Linear Distortions of Periodic Signals

Open Applet in new Tab

Applet description

This applet illustrates the effects of linear distortions (attenuation distortions and phase distortions) with

Meanings of the signals used

• the input signal $x(t)$   ⇒   power $P_x$:

$$x(t) = x_1(t) + x_2(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)+A_2\cdot \cos\left(2\pi f_2\cdot t- \varphi_2\right),$$

• the output signal $y(t)$   ⇒   power $P_y$:

$$y(t) = \alpha_1 \cdot x_1(t-\tau_1) + \alpha_2 \cdot x_2(t-\tau_2),$$

• the matching output signal $z(t)$   ⇒   power $P_z$:

$$z(t) = k_{\rm M} \cdot y(t-\tau_{\rm M}) + \alpha_2 \cdot x_2(t-\tau_2),$$

• the difference signal   $\varepsilon(t) = z(t) - x(t)$   ⇒   power $P_\varepsilon$.

The next block in the above model is Matching: The output signal $y(t)$ is adjusted in amplitude and phase with uniform quantities $k_{\rm M}$ and $\tau_{\rm M}$ for all frequencies which means that this is not a frequency-dependent distortion. Using the signal $z(t)$, a differentiation can be made between:

• attenuation distortion and frequency–independant attenuation, as well as
• phase distortion and pure frequency–independant delay.

The Distortion Power $P_{\rm D}$ is used to measure the strength of the linear distortion and is defined as:

$$P_{\rm D} = \min_{k_{\rm M}, \ \tau_{\rm M}} P_\varepsilon.$$

Theoretical background

Distortions refer to generally unwanted alterations of a message signal through a transmission system. Together with the strong stochastic effects (noise, crosstalk, etc.), they are a crucial limitation for the quality and rate of transmission.

Just as the intensity of noise can be assessed through

• the Noise Power $P_{\rm N}$ and
• the Signal–to–Noise Ratio (SNR) $\rho_{\rm N}$,

distortions can be quantified through

• the Distortion Power $P_{\rm D}$ and
• the Signal–to–Distortion Ratio (SDR)

$$\rho_{\rm D}=\frac{\rm Signal \ Power}{\rm Distortion \ Power} = \frac{P_x}{P_{\rm D} }.$$

Linear and nonlinear distortions

A distinction is made between linear and nonlinear distortions:

• Nonlinear distortions occur, if at all times $t$ the nonlinear correlation $y = g(x) \ne {\rm const.} \cdot x$ exists between the signal values $x = x(t)$ at the input and $y = y(t)$ at the output, whereby $y = g(x)$ is defined as the system's nonlinear characteristic. By creating a cosine signal at the input with frequency $f_0$ the output signal includes $f_0$ as well as multiple harmonic waves. We conclude that new frequencies arise through nonlinear distortion.

For clarification of nonlinear distortions

Description of a linear system

• Linear distortions occur, if the transmission channel is characterized by a frequency response $H(f) \ne \rm const.$ Various frequencies are attenuated and delayed differently. Characteristic of this is that although frequencies can disappear (for example, through a low–pass, a high–pass, or a band–pass), no new frequencies can arise.

In this applet only linear distortions are considered.

Description forms for the frequency response

The generally complex valued frequency response can be represented as follows:

$$H(f) = |H(f)| \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} b(f)} = {\rm e}^{-a(f)}\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} b(f)}.$$

This results in the following description variables:

• The absolute value $|H(f)|$ is called amplitude response and in logarithmic form attenuation function:

$$a(f) = - \ln |H(f)|\hspace{0.2cm}{\rm in \hspace{0.1cm}Neper \hspace{0.1cm}(Np) } = - 20 \cdot \lg |H(f)|\hspace{0.2cm}{\rm in \hspace{0.1cm}Decibel \hspace{0.1cm}(dB) }.$$

• The phase function $b(f)$ indicates the negative frequency–dependent angle of $H(f)$ in the complex plane based on the real axis:

$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) \hspace{0.2cm}{\rm in \hspace{0.1cm}Radian \hspace{0.1cm}(rad)}.$$

Low–pass of order N

Attenuation function $a(f)$ and phase function $b(f)$ of a low–Pass of order $N$

The frequency response of a realizable low–pass (LP) of order $N$ is:

$$H(f) = \left [\frac{1}{1 + {\rm j}\cdot f/f_0 }\right ]^N\hspace{0.05cm}.$$

For example the RC low–pass is a first order low–pass. Consequently we can obtain

• the attenuation function:

$$a(f) =N/2 \cdot \ln [1+( f/f_0)^2] \hspace{0.05cm},$$

• the phase function:

$$b(f) =N \cdot \arctan( f/f_0) \hspace{0.05cm},$$

• the attenuation factor for the frequency $f=f_i$:

$$\alpha_i =|H(f = f_i)| = [1+( f_i/f_0)^2]^{-N/2}$$

$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)= \alpha_i \cdot A_i\cdot \cos(2\pi f_i t)\hspace{0.05cm},$$

• the phase delay for the frequency $f=f_i$:

$$\tau_i =\frac{b(f_i)}{2 \pi f_i} = \frac{N \cdot \arctan( f_i/f_0)}{2 \pi f_i}$$

$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)=A_i\cdot \cos(2\pi f_i (t- \tau_i))\hspace{0.05cm}.$$

High–pass of order N

Attenuation function $a(f)$ and phase function $b(f)$ of a high–pass of order $N$

The frequency response of a realizable high–pass (HP) of order $N$ is:

$$H(f) = \left [\frac{ {\rm j}\cdot f/f_0 }{1 + {\rm j}\cdot f/f_0 }\right ]^N\hspace{0.05cm}.$$

For example the LC high pass is a first grade high pass. Consequently we can obtain

• the attenuation function:

$$a(f) =N/2 \cdot \ln [1+( f_0/f)^2] \hspace{0.05cm},$$

• the phase function:

$$b(f) =-N \cdot \arctan( f_0/f) \hspace{0.05cm},$$

• the attenuation factor for the frequency $f=f_i$:

$$\alpha_i =|H(f = f_i)| = [1+( f_0/f_i)^2]^{-N/2}$$

$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)= \alpha_i \cdot A_i\cdot \cos(2\pi f_i t)\hspace{0.05cm},$$

• the phase delay for the frequency $f=f_i$:

$$\tau_i =\frac{b(f_i)}{2\pi f_i} = \frac{-N \cdot \arctan( f_0/f_i)}{2\pi f_i}$$

$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)=A_i\cdot \cos(2\pi f_i (t- \tau_i))\hspace{0.05cm}.$$

Phase function $b(f)$ of high–pass and low–pass

$\text{Example:}$  This graphic shows the phase function $b(f)$ with the cut–off frequency $f_0 = 1\ \rm kHz$ and order $N=1$

• of a low–pass (green curve),
• of a high–pass (violet curve).

The input signal is sinusoidal with frequency $f_{\rm S} = 1.25\ {\rm kHz}$ whereby this signal is only turned on at $t=0$:

$$x(t) = \left\{ \begin{array}{l} \hspace{0.75cm}0 \\ \sin(2\pi \cdot f_{\rm S} \cdot t ) \\ \end{array} \right.\quad\begin{array}{l} (t < 0), \\ (t>0). \\ \end{array}$$

The left graphic shows the signal $x(t)$. The dashed line marks the first zero at $t = T_0 = 0.8\ {\rm ms}$. The other two graphics show the output signals $y_{\rm LP}(t)$ und $y_{\rm HP}(t)$ of low–pass and high–pass, whereby the change in amplitude was balanced in both cases.

Input signal $x(t)$ (enframed in blue) as well as output signals $y_{\rm LP}(t)$ ⇒   green and $y_{\rm HP}(t)$ ⇒   violet

• The first zero of the signal $y_{\rm LP}(t)$ after the low–pass is delayed by $\tau_{\rm LP} = 0.9/(2\pi) \cdot T_0 \approx 0.115 \ {\rm ms}$ compared to the first zero of $x(t)$   ⇒   marked with green arrow, whereby $b_{\rm LP}(f/f_{\rm S} = 0.9 \ {\rm rad})$ was considered.
• In contrast, the phase delay of the high–pass is negative: $\tau_{\rm HP} = -0.67/(2\pi) \cdot T_0 \approx -0.085 \ {\rm ms}$ and therefore the first zero of $y_{\rm HP}(t)$ occurs before the dashed line.
• Following this transient response, in both cases the zero crossings again come in the raster of the period duration $T_0 = 0.8 \ {\rm ms}.$

Annotation: The shown signals were created using the interactive applet Causal systems – Laplace transform.

Attenuation distortions and phase distortions

Requirements for a non–distorting channel

The adjacent figure shows

• the even attenuation function $a(f)$   ⇒   $a(-f) = a(f)$, and
• the uneven function curve $b(f)$   ⇒   $b(-f) = -b(- f)$

of a non–distorting channel. One notices:

• In a distortion–free system the attenuation function $a(f)$ must be constant between$f_{\rm U}$ and $f_{\rm O}$ around the carrier frequency $f_{\rm T}$, where the input signal exist   ⇒   $x(t) \ne 0$.
• From the specified constant attenuation value $6 \ \rm dB$ follows for the amplitude response $|H(f)| = 0.5$   ⇒   the signal values of all frequencies are thus halved by the system   ⇒   no attenuation distortions.
• In addition, in such a system, the phase function $b(f)$ between $f_{\rm U}$ and $f_{\rm O}$ must increase linearly with the frequency. As a result, all frequency components are delayed by the same phase delay $τ$   ⇒   no phase delay.
• The delay $τ$ is fixed by the slope of $b(f)$. The phase function $b(f) \equiv 0$ would result in a delay–less system   ⇒   $τ = 0$.

The following summary considers that in this applet the input signal is always the sum of two harmonic oscillations,

$$x(t) = x_1(t) + x_2(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)+A_2\cdot \cos\left(2\pi f_2\cdot t- \varphi_2\right),$$

and therefor the channel influence is fully described by the attenuation factors $\alpha_1$ and $\alpha_2$ as well as the phase delays $\tau_1$ and $\tau_2$:

$$y(t) = \alpha_1 \cdot x_1(t-\tau_1) + \alpha_2 \cdot x_2(t-\tau_2).$$

$\text{Summary:}$ 

• A signal $y(t)$ is only distortion–free compared to $x(t)$ if $\alpha_1 = \alpha_2= \alpha$   and   $\tau_1 = \tau_2= \tau$   ⇒   $y(t) = \alpha \cdot x(t-\tau)$.
• Attenuation distortions occur when $\alpha_1 \ne \alpha_2$. If $\alpha_1 \ne \alpha_2$ and $\tau_1 = \tau_2$, then there are exclusively attenuation distortions.
• Phase distortions occur when $\tau_1 \ne \tau_2$. If $\tau_1 \ne \tau_2$ and $\alpha_1 = \alpha_2$, then there are exclusively phase distortions.

Exercises

BlaBla

(1)   We set the parameters for the Input signal $x(t)$ to $A_1 = 0.8\ {\rm V}, \ A_2 = 0.6\ {\rm V}, \ f_1 = 0.5\ {\rm kHz}, \ f_2 = 1.5\ {\rm kHz}, \ \varphi_1 = 90^\circ, \ \varphi_2 = 30^\circ$.

Calculate the signal's cycle duration $T_0$ and power $P_x$. Can you read the value for $P_x$ off the applet?

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}T_0 = \big [\hspace{-0.1cm}\text{ greatest common divisor }(0.5 \ {\rm kHz}, \ 1.5 \ {\rm kHz})\big ]^{-1}\hspace{0.15cm}\underline{ = 2.0 \ {\rm ms}};$

$\hspace{1.85cm} P_x = A_1^2/2 + A_2^2/2 \hspace{0.15cm}\underline{= 0.5 \ {\rm V^2}} = P_\varepsilon\text{, if }\hspace{0.15cm}\underline{k_{\rm M} = 0} \ \Rightarrow \ z(t) \equiv 0$.

(2)  Vary $\varphi_2$ between $\pm 180^\circ$ while assuming the other parameters from Exercise '''(1)'''. How does the value of $T_0$ and $P_x$ change?

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{No changes:}\hspace{0.2cm}\hspace{0.15cm}\underline{ T_0 = 2.0 \ {\rm ms}; \hspace{0.2cm} P_x = 0.5 \ {\rm V^2}}$.

(3)   Vary $f_2$ between $0 \le f_2 \le 10\ {\rm kHz}$ while assuming the other parameters from Exercise '''(1)'''. How does the value of $P_x$ change?

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{No changes if }f_2 \ne 0\text{ oder } f_2 \ne f_1\text{:}\hspace{0.3cm} \hspace{0.15cm}\underline{P_x = 0.5 \ {\rm V^2}}\text{.} \hspace{0.2cm} T_0 \text{ changes if }f_2\text{is not a multiple of }f_1$.

$\hspace{1.85cm}\text{If }f_2 = 0\text{:}\hspace{0.2cm} P_x = A_1^2/2 + A_2^2\hspace{0.15cm}\underline{ = 0.68 \ {\rm V^2}}$. $\hspace{3cm}$

$\hspace{1.85cm}\text{If }f_2 = f_1\text{:}\hspace{0.2cm} P_x = [A_1\cos(\varphi_1) + A_2\cos(\varphi_2)]^2/2 + [A_1\sin(\varphi_1) + A_2\sin(\varphi_2)]^2/2 \text{.}$

$\hspace{1.85cm}\text{Mit } \varphi_1 = 90^\circ, \ \varphi_2 = 30^\circ\text{:}\hspace{0.3cm}\hspace{0.15cm}\underline{ P_x = 0.74 \ {\rm V^2}}\text{.}$

(4)   Going by the previous output signal $x(t)$ we set following parameters to: $\alpha_1 = \alpha_2 = 0.5, \ \tau_1 = \tau_2 = 0.5\ {\rm ms}$, $k_{\rm M} = 1 \text{ and } \tau_{\rm M} = 0$ .

Are there linear distortions? Calculate the reception power $P_y$ and the power $P_\varepsilon$ of the differential signal $\varepsilon(t) = z(t) - x(t)$.

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{ y(t) = 0.5 \cdot x(t- 1\ {\rm ms})}\text{ is only attenuated and delayed, but not distorted.}$

$\hspace{1.85cm}\text{Reception power:}\hspace{0.2cm} P_y = (A_1/2)^2/2 + (A_2/2)^2/2\hspace{0.15cm}\underline{ = 0.125 \ {\rm V^2}}\text{. } P_\varepsilon \text{ is significantly greater:} \hspace{0.1cm} \hspace{0.15cm}\underline{P_\varepsilon = 0.625 \ {\rm V^2}}.$

(5)   With otherwise the same settings as In exercise '''(4)''', vary the matching parameters $k_{\rm M} \text{ and } \tau_{\rm M}$. How big is the distortion power $P_{\rm D}$?

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D}\text{is equal to }P_\varepsilon \text{ when using the ideal matching parameters:} \hspace{0.2cm}k_{\rm M} = 2 \text{ und } \tau_{\rm M}=T_0 - 0.5\ {\rm ms} = 1.5\ {\rm ms}$

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm}z(t) = x(t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\varepsilon(t) = 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm}P_{\rm D}\hspace{0.15cm}\underline{ = P_\varepsilon = 0} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\text{Neither attenuation nor phase distortion.}$

(6)   The channel parameters are now set to: $\alpha_1 = 0.5, \hspace{0.15cm}\underline{\alpha_2 = 0.2}, \ \tau_1 = \tau_2 = 0.5\ {\rm ms}$. Calculate the distortion power $P_{\rm D}$ and The Signal-to-Distortion ratio $(\rm SDR) \ \rho_{\rm D}$.

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D} = P_\varepsilon \text{ when using the best matching parameters:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 2.24} \text{ und } \hspace{0.15cm}\underline{\tau_{\rm M} = 1.5\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.059 \ {\rm V^2}}$.

$\hspace{1.85cm}\text{Attenuation distortions only.} \hspace{0.3cm}\text{Signal-to-Distortion-Ratio}\ \hspace{0.15cm}\underline{\rho_{\rm D} = P_x/P_\varepsilon \approx 8.5}$.

(7)   The channel parameters are now set to: $\alpha_1 = \alpha_2 = 0.5, \ \tau_1 \hspace{0.15cm}\underline{= 2\ {\rm ms} }, \ \tau_2 = 0.5\ {\rm ms}$. Calculate the distortion power $P_{\rm D}$ and the SDR $\rho_{\rm D}$?

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D} = P_\varepsilon \text{when using the best matching parameters:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 1.84} \text{ and } \tau_{\rm M}\hspace{0.15cm}\underline{ = 0.15\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.071 \ {\rm V^2}}$.

$\hspace{1.85cm}\text{Phase distortions only.} \hspace{0.3cm}\text{Signal-to-Distortion-Ratio}\ \hspace{0.15cm}\underline{\rho_{\rm D} = P_x/P_\varepsilon \approx 7}$.

(8)   The channel parameters are now set to: $\hspace{0.15cm}\underline{\alpha_1 = 0.5} , \hspace{0.15cm}\underline{\alpha_2 = 0.2} , \ \hspace{0.15cm}\underline{\tau_1= 0.5\ {\rm ms} }, \ \hspace{0.15cm}\underline{\tau_2 = 0.3\ {\rm ms} }$. Are there attenuation distortions? Are there phase distortions? How can $y(t)$ be approximated? How can $y(t)$ be approximated? Annotation: $\cos(3x) = 4 \cdot \cos(x)^3 - 3\cdot \cos(x).$

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm} \text{Both attenuation and phase distortions, because }\alpha_1 \ne \alpha_2\text{ and }\tau_1 \ne \tau_2$.

$\hspace{1.85cm}y(t) = y_1(t) + y_2(t)\ \Rightarrow \ y_1(t) = A_1 \cdot \alpha_1 \cdot \sin[2\pi f_1\ (t- 0.5\ \rm ms)] = -0.4 \ {\rm V} \cdot \cos(2\pi f_1 t)$

$\hspace{1.85cm} y_2(t) = \alpha_2 \cdot x_2(t- \tau_2) \text{ mit }x_2(t) = A_2 \cdot \cos[2\pi f_2\ (t- 30^\circ)] \approx A_2 \cdot \cos[2\pi f_2\ (t- 1/36 \ \rm ms)]$

$\hspace{1.85cm} \Rightarrow \ y_2(t) = 0.12 \ {\rm V} \cdot \cos[2\pi f_2\ (t- 0.328 \ {\rm ms})] \approx -0.12 \ { \rm V} \cdot \cos[2\pi f_2t]$.

$\hspace{1.85cm} \Rightarrow \ y(t) = y_1(t) + y_2(t) \approx -0.4 \ {\rm V} \cdot [\cos(2\pi \cdot f_1\cdot t) + 1/3 \cdot \cos(2\pi \cdot 3 f_1 \cdot t) = -0.533 \ {\rm V} \cdot \cos^3(2\pi f_1 t)$.

(9)   Assuming the parameters from 88Exercize (8)]]. Calculate the distortion power $P_{\rm D}$ and the SDR $\rho_{\rm D}$?

$\hspace{1.0cm}\text{Best possible adaptation:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 1.96} \text{, } \hspace{0.15cm}\underline{\tau_{\rm M} = 1.65\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.156 \ {\rm V^2} },\hspace{0.1cm}\hspace{0.15cm}\underline{\rho_{\rm D} = 0.500/0.156 \approx 3.2}$.

(10)  Now we set $A_2 = 0$ and $A_1 = 1\ {\rm V}, \ f_1 = 1\ {\rm kHz}, \ \varphi_1 = 0^\circ$. The channel is a Low-pass of order 1 $\underline{(f_0 = 1\ {\rm kHz})}$.
Are there any attenuation or phase distortions? Calculate the channel coefficients $\alpha_1$ and $\tau_1$

$\hspace{1.0cm}\text{At only one frequency there are neither attenuation nor phase distortions.}$ $\hspace{1.0cm}\text{Attenuation factor for }f_1=f_0\text{ and }N=1\text{: }\alpha_1 =|H(f = f_1)| = [1+( f_1/f_0)^2]^{-N/2} = 2^{-1/2}= 1/\sqrt{2}\hspace{0.15cm}\underline{=0.707},$ $\hspace{1.0cm}\text{Phase factor for }f_1=f_0\text{ and }N=1\text{: }\tau_1 = N \cdot \arctan( f_1/f_0)/(2 \pi f_i)=\arctan( 1)/(2 \pi f_i) =1/(8f_1) \hspace{0.15cm}\underline{=0.125 \ \rm ms}.$

Zur Handhabung des Applets

    (A)     Parametereingabe per Slider

    (B)     Bereich der graphischen Darstellung

    (C)     Variationsmöglichkeit für die graphische Darstellung

    (D)     Abspeichern und Zurückholen von Parametersätzen

    (E)     Numerikausgabe des Hauptergebnisses $T_0$; graphische Verdeutlichung durch rote Linie

    (F)     Ausgabe von $x_{\rm max}$ und der Signalwerte $x(t_*) = x(t_* + T_0)= x(t_* + 2T_0)$

    (G)     Darstellung der Signalwerte $x(t_*) = x(t_* + T_0)= x(t_* + 2T_0)$ durch grüne Punkte

    (H)     Einstellung der Zeit $t_*$ für die Signalwerte $x(t_*) = x(t_* + T_0)= x(t_* + 2T_0)$

Details zum obigen Punkt (C)

    (*)   Zoom–Funktionen „$+$” (Vergrößern), „$-$” (Verkleinern) und $\rm o$ (Zurücksetzen)

    (*)   Verschieben mit „$\leftarrow$” (Ausschnitt nach links, Ordinate nach rechts), „$\uparrow$” „$\downarrow$” und „$\rightarrow$”

Andere Möglichkeiten:

    (*)   Gedrückte Shifttaste und Scrollen: Zoomen im Koordinatensystem,

    (*)   Gedrückte Shifttaste und linke Maustaste: Verschieben des Koordinatensystems.

Über die Autoren

Dieses interaktive Berechnungstool wurde am Lehrstuhl für Nachrichtentechnik der Technischen Universität München konzipiert und realisiert.

• Die erste Version wurde 2005 von Bettina Hirner im Rahmen ihrer Diplomarbeit mit „FlashMX–Actionscript” erstellt (Betreuer: Günter Söder ).
• 2018 wurde dieses Programm von Jimmy He im Rahmen seiner Bachelorarbeit (Betreuer: Tasnád Kernetzky) auf „HTML5” umgesetzt und neu gestaltet.

Nochmalige Aufrufmöglichkeit des Applets in neuem Fenster

Open Applet in new Tab