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7.1 Laplace Transform

The Laplace transform of \( f(t) \) is defined as:

\( e^{-st} \) is the kernel

\[ \mathcal{L} \{ f(t) \} = F(s) = \int_{0}^{\infty} f(t) e^{-st} dt \]

This is one example of integral transforms (another commonly used one is the Fourier transform).

Let's look at the transforms of simple functions

\[ \begin{aligned} \mathcal{L} \{ 1 \} = F(s) &= \int_{0}^{\infty} 1 \cdot e^{-st} dt \\ &= \lim_{a \to \infty} \int_{0}^{a} e^{-st} dt \quad \text{(treat } s \text{ as a "constant", } t \text{ is variable)} \\ &= \lim_{a \to \infty} \left. -\frac{1}{s} e^{-st} \right|_{t=0}^{t=a} \\ &= \lim_{a \to \infty} \left( -\frac{1}{s} e^{-sa} + \frac{1}{s} \right) \end{aligned} \]

\( -\frac{1}{s} e^{-sa} \) must go to 0 for integral to converge.

\( s > 0 \)

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\[ \mathcal{L} \{ 1 \} = \frac{1}{s}, \quad s > 0 \]
\[ \mathcal{L} \{ t \} = \int_{0}^{\infty} t e^{-st} dt = \lim_{a \to \infty} \int_{0}^{a} t e^{-st} dt \]
By parts:
\( u = t \quad dv = e^{-st} dt \)
\( du = dt \quad v = -\frac{1}{s} e^{-st} \)
\( uv - \int v du \)
\[ \begin{aligned} &= \lim_{a \to \infty} \left( \left. -\frac{t}{s} e^{-st} \right|_{0}^{a} + \int_{0}^{a} \frac{1}{s} e^{-st} dt \right) \\ &= \lim_{a \to \infty} \left( \left. -\frac{t}{s} e^{-st} \right|_{0}^{a} - \left. \frac{1}{s^2} e^{-st} \right|_{0}^{a} \right) \\ &= \lim_{a \to \infty} \left( -\frac{a}{s} e^{-sa} - \frac{1}{s^2} e^{-sa} + \frac{1}{s^2} \right) \end{aligned} \]

Terms with \( e^{-sa} \) must go to 0.

So, \( s > 0 \)

\[ \mathcal{L} \{ t \} = \frac{1}{s^2}, \quad s > 0 \]

Similarly, we can show that:

\[ \begin{aligned} \mathcal{L} \{ t^2 \} &= \frac{2}{s^3}, \quad s > 0 \\ \mathcal{L} \{ t^n \} &= \frac{n!}{s^{n+1}}, \quad s > 0 \end{aligned} \]
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Laplace Transform of Exponential Functions

We begin by calculating the Laplace transform of the exponential function \( e^{ct} \):

\[ \begin{aligned} \mathcal{L} \{ e^{ct} \} &= \int_{0}^{\infty} e^{ct} e^{-st} dt \\ &= \lim_{a \to \infty} \int_{0}^{a} e^{(c-s)t} dt \\ &= \lim_{a \to \infty} \left. \frac{1}{c-s} e^{(c-s)t} \right|_{0}^{a} = \lim_{a \to \infty} \left( \frac{1}{c-s} e^{(c-s)a} - \frac{1}{c-s} \right) \end{aligned} \]

Note: For the limit to exist, the term \( \frac{1}{c-s} e^{(c-s)a} \) must go to 0 as \( a \to \infty \). This requires \( c-s < 0 \), so \( s > c \).

\[ \mathcal{L} \{ e^{ct} \} = \frac{1}{s-c} , \quad s > c \]

Linearity of the Laplace Transform

The Laplace transform is linear because integration is linear.

\[ \begin{aligned} \mathcal{L} \{ f(t) + g(t) \} &= \int_{0}^{\infty} [f(t) + g(t)] e^{-st} dt \\ &= \int_{0}^{\infty} [f(t)e^{-st} + g(t)e^{-st}] dt \\ &= \int_{0}^{\infty} f(t)e^{-st} dt + \int_{0}^{\infty} g(t)e^{-st} dt \\ &= \mathcal{L} \{ f(t) \} + \mathcal{L} \{ g(t) \} \end{aligned} \]
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Linearity also applies to scalar multiplication:

\[ \begin{aligned} \mathcal{L} \{ c \cdot f(t) \} &= \int_{0}^{\infty} c \cdot f(t) e^{-st} dt \quad \text{(where } c \text{ is a constant)} \\ &= c \cdot \int_{0}^{\infty} f(t) e^{-st} dt = c \cdot \mathcal{L} \{ f(t) \} \end{aligned} \]

Example Calculation

Calculate the transform for a sum of functions:

\[ \begin{aligned} &\mathcal{L} \{ 3 + 5e^{-2t} - 10t^3 \} \\ &= \mathcal{L} \{ 3 \} + \mathcal{L} \{ 5e^{-2t} \} - \mathcal{L} \{ 10t^3 \} \\ &= 3 \mathcal{L} \{ 1 \} + 5 \mathcal{L} \{ e^{-2t} \} - 10 \mathcal{L} \{ t^3 \} \\ &= 3 \cdot \frac{1}{s} + 5 \cdot \frac{1}{s+2} - 10 \cdot \frac{3!}{s^4} , \quad s > 0 \end{aligned} \]

Existence of Laplace Transforms

Do all functions \( f(t) \) have Laplace transforms?

\[ \mathcal{L} \{ f(t) \} = \int_{0}^{\infty} f(t) e^{-st} dt \]
  • The existence depends on whether we can perform the integral.
  • The transform exists as long as the function is piecewise continuous.
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Laplace Transform of Piecewise Continuous Functions

(finite number of discontinuities)

\[ \int_{0}^{\infty} f(t) e^{-st} dt = \int_{0}^{a} \dots + \int_{a}^{b} \dots + \int_{b}^{c} \dots + \int_{c}^{\infty} \dots \]
Graph of a piecewise function f(t) with jump discontinuities at points a, b, and c on the t-axis.

for example,

\[ f(t) = \begin{cases} 2t + 1 & 0 \le t < 3 \\ e^t & 3 \le t < \infty \end{cases} \]
\[ F(s) = \int_{0}^{3} (2t + 1) e^{-st} dt + \int_{3}^{\infty} e^t e^{-st} dt \] \[ \vdots \] \[ = \frac{1}{s} - \frac{7e^{-3s}}{s} + \frac{2}{s^2} - \frac{2e^{-3s}}{s^2} + \frac{e^{-3(s-1)}}{s-1}, \quad s > 1 \]
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Laplace and Inverse Laplace Transforms

Laplace transform:

\[ \mathcal{L} \{f(t)\} = F(s) = \int_{0}^{\infty} f(t) e^{-st} dt \]

Inverse Laplace transform:

\[ \mathcal{L}^{-1} \{F(s)\} = f(t) = \frac{1}{2\pi i} \lim_{T \to \infty} \int_{\gamma - iT}^{\gamma + iT} e^{st} F(s) ds \]

usually not used in practice

usually, we do table look up for both transforms