# Dragon Notes

UNDER CONSTRUCTION
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# Laplace Transform

$$\large{}\boxed{\bn{X}(\bn{s})=\mathcal{L}[x(t)]=\int_{0^-}^{\infty}x(t)e^{-\bn{s}t}dt}$$
$$\vpL^{\vpl}\large{}\boxed{x(t)=\mathcal{L}^{-1}[\bn{X}(\bn{s})]=\frac{1}{j2\pi}\int_{\sigma-j\infty}^{\sigma+j\infty}\bn{X}(\bn{s})e^{\bn{s}t}d\bn{s}}$$
$$\vpL \lrg{} \bn{s}=\sigma+j\omega ,\ \ x(t)u(t)\ \t{is implicit,}\ \ \sigma>\sigma_c$$

Unilateral Laplace Transform, applied to causal signals.
$$\vpl$$The $$0^-$$ in the lower limit of integration serves to account for initial conditions.
Region of Convergence: $$\vpl$$$$\sigma>\sigma_c$$, where $$\sigma_c$$ is the smallest value of $$\sigma$$ for which the transform with an absolute value integrand converges.
$\int_{0^-}^{\infty}|x(t)e^{-\bn{s}t}|dt=\int_{0^-}^{\infty}|x(t)||e^{-\sigma t}||e^{-j\omega t}|dt \rightarrow \int_{0^-}^{\infty}|x(t)|e^{-\sigma t}dt<\infty$

### Transform Properties

 Property $$x(t)$$ $$\bn{X}(\bn{s})=\mathcal{L} [x(t)]$$ $$[1]\ \t{Constant multiplication}$$ $$\ds Kx(t)$$ ⬌ $$\ds K\bn{X}(\bn{s})$$ $$[2]\ \t{Linearity}$$ $$\ds K_1x_1(t)+K_2x_2(t)$$ ⬌ $$\ds K_1\bn{X}_1(\bn{s})+K_2\bn{X}_2(\bn{s})$$ $$[3]\ \t{Time scaling}$$ $$\ds x(at),\quad a>0$$ ⬌ $$\ds \frac{1}{a}\bn{X}\lrpar{\frac{\bn{s}}{a}}$$ $$[4]\ \t{Time shift}$$ $$\ds x(t-T)u(t-T)$$ ⬌ $$\ds e^{-T\bn{s}}\nblr{X}{s}$$ $$[5]\ \t{Frequency shift}$$ $$\ds e^{-\bn{a}t}x(t)$$ ⬌ $$\ds \nblr{X}{s+a}$$ $$[6]\ \t{Time 1st-derivative}$$ $$\ds x'=\frac{dx}{dt}$$ ⬌ $$\ds \bn{s} \nblr{X}{s}-x(0^-)$$ $$[7]\ \t{Time 2nd-derivative}$$ $$\ds x''=\frac{d^2x}{dt^2}$$ ⬌ $$\ds \bn{s}^2\nblr{X}{s}-\bn{s}x(0^-)$$ $$[8]\ \t{Time integral}$$ $$\ds \int_{0^-}^t x(t')dt'$$ ⬌ $$\ds \frac{1}{\bn{s}}\nblr{X}{s}$$ $$[9]\ \t{Frequency derivative}$$ $$\ds tx(t)$$ ⬌ $$\ds -\frac{d}{d\bn{s}} \nblr{X}{s}=-\nblr{X'}{s}$$ $$[10]\ \t{Frequency integral}$$ $$\ds \frac{x(t)}{t}$$ ⬌ $$\ds \int_{\bn{s}}^{\infty}\bn{X}(\bn{s'})d\bn{s'}$$ $$[11]\ \t{Initial value}$$ $$\ds \vphantom{\lmt{s}{\infty}}x(0^+)$$ $$\vphantom{\lmt{s}{\infty}}\bf{=}\hspace{1px}$$ $$\ds \lmt{\bn{s}}{\infty}\bn{s} \nblr{X}{s}$$ $$[12]\ \t{Final value}$$ $$\ds \lmt{t}{\infty}x(t)=x(\infty)$$ $$\vphantom{\lmt{s}{\infty}}\bf{=}\hspace{1px}$$ $$\ds \lmt{\bn{s}}{0}\bn{s}\nblr{X}{s}$$ $$[13]\ \t{Convolution}$$ $$\ds x_1(t)*x_2(t)$$ ⬌ $$\ds \bn{X}_1(\bn{s})\bn{X}_2(\bn{s})$$

Dragon Notes,   Est. 2018     About