I have a table of laplace transforms called New Laplace Transform Table produced by dartmouth.edu and I'm having difficulty with the following section: -
Both the above (27 and 27a) appear to be useful when applying a step to a 2nd order under-damped low pass filter yet, if I try and rationalize them I find an anomaly (most probably in my math).
For instance, in "27" I can equate $a=\zeta$ and I get: -
$$\dfrac{1}{s}\cdot \dfrac{1}{s^2 + 2\zeta s+ \zeta^2+b^2}$$
I then make $\zeta^2 + b^2 = 1$ thus, $b = \sqrt{1-\zeta^2}$
That leaves me with the inverse solution of: -
$$\dfrac{1}{\zeta^2+1-\zeta^2}+\dfrac{1}{\sqrt{1-\zeta^2}\cdot(\zeta^2+1-\zeta^2)}\cdot e^{-\zeta t}\cdot\sin(t\sqrt{1-\zeta^2}-\phi)$$
$$=1+\dfrac{1}{\sqrt{1-\zeta^2}}\cdot e^{-\zeta t}\cdot\sin(t\sqrt{1-\zeta^2}-\phi)$$
If I use "27a" and equate $\omega_n$ to 1 (for reasons of normalization), I get: -
$$=1-\dfrac{1}{\sqrt{1-\zeta^2}}\cdot e^{-\zeta t}\cdot\sin(t\sqrt{1-\zeta^2}+\phi)$$
Notice that the sign has changed for the equation following "1" and also the sign has changed for $\phi$. I'm aware that the solution for $\phi$ appears to be different for both examples in that "27" uses
arctan($\dfrac{\sqrt{1-\zeta^2}}{\zeta}$) and
"27a" uses arccos($\zeta$) but these are mathematically the same as far as I can tell.
I also wrote some java code to try-out both solutions and they give similar results (certainly within 1 %) but noticeably not exactly the same. I also used an elecrical engineering sim tool to model the step-response of the filter and got results that are the same as "27a" implemented in java.
Question: Is it me that is wrong in my analysis/understanding or is the table (27) actually wrong?
Both formulas are fine. We have $$\phi_1 = \arctan(\sqrt {1 - \zeta^2}, -\zeta), \\ \phi_2 = \arccos \zeta, \\ \phi_1 = \pi - \phi_2, \\ \sin(A - \phi_1) = -\sin(A + \phi_2).$$ $\arctan(y, x)$ is the value of $\arg (x+i y)$ in $(-\pi, \pi]$, not the same as $\arctan(y/x)$.