Exercise 4.6: k-parameters and alpha-parameters

Attenuation function per unit length,
valid for "copper twin wire" (0.5 mm)

For symmetrical copper twisted pairs, the following empirical formula can be found in [PW95], which is valid for the frequency range $0 \le f \le 30 \ \rm MHz$:

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

In contrast, the attenuation function per unit length of a coaxial cable is usually given in the following form:

$$\alpha_{\rm II}(f) = \alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f}\hspace{0.05cm}.$$

Especially for the calculation of impulse response and rectangular response it is advantageous also for the copper twisted pairs to choose the second representation form with the cable parameters $\alpha_0$, $\alpha_1$ and $\alpha_2$ instead of the representation with $k_1$, $k_2$ and $k_3$.

For the conversion, one proceeds as follows:

From above equations, it is obvious that the coefficient characterizing the DC signal attenuation is $\alpha_0 = k_1$.
To determine $\alpha_1$ and $\alpha_2$, it is assumed that the mean square error should be minimum in the range of a given bandwidth $B$:

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

The difference $\varepsilon^2(f)$ and the mean square error ${\rm E}\big[\varepsilon^2(f)\big]$ are obtained as follows:

$$\varepsilon^2(f) = \big [ \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f} - k_2 \cdot (f/f_0)^{k_3}\big ]^2 =\alpha_1^2 \hspace{0.05cm}\cdot\hspace{0.05cm} f^2 + 2 \alpha_1 \alpha_2 \hspace{0.05cm}\cdot\hspace{0.05cm} f^{1.5} + \alpha_1^2 \hspace{0.05cm}\cdot\hspace{0.05cm} f + k_2^2\hspace{0.05cm}\cdot\hspace{0.05cm} \frac{f^{2k_3}}{f_0^{2k_3}} - 2 k_2 \alpha_1 \hspace{0.05cm}\cdot\hspace{0.05cm} \frac{f^{k_3+1}} {f_0^{k_3}}-{2 k_2 \alpha_2} \hspace{0.05cm}\cdot\hspace{0.05cm} \frac{f^{k_3+0.5}}{f_0^{k_3}}$$

$$\Rightarrow \hspace{0.3cm}{\rm E}\big[\varepsilon^2(f)\big] = \alpha_1^2 \hspace{0.05cm}\cdot\hspace{0.05cm}\frac{B^3}{3} + \frac{4}{5} \hspace{0.05cm}\cdot\hspace{0.05cm}\alpha_1 \alpha_2 \hspace{0.05cm}\cdot\hspace{0.05cm}B^{2.5} + \alpha_1^2 \hspace{0.05cm}\cdot\hspace{0.05cm} \frac{B^2}{2} + \frac{k_2^2}{2k_3 +1} \hspace{0.05cm}\cdot\hspace{0.05cm} \frac{B^{2k_3+1}}{f_0^{2k_3}} - \hspace{0.15cm} \frac{2 k_2 \alpha_1}{k_3 + 2} \hspace{0.05cm}\cdot\hspace{0.05cm} $$

This equation contains the cable parameters $\alpha_1$, $\alpha_2$, $k_2$ and $k_3$ to be calculated as well as the bandwidth $B$, within which the approximation should be valid.

By setting the derivatives of ${\rm E}\big[\varepsilon^2(f)\big]$ to $\alpha_1$ and $\alpha_2$ to zero, two equations are obtained for the best possible coefficients $\alpha_1$ and $\alpha_2$ that minimize the mean square error. These can be represented in the following form:

$$\frac{{\rm d}\,{\rm E}\big[\varepsilon^2(f)\big]}{{\rm d}\,{\alpha_1}} = 0 \hspace{0.2cm} \Rightarrow \hspace{0.2cm} \alpha_1 + C_1 \cdot \alpha_2 + C_2 = 0 \hspace{0.05cm} ,$$

$$\frac{{\rm d}\,{\rm E}\big[\varepsilon^2(f)\big]}{{\rm d}\,{\alpha_2}} = 0 \hspace{0.2cm} \Rightarrow \hspace{0.2cm} \alpha_1 + D_1 \cdot \alpha_2 + D_2 = 0 \hspace{0.05cm} . $$

From the equation $C_1 \cdot \alpha_2 + C_2 = D_1 \cdot \alpha_2 + D_2$, the coefficient $\alpha_2$ can be calculated and then the coefficient $\alpha_1$ can be calculated from each of the two equations above.

The graph shows the attenuation function per unit length for a copper twin wire with $\text{0.5 mm}$ diameter, whose $k$–parameters are:

$$k_1 = 4.4\, {\rm dB}/{\rm km} \hspace{0.05cm}, \hspace{0.2cm} k_2 = 10.8\, {\rm dB}/{\rm km}\hspace{0.05cm}, \hspace{0.2cm}k_3 = 0.60\hspace{0.05cm} \hspace{0.05cm}.$$

The red curve shows the function $\alpha(f)$ calculated with this parameters. For $f = 30 \ \rm MHz$ the attenuation function per unit length is $\alpha(f)= 87.5 \ \rm dB/km$.
The blue curve gives the approximation with the $\alpha$–coefficients. This is almost indistinguishable from the red curve within the drawing accuracy.

Notes:

The exercise belongs to the chapter Properties of Balanced Copper Pairs.

You can use the (German language) interactive SWF applet "Dämpfung von Kupferkabeln" ⇒ "Attenuation of copper cables" .
[PW95] denotes the following literature reference: Pollakowski, P.; Wellhausen, H.-W.: Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz. Deutsche Telekom AG, Forschungs- und Technologiezentrum Darmstadt, 1995.

Questions

Solution

(1) Solutions 1 and 6 are correct:

The derivative of the given expected value with respect to $\alpha_1$ gives:

$$\frac{{\rm d}\,{\rm E}[\varepsilon^2(f)]}{{\rm d}\,{\alpha_1}} = \frac{2}{3}\cdot B^3 \cdot \alpha_1 + \frac{4}{5}\cdot B^{2.5} \cdot \alpha_2 - \frac{2 k_2 }{k_3 + 2} \cdot \frac{B^{k_3+2}}{f_0^{k_3}}= 0 \hspace{0.05cm} .$$

By setting it to zero and dividing by $2B^2/3$, we obtain:

$$\alpha_1 + \frac{6}{5}\cdot B^{-0.5} \cdot \alpha_2 - \frac{3 k_2 }{k_3 +2} \cdot \frac{B^{k_3-1}}{f_0^{k_3}}= 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} C_1 = \frac{6}{5}\cdot B^{-0.5} \hspace{0.05cm} , \hspace{0.5cm} C_2 = - \frac{3 k_2 }{k_3 +2} \cdot \frac{B^{k_3-1}}{f_0^{k_3}} \hspace{0.05cm} .$$

(2) Solutions 2 and 5 are correct:

Using the same procedure as in subtask (1), we obtain:

$$\frac{{\rm d}\,{\rm E}[\varepsilon^2(f)]}{{\rm d}\,{\alpha_2}} = \frac{4}{5}\cdot B^{2.5} \cdot \alpha_1 + B^{2} \cdot \alpha_2 - \frac{2 k_2 }{k_3 + 1.5} \cdot \frac{B^{k_3+1.5}}{f_0^{k_3}}= 0$$

$$\Rightarrow \hspace{0.3cm} \alpha_1 + \frac{5}{4}\cdot B^{-0.5} \cdot \alpha_2 - \frac{2.5 \cdot k_2 }{k_3 +1.5} \cdot \frac{B^{k_3-1}}{f_0^{k_3}}= 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}D_1 = \frac{5}{4}\cdot B^{-0.5} \hspace{0.05cm} , \hspace{0.3cm}D_2 = - \frac{2.5 \cdot k_2 }{k_3 +1.5} \cdot \frac{B^{k_3-1}}{f_0^{k_3}} \hspace{0.05cm} .$$

(3) Both solutions are correct:

From $C_1 \cdot \alpha_2 + C_2 = D_1 \cdot \alpha_2 + D_2$ we obtain a linear equation for $\alpha_2$. With the result from (2) we can write:

$$\alpha_2 = \frac{D_2 - C_2}{C_1 - D_1} = \frac{- \frac{2.5 \cdot k_2 }{k_3 +1.5} \cdot \frac{B^{k_3-1}}{f_0^{k_3}} + \frac{3 k_2 }{k_3 +2} \cdot \frac{B^{k_3-1}}{f_0^{k_3}}}{{6}/{5}\cdot B^{-0.5} - {5}/{4}\cdot B^{-0.5}} = \frac{- {2.5 \cdot k_2 }\cdot(k_3 +2) + {3 k_2 }\cdot (k_3 +1.5) }{({6}/{5} - {5}/{4})(k_3 +1.5)(k_3 +2)} \cdot \frac{B^{k_3-0.5}}{f_0^{k_3}}$$

$$ \Rightarrow \hspace{0.3cm}\alpha_2 = 10 \cdot (B/f_0)^{k_3 -0.5}\cdot \frac{1-k_3}{(k_3 + 1.5)(k_3 + 2)}\cdot \frac {k_2}{\sqrt{f_0}} \hspace{0.05cm} .$$

For the parameter $\alpha_1$ then holds:

$$\alpha_1 = - C_1 \cdot \alpha_2 - C_2 = -\frac{6}{5}\cdot B^{-0.5} \cdot 10 \cdot (B/f_0)^{k_3 -0.5}\cdot \frac{1-k_3}{(k_3 + 1.5)(k_3 + 2)}\cdot \frac {k_2}{\sqrt{f_0}} +\frac{3 k_2 }{k_3 +2} \cdot \frac{B^{k_3-1}}{f_0^{k_3}}$$

$$ \Rightarrow \hspace{0.3cm}\alpha_1 = (B/f_0)^{k_3 -1}\cdot \frac{-12 \cdot (1-k_3) + 3 \cdot (k_3 + 1.5)}{(k_3 + 1.5)(k_3 + 2)} \cdot \frac {k_2}{f_0} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\alpha_1 =15 \cdot (B/f_0)^{k_3 -1}\cdot \frac{k_3 -0.5}{(k_3 + 1.5)(k_3 + 2)}\cdot \frac {k_2}{f_0}\hspace{0.05cm} .$$

Regardless of the bandwidth, we obtain for $k_3 = 1$:

$$\alpha_1 = (B/f_0)^{k_3 -1}\cdot \frac{15 \cdot (k_3 -0.5)}{(k_3 + 1.5)(k_3 + 2)}\cdot \frac {k_2}{f_0} = \frac{15 \cdot 0.5}{2.5 \cdot 3}\cdot \frac {k_2}{f_0} \hspace{0.15cm}\underline{ = {k_2}/{f_0}}\hspace{0.05cm} ,$$

$$ \alpha_2 = (B/f_0)^{k_3 -0.5}\cdot \frac{10 \cdot (1-k_3)}{(k_3 + 1.5)(k_3 + 2)}\cdot \frac {k_2}{\sqrt{f_0}}\hspace{0.15cm}\underline{= 0} \hspace{0.05cm} .$$

In contrast, for $k_3 = 0.5$:

$$\alpha_1 = (B/f_0)^{k_3 -1}\cdot \frac{15 \cdot (k_3 -0.5)}{(k_3 + 1.5)(k_3 + 2)}\cdot \frac {k_2}{f_0} \hspace{0.15cm}\underline{= 0}\hspace{0.05cm} ,$$

$$ \alpha_2 = (B/f_0)^{k_3 -0.5}\cdot \frac{10 \cdot (1-k_3)}{(k_3 + 1.5)(k_3 + 2)}\cdot \frac {k_2}{\sqrt{f_0}}= \frac{10 \cdot 0.5}{2 \cdot 2.5}\cdot \frac {k_2}{\sqrt{f_0}} = \hspace{0.15cm}\underline{ {k_2}/{\sqrt{f_0}}} \hspace{0.05cm} .$$

(4) For the two coefficients, with $k_2 = 10.8 \ \rm dB/km$, $k_3 = 0.6 \ \rm dB/km$ and $B/f_0 = 30$:

$$\alpha_1 = (B/f_0)^{k_3 -1}\cdot \frac{15 \cdot (k_3 -0.5)}{(k_3 + 1.5)(k_3 + 2)}\cdot \frac {k_2}{f_0} = 30^{-0.4}\cdot \frac{15 \cdot 0.1}{2.1 \cdot 2.6}\cdot \frac {10.8 \, {\rm dB/km} }{1 \, {\rm MHz}} \hspace{0.15cm}\underline{ \approx 0.761\, {{\rm dB} }/{({\rm km \cdot MHz})}} \hspace{0.05cm} ,$$

$$ \alpha_2 = (B/f_0)^{k_3 -0.5}\cdot \frac{10 \cdot (1-k_3)}{(k_3 + 1.5)(k_3 + 2)}\cdot \frac {k_2}{\sqrt{f_0}}= \frac {k_2}{\sqrt{f_0}} = 30^{0.1}\cdot \frac{10 \cdot 0.4}{2.1 \cdot 2.6}\cdot \frac {10.8 \, {\rm dB/km} }{1 \, {\rm MHz^{0.5}}} \hspace{0.15cm}\underline{ \approx 11.1\, {{\rm dB} }/{({\rm km \cdot \sqrt{MHz}}})}\hspace{0.05cm} .$$

(5) According to the given equation $\alpha_{\rm II}(f)$ thus also holds:

$$\alpha_{\rm II}(f = 30 \, {\rm MHz}) = \alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f} = \big [ \hspace{0.05cm} 4.4 + 0.761 \cdot 30 + 11.1 \cdot \sqrt {30}\hspace{0.05cm} \big ]\frac {\rm dB}{\rm km } \hspace{0.15cm}\underline{\approx 88.1\, {\rm dB}/{\rm km }} \hspace{0.05cm}.$$

	$C_1 = 6/5 \cdot B^{-0.5}$,
	$C_1 = 5/4 \cdot B^{-0.5}$,
	$C_1 = 4/3 \cdot B^{2}$,
	$C_2 = -4/3 \cdot B^{-2$}$,
	$C_2 = -5/2 \cdot k_2/(k_3 +1.5) \cdot B^{k_3 -1} \cdot f_0^{-k_3}$,
	$C_2 = -3 \cdot k_2/(k_3 +2) \cdot B^{k_3 -1} \cdot f_0^{-k_3}$.

	$D_1 = 6/5 \cdot B^{-0.5}$,
	$D_1 = 5/4 \cdot B^{-0.5}$,
	$D_1 = 4/3 \cdot B^{2}$,
	$D_2 = -4/3 \cdot B^{-2}$,
	$D_2 = -5/2 \cdot k_2/(k_3 +1.5) \cdot B^{k_3 -1} \cdot f_0^{-k_3}$,
	$D_2 = -3 \cdot k_2/(k_3 +2) \cdot B^{k_3 -1} \cdot f_0^{-k_3}$.

	For $k_3=1.0$, $\alpha_1 = k_2/f_0$ and $\alpha_2 = 0$.
	For $k_3=0.5$, $\alpha_1 = 0$ and $\alpha_2 = k_2/f_0^{0.5}$.