Difference between revisions of "Applets:Attenuation of Copper Cables"

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Applet Description

Theoretical Background

Magnitude Frequency Response and Attenuation Function

Following relationship exists between the magnitude frequency response and the attenuation function:

$$\left | H_{\rm K}(f)\right |=10^{-a_\text{K}(f)/20} = {\rm e}^{-a_\text{K, Np}(f)}.$$

The index „K” makes it clear, that the considered LTI system is a cable(Ger: Kabel).
For the first calculation rule, the damping function $a_\text{K}(f)$ must be used in $\rm dB$ (decibel).
For the first calculation rule, the damping function $a_\text{K, Np}(f)$ must be used in $\rm Np$ (Neper).
The following conversions apply: $\rm 1 \ dB = 0.05 \cdot \ln (10) \ Np= 0.1151 \ Np$ or $\rm 1 \ Np = 20 \cdot \lg (e) \ dB= 8.6859 \ dB$.
This applet exclusively uses dB values.

Attenuation Function of a Coaxial Cable

According to [Wel77]^[1] the Attenuation Function of a Coaxial Cable of length $l$ is given as follows:

$$a_{\rm K}(f)=(\alpha_0+\alpha_1\cdot f+\alpha_2\cdot \sqrt{f}) \cdot l.$$

It is important to note the difference between $a_{\rm K}(f)$ in $\rm dB$ and the „alpha” coefficient with other pseudo–units.
The attenuation function $a_{\rm K}(f)$ is directly proportional to the cable length $l$; $a_{\rm K}(f)/l$ is referred to as the „attenuation factor” or „kilometric attenuation”.
The frequency-independent component $α_0$ of the attenuation factor takes into account the Ohmic losses.
The frequency proportional portion $α_1 · f$ of the attenuation factor is due to the derivation losses („crosswise loss”) .
the dominant portion $α_2$ goes back to Skineffekt, which causes a lower current density inside the conductor compared to its surface. As a result, the resistance of an electric line increases with the square root of the frequency.

The constants for the standard coaxial cable with a 2.6 mm inner diameter and a 9.5 mm outer diameter ⇒ short Coax (2.6/9.5 mm) are:

$$\alpha_0 = 0.014\, \frac{ {\rm dB} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_1 = 0.0038\, \frac{ {\rm dB} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 2.36\, \frac{ {\rm dB} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}.$$

The same applies to the coaxial coaxial cable' ⇒ short Coax (1.2/4.4 mm):

$$\alpha_0 = 0.068\, \frac{ {\rm dB} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_1 = 0.0039\, \frac{ {\rm dB} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_2 =5.2\, \frac{ {\rm dB} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}.$$

These values can be calculated from the cables' geometric dimensions and have been confirmed by measurements at the Fernmeldetechnisches Zentralamt in Darmstadt – see [Wel77]^[1] . They are valid for a temperature of 20 ° C (293 K) and frequencies greater than 200 kHz.

Attenuation Function of a Two–wired Line

According to [PW95]^[2] the attenuation function of a Two–wired Line of length $l$ is given as follows:

$$a_{\rm K}(f)=(k_1+k_2\cdot (f/{\rm MHz})^{k_3}) \cdot l.$$

This function is not directly interpretable, but is a phenomenological description.

[PW95]^[2]also provides the constants determined by measurement results:

$d = 0.35 \ {\rm mm}$: $k_1 = 7.9 \ {\rm dB/km}, \hspace{0.2cm}k_2 = 15.1 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.62$,
$d = 0.40 \ {\rm mm}$: $k_1 = 5.1 \ {\rm dB/km}, \hspace{0.2cm}k_2 = 14.3 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.59$,
$d = 0.50 \ {\rm mm}$: $k_1 = 4.4 \ {\rm dB/km}, \hspace{0.2cm}k_2 = 10.8 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.60$,
$d = 0.60 \ {\rm mm}$: $k_1 = 3.8 \ {\rm dB/km}, \hspace{0.2cm}k_2 = \hspace{0.25cm}9.2 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.61$.

From these numerical values one recognizes:

The attenuation factor $α(f)$ and the attenuation function $a_{\rm K}(f) = α(f) · l$ depend significantly on the pipe diameter. The cables laid since 1994 with $d = 0.35 \ \rm (mm)$ and $d = 0.5$ mm have a 10% greater attenuation factor than the older lines with $d = 0.4$ or $d= 0.6$.
However, this smaller diameter, which is based on the manufacturing and installation costs, significantly reduces the range $l_{\rm max}$ of the transmission systems used on these lines, so that in the worst case scenario expensive intermediate generators have to be used.
The current transmission methods for copper lines prove only a relatively narrow frequency band, for example $120\ \rm kHz$ with ISDN and ca. $1100 \ \rm kHz$ with DSL. For $f = 1 \ \rm MHz$ the attenuation factor of a 0.4 mm cable is around $20 \ \rm dB/km$, so that even with a cable length of $l = 4 \ \rm km$ the Attenuation does not exceed $80 \ \rm dB$.

Conversion Between $k$ and $\alpha$ parameters

The $k$–parameters of the attenuation factor ⇒ $\alpha_{\rm I} (f)$ can be converted into corresponding $\alpha$–parameters ⇒ $\alpha_{\rm II} (f)$:

$$\alpha_{\rm I} (f) = k_1 + k_2 \cdot (f/f_0)^{k_3}\hspace{0.05cm}, \hspace{0.2cm}{\rm mit} \hspace{0.15cm} f_0 = 1\,{\rm MHz},$$

$$\alpha_{\rm II} (f) = \alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f}.$$

As a criterion of this conversion, we assume that the quadratic deviation of these two functions is minimal within a bandwidth $B$:

$$\int_{0}^{B} \left [ \alpha_{\rm I} (f) - \alpha_{\rm II} (f)\right ]^2 \hspace{0.1cm}{\rm d}f \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Minimum} \hspace{0.05cm} .$$

It is obvious that $α_0 = k_1$. The parameters $α_1$ and $α_2$ are dependent on the underlying bandwidth $B$ and are:

$$\begin{align*}\alpha_1 & = 15 \cdot (B/f_0)^{k_3 -1}\cdot \frac{k_3 -0.5}{(k_3 + 1.5)(k_3 + 2)}\cdot {k_2}/{ {f_0} }\hspace{0.05cm} ,\\ \alpha_2 & = 10 \cdot (B/f_0)^{k_3 -0.5}\cdot \frac{1-k_3}{(k_3 + 1.5)(k_3 + 2)}\cdot {k_2}/{\sqrt{f_0} }\hspace{0.05cm} .\end{align*}$$

$\text{Example 1:}$

For $k_3 = 1$ (frequency proportional attenuation factor) we get $\alpha_0 = k_0\hspace{0.05cm} ,\hspace{0.2cm} \alpha_1 = {k_2}/{ {f_0} }\hspace{0.05cm} ,\hspace{0.2cm} \alpha_2 = 0\hspace{0.05cm} .$
For $k_3 = 0.5$ (Skin effect) we get the coefficients: $\alpha_0 = k_0\hspace{0.05cm} ,\hspace{0.2cm}\alpha_1 = 0\hspace{0.05cm} ,\hspace{0.2cm} \alpha_2 = {k_2}/{\sqrt{f_0} }\hspace{0.05cm}.$
For $k_3 < 0.5$ we get a negative $\alpha_1$. Conversion is only possible for $0.5 \le k_3 \le 1$.

Umrechnung in Gegenrichtung

Fehlt noch

Channel Influence on the Binary Nyquistent Equalization

Simplified block diagram of the optimal Nyquistent equalizer

Going by the block diagram: Between the Dirac source and the decider are the frequency responses for the transmitter ⇒ $H_{\rm S}(f)$, Channel ⇒ $H_{\rm K}(f)$ and receiver ⇒ $H_{\rm E}(f)$.

In this applet

we neglect the influence of the transmitted pulse form ⇒ $H_{\rm S}(f) \equiv 1$ ⇒ dirac shaped transmission signal $s(t)$,
presuppose a binary Nyquist system with cosine–roll-off around the Nyquistf requency $f_{\rm Nyq} = [f_1 + f_2]/2 =1(2T)$ :

$$H_{\rm K}(f) · H_{\rm E}(f) = H_{\rm CRO}(f).$$

This means: The first Nyquist criterion is met ⇒
Timely successive impulses do not disturb each other ⇒ there are no Intersymbol Interferences.

In the case of white noise, the transmission quality is thus determined solely by the noise power in front of the receiver:

$$P_{\rm N} =\frac{N_0}{2} \cdot \int_{-\infty}^{+\infty} |H_{\rm E}(f)|^2 \ {\rm d}f\hspace{1cm}\text{mit}\hspace{1cm}|H_{\rm E}(f)|^2 = \frac{|H_{\rm CRO}(f)|^2}{|H_{\rm K}(f)|^2}.$$

The lowest possible noise performance results with an ideal channel ⇒ $H_{\rm K}(f) \equiv 1$ and a rectangular $H_{\rm CRO}(f) \equiv 1$ in $|f| \le f_{\rm Nyq}$:

$$P_\text{N, min} = P_{\rm N} \ \big [\text{optimal system: }H_{\rm K}(f) \equiv 1, \ r=0 \big ] = N_0 \cdot f_{\rm Nyq} .$$

$\text{Definitions:}$

As a quality criterion for a given system we use the total efficiency:

$$\eta_\text{K+R} = \frac{P_{\rm N} \ \big [\text{Given system: Channel }H_{\rm K}(f), \ \text{Roll-off factor }r \big ]}{P_{\rm N} \ \big [\text{optimal system: }H_{\rm K}(f) \equiv 1, \ r=0 \big ]} =\frac{1}{f_{\rm Nyq} } \cdot \int_{0}^{+\infty} \vert H_{\rm E}(f) \vert^2 \ {\rm d}f \le 1.$$

This system size is specified in the applet for both parameter sets in logarithm form: $10 \cdot \lg \ \eta_\text{K+R} \le 0 \ \rm dB$.

Through variation and optimization of the Roll-off factor $r$ we get the Channel efficiency:

$$\eta_\text{K} = \min_{0 \le r \le 1} \ \eta_\text{K+R} .$$

File:Applet Kabeldämpfung 3 version2.png

Frequency response with Cosine–Roll-off

$\text{Beispiel 2:}$ The graph shows the square value frequency response $\left \vert H_{\rm E}(f)\right \vert ^2 $ mit $\left \vert H_{\rm E}(f)\right \vert = H_{\rm CRO}(f) / \left \vert H_{\rm K}(f)\right \vert$ for the following boundary conditions:

Attenuation function of the channel: $a_{\rm K}(f) = 1 \ {\rm dB} \cdot \sqrt{f/\ {\rm MHz} }$,
Nyquist frequency: : $f_{\rm Nyq} = 20 \ {\rm MHz}$, Roll-off factor $r = 0.5$

This results in the following consequences:

In the area up to $f_{1} = 10 \ {\rm MHz: }$ $H_{\rm CRO}(f) = 1$ ⇒ $\left \vert H_{\rm E}(f)\right \vert ^2 = \left \vert H_{\rm K}(f)\right \vert ^{-2}$ (see yellow deposit).
The flank of $H_{\rm CRO}(f)$ is only effective from $f_{1}$ to $f_{2} = 30 \ {\rm MHz}$ and $\left \vert H_{\rm E}(f)\right \vert ^2$ decreases more and more.
The maximum of $\left \vert H_{\rm E}(f_{\rm max})\right \vert ^2$ at $f_{\rm max} \approx 11.5 \ {\rm MHz}$ is twice the value of $\left \vert H_{\rm E}(f = 0)\right \vert ^2 = 1$.
The integral over $\left \vert H_{\rm E}(f)\right \vert ^2$ is a measure of the effective noise power. In the current example this is $4.6$ times bigger than the minimal noise power (for $a_{\rm K}(f) = 0 \ {\rm dB}$ and $r=1$) ⇒ $10 \cdot \lg \ \eta_\text{K+E} \approx - 6.6 \ {\rm dB}.$

Exercises

First choose an exercise number.
An exercise description is displayed.
Parameter values are adjusted to the respective exercises.
Click „Hide solition” to display the solution.
Exercise description and solution in english

Number „0” is a „Reset” button:

Sets parameters to initial values (when loading the page).
Displays a „Reset text” to describe the applet further.

In der folgenden Beschreibung bedeutet

Blau: Verteilungsfunktion 1 (im Applet blau markiert),
Rot: Verteilungsfunktion 2 (im Applet rot markiert).

(1) First set Blue to $\text{Coax (1.2/4.4 mm)}$ and then to $\text{Coax (2.6/9.5 mm)}$. The cable length is $l_{\rm Blue}= 5\ \rm km$.

Interpret $a_{\rm K}(f)$ and $\vert H_{\rm K}(f) \vert$, in particular the functional values $a_{\rm K}(f = f_\star = 30 \ \rm MHz)$ and $\vert H_{\rm K}(f = 0) \vert$.

$\Rightarrow\hspace{0.3cm}\text{The attenuation function increases approximately }\sqrt{f}\text{ and the magnitude frequency response falls similarly to an exponential function};$

$\hspace{1.15cm}\text{Coax (1.2/4.4 mm): }a_{\rm K}(f = f_\star) = 143.3\text{ dB;}\hspace{0.5cm}\vert H_{\rm K}(f = 0) \vert = 0.96.$

$\hspace{1.15cm}\text{Coax (2.6/9.5 mm): }a_{\rm K}(f = f_\star) = 65.3\text{ dB;}\hspace{0.5cm}\vert H_{\rm K}(f = 0) \vert = 0.99;$

(2) Set Blue to $\text{Coax (1.2/4.4 mm)}$ and $l_{\rm Blue} = 3\ \rm km$. How is $a_{\rm K}(f =f_\star = 30 \ \rm MHz)$ affected by $\alpha_0$, $\alpha_1$ und $\alpha_2$?

$\Rightarrow\hspace{0.3cm}\alpha_2\text{is crucial (Skin effect). The contributions of } \alpha_0\text{ (ca. 0.1 dB) and }\alpha_1 \text{ (ca. 0.6 dB) are comparatively small.}$

(3) Additionally, set Red to $\text{Two–wired Line (0.5 mm)}$ and $l_{\rm Red} = 1\ \rm km$. What is the resulting value for $a_{\rm K}(f =f_\star= 30 \ \rm MHz)$?

Up to what length $l_{\rm Red}$ does the red attenuation function go under the blue one?

$\Rightarrow\hspace{0.3cm}\text{Red curve: }a_{\rm K}(f = f_\star) = 87.5 {\ \rm dB} \text{. The above condition is fulfilled for }l_{\rm Red} = 0.7\ {\rm km} \ \Rightarrow \ a_{\rm K}(f = f_\star) = 61.3 {\ \rm dB}.$

(4) Set Red to ${k_1}' = 0, {k_2}' = 10, {k_3}' = 0.75, {l_{\rm red} } = 1 \ \rm km$ and vary the Parameter $0.5 \le k_3 \le 1$.

What observations can be made based on $a_{\rm K}(f)$ and $\vert H_{\rm K}(f) \vert$?

$\Rightarrow\hspace{0.3cm}\text{With }k_2\text {being constant, }a_{\rm K}(f)\text{ increases with bigger values of }k_3\text{ and }\vert H_{\rm K}(f) \vert \text{ decreases faster and faster. With }k_3 =1: a_{\rm K}(f)\text{ rises linearly.}$

$\hspace{1.15cm}\text{With, }k_3 \to 0.5\text{ the attenuation function is more and more determined by the skin effect, same as the coaxial cable.}$

(5) Set Red to $\text{Two–wired Line (0.5 mm)}$ and Blue to $\text{Conversion of Red}$. For the length use $l_{\rm Rot} = l_{\rm Blau} = 1\ \rm km$.

Analyse and interpret the displayed functions $a_{\rm K}(f)$ and $\vert H_{\rm K}(f) \vert$.

$\Rightarrow\hspace{0.3cm}\text{Very good approximation of the two-wire line through the blue parameter set, both with regard to }a_{\rm K}(f) \text{, as well as }\vert H_{\rm K}(f) \vert.$

(6) We assume the settings of (5). Which parts of the attenuation function are due to ohmic loss, lateral losses and skin effect?

$\Rightarrow\hspace{0.3cm}\text{Solution based on '''Blue''': }a_{\rm K}(f = f_\star= 30 \ {\rm MHz}) = 88.1\ {\rm dB}, \hspace{0.2cm}\text{without }\alpha_0\text{: }83.7\ {\rm dB}, \hspace{0.2cm}\text{without }\alpha_0 \text{ and } \alpha_1\text{: }60.9\ {\rm dB}.$

$\hspace{1.15cm}\text{With a two-wire cable, the influence of the longitudinal and transverse losses is significantly greater than with a coaxial cable.}$

(7) Set Blue to ${\alpha_0}' = {\alpha_1}' ={\alpha_2}' = 0$ and Red to ${k_1}' = 2, {k_2}' = 0, {l_{\rm red} } = 1 \ \rm km$. Additionally, set ${f_{\rm Nyq} }' =15$ and $r= 0.5$.

How big is the total efficiency $\eta_\text{K+E}$ and the channel efficiency $\eta_\text{K}$?

$\Rightarrow\hspace{0.3cm}10 \cdot \lg \ \eta_\text{K+E} = -0.7\ \ {\rm dB}\text{ (Blue: ideal system) and }10 \cdot \lg \ \eta_\text{K+E} = -2.7\ \ {\rm dB}\text{ (Red: DC signal attenuation only)}$.

$\hspace{0.95cm}\text{The best possible rolloff factor is }r = 1.\text{ Therefore }10 \cdot \lg \ \eta_\text{K} = 0 \ {\rm dB}\text{ (Blue) or }10 \cdot \lg \ \eta_\text{K} = -2\ {\rm dB}\text{ (Red)}.$

(8) The same settings apply as in (7). Under what transmission power $P_{\rm red}$ in respect to $P_{\rm blue}$ do both systems achieve the same error probability?

$\Rightarrow\hspace{0.3cm}\text{It has to apply: }10 \cdot \lg \ P_{\rm red}/P_{\rm blue} =2 \ {\rm dB} \ \ \text{ ⇒ } \ \ P_{\rm red}/P_{\rm blue} = 10^{0.2} = 1.585.$

(9) Set Blue tof ${\alpha_0}' = {\alpha_1}' = 0, \ {\alpha_2}' = 3, \ {l_{\rm blue} }' = 2$ and Red to „Inactive”. Additionally set ${f_{\rm Nyq} }' =15$ and $r= 0.7$.

What course does $\vert H_{\rm E}(f) \vert have$? Calculate the total efficiency $\eta_\text{K+E}$ and the channel efficiency$\eta_\text{K}$

$\Rightarrow\hspace{0.3cm}\text{For} f < 7.5 {\ \rm MHz: } \vert H_{\rm E}(f) \vert = \vert H_{\rm K}(f) \vert ^{-1}.\text{ For }(f > 25 {\ \rm MHz): }\vert H_{\rm E}(f) \vert = 0.\text{ Inbetween is the effect of the CRO–flank.}$

$\hspace{0.95cm}\text{The best possible rolloff factor }r = 0.5\text{is already set: }\Rightarrow \ 10 \cdot \lg \ \eta_\text{K+E} = 10 \cdot \lg \ \eta_\text{K} \approx - 18.1 \ {\rm dB}.$

(10) Set Blue to ${\alpha_0}' = {\alpha_1}' = 0, \ {\alpha_2}' = 3, \ {l_{\rm blue} }' = 8$ and Red to „Inactive”. Additionally, set ${f_{\rm Nyq} }' =15$ and $r= 0.5$.

How big is $\vert H_{\rm E}(f = 0) \vert$? What is the maximum value of $\vert H_{\rm E}(f) \vert$? Calculate the channel efficiency $\eta_\text{K}$

$\Rightarrow\hspace{0.3cm}\vert H_{\rm E}(f = 0) \vert = \vert H_{\rm E}(f = 0) \vert ^{-1}= 1 \text{ and the maximum value } \vert H_{\rm E}(f) \vert \text{ is approximately }37500\text{ for }r=0.7 \Rightarrow 10 \cdot \lg \ \eta_\text{K+E} \approx -89.2 \ {\rm dB},$

$\hspace{0.95cm}\text{because the integral over }\vert H_{\rm E}(f) \vert^2\text{is huge. After the optimization }r=0.17 \text{ we get }10 \cdot \lg \ \eta_\text{K} \approx -82.6 \ {\rm dB}.$

(11) The same settings apply as in (10) and $r= 0.17$. Vary the cable length up to $l_{\rm blue} =10 \ \rm km$.

How much does the maximum value of $\vert H_{\rm E}(f) \vert$, the channel efficiency $\eta_\text{K}$ and the optimal rolloff factor $r_{\rm opt}$ change?

$\Rightarrow\hspace{0.3cm}\text{The maximum value of } \vert H_{\rm E}(f) \vert \text{ increases and }10 \cdot \lg \ \eta_\text{K} \text{ decreases more and more.}$

$\hspace{0.95cm}\text{At 10 km length } 10 \cdot \lg \ \eta_\text{K} \approx -104.9 \ {\rm dB} \text{ and } r_{\rm opt}=0.14\text{. For }f_\star \approx 14.5\ {\rm MHz} \Rightarrow \vert H_{\rm E}(f = f_\star) = 352000 \cdot \approx \vert H_{\rm E}(f =0)\vert$.

Vorgeschlagene Parametersätze

(1)   Nur blauer Parametersatz, $l=1$ km, $B=30$ MHz, $r=0$, $a_0=20$, $a_1=0$, $a_2=0$:
Konstante Werte $a_K=20$ dB und $\left| H_K(f)\right|=0.1$. Nur Ohmsche Verluste werden berücksichtigt.
(2) Parameter wie (1), aber zusätzlich $a_1=1$ dB/(km · MHz):
Linearer Anstieg von $a_K(f)$ zwischen $20$ dB und $50$ dB, $\left| H_K(f)\right|$ fällt beidseitig exponentiell ab.
(3)   Parameter wie (1), aber $a_0=0$, $a_1=0$, $a_2=1$ dB/(km · MHz^1/2).
$a_K(f)$ und $\left| H_K(f)\right|$ werden ausschließlich durch den Skineffekt bestimmt. $a_K(f)$ ist proportional zu $f^{1/2}$.
(4)   Parameter wie (1), aber nun mit der Einstellung „Koaxialkabel $2.6/9.5$ mm“ (Normalkoaxialkabel):
Es überwiegt der Skineffekt; $a_k$ ($f=30$ MHz)$=13.05$ dB; ohne $a_0$: $13.04$ dB, ohne $a_1=12.92$ dB.
(5)   Parameter wie (1), aber nun mit der Einstellung „Koaxialkabel $1.2/4.4$ mm“ (Kleinkoaxialkabel):
Wieder überwiegt der Skineffekt; $a_k$ ($f=30$ MHz)$=28.66$ dB; ohne $a_0$: $28.59$ dB, ohne $a_1=28.48$ dB.
(6)   Nur roter Parametersatz, $l=1 km$, $b=30$ MHz, $r=0$, Einstellung „Zweidrahtleitung $0.4$ mm“.
Skineffekt ist auch hier dominant; $a_k$ ($f=30$ MHz)$=111.4$ dB; ohne $k_1$: $106.3$ dB.
(7)   Parameter wie (6), aber nun Halbierung der Kabellänge ($l=0.5$ km):
Auch die Dämpfungswerte werden halbiert: $a_k$ ($f=30$ MHz)$=55.7$ dB; ohne $k_1$: $53.2$ dB.
(8)   Parameter wie (7), dazu im blauen Parametersatz die umgerechneten Werte der Zweidrahtleitung:
Sehr gute Approximation der $k$-Parameter durch die $a$-Parameter; Abweichung < $0.4$ dB.
(9)   Parameter wie (8), aber nun Approximation auf die Bandbreite $B=20$ MHz:
Noch bessere Approximation der $k$-Parameter durch die $a$-Parameter; Abweichung < $0.15$ dB.
(10)   Nur blauer Parametersatz, $l=1$ km, $B=30$ MHz, $r=0$, $a_0=a_1=a_2=0$; unten Darstellung $\left| H_K(f)\right|^2$:
Im gesamten Bereich ist $\left| H_K(f)\right|^2=1$; der Integralwert ist somit $2B=60$ (in MHz).
(11)   Parameter wie (10), aber nun mit Einstellung „Koaxialkabel $2.6/9.5$ mm“ (Normalkoaxialkabel):
$\left| H_K(f)\right|^2$ ist bei $f=1$ etwa $1$ und steigt zu den Rändern bis ca. $20$. Der Integralwert ist ca. $550$.
(12)   Parameter wie (11), aber nun mit der deutlich größeren Kabellänge $l=5$ km:
Deutliche Verstärkung des Effekts; Anstieg bis ca. $3.35\cdot 10^6$ am Rand und Integralwert $2.5\cdot 10^7$.
(13)   Parameter wie (12), aber nun mit Rolloff-Faktor $r=0.5$:
Deutliche Abschwächung des Effekts; Anstieg bis ca. $5.25\cdot 10^4$ ($f$ ca. $20$ MHz), Integralwert ca. $1.07\cdot 10^6$.
(14)   Parameter wie (13), aber ohne Berücksichtigung der Ohmschen Verluste ($a_0=0$):
Nahezu gleichbleibendes Ergebnis; Anstieg bis ca. $5.15\cdot 10^4$ ($f$ ca. $20$ MHz), Integralwert ca. $1.05\cdot 10^6$.
(15)   Parameter wie (14), aber auch ohne Berücksichtigung der Querverluste ($a_1=0$):
Ebenfalls kein großer Unterschied; Anstieg bis ca. $4.74\cdot 10^4$ ($f$ ca. $20$ MHz), Integralwert ca. $0.97\cdot 10^6$.
(16)   Nur roter Parametersatz, $l=1$ km, $B=30$ MHz, $r=0.5$, Einstellung „Zweidrahtleitung $0.4$ mm“:
Anstieg bis ca. $3\cdot 10^8$ ($f$ ca. $23$ MHz), Integralwert ca. $4.55\cdot 10^9$; ohne $k_1$: $0.93\cdot 10^8$ ($f$ ca. $23$ MHz) bzw. $1.41\cdot 10^9$.

Quellenverzeichnis