathcal{L} \left[ g(t) \right] \\ &= a F(s) + b G(s) \end{aligned} } \]

\[ \boxed{ \begin{aligned} \mathcal{L}\left[ 6+4e^{-3t} \right] &= \mathcal{L}[6] + \mathcal{L}\left[ 4e^{-3t}\right] \\ &= 2 \cdot \mathcal{L}[3] + 4 \cdot \mathcal{L}\left[ e^{-3t}\right] \\ &= 2 \cdot \underbrace{\frac{3}{s}}_{\mathcal{L}[3]} + 4 \cdot \underbrace{\frac{1}{s+3}}_{\mathcal{L}\left[ e^{-3t}\right]} = \frac{6(s+3) + 4s}{s(s+3)} \end{aligned} } \]

The Derivative Property of Laplace Transforms

  • Even without knowing exactly what \(x(t)\) is, we can transform first derivatives: \[\mathcal{L} \left[ \frac{dx}{dt} \right] = s \mathcal{L} \left[ x(t) \right] - x(0)\]
  • and second derivatives: \[\mathcal{L} \left[ \frac{d^2x}{dt^2} \right] = s^2 \mathcal{L} \left[ x(t) \right] - s x(0) - \dot{x}(0)\]

Derivation (without much nuance re: limits): \[ \boxed{ \begin{aligned} \mathcal{L}\left[ \frac{dx}{dt}\right] = \int_0^{\infty} \frac{dx}{dt} e^{-st} dt &= \left. x(t) e^{-st} \right|_0^{\infty} + s \underbrace{\int_0^{\infty} x(t) e^{-st} dt}_{\mathcal{L}[x(t)]} \\ &= \left(x(\infty) e^{-\infty} - x(0) e^{0} \right) + s X(s) \\ &= sX(s)-x(0) \end{aligned} } \] and repeated application of the above procedure gives us \(\displaystyle \mathcal{L} \left[ \frac{d^n x}{dt^n} \right]\)

Laplace Transform to solve 1st-order system

  • Consider the general first-order system \[ \dot{x} + a x = f(t)\]
  • Apply \(\mathcal{L}[\cdot]\) to both sides

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

  • Collect \(X(s)\) terms together and solve for \(X\):

\[ \begin{aligned} X(s) \cdot (s+a) &= F(s) + x(0) \\ X(s) &=\frac{F(s)}{s+a} + \frac{x(0)}{s+a} \end{aligned} \]

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

A familiar problem but with Laplace

\[RC \dot{v} + v = v_s \text{(const.)}, \quad v(0) = 0\]

Using the Laplace Transform to solve IVPs

Solve \[\dot{x} + 2x = 6+4 e^{-3t}, \quad x(0) = 0\] using Laplace Transforms:

  • first find \(X(s)\)
  • then use \(\mathcal{L}^{-1}\) to find \(x(t)\)

Two new functions: The Heaviside Step function

a.k.a. the “unit step function”:

\[u_s(t) = \begin{cases} 0 & {t < 0} \\ 1 & {t>0} \end{cases}\]

Two new functions: The Dirac Delta function

a.k.a. the “impulse”:

\[\delta(t) = \begin{cases} 0 & {t \neq 0} \\ \infty & {t = 0} \end{cases}\]

such that \(\displaystyle \int_{-\infty}^{+\infty} \delta(t) = 1\)

The Step function and the Impulse Function

\[\boxed{\frac{d}{dt} \left( u_s(t) \right) = \delta(t)}\]

Rational Functions & Partial Fractions Expansion

Definition

A rational function is any function that can be expressed as a fraction in which the numerator and denominator are both polynomials.

Express the following rational function as a sum of fractions:

\[\frac{10s+18}{s(s+3)^2} = \frac{A}{s} + \frac{B}{s+3} + \frac{C}{(s+3)^2}\]

Partial Fractions Expansion

  • For a given rational function \[\frac{p(s)}{q(s)} \] where \(p(s)\) is of lower or equal degree than \(q(s)\),

  • write the function as \[\frac{p(s)}{q(s)} = \frac{(s-p_1)(s-p_2)(...)(s-p_m)}{(s-q_1)(s-q_2)(s-q_3)... (s-q_n)}\]

  • where \(p_i\) is the \(i\) ’th root of \(p(s)\) and \(q_i\) is the \(i\) ’th root of \(q(s)\).

  • It is possible to find constants \(A\), \(B\), \(C\), … such that \[\frac{p(s)}{q(s)} = \frac{A}{s-q_1} + \frac{B}{s-q_2} + \frac{C}{s-q_3} + ...\]

  • If any roots are repeated, the corresponding term appears once for each time the root is repeated.

  • For example, if \(q(s) = s (s+3)^2,\) then the partial fraction expansion will be \[\frac{A}{s} + \frac{B}{s+3} + \frac{C}{(s+3)^2}\]

  • For example, if \(q(s) = s^2 (s+3),\) then the partial fraction expansion will be \[\frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+3} \]