doc/Rosenfalck.md

In ime-luebeck/semgsim: Simulating surface electromyographic (sEMG) and force signals

The original equation due to Rosenfalck, eq (16) in Farina et al.: $$V(z [mm]) = \begin{cases} A z^3 e^{-z} + B & \text{if } z \gt 0 \ 0 & \text{else} \end{cases} $$

The same function, taking a variable in [m]: $$V(z [m]) = \begin{cases} A (1000z)^3 e^{-1000z} + B & \text{if } z \gt 0 \ 0 & \text{else} \end{cases} $$

Its first derivative in the negative direction $$ \psi(z) = \frac{d}{dz} V(-z)$$ as defined by Farina et al.: $$ \psi(z [mm]) = -A e^{z} \cdot [3 z^2 + z^3] \cdot H(-z) $$ where $H(z)$ denotes the Heaviside step function.

The same function but taking an argument in [m]: $$ \psi(z [m]) = \frac{d}{dz} [-A z^3 10^9 e^{1000z} + B] \quad \text{if } z \lt 0 \text{ else } 0 \ = -3A z^2 10^9 e^{1000z} -A z^3 10^{12} e^{1000z} \quad \text{if } z \lt 0 \text{ else } 0 \ = -A e^{1000z} \cdot [3 z^2 10^9 + z^3 10^{12}] \cdot H(-z) $$

The Wolfram Alpha query integral from -inf to inf of ((- A e^(1000t) * (3*t^2*10^9 + t^3*10^12) * Heaviside(-t))*e^(-j omega t)) [^1] yields $$ \mathcal{F}{\psi(z [m])}(f) = \frac{6 \cdot 10^9 A j \omega}{(j\omega- 1000)^4} \ = \frac{12j \cdot 10^9 A \pi f}{(2 j \pi f- 1000)^4} $$ which is the quantity called $\Psi$ in the Farina paper.

Let's try to derive the same thing analytically.

We know that (for a non-unitary Fourier transform, cf. Fourier-transforms.Rmd) $$ x^n f(x) \mapsto i^n \frac{d^n \hat{f}(\omega)}{d\omega^n}. $$

Our expression for $\psi$ above can be decomposed as $$ \psi(z[m]) = -A e^{1000z} \cdot [3 z^2 10^9 + z^3 10^{12}] \cdot H(-z) \ = - 3A z^2 e^{1000z} 10^9 H(-z) - A z^3 e^{1000z} 10^{12} H(-z). $$

...

[^1]: Notice that that query is not the same as simply calling fourier transform of (- A e^(1000t) * (3*t^2*10^9 + t^3*10^12) * Heaviside(-t), because that will use another type of Fourier transform! (Probably a unitary one?) The difference should be a factor of $\sqrt{2 \pi}$.

ime-luebeck/semgsim documentation built on April 14, 2022, 11:02 p.m.