Linear Distortions of Periodic Signals

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 $P_\varepsilon$.

Beginn Anpassen 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:

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

Ende Anpassen

Beginn Änderungen im deutschen Text:

Als nächster Block im obigen Modell folgt das „Matching”: Dabei wird das Ausgangssignal $y(t)$ mit für alle Frequenzen einheitlichen Größen $k_{\rm M}$ und $\tau_{\rm M}$ in Amplitude bzw. Phase angepasst. Dies ist also keine frequenzabhängige Entzerrung. Anhand des Signals $z(t)$ kann unterschieden werden

zwischen einer Dämpfungsverzerrung und einer frequenzunabhängigen Dämpfung, sowie
zwischen einer Phasenverzerrung und einer für alle Frequenzen gleichen Laufzeit.

Als Maß für die Stärke der linearen Verzerrungen wird die Verzerrungsleistung (englisch: Distortion Power) $P_{\rm D}$ verwendet. Für diese gilt:

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

'Ende Änderungen im deutschen Text:

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 „Stärke” 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 value 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 or a High–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 curve:

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

The Phase response $b(f)$ 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

Dämpfungsverlauf und Phasenverlauf eines Tiefpasses N–ter Ordnung

The frequency response of a realizable N grade low pass 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 grade low pass. Consequently we can obtain

the attenuation curve:

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

the phase curve:

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

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

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

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

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

Dämpfungsverlauf und Phasenverlauf eines Hochpasses N–ter Ordnung

The frequency response of a realizable N grade high pass 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 curve:

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

the phase curve:

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

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

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

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

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

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

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

Beginn Änderungen im deutschen Text:

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 the order $N=1$

of a Low–pass as a green graph and
of a High–pass as a violet graph.

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

Formel überprüfen!

$$x(t) = \left\{ \begin{array}{l} \hspace{0.75cm}0 \\ \sin(2\pi \cdot f_{\rm S} \cdot t ) \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r} } \\ {\rm{f\ddot{u}r} } \\ \end{array}\begin{array} t < 0, \\ t>0. \\ \end{array}$$

The left (outlined in blue) 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 $y_{\rm TP}(t)$ und $y_{\rm HP}(t)$of High–pass and Low–pass, whereby the change in amplitude was balanced in both cases.

Input signal $x(t)$ as well as output signals $y_{\rm TP}(t)$ und $y_{\rm HP}(t)$

The first zero of the signal $y_{\rm TP}(t)$ after the Low–pass is delayed by $\tau_{\rm TP} = 0.9/(2\pi) \cdot T_0 \approx 0.115 \ {\rm ms}$ compared to the first zero of $x(t)$ ⇒ marked with green arrow, whereby $b_{\rm TP}(f/f_{\rm S} = 0.9 \ {\rm rad})$ 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 therefor the first zero $y_{\rm HP}(t)$ 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 signal paths 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 curve $a(f)$ ⇒ $a(-f) = a(f)$, and
the uneven phase 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 proportions of $x(t)$ exist.
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 curve $b(f)$ 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 $b(f)$. $b(f) = 0$ 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 $\tau_1 = \tau_2$:

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

$\text{Summary:}$

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.
A signal $y(t)$ is only distortion–free compared to $x(t)$ if $\alpha_1 = \alpha_2= \alpha$ and $\tau_1 = \tau_2= \tau$ ⇒ $y(t) = \alpha \cdot x(t-\tau)$.

Ende Änderungen im deutschen Text:

Vorschlag für die Versuchsdurchführung

BlaBla

(1) We set the parameters for the transmitter signal $x(t)$ to $A_1 = 0.8\ {\rm V}, \ A_2 = 0.6\ {\rm V}, \ f_1 = 0.5\ {\rm kHz}, \ f_2 = 1.5\ {\rm kHz}, \ \varphi_1 = 90^\circ, \ \varphi_2 = 0^\circ$.

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 (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 (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{. Mit } \varphi_1 = 90^\circ, \ \varphi_2 = 0^\circ\text{:}\hspace{0.3cm}\hspace{0.15cm}\underline{ P_x = 0.5 \ {\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 (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 $\rm SDR$ $\rho_{\rm D}$?

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

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

(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 $\rho_{\rm D}$?

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D} = P_\varepsilon \text{when using the best matching parameters:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 1.82} \text{ and } \tau_{\rm M}\hspace{0.15cm}\underline{ = 0.15\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.072 \ {\rm V^2}}$.

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

(8) `The channel parameters are now set to: $\hspace{0.15cm}\underline{\alpha_1 = 0.5} , \hspace{0.15cm}\underline{\alpha_2 = 0.2} , \ \hspace{0.15cm}\underline{\tau_1= 2\ {\rm ms} }, \ \hspace{0.15cm}\underline{\tau_2 = 0.5\ {\rm ms} }$. Calculate the distortion power $P_{\rm D}$ and the SDR $\rho_{\rm D}$? How can $y(t)$ be approximated?

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm} \text{Both attenuation and phase distortions. Best possible adaptation:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 2.06} \text{, } \hspace{0.15cm}\underline{\tau_{\rm M} = 0.15\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.136 \ {\rm V^2}},\hspace{0.1cm}\hspace{0.15cm}\underline{\rho_{\rm D} \approx 3.7}$.

$\hspace{1.85cm}\text{Combining }\varphi \text{- und } \tau\text{-Parameter: } y(t) = 0.4 \ {\rm V} \cdot \sin\ (2\pi f_1 t) - 0.12 \ {\rm V} \cdot \sin\ (2\pi \cdot 3f_1\cdot t) \hspace{0.15cm}\underline{\approx 0.52 \ {\rm V} \cdot \sin^3(2\pi f_1 t)}$.

(9) Nun gelte $\underline{A_1 = A_2 = 1\ {\rm V}, \ f_1 = 1\ {\rm kHz}, \ f_2 = 1\ {\rm kHz}, \ \varphi_1 = 0^\circ, \ \varphi_2 = 0^\circ}$. Der Kanal sei ein Tiefpass erster Ordnung $\underline{(f_0 = 1\ {\rm kHz})}$.

Gibt es Dämpfungsverzerrungen? Gibt es Phasenverzerrungen? Wie groß ist nun die Verzerrungsleistung $P_{\rm D}$?

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm} \text{Dämpfungsverzerrungen, da }\hspace{0.15cm}\underline{\alpha_1 = 0.71 \ne \alpha_2 = 0.45} \text{; geringere Phasenverzerrungen, da }\hspace{0.15cm}\underline{ \tau_1 = 0.13 \ {\rm ms} \approx \tau_2 = 0.09 \ {\rm ms}}$.

$\hspace{1.85cm}\text{ Verzerrungsleistung }\hspace{0.15cm}\underline{P_{\rm D} = 0.074 \ {\rm V^2}} \text{ bei bestmöglicher Anpassung:} \hspace{0.2cm}k_{\rm M}\hspace{0.15cm}\underline{ = 1.6} \text{ und } \tau_{\rm M}\hspace{0.15cm}\underline{ = 0.9\ {\rm ms} }$.

(10) Wie ändern sich die Kanalparameter durch einen Tiefpass zweiter Ordnung gegenüber einem Tiefpass erster Ordnung $(f_0 = 1\ {\rm kHz})$.

Wie groß ist nun die Verzerrungsleistung $P_{\rm D}$? Wie groß ist nun die Verzerrungsleistung $P_{\rm D}$?

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm} \text{Es gilt }\hspace{0.15cm}\underline{\alpha_1 = 0.71^2 \approx 0.5, \alpha_2 = 0.45^2 \approx 0.5, \tau_1 = 2 \cdot 0.13 \approx 0.25 \ {\rm ms} \tau_2 = 2 \cdot 0.09 \ {\rm ms} \approx 0.18 \ {\rm ms}} $.

$\hspace{1.85cm}P_{\rm D} = 0.228 \ {\rm V^2} \text { ist größer und der 2 kHz-Anteil wird im Vergleich zum 2 kHz-Anteil noch mehr unterdrückt}$.

(11) Welche Unterschiede ergeben sich bei einem Hochpass zweiter Ordnung gegenüber einem Tiefpass zweiter Ordnung $(f_0 = 1\ {\rm kHz})$.

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm} \text{???????????????}$

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.

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