Laplace Transform

The Laplace transform maps a function to a unary operator on complex numbers . A sufficient condition for to exist is that be measurable and be integrable.

A function is exponential type if there are constants such that for all .

Define the operations:

1. [differentiation]
2. [integration] .
3. [convolution]
4. [complex conjugation] .

Then .

The transform has several properties.

1. 
2. 
3. 
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5. 
6. 
7. 
8. 
9. is -periodic,
10. 
11. 

Linearity is a consequence of the linearity of the integral. Provided the derivative is of exponential type, the second result follows from

$$\begin{array}{ccc}L\left(Df\right)\left(s\right)& =& {\int }_0^{\infty }Df\left(t\right)e^{–st}dt\\ & =& {.f\left(t\right)e^{–st}|}_0^{\infty }–{\int }_0^{\infty }f\left(t\right)\left(–s{e}^{–st}\right)dt\\ & =& sL\left(f\right)\left(s\right)–f\left(0+\right)\end{array}$$

The third result is an inductive application of the second result:

$$\begin{array}{ccc}L\left(D^{n}f\right)\left(s\right)& =& sL\left(D^{n–1}f\right)\left(s\right)–\left(D^{n–1}f\right)\left(0+\right)\\ & =& s\left(sL\left(D^{n–2}f\right)\left(s\right)–\left(D^{n–2}f\right)\left(0+\right)\right)–\left(D^{n–1}f\right)\left(0+\right)\\ & =& s^{n}L\left(f\right)\left(s\right)–\sum _{j=0}^{n–1}{s}^j\left(D^{n–1–j}f\right)\left(0+\right)\end{array}$$

The fourth result uses the second result and applied to ,

$$\begin{array}{ccc}sL\left(If\right)\left(s\right)& =& \left(If\right)\left(0+\right)+L\left(DIf\right)\left(s\right)\\ L\left(If\right)\left(s\right)& =& s^{–1}L\left(f\right)\left(s\right)\end{array}$$

The fifth result is a consequence of aggegration of the exponential term:

$$\begin{array}{ccc}L\left(f\left(t\right)e^{at}\right)\left(s\right)& =& {\int }_0^{\infty }f\left(t\right)e^{at}{e}^{–st}dt\\ & =& {\int }_0^{\infty }f\left(t\right)e^{–\left(s–a\right)t}dt\\ & =& L\left(f\right)\left(s–a\right)\end{array}$$The sixth result is a consequence of rescaling

$$\begin{array}{ccc}L\left(f\left(at\right)\right)\left(s\right)& =& {\int }_0^{\infty }f\left(at\right)e^{–st}dt\\ & =& {\int }_0^{\infty }f\left(u\right)e^{–s\left(u/a\right)}d\left(u/a\right)\\ & =& a^{–1}{\int }_0^{\infty }f\left(u\right)e^{–\left(s/a\right)u}du\\ & =& a^{–1}L\left(f\right)\left(a^{–1}s\right)\end{array}$$

The seventh result is an application of Fubini’s theorem

$$\begin{array}{ccc}L\left(f★g\right)\left(s\right)& =& {\int }_0^{\infty }[{\int }_0^{t}f\left(u\right)g\left(t–u\right)du]e^{–st}dt\\ & =& {\int }_0^{\infty }{\int }_0^{\infty }f\left(u\right)g\left(t–u\right)e^{–s\left(t–u+u\right)}I\left(u<t\right)dudt\\ & =& {\int }_0^{\infty }{\int }_0^{\infty }f\left(u\right)e^{–su}g\left(t–u\right)e^{–s\left(t–u\right)}I\left(0<t–u\right)dudt\\ & =& {\int }_0^{\infty }f\left(u\right)e^{–su}du{\int }_0^{\infty }g\left(w\right)e^{–sw}dw\\ & =& L\left(f\right)\left(s\right)\cdot L\left(g\right)\left(s\right)\end{array}$$

The eigth result

$$\begin{array}{ccc}L\left(Cf\right)\left(s\right)& =& {\int }_0^{\infty }Cf\left(t\right)e^{–st}dt\\ & =& C[{\int }_0^{\infty }f\left(t\right)C\left(e^{–st}\right)dt]\\ & =& C[{\int }_0^{\infty }f\left(t\right)e^{–C\left(s\right)t}dt]\\ & =& C\left(L\left(f\right)\left(C\left(s\right)\right)\right)\end{array}$$The ninth result

$$\begin{array}{ccc}L\left(f\right)\left(s\right)& =& {\int }_0^{\infty }f\left(t\right)e^{–st}dt\\ & =& \sum _{j=0}^{\infty }{\int }_{jT}^{\left(j+1\right)T}f\left(t\right)e^{–st}dt\\ & =& \sum _{j=0}^{\infty }{\int }_0^{T}f\left(u\right)e^{–s\left(u–jT\right)}du\\ & =& \sum _{j=0}^{\infty }{\left(e^{–sT}\right)}^j{\int }_0^T{e}^{–st}f\left(t\right)dt\\ & =& \frac{1}{1–e^{–sT}}{\int }_0^T{e}^{–st}f\left(t\right)dt\end{array}$$

The tenth result

$$\begin{array}{ccc}L\left(tf\left(t\right)\right)\left(s\right)& =& {\int }_0^{\infty }tf\left(t\right)e^{–st}dt\\ & =& {\int }_0^{\infty }f\left(t\right)\left(–\frac{\partial }{\partial s}\right)e^{–st}dt\\ & =& \left(–\frac{\partial }{\partial s}\right){\int }_0^{\infty }f\left(t\right)e^{–st}dt\end{array}$$

The inverse transform is defined as

such that the constant is chosen to ensure the line integral is within the region of convergence of with the limit interpreted in the weak-* topology (to be examined at a future date).

The eleventh result

$$\begin{array}{ccc}L\left(f\left(t\right)g\left(t\right)\right)\left(s\right)& =& {\int }_0^{\infty }f\left(t\right)g\left(t\right)e^{–st}dt\\ & =& {\int }_0^{\infty }\left[\frac{1}{2\pi i}\underset{T\to \infty }{lim}{\int }_{c–iT}^{c+iT}F\left(u\right)e^{ut}du\right]g\left(t\right)e^{–st}dt\\ & =& \frac{1}{2\pi i}\underset{T\to \infty }{lim}{\int }_{c–iT}^{c+iT}F\left(u\right)[{\int }_0^{\infty }g\left(t\right)e^{–\left(s–u\right)t}dt]du\\ & =& \frac{1}{2\pi i}\underset{T\to \infty }{lim}{\int }_{c–iT}^{c+iT}F\left(u\right)G\left(s–u\right)du\end{array}$$

Laplace Transforms of Common Signals

Define the delay operator .

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