If there are any errors or clarifications needed, please email me at
http://www.reading.ac.uk/~shshawin/
$$ F(s)=\int_0^\infty f(t) e^{-st}dt $$
$$ f(t)={1 \over 2\pi i}\int_{c-i\infty}^{c+i\infty} e^{st} F(s)ds $$
Residue Theorem: If $f(z)$ is analytic within and on $C$ except for a finite number of poles $$ \int_C f(z) dz = 2\pi i S $$ Where $S$ is the sum of the residues at the poles within $C$
Residue $$ a_{-1}=\lim_{ z\to a }{1 \over (n-1)!} {d^{n-1}\over dz^{n-1}}\bigl((z-a)^n f(z)\bigr)$$
Final value Theorem: $$ f(\infty)=\lim_{s\to 0} sF(s) $$ Initial value Theorem: $$ f(0^+)=\lim_{s\to \infty} sF(s) $$
| $\displaystyle f(t)$ | $\displaystyle F(s)$ |
|---|---|
| unit impulse $\displaystyle\delta(t)$ | 1 |
| unit impulse $\displaystyle\delta(t-t_0)$ | $\displaystyle e^{-st_0}$ |
| Unit step u(t) | $\displaystyle{1\over s}$ |
| $u(t_0-t)$ | $\displaystyle{{e^{st_0}\over s}}$ |
| ramp of $kt$ | $\displaystyle{{k\over s^2}}$ |
| ${df(t)\over dt}$ | $sF(s)-f(0)$ |
| $e^{-at}$ | ${1 \over s+a}$ |
| $e^{-t/T}$ | ${T \over sT+1}$ |
| $f(at)$ | $ {1\over a}F({s\over a})$ |
| $f*g$ | $ FG$ |
| $A\sin\omega t$ | $ {A\omega\over s^2+\omega^2}$ |
| $A\cos\omega t$ | $ {As\over s^2+\omega^2}$ |
| $\displaystyle \frac{{e}^{-a\,t}}{b-a}-\frac{{e}^{-b\,t}}{b-a} $ | $\displaystyle \frac{1}{\left( a+s\right) \,\left( b+s\right) }$ |
| $\displaystyle 1-{e}^{-a\,t}$ | $\displaystyle \frac{a}{s\,\left( a+s\right) }$ |
| Cosine arch $\displaystyle 1-\cos(2\pi t/T)$ | $\displaystyle \frac{1}{s}- \frac{e^{-sT}}{s} -\frac{s }{ s^2+\omega^2} + \frac{s e^{-sT} }{ s^2+\omega^2}$ |
| sinc function $\displaystyle \frac{\mathrm{sin}\left( t\right) }{t}$ | $\displaystyle \frac\pi2-\mathrm{atan}\left( s\right)$ |
| $\displaystyle \mathrm{e}^{-\beta\,t}\,\frac{1}{\sqrt{\beta^2-k}} \sinh\left(t\,\sqrt{\beta^2-k}\right)$ | $\displaystyle \frac1{(s^2+2\beta s+k)}$ |
| $\displaystyle \frac{1}{k}\left(1-{\mathrm{e}}^{-\beta\,t}\,\left(\cosh\left(t\,\sqrt{\beta^2-k}\right)+\frac{\beta}{\sqrt{\beta^2-k}} \sinh\left(t\,\sqrt{\beta^2-k}\right) \right)\right)$ | $\displaystyle \frac1{s(s^2+2\beta s+k)}$ |
| $\displaystyle \frac{{e}^{-\omega\,t\,\zeta}\,\mathrm{sin}\left( \omega\,t\,\sqrt{1-{\zeta}^{2}}\right) }{\sqrt{1-{\zeta}^{2}}}$ | $\displaystyle \frac{\omega}{{\omega}^{2}+2\,\zeta\,\omega\,s+{s}^{2}} $ |
| $\displaystyle 1- {e}^{-\omega\,t\,\zeta}\,\left(\frac{\zeta\,\mathrm{sin}\left( \omega\,t\,\sqrt{1-{\zeta}^{2}}\right) }{\sqrt{1-{\zeta}^{2}}}+\mathrm{cos}\left( \omega\,t\,\sqrt{1-{\zeta}^{2}}\right) \right)$ | $\displaystyle \frac{{\omega}^{2}}{s\,\left( {\omega}^{2}+2\,\zeta\,\omega\,s+{s}^{2}\right) }$ |
For the second order systems $\omega$ is the natural frequency $\omega=\sqrt{k/m}$ and $\zeta$ is the damping where $2\zeta\omega={b/m}$. $2\beta=2\zeta\omega=b$
if $x(t)$ then Laplace transform of $\frac{dx}{dt}$ is \[ L(\frac{dx}{dt})=\int_0^\infty \frac{dx}{dt} e^{-st}dt=sx(s)+x(0) \]
By recursion it is then possible to compute the Laplace transform of $\frac{d^2x}{dt^2}$
$$ F(z)=\sum_{n=-\infty}^\infty f[n] z^{-n} $$ where $z=e^{j\omega t}$
Shifting Theorem. if $f[n] \leftrightarrow F(z)$
then $f[n-m] \leftrightarrow z^{-m}F(z)$
Convolution Theorem. if $f_1[n] \leftrightarrow F_1(z)$ and $f_2[n] \leftrightarrow F_2(z)$ then $f_1 \otimes f_2 \leftrightarrow F_1F_2$
Final value Theorem: $$ f[\infty]=\lim_{z\to 1} (z-1)F(z) $$ Initial value Theorem: $$ f[0]=\lim_{z\to \infty} F(z) $$