time | Laplace | |
impulse | \delta \left ( t\right ) | 1 |
delayed impulse | \delta \left ( t-a\right ) | e^{-as} |
unit step | u\left ( t\right ) | \frac{1}{s} |
delayed unit impulse | u\left ( t-a\right ) | \frac{1}{s}e^{-as} |
ramp | t | \frac{1}{s^{2}} |
parabolic | t^{2} | \frac{2}{s^{3}} |
tf\left ( t\right ) | -F^{\prime }\left ( s\right ) | |
scaled | f\left ( at\right ) | \frac{1}{a}F\left ( \frac{s}{a}\right ) |
e^{at}f\left ( t\right ) | F\left ( s-a\right ) | |
f\left ( t\right ) | F\left ( s\right ) | |
derivative | f^{\prime }\left ( t\right ) | sF\left ( s\right ) -f\left ( 0\right ) |
second derivative | f^{\prime \prime }\left ( t\right ) | s^{2}F\left ( s\right ) -sf\left ( 0\right ) -f^{\prime }\left ( 0\right ) |
integrator | \int _{0}^{t}f\left ( \tau \right ) d\tau | \frac{1}{s}F\left ( s\right ) |
delay in time | f\left ( t-a\right ) u\left ( t-a\right ) | e^{-as}F\left ( s\right ) |
convolution | f\left ( t\right ) \circledast g\left ( t\right ) | F\left ( s\right ) G\left ( s\right ) |
e^{-at}u\left ( t\right ) | \frac{1}{s+a} | |
e^{-a\left \vert t\right \vert } | \frac{2a}{a^{2}-s^{2}} | |
\left ( 1-e^{-at}\right ) u\left ( t\right ) | \frac{a}{s\left ( s+a\right ) } | |
\sin \left ( at\right ) u\left ( t\right ) | \frac{a}{s^{2}+a^{2}} | |
\cos \left ( at\right ) u\left ( t\right ) | \frac{s}{s^{2}+a^{2}} | |
e^{-bt}\sin \left ( at\right ) u\left ( t\right ) | \frac{a}{\left ( s+b\right ) ^{2}+a^{2}} | |
e^{-bt}\cos \left ( at\right ) u\left ( t\right ) | \frac{s+b}{\left ( s+b\right ) ^{2}+a^{2}} | |
For B,C, expand now and compare coefficients (but there should be faster way)
Suppose F\left ( s\right ) =\frac{N\left ( s\right ) }{D\left ( s\right ) } is stable, then \lim _{t\rightarrow \infty }f\left ( t\right ) =\lim _{s\rightarrow 0}sF\left ( s\right )
F\left ( s\right ) is allowed to have only one pole at origin and still use FVT. But if F\left ( s\right ) has more than one pole at the origin, or unstable, we can’t use FVT to determine \lim _{t\rightarrow \infty }f\left ( t\right ) .
\frac{E\left ( s\right ) }{R\left ( s\right ) }=\frac{1}{1+GH}. To have \lim _{t\rightarrow \infty }e\left ( t\right ) =0 when the input R\left ( s\right ) is step, we need to have integrator in G\left ( s\right ) H\left ( s\right ) . Since we want GH to be very large for s=0. And integrator is \frac{1}{s}. If the input is ramp t, then we need \frac{1}{s^{2}} in GH. If the input is t^{2} then we need \frac{1}{s^{3}} in the controller. and so on.
G\left ( s\right ) =\frac{\omega _{n}^{2}}{s^{2}+2\zeta \omega _{n}s+\omega _{n}^{2}}, y_{step}\left ( t\right ) =1-\frac{e^{-\zeta \omega _{n}t}}{\sqrt{1-\zeta ^{2}}}\left ( \sin \omega _{n}\sqrt{1-\zeta ^{2}}t+\phi \right ) where \phi =\cos ^{-1}\zeta . Maximum overshoot is e^{\frac{-\pi \zeta }{\sqrt{1-\zeta ^{2}}}}. Resonance frequency \omega _{r}=\omega _{n}\sqrt{1-2\zeta ^{2}}, and \left \vert G\left ( \omega _{r}\right ) \right \vert =\frac{1}{2\zeta \sqrt{1-\zeta ^{2}}} l.3981 — TeX4ht warning — “SaveEverypar’s: 2 at “begindocument and 3 “enddocument —