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7.1 Laplace Transform

the definition of Laplace transform of \(f(t)\) is

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

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

Laplace transform is one type of integral transform

(another commonly used transform is the Fourier transform)

let's look at the transforms of some simple functions

\(f(t) = 1\)

\[ \mathcal{L}\{1\} = F(s) = \int_{0}^{\infty} 1 \cdot e^{-st} dt = \lim_{a \to \infty} \int_{0}^{a} 1 \cdot e^{-st} dt \]

Note: \(s\) is "constant"

Note: \(t\) is variable

\[ = \lim_{a \to \infty} -\frac{1}{s} e^{-st} \Big|_{t=0}^{t=a} \]
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\[ = \lim_{a \to \infty} \left( -\frac{1}{s} e^{-sa} + \frac{1}{s} \right) \]

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

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

\(\mathcal{L}\{t\} = F(s) = \int_{0}^{\infty} t \cdot e^{-st} dt = \lim_{a \to \infty} \int_{0}^{a} t e^{-st} dt \)

by parts:

  • \(u = t\)
  • \(du = dt\)
  • \(dv = e^{-st} dt\)
  • \(v = -\frac{1}{s} e^{-st}\)

\(uv - \int v du\)

\[ = \lim_{a \to \infty} \left( -\frac{t}{s} e^{-st} \Big|_{0}^{a} + \int_{0}^{a} \frac{1}{s} e^{-st} dt \right) \]
\[ = \lim_{a \to \infty} \left( -\frac{t}{s} e^{-st} \Big|_{0}^{a} - \frac{1}{s^2} e^{-st} \Big|_{0}^{a} \right) \]
\[ = \lim_{a \to \infty} \left( -\frac{a}{s} e^{-sa} - \frac{1}{s^2} e^{-sa} + \frac{1}{s^2} \right) \]

\(-\frac{a}{s} e^{-sa}\) and \(-\frac{1}{s^2} e^{-sa}\) must go to zero for integral to converge, so \(s > 0\)

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

Using the same process, we find \[ \mathcal{L}\{t^2\} = \frac{2}{s^3} \]

\[ \begin{aligned} \mathcal{L}\{t^3\} &= \int_{0}^{\infty} t^3 e^{-st} dt \\ &= \lim_{a \to \infty} \int_{0}^{a} t^3 e^{-st} dt \end{aligned} \]
Integration by parts:
\[ \begin{aligned} u &= t^3 & dv &= e^{-st} dt \\ du &= 3t^2 dt & v &= -\frac{1}{s} e^{-st} \end{aligned} \]
\[ \begin{aligned} &= \lim_{a \to \infty} \left( \left. -\frac{t^3}{s} e^{-st} \right|_0^a + \int_{0}^{a} \frac{3t^2}{s} e^{-st} dt \right) \\ &\quad \text{Note: } \left. -\frac{t^3}{s} e^{-st} \right|_0^a \to 0 \text{ if } s > 0 \\ &= \frac{3}{s} \int_{0}^{\infty} t^2 e^{-st} dt = \frac{3}{s} \cdot \frac{2}{s^3} = \frac{3 \cdot 2}{s^4} \end{aligned} \]

Note: The integral above is \( \mathcal{L}\{t^2\} \).

Repeat and eventually we get

\[ \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, \quad s > 0 \]
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\[ \begin{aligned} \mathcal{L}\{e^t\} &= \lim_{a \to \infty} \int_{0}^{a} e^t e^{-st} dt \\ &= \lim_{a \to \infty} \int_{0}^{a} e^{(1-s)t} dt \\ &= \lim_{a \to \infty} \left. \frac{1}{1-s} e^{(1-s)t} \right|_0^a \\ &= \lim_{a \to \infty} \frac{1}{1-s} e^{(1-s)a} - \frac{1}{1-s} = \frac{1}{s-1} \end{aligned} \]

Condition for convergence:

\( \frac{1}{1-s} e^{(1-s)a} \) must go to 0

\( 1 - s < 0 \)

\( s > 1 \)

\[ \mathcal{L}\{e^{at}\} = \frac{1}{s-a} \iff s > a \]

Linearity of the Laplace Transform

Laplace transform is linear because integration is linear.

\[ \begin{aligned} \mathcal{L}\{f(t) + g(t)\} &= \int_{0}^{\infty} [f(t) + g(t)] e^{-st} dt \\ &= \int_{0}^{\infty} f(t) e^{-st} dt + \int_{0}^{\infty} g(t) e^{-st} dt \end{aligned} \]
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Properties of the Laplace Transform

Linearity Properties

\[ \mathcal{L} \{ f(t) + g(t) \} = \mathcal{L} \{ f(t) \} + \mathcal{L} \{ g(t) \} \]\[ \mathcal{L} \{ c f(t) \} = c \cdot \mathcal{L} \{ f(t) \} \]

Example

For example,

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

Existence of the Laplace Transform

Any \( f(t) \) that is piecewise continuous has a Laplace transform.

Piecewise continuous: finite number of discontinuities on \( 0 < t < \infty \)

Because:

\[ \int_0^\infty f(t) e^{-st} dt = \int_0^a \dots + \int_a^b \dots + \int_b^c \dots + \int_c^\infty \dots \]
Graph of a piecewise continuous function f(t) with jump discontinuities at points a, b, and c on the t-axis.
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Example: Piecewise Function

For example,

\[ f(t) = \begin{cases} 2t+1 & 0 \le t < 3 \\ e^t & 3 \le t < \infty \end{cases} \]
\[ \begin{aligned} F(s) &= \int_0^\infty f(t) e^{-st} dt \\ &= \int_0^3 (2t+1) e^{-st} dt + \int_3^\infty e^t e^{-st} dt \\ &\vdots \\ &= \frac{1}{s} - \frac{7e^{-3s}}{s} + \frac{2}{s^2} - \frac{2e^{-3s}}{s^2} + \frac{e^{-3(s-1)}}{s-1}, \quad s > 1 \end{aligned} \]

Definitions

Laplace transform:

\[ \mathcal{L} \{ f(t) \} = F(s) = \int_0^\infty f(t) e^{-st} dt \]

Inverse Laplace transform:

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

not used in practice

usually we use a table of transforms