Contents
Programmbeschreibung
The applet should clarify the properties of digital filters, whereby we confine ourselves to filters of the order $M=2$. Both non-recursive filters $\rm (FIR$, Finite Impulse Response$)$ as well as recursive filters $\rm (IIR$, Infinite Impulse Response$)$.
The input signal $x(t)$ is represented by the sequence $〈x_ν〉$ of its samples, where $x_ν$ stands for $x(ν · T_{\rm A})$. The output sequence $〈y_ν〉$is calculated, i.e. the discrete-time representation of the output signal $y(t)$.
- $T_{\rm A}$ denotes the time interval between two samples.
- We also limit ourselves to causal signals and systems, which means that $x_ν \equiv 0$ and $y_ν \equiv 0$ for $ν \le 0$.
It should also be noted that we denote the initial sequence $〈y_ν〉$ as
(1) the discrete-time impulse response $〈h_ν〉$ if the “discrete-time Dirac function” is present at the input: $〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉,$
(2) the time-discrete step response $〈\sigma_ν〉$ if the “time-discrete step function” is present at the input: $〈x_ν〉= 〈1,\ 1,\ 1,\ 1,\ 1,\ 1,\ 1, \text{...}〉,$
(3) the discrete-time rectangle response $〈\rho_ν^{(2, 4)}〉$ if the “discrete-time rectangle function” is present at the input: $〈x_ν〉= 〈0,\ 0,\ 1,\ 1,\ 1,\ 0,\ 0, \text{...}〉;$
In quotation marks are the beginning of the ones $(2)$ and the position of the last ones $(4)$.
Theoretical background
General block diagram
Each signal $x(t)$ can only be represented on a computer by the sequence $〈x_ν〉$ of its samples, where $x_ν$ stands for $x(ν · T_{\rm A})$. right |frame| Block diagram of a digital (IIR–) filter $M$–order
- The time interval $T_{\rm A}$ between two samples is limited by the sampling theorem.
- We limit ourselves here to causal signals and systems, which means that $x_ν \equiv 0$ for $ν \le 0$.
- In order to determine the influence of a linear filter with frequency response $H(f)$ on the time-discrete input signal $〈x_ν〉$, it is advisable to describe the filter discrete-time. In the time domain, this happens with the discrete-time impulse response $〈h_ν〉$.
- On the right you can see the corresponding block diagram. The following therefore applies to the samples of the output signal $〈y_ν〉$ thus holds:
- $$y_\nu = \sum\limits_{\mu = 0}^M {a_\mu } \cdot x_{\nu - \mu } + \sum\limits_{\mu = 1}^M {b_\mu } \cdot y_{\nu - \mu } .$$
The following should be noted here:
- The index $\nu$ refers to sequences, for example at the input $〈x_ν〉$ and output $〈y_ν〉$.
- On the other hand, we use the index $\mu$ to identify the $a$ and $b$ filter coefficients.
- The first sum describes the dependency of the current output $y_ν$ on the current input $x_ν$ and on the $M$ previous input values $x_{ν-1}$, ... , $x_{ν-M}$.
- The second sum indicates the influence of $y_ν$ by the previous values $y_{ν-1}$, ... , $y_{ν-M}$ at the filter output. It specifies the recursive part of the filter.
- The integer parameter $M$ is called the order of the digital filter. In the program, this value is limited to $M\le 2$.
$\text{Definitions:}$
(1) The output sequence $〈y_ν〉$ is called the discrete-time impulse response $〈h_ν〉$ if the “discrete-time Dirac function” is present at the input:
- $$〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉 .$$
(2) The output sequence $〈y_ν〉$ is called the time-discrete step response $〈\sigma_ν〉$ if the “time-discrete step function” is present at the input:
- $$〈x_ν〉= 〈1,\ 1,\ 1,\ 1,\ 1,\ 1,\ 1, \text{...}〉 .$$
(3) The output sequence $〈y_ν〉$ is called the discrete-time rectangle response $〈\rho_ν^{(2, 4)}〉$ if the “discrete-time rectangular function” is present at the input:
- $$〈x_ν〉= 〈0,\ 0,\ 1,\ 1,\ 1,\ 0,\ 0, \text{...}〉 .$$
- The beginning of ones $(2)$ and the position of the last ones $(4)$ are given in single quotes.
Non-recursive filter ⇒ FIR–filter
right |frame| Non-recursive digital filter $($FIR filter$)$ $M$ order $\text{Definition:}$ If all feedback coefficients $b_{\mu} = 0$ , one speaks of one non-recursive filter. In the English language literature, the term FIR filter (Finite Impulse Response) is also used for this.
The following applies to the order $M$ applies:
- The output value $y_ν$ depends only on the current and the previous $M$ input values:
- $$y_\nu = \sum\limits_{\mu = 0}^M {a_\mu \cdot x_{\mu - \nu } } .$$
- Time-discrete impulse response with $〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉$:
- $$〈h_\mu〉= 〈a_0,\ a_1,\ \text{...},\ a_M〉 .$$
$\text{Example 1:}$ A two-way channel where
- the signal on the main path arrives undamped compared to the input signal but is delayed by $2\ \rm µ s$ arrives with a delay, and
- at $4\ \rm µ s$ distance – so absolutely at time $t = 6\ \rm µ s$ – follows an echo with half the amplitude,
can be simulated by a non-recursive filter according to the sketch above, whereby the following parameter values must be set:
- $$M = 3,\quad T_{\rm A} = 2\;{\rm{µ s} },\quad a_{\rm 0} = 0,\quad a_{\rm 1} = 1, \quad a_{\rm 2} = 0, \quad a_{\rm 3} = 0.5.$$
$\text{Example 2:}$ Consider a non-recursive filter with the filter coefficients $a_0 = 1,\hspace{0.5cm} a_1 = 2,\hspace{0.5cm} a_2 = 1.$ right|frame|Nichtrekursives Filter
(1) The conventional impulse response is: $h(t) = \delta (t) + 2 \cdot \delta ( {t - T_{\rm A} } ) + \delta ( {t - 2T_{\rm A} } ).$
⇒ discrete-time impulse response: $〈h_\mu〉= 〈1,\ 2,\ 1〉 .$
(2) The frequency response $H(f)$ is the Fourier transform of $h(t)$. By applying the displacement theorem:
- $$H(f) = 2\big [ {1 + \cos ( {2{\rm{\pi }\cdot }f \cdot T_{\rm A} } )} \big ] \cdot {\rm{e} }^{ - {\rm{j} }2{\rm{\pi } }fT_{\rm A} }\hspace{0.5cm}\Rightarrow \hspace{0.5cm}H(f = 0) = 4.$$
(3) It follows that the time-discrete step response $〈\sigma_ν〉$ tends to become $4$ for large $\nu$.
(4) The discrete-time convolution of the input sequence $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;0,\;0,\;0,\;1,\;0,\;0,\;0,\;\text{...} } \hspace{0.05cm} \right\rangle$ with $\left\langle \hspace{0.05cm}{h_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{1, \ 2,\ 1 } \hspace{0.05cm}\right\rangle$ results
- $$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;2,\;1,\;0,\;1,\;2,\;1,\;0,\;0,\;0,\;0,\; \text{...} \;} \right\rangle. $$
(5) The discrete-time convolution of the input sequence $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;1,\;0,\;0,\;1,\;0,\;0,\;0,\;\text{...} } \hspace{0.05cm} \right\rangle$ with $\left\langle \hspace{0.05cm}{h_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{1, \ 2,\ 1 } \hspace{0.05cm}\right\rangle$ results
- $$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;3,\;3,\;2,\;2,\;1,\;0,\;0,\;0,\;0,\;0,\; \text{...} \;} \right\rangle. $$
Recursive filter ⇒ IIR filter
right|frame|First order recursive filter $\text{Definition:}$
- If at least one of the feedback coefficients is $b_{\mu} \ne 0$, then this is referred to as a recursive filter (see graphic on the right). The term IIR filter (Infinite Impulse Response) is also used for this, particularly in the English-language literature. This filter is dealt with in detail in the trial implementation.
- If all forward coefficients are also identical $a_\mu = 0$ with the exception of $a_0$, a purely recursive filter is available (see graphic on the left).
left|frame| Purely recursive first order filter
In the following we restrict ourselves to the special case “purely recursive filter of the first order”. This filter has the following properties:
- The output value $y_ν$ depends (indirectly) on an infinite number of input values:
- $$y_\nu = \sum\limits_{\mu = 0}^\infty {a_0 \cdot {b_1} ^\mu \cdot x_{\nu - \mu } .}$$
- This shows the following calculation:
- $$y_\nu = a_0 \cdot x_\nu + b_1 \cdot y_{\nu - 1} = a_0 \cdot x_\nu + a_0 \cdot b_1 \cdot x_{\nu - 1} + {b_1} ^2 \cdot y_{\nu - 2} = a_0 \cdot x_\nu + a_0 \cdot b_1 \cdot x_{\nu - 1} + a_0 \cdot {b_1} ^2 \cdot x_{\nu - 2} + {b_1} ^3 \cdot y_{\nu - 3} = \text{...}. $$
- By definition, the discrete-time impulse response is the same as the output sequence if there is a single "one" at $t =0$ at the input.
- $$h(t)= \sum\limits_{\mu = 0}^\infty {a_0 \cdot {b_1} ^\mu \cdot \delta ( {t - \mu \cdot T_{\rm A} } )}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}〈\hspace{0.05cm}h_\mu\hspace{0.05cm}〉= 〈\hspace{0.05cm}a_0, \ a_0\cdot {b_1}, \ a_0\cdot {b_1}^2 \ \text{...} \hspace{0.05cm}〉.$$
$\text{Conclusion:}$ With a recursive filter, the (time-discrete) impulse response extends to infinity with $M = 1$:
- For reasons of stability, $b_1 < 1$ must apply.
- With $b_1 = 1$ the impulse response $h(t)$ would extend to infinity and with $b_1 > 1$ the variable $h(t)$ would even continue to infinity.
- With such a recursive filter of the first order, each individual Dirac line is exactly the factor $b_1$ smaller than the previous Dirac line:
- $$h_{\mu} = h(\mu \cdot T_{\rm A}) = {b_1} \cdot h_{\mu -1}.$$
frame| Discrete-time impulse response | rechts $\text{Example 3:}$ The graphic opposite shows the discrete-time impulse response $〈\hspace{0.05cm}h_\mu\hspace{0.05cm}〉of a recursive filter of the first order with the parameters $a_0 = 1$ and $b_1 = 0.6$. *The (time-discrete) course is exponentially falling and extends to infinity. *The ratio of the weights of two successive Diracs is $b_1 = 0.6$. <div style="clear:both;"> </div> </div> ==='"`UNIQ--h-5--QINU`"'Recursive filter as a sine generator=== [[Datei:P_ID622__Sto_A_5_4.png|right|frame|'''Changing the''' proposed filter structure '''to''' $T_{\rm A}$]] The graphic shows a second-order digital filter that is suitable for generating a time-discrete sine function on a digital signal processor (DSP) if the input sequence $\left\langle \hspace{0.05cm} {x_\nu } \hspace{0.05cm}\right\rangle$ a (time-discrete) Dirac function is: :'"`UNIQ-MathJax16-QINU`"' The five filter coefficients result from the: [https://de.wikipedia.org/wiki/Z-Transformation $Z$ transformation]: :'"`UNIQ-MathJax17-QINU`"' After implementing this equation using a second-order recursive filter, the following filter coefficients are obtained: :'"`UNIQ-MathJax18-QINU`"' *The filter coefficients $a_0$ and $a_2$ can be omitted and $b_2=-1$ has a fixed value. *The angular frequency $\omega_0$ of the sine wave is therefore only determined by $a_0$ and $a_0$. <div class="greybox"> $\text{Example 3:}$ Let $a_1 = 0.5$, $b_1 = \sqrt 3$, $x_0 = 1$ and $x_{\nu \hspace{0.05cm}\ne\hspace{0.05cm} 0} = 0$. '''(1)''' Then the following applies to the initial values $y_\nu$ at times $\nu \ge 0$:<br> :* $y_0 = 0;$ :* $y_1 = 0.5$ ⇒ the „$1$” at the input only has an effect at time $\nu = 1$ because of $a_0= 0$ at the output; :* $y_2 = b_1 \cdot y_1 - y_0 = {\sqrt 3 }/{2} \approx 0.866$ ⇒ with $\nu = 2$ the recursive part of the filter also takes effect; :* $y_3 = \sqrt 3 \cdot y_2 - y_1 = \sqrt 3 \cdot {\sqrt 3 }/{2} - {1}/{2} = 1$ ⇒ for $\nu \ge 2$ the filter is purely recursive: $y_\nu = b_1 \cdot y_{\nu - 1} - y_{\nu - 2}$; :* $y_4 = \sqrt 3 \cdot y_3 - y_2 = \sqrt 3 \cdot 1 - {\sqrt 3 }/{2} = {\sqrt 3 }/{2};$ :* $y_5 = \sqrt 3 \cdot y_4 - y_3 = \sqrt 3 \cdot {\sqrt 3 }/{2} - 1 = 0.5;$ :* $y_6 = \sqrt 3 \cdot y_5 - y_4 = \sqrt 3 \cdot {1}/{2} - {\sqrt 3 }/{2} = 0;$ :* $y_7 = \sqrt 3 \cdot y_6 - y_5 = \sqrt 3 \cdot 0 - {1}/{2} = - 0.5.$ '''(2)''' By continuing the recursive algorithm one gets for large $\nu$–values: $y_\nu = y_{\nu - 12}$ ⇒ $T_0/T_{\rm A}= 12.$ <div style="clear:both;"> </div> </div> =='"`UNIQ--h-6--QINU`"'Test execution== [[Datei:Exercises_binomial_fertig.png|right]] *First select the number '''1''' ... '''10''' of the task to be processed. *A task description is displayed. The parameter values are adjusted. *Solution after pressing "sample solution". *The number '''0''' corresponds to a "reset": Same setting as when the program was started. <br clear="all"> <div class="bluebox"> '''(1)''' The filter coefficients are $a_0=0.25$, $a_1=0.5$, $a_2=0.25$, $b_1=b_2=0$. Which filter is it?<br> Interpret the impulse response $〈h_ν〉$, the step response $〈\sigma_ν〉$ and the rectangular response $〈\rho_ν^{(2, 8)}〉$ each in a time-discrete representation. <div style="clear:both;"> </div> </div> :* Due to the missing $b$ coefficients, it is a non-recursive digital filter ⇒ '''FIR filter''' (''Finite Impulse Response''). :* The impulse response consists of $M+1=3$ Dirac lines according to the $a$ coefficients: $〈h_ν〉= 〈a_0, \ a_1,\ a_2〉= 〈0.25, \ 0.5,\ 0.25,\ 0, \ 0, \ 0,\text{...}〉 $. :* The step response is: $〈\sigma_ν〉= 〈0.25, \ 0.75,\ 1,\ 1, \ 1, \ 1,\text{...}〉 $. The final value is equal to the DC signal transfer factor $H(f=0)=a_0+a_1+a_2 = 1$. :* The distortions with rise and fall can also be seen from the rectangular response $〈\rho_ν^{(2, 8)}〉= 〈0,\ 0, 0.25, \ 0.75,\ 1,\ 1, \ 1, \ 1, \ 1, \ 0.75, \ 0.25, \ \text{...}〉$. <div class="bluebox"> '''(2)''' How do the results differ with $a_2=-0.25$? <div style="clear:both;"> </div> </div> :* Taking into account $H(f=0)= 0.5$ there are comparable consequences ⇒ Step response: $〈\sigma_ν〉= 〈0.25, \ 0.75,\ 0.5,\ 0.5, \ 0.5, \ 0.5,\text{...}〉 $. <div class="bluebox"> '''(3)''' Now let the filter coefficients $a_0=1$, $b_1=0.9$ and $a_1=a_2= b_2=0$. Which filter is it? Interpret the impulse response $〈h_ν〉$. <div style="clear:both;"> </div> </div> :* It is a recursive digital filter ⇒ '''IIR filter''' (''Infinite Impulse Response'') of the first order. It is the discrete-time analogue of the RC low pass. :* Starting from $h_0= 1$ is $h_1= h_0 \cdot b_0= 0.9$, $h_2= h_1 \cdot b_0= b_0^2=0.81$, $h_3= h_2 \cdot b_0= b_0^3=0.729$, and so on ⇒ $〈h_ν〉$ extends to infinity. :* Impulse response $h(t) = {\rm e}^{-t/T}$ with $T$: intersection $($Tangente bei $t=0$, Abscissa$)$ ⇒ $h_\nu= h(\nu \cdot T_{\rm A}) = {\rm e}^{-\nu/(T/T_{\rm A})}$ with $T/T_{\rm A} = 1/(h_0-h_1)= 10$. :* So: The values of the continuous time differ from the discrete-time impulse response. This results in the values $1.0, \ 0.9048,\ 0.8187$ ... <div class="bluebox"> '''(4)''' The filter setting is retained. Interpret the step response $〈h_ν〉$ and the rectangular response $〈\rho_ν^{(2, 8)}〉$. What is the value for $H(f=0)$? <div style="clear:both;"> </div> </div> :* The step response is the integral over the impulse response: $\sigma(t) = T \cdot (1-{\rm e}^{-t/T}) ]$ ⇒ $\sigma_\nu= 10 \cdot (1-{\rm e}^{-\nu/10})$ ⇒ $\sigma_0=1$, $\sigma_1=1.9$, $\sigma_2=2.71$, ... :* For large $\nu$ values, the (time-discrete) step response tends to the DC signal transmission factor $H(f=0)= 10$: $\sigma_{40}=9.867$, $\sigma_{50}=9.954$, $\sigma_\infty=10$. :* The rectangular response $〈\rho_ν^{(2, 8)}〉$ increases with a delay of $2$ in the same way as $〈\sigma_ν〉$. In the area $\nu \ge 8$ the $\rho_ν$ values decrease exponentially. <div class="bluebox"> '''(5)''' Wir betrachten weiterhin das Filter mit $a_0=1$, $b_1=0.9$, $a_1=a_2= b_2=0$. Wie lautet die Ausgangsfolge $〈y_ν〉$ für die Eingangsfolge $〈x_ν〉= 〈1,\ 0,\ -0.5〉$? <br> ''Hinweis'': Die Aufgabe lässt sich ebenfalls mit diesem Programm lösen, obwohl die hier betrachtete Konstellation nicht direkt einstellbar ist. <div style="clear:both;"> </div> </div> :* Man behilft sich, indem man den Koeffizienten $a_2=-0.5$ setzt und dafür die Eingangsfolge auf $〈x_ν〉= 〈1,\ 0,\ 0,\ \text{ ...}〉$ ⇒ „Diracfunktion” reduziert. :* Die tatsächliche Impulsantwort dieses Filters $($mit $a_2=0)$ wurde in Aufgabe '''(3)''' ermittelt: $h_0= 1$, $h_1= 0.9$, $h_2= 0.81$, $h_3= 0.729$, $h_4= 0.646$. :* Die Lösung dieser Aufgabe lautet somit: $y_0 = h_0= 1$, $y_1= h_1= 0.9$, $y_2 =h_2-h_0/2= 0.31$, $y_3 =h_3-h_1/2= 0.279$, $y_4 =h_4-h_2/2= 0.251$. :* Vorsicht: Sprungantwort und Rechteckantwort beziehen sich nun auf das fiktive Filter $($mit $a_2=-0.5)$ und nicht auf das eigentliche Filter $($mit $a_2=0)$. <div class="bluebox"> '''(6)''' Betrachten und interpretieren Sie die Impulsanwort und die Sprungantwort für die Filterkoeffizienten $a_0=1$, $b_1=1$, $a_1=a_2= b_2=0$. <div style="clear:both;"> </div> </div> :* '''Das System ist instabil''': Eine zeitdiskrete Diracfunktion am Eingang $($zur Zeit $t=0)$ bewirkt im Ausgangsignal unendlich viele Diracs gleicher Höhe. :* Eine zeitdiskrete Sprungfunktion am Eingang bewirkt im Ausgangsignal unendlich viele Diracs mit monoton ansteigenden Gewichten (bis ins Unendliche). <div class="bluebox"> '''(7)''' Betrachten und interpretieren Sie Impulsanwort und Sprungantwort für die Filterkoeffizienten $a_0=1$, $b_1=-1$, $a_1=a_2= b_2=0$. <div style="clear:both;"> </div> </div> :* Im Gegensatz zur Aufgabe '''(6)''' sind hier die Gewichte der Impulsantwort $〈h_ν〉$ nicht konstant gleich $1$, sondern alternierend $\pm 1$. Das System ist ebenfalls instabil. :* Bei der Sprunganwort $〈\sigma_ν〉$ wechseln sich dagegen die Gewichte alternierend zwischen $0$ $($bei geradem $\nu)$ und $1$ $($bei ungeradem $\nu)$ ab. <div class="bluebox"> '''(8)''' Wir betrachten den „Sinusgenerator”: $a_1=0.5$, $b_1=\sqrt{3}= 1.732$, $b_2=-1.$ Vergleichen Sie die Impulsantwort mit den berechneten Werten in $\text{Beispiel 4}$. <br> Wie beinflussen die Parameter $a_1$ und $b_1$ die Periodendauer $T_0/T_{\rm A}$ und die Amplitude $A$ der Sinusfunktion? <div style="clear:both;"> </div> </div> :* $〈x_ν〉=〈1, 0, 0, \text{...}〉$ ⇒ $〈y_ν〉=〈0, 0.5, 0.866, 1, 0.866, 0.5, 0, -0.5, -0.866, -1, -0.866, -0.5, 0, \text{...}〉$ ⇒ '''Sinus''', Periode $T_0/T_{\rm A}= 12$, Amplitude $1$. :* Die Vergrößerung/Verkleinerung von $b_1$ führt zur größeren/kleineren Periodendauer $T_0/T_{\rm A}$ und zur größeren/kleineren Amplitude $A$. Es muss $b_1 < 2$ gelten. :* $a_1$ beinflusst nur die Amplitude, nicht die Periodendauer. Für $a_1$ gibt es keine Wertebegrenzumg. Bei negativem $a_1$ ergibt sich die Minus–Sinusfunktion. :* '''Gibt es hier keine Diskrepanz zu h(t) wertkontinuierlich ???''' <div class="bluebox"> '''(9)''' Die Grundeinstellung bleibt erhalten. Mit welchen $a_1$ und $b_1$ ergibt sich eine Sinusfunktion mit Periodendauer $T_0/T_{\rm A}=16$ und Amplitude $A=1$? <div style="clear:both;"> </div> </div> :* Durch Probieren erreicht man mit $b_1= 1.8478$ tatsächlich die Periodendauer $T_0/T_{\rm A}=16.$ Allerdings erhöht sich dadurch die Amplitude auf $A=1.307$. :* Die Anpassung des Parameters $a_1= 0.5/1.307=0.3826$ führt dann zur gewünschten Amplitude $A=1$. :* Oder man kann das auch wie im Beispiel berechnen: $b_1 = 2 \cdot \cos ( {2{\rm{\pi }}\cdot{T_{\rm A}}/{T_0 }})= 2 \cdot \cos (\pi/8)=1.8478$, $a_1 = \sin (\pi/8)=0.3827$. <div class="bluebox"> '''(10)''' Wir gehen weiter vom „Sinusgenerator” aus. Welche Modifikationen muss man vornehmen, um damit einen „Cosinus” zu generieren? <div style="clear:both;"> </div> </div> :* Mit $a_1=0.5$, $b_1=\sqrt{3}= 1.732$, $b_2=-1$ sowie $〈x_ν〉=〈1, 1, 1, \text{...}〉$ ist die Ausgangsfolge $〈y_ν〉$ das zeitdiskrete Analogon der Sprungantwort $\sigma(t)$.
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Zur Handhabung des Applets
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).
- 2020 wurde das Programm von Andre Schulz (Bachelorarbeit LB, Betreuer: Benedikt Leible und Tasnád Kernetzky ) unter „HTML5” neu gestaltet.
