Applets:Attenuation of Copper Cables

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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] 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] . 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] 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]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   ⇒   $\alpha_{\rm II} (f)$:

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

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

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:

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

In the opposite direction the conversion rule for the exponent is:

$$k_3 = \frac{A + 0.5} {A +1}, \hspace{0.2cm}\text{Auxiliary variable: }A = \frac{2} {3} \cdot \frac{\alpha_1 \cdot \sqrt{f_0}}{\alpha_2} \cdot \sqrt{B/f_0}.$$

With this result you can specify $ k_2 $ with each of the above equations:

$\text{Example 1:}$ 

• 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 $0.5 \le k_3 \le 1$.
• For $0.5 \le k_3 \le$ we get the coefficients $\alpha_1 > 0$ and $\alpha_2 > 0$, which are also dependent on $B/f_0$.
• From $\alpha_1 = 0.3\, {\rm dB}/ ({\rm km \cdot MHz}) \hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 3\, {\rm dB}/ ({\rm km \cdot \sqrt{MHz} })\hspace{0.05cm},\hspace{0.2cm}B = 30 \ \rm MHz$ folgt $k_3 = 0.63$ und $k_2 = 2.9 \ \rm dB/km$.

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:

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

Datei:Applet Kabeldämpfung 3 version2.png

Frequency response with Cosine–Roll-off

$\text{Example 2:}$  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 „Show solution” to display the solution.
• Exercise description and solution are in english.

Number „0” is a „Reset” button:

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

In the following desctiption:

• Blue means   Distribution 1 (blau in the applet),
• Red means     Distributeion 2 (red in the applet).

(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 $\vert H_{\rm K}(f = 0) \vert$.

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

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

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

(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 $\alpha_2$?

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

(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 stay under the blue one?

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

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

How do the parameters affect $a_{\rm K}(f)$ and $\vert H_{\rm K}(f) \vert$?

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

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

(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?

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

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

(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 are 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 $P_{\rm red}$ with respect to $P_{\rm blue}$ do both systems achieve the same error probability?

$\Rightarrow\hspace{0.3cm}\text{We need to achieve }10 \cdot \lg \frac{P_{\rm red}}{P_{\rm blue}} = 2 \ {\rm dB} \ \ \Rightarrow \ \ \frac{P_{\rm red}}{P_{\rm blue}} = 10^{0.2} = 1.585.$

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

How does $\vert H_{\rm E}(f) \vert$ look like? Calculate the total efficiency $\eta_\text{K+E}$ and the channel efficiency$\eta_\text{K}$

.

$\Rightarrow\hspace{0.3cm}\text{For} f < 7.5 {\ \rm MHz: } \vert H_{\rm E}(f) \vert = \vert H_{\rm K}(f) \vert ^{-1}.\text{ For } f > 25 {\ \rm MHz: }\vert H_{\rm E}(f) \vert = 0.\text{ In between, the effect of the CRO edge can be observed.}$

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

(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 do 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?

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

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

References

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