Aufgaben:Exercise 4.1Z: Transmission Behavior of Short Cables

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Short line section

We assume a homogeneous and reflection-free terminated line of length  $l$  so that the following applies to the spectral function at the output:

$$U_2(f) = U_1(f) \cdot {\rm e}^{-\hspace{0.02cm}\gamma(f) \hspace{0.05cm} \cdot \hspace{0.05cm}l} \hspace{0.05cm}.$$

Here  $\gamma(f)$  describes the  complex propagation function  of an extremely short line of infinitesimal length  $dx$,  which can be represented with the parameters  $R\hspace{0.05cm}'$,  $L\hspace{0.05cm}'$,  $G\hspace{0.08cm}'$ and  $C\hspace{0.08cm}'$ (see diagram) as follows:

$$\gamma(f) = \sqrt{(R\hspace{0.05cm}' + {\rm j} \cdot 2\pi f \cdot L\hspace{0.05cm}') \cdot (G\hspace{0.08cm}' + {\rm j} \cdot 2\pi f \cdot C\hspace{0.08cm}')} = \alpha (f) + {\rm j} \cdot \beta (f)\hspace{0.05cm}.$$

The real part of  $\gamma(f)$  results in

• The real part of  $\gamma(f)$  results in the attenuation function $\alpha(f)$  (per unit length).
• The imaginary part of  $\gamma(f)$  results in the phase function  $\beta(f)$ (per unit length).

After some calculation one can write for these sizes:

$$\alpha(f) = \sqrt{{1}/{2}\cdot \left (R\hspace{0.05cm}' \cdot G\hspace{0.08cm}' - \omega^2 \cdot L\hspace{0.05cm}' \cdot C\hspace{0.08cm}'\right)+ {1}/{2}\cdot \sqrt{(R\hspace{0.05cm}'\hspace{0.05cm}^2 + \omega^2 \cdot L\hspace{0.05cm}'\hspace{0.05cm}^2) \cdot (G\hspace{0.08cm}'\hspace{0.05cm}^2 + \omega^2 \cdot C\hspace{0.08cm}'\hspace{0.05cm}^2)}} \hspace{0.1cm}\bigg |_{\hspace{0.05cm}\omega \hspace{0.05cm}= \hspace{0.05cm}2\pi f},$$

$$\beta(f) = \sqrt{{1}/{2}\cdot \left (-R\hspace{0.05cm}' \cdot G\hspace{0.08cm}' + \omega^2 \cdot L\hspace{0.05cm}' C\hspace{0.08cm}'\right)+ {1}/{2}\cdot \sqrt{(R\hspace{0.05cm}'\hspace{0.05cm}^2 + \omega^2 \cdot L\hspace{0.05cm}'\hspace{0.05cm}^2) \cdot (G\hspace{0.08cm}'\hspace{0.05cm}^2 + \omega^2 \cdot C\hspace{0.08cm}'\hspace{0.05cm}^2)}} \hspace{0.1cm}\bigg |_{\hspace{0.05cm}\omega \hspace{0.05cm}= \hspace{0.05cm}2\pi f}.$$

• For the attenuation function  $a(f)$  the pseudo unit  "Neper"  (Np)  has to be added additionally and for the phase function  $b(f)$  "Radian"  (rad).  
• Since the primary line parameters are each related to the line length,  $\alpha(f)$  and  $\beta(f)$  have the units  "Np/km"  and  "rad/km",  respectively.

Another important descriptive quantity besides  $\gamma(f)$  is the  wave impedance  $Z_{\rm W}(f)$,  which gives the relationship between voltage and current of the two running waves at each location.  It holds:

$$Z_{\rm W}(f) = \sqrt{\frac {R\hspace{0.05cm}' + {\rm j} \cdot \omega L\hspace{0.05cm}'}{G\hspace{0.08cm}' + {\rm j} \cdot \omega C\hspace{0.08cm}'}} \hspace{0.1cm}\bigg |_{\hspace{0.05cm} \omega \hspace{0.05cm}= \hspace{0.05cm}2\pi f}.$$

Notes:

• The exercise belongs to the chapter  Some Results from Line Transmission Theory.

• Use the following values for the numerical calculations:

$$R\hspace{0.05cm}' = 100\,\,{\rm \Omega}/{ {\rm km} }\hspace{0.05cm},\hspace{0.3cm} G\hspace{0.08cm}' = 1\,\,{\rm µ S}/{ {\rm km}}\hspace{0.05cm},\hspace{0.3cm} 2\pi L\hspace{0.03cm}' = 2\,\,{\rm mH}/{ {\rm km}} \hspace{0.05cm},\hspace{0.3cm} 2\pi C\hspace{0.08cm}' = 200\,\,{\rm nF}/{ {\rm km}} \hspace{0.05cm}.$$

Questions

1

Specify  $\alpha(f)$,  $\beta(f)$ and  $Z_{\rm W}(f)$  for frequency  $f = 0$  ("direct current").

$\alpha(f =0) \ =$ $\ \rm Np/km$

$\beta(f = 0) \ =$ $\ \rm rad/km$

$Z_{\rm W}(f = 0) \ =$ $\ \rm \Omega$

2

Calculate the attenuation function  $\alpha(f)$  (per unit length)  for  $f = 100\ \rm kHz$.

$\alpha(f = 100\ \rm kHz) \ = \ $ $\ \rm Np/km$

3

Give the approximations of  $Z_{\rm W}(f)$  and  $\alpha(f)$,  valid for  $f → \infty$ .

$ Z_{\rm W}(f → \infty) \ = \ $ $\ \rm \Omega$

$\alpha(f → \infty) \ = \ $ $\ \rm Np/km$

4

Use  $\omega L\hspace{0.03cm}' \ll R\hspace{0.05cm}'$  and  $\omega C\hspace{0.08cm}' \gg G\hspace{0.08cm}'$  to derive an  $\alpha(f)$  approximation for  (not too)  small frequencies.
What is the attenuation function per unit length for  $ f = 1 \ \rm kHz$  and  $ f = 4 \ \rm kHz$?

$\alpha(f = 1\  \rm kHz) \ = \ $ $\ \rm Np/km$

$\alpha(f = 4\ \rm kHz) \ = \ $ $\ \rm Np/km$

5

For the same frequency range,  give a suitable approximation for the wave impedance  $Z_{\rm W}(f)$ .
What value results for  $ f = 1 \ \rm kHz$?

${\rm Re}\{Z_{\rm W}(f = 1\ \rm kHz)\} \ = \ $ $\ \rm \Omega$

${\rm Im}\{Z_{\rm W}(f = 1\ \rm kHz)\} \ = \ $ $\ \rm \Omega$

Solution

Solution

(1)  If you insert the frequency  $f = 0$  into the given equations, we obtain

$$\alpha(f = 0) = [1\,{\rm Np}] \hspace{-0.03cm}\cdot \hspace{-0.03cm} \sqrt{{1}/{2}\hspace{-0.03cm}\cdot \hspace{-0.03cm} R\hspace{0.03cm}' \hspace{-0.03cm}\cdot \hspace{-0.03cm} G\hspace{0.03cm}'+ {1}/{2}\cdot R\hspace{0.03cm}' \hspace{-0.03cm}\cdot \hspace{-0.03cm} G\hspace{0.03cm}'} = [1\,{\rm Np}] \hspace{-0.03cm}\cdot \hspace{-0.03cm} \sqrt{ R\hspace{0.03cm}' \hspace{-0.03cm}\cdot \hspace{-0.03cm} G\hspace{0.03cm}'} = [1\,{\rm Np}] \hspace{-0.03cm}\cdot \hspace{-0.03cm} \sqrt{ 100\,{\rm \Omega/km} \hspace{-0.03cm}\cdot \hspace{-0.03cm} 10^{-6}\,{\rm (\Omega \hspace{-0.03cm}\cdot \hspace{-0.03cm} km})^{-1}} \hspace{0.15cm}\underline{= 0.01\,{\rm Np}/{ {\rm km}} }\hspace{0.05cm},$$

$$\beta(f = 0) = [1\,{\rm rad}] \cdot \sqrt{-{1}/{2}\cdot R\hspace{0.03cm}' \cdot G\hspace{0.03cm}'+ {1}/{2}\cdot R\hspace{0.03cm}' \cdot G\hspace{0.03cm}'} \hspace{0.15cm}\underline{= 0 }\hspace{0.05cm},$$

$$Z_{\rm W}(f = 0) = \sqrt{\frac {R\hspace{0.03cm}'}{G\hspace{0.03cm}'}} = \sqrt{\frac {100\,{\rm \Omega/km}}{{\rm 10^{-6}/(\Omega \cdot km})}}\hspace{0.15cm}\underline{= 10\, {\rm k \Omega}}\hspace{0.05cm}.$$

The DC signal attenuation becomes relevant,

• if the useful signal is to be transmitted in the baseband and has a DC component, or
• if the network termination at the participant must be supplied with power from the local exchange (remote power supply).

(2)  With  $f = 10^{5} \ \rm Hz$  and the specified values, the following holds:

$$f \cdot 2\pi L' = 10^5\,\frac{1}{ {\rm s}} \cdot 2 \cdot 10^{-3}\,\frac{\rm \Omega \cdot s}{ {\rm km}}= 200 \,\frac{\rm \Omega }{ {\rm km}} \hspace{0.05cm},\hspace{1.05cm} f \cdot 2\pi C' = 10^5\,\frac{1}{ {\rm s}} \cdot 2 \cdot 10^{-7}\,\frac{\rm s}{ {\rm \Omega \cdot km}}= 0.02 \,\frac{\rm 1 }{ {\rm \Omega \cdot km}} \hspace{0.05cm}.$$

This results in the following for the attenuation function in "Np/km":

$$\alpha(f = 100\,{\rm kHz}) = \sqrt{ {1}/{2}\cdot \left (100 \cdot 10^{-6} - 200 \cdot 0.02 \right)+ {1}/{2} \cdot \sqrt{(100^2 + 200^2) \cdot (10^{-12} + 0.02^2)}} $$

$$ \Rightarrow \; \; \alpha(f = 100\,{\rm kHz}) \approx \sqrt{{1}/{2}\cdot \left (10^{-4} - 4 \right)+ {1}/{2}\cdot \sqrt{5 \cdot 10^{4} \cdot 4 \cdot 10^{-4}}} \approx \sqrt {-2 + \frac{\sqrt{20}}{ 2}} \hspace{0.15cm}\underline{\approx 0.486 \ {\rm Np/km}} \hspace{0.05cm}.$$

(3)  The limit for  $f → \infty$  results if one neglects the second term in the numerator  $R\hspace{0.03cm}'$  and in the denominator  $G\hspace{0.08cm}'$ :

$$\lim_{f \rightarrow \infty} \hspace{0.1cm} Z_{\rm W}(f) = \lim_{\omega \rightarrow \infty} \hspace{0.1cm} \sqrt{\frac {R\hspace{0.03cm}' + {\rm j} \cdot \omega L'}{G' + {\rm j} \cdot \omega C\hspace{0.03cm}'}} =\sqrt{\frac {2 \pi L\hspace{0.03cm}' }{2 \pi C\hspace{0.03cm}'}}=\sqrt{\frac {2 \cdot 10^{-3}\,{\rm \Omega \cdot s} } {2 \cdot 10^{-73}\,{\rm s/\Omega} }} \hspace{0.15cm}\underline{= 100\,{\rm \Omega }}\hspace{0.05cm}.$$

• The approximation for the attenuation function is more difficult to derive. Starting from

$$\alpha(\omega) = \sqrt{ {1}/{2}\cdot \left (R\hspace{0.03cm}' G\hspace{0.03cm}' - \omega^2 \cdot L\hspace{0.03cm}' C\hspace{0.03cm}'\right)+ {1}/{2}\sqrt{(R\hspace{0.03cm}'\hspace{0.03cm}^2 + \omega^2 \cdot L\hspace{0.03cm}'\hspace{0.03cm}^2) \cdot (G\hspace{0.03cm}'\hspace{0.03cm}^2 + \omega^2 \cdot C\hspace{0.03cm}\hspace{0.03cm}'^2)}}$$

then also the following applies:

$$2 \cdot \alpha^2(\omega) = R\hspace{0.03cm}' G\hspace{0.03cm}' + \omega^2 \cdot L' C'\cdot \left [-1 +\sqrt{(1 + \frac{R\hspace{0.03cm}'^2}{ \omega^2 \cdot L\hspace{0.03cm}'^2}) \cdot (1 + \frac{G\hspace{0.03cm}'^2}{ \omega^2 \cdot C\hspace{0.03cm}'^2})} \hspace{0.1cm} \right]$$

$$\Rightarrow \; \; 2 \cdot \alpha^2(\omega) \approx R\hspace{0.03cm}' G\hspace{0.03cm}' + \omega^2 \cdot L' C\hspace{0.03cm}'\cdot \left [-1 +\sqrt{1 + \frac{R'^2}{ \omega^2 \cdot L'^2}+ \frac{G\hspace{0.03cm}'^2}{ \omega^2 \cdot C\hspace{0.03cm}'^2}} \hspace{0.1cm} \right].$$

• Using the approximation $\sqrt{1 + x}\approx 1+x/2$  valid for small  $x$ , one arrives at the intermediate result for (infinitely) large frequencies:

$$2 \cdot \alpha^2(\omega \rightarrow \infty) = R\hspace{0.03cm}' G\hspace{0.05cm}' + \omega^2 \cdot L' C\hspace{0.05cm}'\cdot \left [ -1 +1 + {1}/{2} \cdot \left ( \frac{R\hspace{0.03cm}'\hspace{0.03cm}^2}{ \omega^2 \cdot L\hspace{0.03cm}'\hspace{0.03cm}^2}+ \frac{G\hspace{0.03cm}'\hspace{0.03cm}^2}{ \omega^2 \cdot C\hspace{0.03cm}'\hspace{0.03cm}^2} \right) \hspace{0.1cm} \right] $$

$$\Rightarrow \hspace{0.3cm} 2 \cdot \alpha^2(\omega \rightarrow \infty) = \frac{2 \cdot R\hspace{0.03cm}' G\hspace{0.03cm}' C\hspace{0.03cm}' L'+ R\hspace{0.03cm}'\hspace{0.03cm}^2 C\hspace{0.03cm}'\hspace{0.03cm}^2+ G\hspace{0.03cm}'\hspace{0.03cm}^2 L\hspace{0.03cm}'\hspace{0.03cm}^2}{2 \cdot C\hspace{0.03cm}' L\hspace{0.03cm}' }= \frac{(R\hspace{0.03cm}' C\hspace{0.03cm}' + G\hspace{0.03cm}' L\hspace{0.03cm}')^2}{2 \cdot C\hspace{0.03cm}' L\hspace{0.03cm}' }$$

$$\Rightarrow \hspace{0.3cm} \alpha(\omega \rightarrow \infty) = {1}/{2}\cdot \frac{R\hspace{0.03cm}' C\hspace{0.03cm}' + G\hspace{0.03cm}' L\hspace{0.03cm}'}{\sqrt{ C\hspace{0.03cm}' L\hspace{0.03cm}' }}= {1}/{2}\cdot \left [R\hspace{0.03cm}' \cdot \sqrt{\frac{C\hspace{0.03cm}'}{L\hspace{0.03cm}'}}+G\hspace{0.03cm}' \cdot \sqrt{\frac{L\hspace{0.03cm}'}{C\hspace{0.03cm}'}}\right]\hspace{0.05cm}.$$

• With the numerical values inserted, we get

$$\alpha(f \rightarrow \infty) = \alpha(\omega \rightarrow \infty) = {0.5\,{\rm Np/km}}\cdot \left [100 \cdot \sqrt{\frac{2 \cdot 10^{-7}}{2 \cdot10^{-3}}}+10^{-6} \cdot \sqrt{\frac{2 \cdot10^{-3}}{2 \cdot10^{-7}}}\right] \hspace{0.15cm}\underline{\approx 0.5 \, {\rm Np}/{\rm km}}\hspace{0.05cm}.$$

(4)  For small frequencies,  $\omega L\hspace{0.03cm}' \ll R\hspace{0.03cm}'$  and  $ \omega C\hspace{0.03cm}' \gg G\hspace{0.03cm}'$ apply.

• Neglecting the  $\omega^2$–part, one obtains:

$$\alpha(f) = \sqrt{\frac {1}{2}\cdot \left (R\hspace{0.03cm}' G\hspace{0.03cm}' - \omega^2 \cdot L\hspace{0.03cm}' C\hspace{0.03cm}'\right)+ \frac {1}{2}\sqrt{(R\hspace{0.03cm}'\hspace{0.03cm}^2 + \omega^2 \cdot L\hspace{0.03cm}'\hspace{0.03cm}^2) \cdot (G\hspace{0.03cm}'\hspace{0.03cm}^2+ \omega^2 \cdot C\hspace{0.03cm}'\hspace{0.03cm}^2)}} \hspace{0.1cm}\bigg |_{\hspace{0.05cm}\omega \hspace{0.05cm}= \hspace{0.05cm}2\pi f}$$

$$ \Rightarrow \hspace{0.3cm} \alpha(f) \approx \sqrt{\frac {R\hspace{0.03cm}' G\hspace{0.03cm}'}{2}+ \frac {R\hspace{0.03cm}' \cdot \omega C\hspace{0.03cm}'}{2}} \hspace{0.1cm}\bigg |_{\hspace{0.03cm}\omega \hspace{0.05cm}= \hspace{0.05cm}2\pi f} \approx \sqrt{ {1}/{2} \cdot f \cdot R\hspace{0.03cm}' \cdot 2 \pi C\hspace{0.03cm}'} \hspace{0.05cm}.$$

• Here it is considered that the first part can be neglected according to subtask  (1)  except for the frequency  $f = 0$ .
• For the frequency  $f = 1 \ \rm kHz$  we get the approximation

$$\alpha(f = 1\,{\rm kHz}) = \sqrt{ {1}/{2} \cdot 10^{3}\,{\rm Hz} \cdot 100\,\frac{\rm \Omega }{ {\rm km}} \cdot 2 \cdot 10^{-7} \,\frac{\rm s }{ {\rm \Omega \cdot km}}} \hspace{0.15cm}\underline{= 0.1\,{\rm Np }/{ {\rm km}}} \hspace{0.05cm}.$$

• For frequency  $f = 4 \ \rm kHz$  the attenuation function per unit length is twice as large:

$$\alpha(f = 4\,{\rm kHz}) \hspace{0.15cm}\underline{ = 0.2\,{\rm Np }/{ {\rm km}}} \hspace{0.05cm}.$$

(5)  The wave impedance at low frequencies is approximated by:

$$Z_{\rm W}(f) = \sqrt{\frac {R\hspace{0.03cm}' + {\rm j} \cdot f \cdot 2 \pi L\hspace{0.03cm}'}{G\hspace{0.03cm}' + {\rm j} \cdot f \cdot 2 \pi C\hspace{0.03cm}'}} \approx \sqrt\frac{1 }{ {\rm j}} \cdot \sqrt{\frac {R\hspace{0.03cm}' }{ f \cdot 2 \pi C\hspace{0.03cm}'}}= (1 - {\rm j})\cdot \sqrt{\frac {R\hspace{0.03cm}' }{ 2 \cdot f \cdot 2 \pi C\hspace{0.03cm}'}}\hspace{0.05cm}.$$

• With the specified line fittings we obtain:

$${\rm Re}\{Z_{\rm W}(f= 1\,{\rm kHz})\} = \sqrt{\frac {100\,{\rm \Omega/km }}{ 2 \cdot 10^{3}\,{\rm Hz} \cdot 2 \cdot 10^{-7} \,{\rm s/(\Omega \cdot km) }}} \hspace{0.15cm}\underline{= 500\,{\rm \Omega}}\hspace{0.05cm},$$

$$ {\rm Im}\{Z_{\rm W}(f= 1\,{\rm kHz})\} = -{\rm Re}\{Z_{\rm W}(f= 1\,{\rm kHz})\}\hspace{0.15cm}\underline{= -500\,{\rm \Omega}}\hspace{0.05cm}.$$