Theory inside nested-shaper

Definitions of Convolution

\[ h(t) = \int_{-\infty}^{\infty}f(x)g(t-x)dx=f(t)*g(t) \tag{continuous domain} \]
\[ h[n] = (f*g)[n]=\sum_{i=0}^{K-1}f[k] \cdot g[n-k] \tag{discrete domain} \]

The Convolution Theorem

\[ h = f * g \iff H[m] = F[m] \cdot G[m] \iff H(s)= F(s)G(s) \]

Step and SMA function in frequency domain

Step function, defined by

\[ u(t)=\begin{cases} 0 & \text{if } t < 0 \\ A & \text{if } t \geq 0 \\ \end{cases} \to U(s)=\mathcal{L}\{u(t)\}=\frac{A}{s} \tag{1} \]

SMA function, defined by

\[ \text{SMA}_k(t)=\frac{1}{T_k}(H(t)-H(t-T_k)) \to \text{SMA}_k(s) = \mathcal{L}\{SMA_k(t)\} = \frac{1}{T_k}(\frac{1-e^{T_ks}}{s}) \tag{2} \]

Heaviside unit step, defined by

\[ H(t)=\begin{cases} 1 & \text{if } t \geq 0 \\ 0 & \text{if } t < 0 \\ \end{cases} \]

Laplace transforms used in this theory

\[ \mathcal{L}\{t^{n}\}=\frac{n!}{s^{n+1}} \tag{3} \]
\[ \mathcal{L}\{H(t-a)\}=\frac{e^{-as}}{s}\ (a \geq 0) \tag{4} \]
\[ \mathcal{L}\{f(t-a)H(t-a)\}=e^{-as}\mathcal{L}\{f(t)\} \tag{5} \]

Step function with nested SMA.

Based on the convolution theorem, we apply nested SMAs(total n) on step function. $$ R(s)=U(s)\text{SMA}_1(s)\text{SMA}_2(s)\cdots\text{SMA}_n(s) $$

\[ =\frac{A}{s^{n+1}}\frac{1}{T_1T_2\cdots T_n}(1-e^{T_1s})(1-e^{T_2s})\cdots (1-e^{T_ns})=\frac{A}{n!}\frac{n!}{s^{n+1}}\frac{1}{\prod^{n}_{k=1}T_{k}}\prod^{n}_{k=1}(1-e^{T_ks}) \]
\[ =\frac{A}{n!}\frac{n!}{s^{n+1}}\frac{1}{\prod^{n}_{k=1}T_{k}}\sum^{n}_{k=0}(-1)^k\sum_{\stackrel{I\subseteq\{1,2,\cdots,n\}}{|I|=k}}e^{-s\sum_{i\in I}T_i} \tag{6} \]

If SMAs have same amplitude,

\[ R(s)=U(s)\text{SMA}^n(s)=\frac{A}{n!}\frac{n!}{s^{n+1}}\frac{1}{T^n}\sum^n_{k=0}(-1)^k\binom{n}{k}e^{-kTs} \tag{7} \]

Apply inverse laplace transform on (7), in case of all SMAs have same amplitude, for simplicity.

\[ r(t)=\mathcal{L^{-1}}\{R(s)\}=\frac{A}{n!}\frac{1}{T^n}\sum^n_{k=0}(-1)^k\binom{n}{k}(t-kT)^nH(t-kT) \tag{8} \]

Let's investigate first Heaviside section.

\[ r(t)=\frac{A}{n!}\frac{1}{T^n}t^n\ (t \leq T) \tag{9} \]

And we can prove important characteristics about derivatives.

\[ \therefore \frac{d^k r(t)}{dt^k} = \frac{A}{(n-k)!T^n}t^{n-k}\ (\leq \frac{A}{T^k}) \tag{10} \]

(10) implies the derivative of results, is always smaller than initial amplitude / nested SMA's amplitude.

If we generalize the results in case of SMAs have different amplitude, and first sections,

\[ \frac{d^k r(t)}{dt^k} \leq \frac{A}{\prod_{n=1}^{k} T_{n}} \]

Here is an example, step input with 10.0,

applied with 5 filters,

SMA amplitude with 2.0s, 2.0s, 1.0s, 0.5s, 0.50

example

You can see that, (first section, see t < 2.0s)

Velocity <= 10.0 / 2.0 = 5.0

Acceleration <= 10.0 / (2.0*2.0) = 2.5

Jerk <= 10.0 / (2.0*2.0*1.0) = 2.5

snap <= 10.0 / (2.0*2.0*1.0*0.5) = 5.0

Implementation details about SMA

The mean over last N data-points calculated as

\[ y[n] = \frac{1}{N}\sum_{i=0}^{N-1}x[n-i] \tag{11} \]

Iterating all over the N data points is computationally expensive. However, averaging all data at each time will provide a numerically stable results in long time calculation. Other way to calculate SMA is (called as recursive/cumulative),

\[ y[n+1] = y[n] + \frac{1}{N}(x[n+1]-x[n-k+1]) \]

Calculating SMA by cumulative method suffers from numerical error on long time operation. In order to minimize numerical error, i implemented Kahan summation algorithm.

averaging angles.

What if samples are on the different algebra space? For example, we should use circular mean for angle.

\[ \bar{\alpha} = \text{atan2}(\frac{1}{n}\sum^n_{j=1} \text{sin} (\alpha_{j}), \sum^n_{j=1} \text{cos} (\alpha_{j})) \]