Aufgaben:Exercise 4.1: Attenuation Function

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Attenuation function per unit length   ⇒   $\alpha(f)$  and two bounds

The attenuation function per unit length   ⇒   $\alpha(f)$  – pronounced  "alpha"  – of a line indicates the attenuation related to the line length.  This quantity is determined by the primary line parameters  $R\hspace{0.05cm}'$,  $L\hspace{0.05cm}'$,  $G\hspace{0.08cm}'$ and  $C\hspace{0.08cm}'$  where the exact equation is somewhat complicated.   Therefore, two more manageable approximations have been developed:

$$\frac{\alpha_{_{\rm I}}(f)}{\rm Np} = {1}/{2} \cdot \left [R\hspace{0.05cm}' \cdot \sqrt{{C\hspace{0.08cm}'}/{ L\hspace{0.05cm}'} } + G\hspace{0.08cm}' \cdot \sqrt{{L\hspace{0.05cm}'}/{ C\hspace{0.08cm}'} }\hspace{0.05cm}\right ]\hspace{0.05cm},$$

$$\frac{\alpha_{_{\rm II}}(f)}{\rm Np} = \sqrt{1/2 \cdot \omega \cdot {R\hspace{0.05cm}' \cdot C\hspace{0.08cm}'} }\hspace{0.1cm}\bigg |_{\omega \hspace{0.05cm}= \hspace{0.05cm}2\pi f}\hspace{0.05cm}.$$

These two approximations are shown in the graph together with the actual  $\alpha(f)$  curve.  The intersection of  $\alpha_{\rm I}(f)$  and  $\alpha_{\rm II}(f)$  gives the characteristic frequency  $f_∗$  with the following meaning:

• For  $f \gg f_∗$  holds  $α(f) ≈ α_{\rm I}(f)$.
• For  $f \ll f_∗$  holds  $α(f) ≈ α_{\rm II}(f)$.

These approximations are used to determine the function  $\alpha(f)$  for a message signal of frequency  $f_0 = 2 \ \rm kHz$,  whereby the following transmission media are to be considered:

• a copper cable with  $0.6 \ \rm mm$  diameter:

$$R\hspace{0.05cm}' = 130\,\,{\rm \Omega}/{ {\rm km} }\hspace{0.05cm},\hspace{0.3cm}L\hspace{0.03cm}' = 0.6\,\,{\rm mH}/{ {\rm km}} \hspace{0.05cm},\hspace{0.3cm}G\hspace{0.08cm}' = 1\,\,{\rm µ S}/{ {\rm km}} \hspace{0.05cm},\hspace{0.3cm}C\hspace{0.08cm}' = 35\,\,{\rm nF}/{ {\rm km}} \hspace{0.05cm},$$

• a bronze overhead line with  $5 \ \rm mm$  diameter:

$$R\hspace{0.05cm}' = 2.2\,\,{\rm \Omega}/{ {\rm km} }\hspace{0.05cm},\hspace{0.3cm}L\hspace{0.03cm}' = 1.8\,\,{\rm mH}/{ {\rm km}} \hspace{0.05cm},\hspace{0.3cm}G\hspace{0.08cm}' = 0.5\,\,{\rm µ S}/{ {\rm km}} \hspace{0.05cm},\hspace{0.3cm}C\hspace{0.08cm}' = 6.7\,\,{\rm nF}/{ {\rm km}}\hspace{0.05cm}.$$

Notes:

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

• The reference unit in the above equations for  $α_{\rm I}(f)$  and  $α_{\rm II}(f)$  and thus also for the total attenuation function (per unit length)  $α(f)$  results from the fact that the magnitude frequency response is defined as  $|H(f)| = {\rm e}^{-a}$ .
• From this follows for the attenuation  $ a = - {\rm ln} \; |H(f)|$, where the relationship via the natural logarithm is denoted by "Neper" (Np).
• The unit of the attenuation function per unit length  $α = a/l$  is thus  "Np/km".

Questions

1 Calculate for the copper cable and the bronze cable the given approximation  $\alpha_{\rm I}$ .

${\rm Copper}\hspace{-0.1cm}:\hspace{0.2cm} \alpha_{\rm I} \ = \ $ $\ \rm Np/km$

${\rm Bronze}\hspace{-0.1cm}:\hspace{0.2cm} \alpha_{\rm I} \ = \ $ $\ \rm Np/km$

2 Specify the respective characteristic frequency  $f_*$  that bounds the ranges of validity of the two approximations.

${\rm Copper}\hspace{-0.1cm}:\hspace{0.2cm} f_* \ = \ $ $\ \rm kHz$

${\rm Bronze}\hspace{-0.1cm}:\hspace{0.2cm} f_* \ = \ $ $\ \rm kHz$

3 Using the two approximations,  give the attenuation function for frequency  $f_0 = 2 \ \rm kHz$ .

${\rm Copper}\hspace{-0.1cm}: \hspace{0.2cm} \alpha (f = f_0) \ = \ $ $\ \rm Np/km$

${\rm Bronze}\hspace{-0.1cm}:\hspace{0.2cm} \alpha (f = f_0) \ = \ $ $\ \rm Np/km$

Solution

Solution

(1)  For the copper cable,  $R\hspace{0.03cm}' = 130\,\,{\rm \Omega}/{ {\rm km} }\hspace{0.05cm},\hspace{0.3cm}

L' = 0.6\,\,{\rm mH}/{ {\rm km}} \hspace{0.05cm},\hspace{0.3cm}
G\hspace{0.03cm}' = 1\,\,{\rm µ S}/{ {\rm km}} \hspace{0.05cm},\hspace{0.3cm}
C\hspace{0.03cm}' = 35\,\,{\rm nF}/{ {\rm km}} \hspace{0.05cm}$:

$${\alpha_{_{\rm I}}(f)} = \frac{1 \,\rm Np/km}{2} \cdot\left [130\,{\rm \Omega} \cdot \sqrt{\frac{35 \cdot 10^{-9}\,{\rm s/\Omega}}{ 0.6 \cdot 10^{-3}\,{\rm \Omega \,s}} }+ 10^{-6}\,{\rm \Omega^{-1}} \cdot \sqrt{\frac{0.6 \cdot 10^{-3}\,{\rm \Omega \,s}}{ 35 \cdot 10^{-9}\,{\rm s/\Omega}} }\hspace{0.1cm}\right] $$

$$ \Rightarrow \; \alpha_{\rm I}(f) = 1/2 \cdot\left [130 \cdot 7.638 \cdot 10^{-3}+ 10^{-6} \cdot 0.131 \cdot 10^{3}\right] {\rm Np/km} \hspace{0.15cm}\underline{= 0.496\,{\rm Np/km}}\hspace{0.05cm}.$$

For the bronze line the result is  $R\hspace{0.03cm}' = 2.2\,\,{\rm \Omega}/{ {\rm km} }\hspace{0.05cm},\hspace{0.3cm}

L' = 1.8\,\,{\rm mH}/{ {\rm km}} \hspace{0.05cm},\hspace{0.3cm}
G\hspace{0.03cm}' = 0.5\,\,{\rm µ S}/{ {\rm km}} \hspace{0.05cm},\hspace{0.3cm}
C\hspace{0.03cm}' = 6.7\,\,{\rm nF}/{ {\rm km}}$:

$$\alpha_{\rm I}(f) = 1/2 \cdot\left [2.2 \cdot \sqrt{\frac{6.7 \cdot 10^{-9}}{ 1.8 \cdot 10^{-3}} }+ 0.5 \cdot 10^{-6} \cdot \sqrt{\frac{ 1.8 \cdot 10^{-3}} {6.7 \cdot 10^{-9}}}\hspace{0.1cm}\right] $$

$$ \Rightarrow \; \alpha_{\rm I}(f) = \frac{1 \,\rm Np/km}{2} \cdot\big [4.244 \cdot 10^{-3}+ 0.259 \cdot 10^{-3}\big] {\rm Np/km}\hspace{0.15cm}\underline{= 0.0023\,{\rm Np}/{ {\rm km} }}\hspace{0.05cm}.$$

(2)  The bound  $α_{\rm I}(f)$  calculated in subtask  (1)  is valid only for  $f \gg f_∗$,  while the bound  $α_{\rm II}(f)$  is valid for  $f \ll f_∗$.

• The characteristic frequency is obtained as the intersection of the two approximations:

$$\alpha_{\rm II}(f = f_{\star}) = \sqrt{1/2 \cdot \omega_{\star} \cdot R' \cdot C' }\hspace{0.1cm}\bigg |_{\omega_{\star} \hspace{0.05cm}= \hspace{0.05cm}2\pi f_{\star}} = \alpha_{\rm I}(f = f_{\star})$$

• For the copper cable with  $\text{0.6 mm}$  diameter,  the following equation holds:

$$f_{\star} = \frac {{\alpha^2_{_{\rm I}}(f = f_{\star})}}{\pi \cdot R' \cdot C'}=\frac {0.496^2 \, {\rm 1/km^2}}{\pi \cdot 130\,{\rm \Omega/km} \cdot 35 \cdot 10^{-9}\,{\rm s/(\Omega \cdot km)}}\hspace{0.15cm}\underline{= 17.2\,{\rm kHz}}\hspace{0.05cm}.$$

• In contrast,  for the bronze line with diameter  $\text{5 mm}$ :

$$f_{\star} =\frac {(2.25 \cdot 10^{-3})^2 }{\pi \cdot 2.2 \cdot 6.7 \cdot 10^{-9}}\,{\rm kHz}\hspace{0.15cm}\underline{= 0.109\,{\rm kHz}}\hspace{0.05cm}.$$

(3)  For the copper cable  $f_0 \ll f_∗$ holds.

• Therefore,  the approximation is  $α_{\rm II}(f)$   ⇒   "strong attenuation"  should be used:

$$\alpha(f = f_0) \approx \sqrt{\pi \cdot f_0 \cdot R' \cdot C'}= \sqrt{\pi \cdot 2 \cdot 10^{3} \cdot 130 \cdot 35 \cdot 10^{-9}}\hspace{0.1cm}{\rm Np}/{ {\rm km} }\hspace{0.15cm}\underline{ = 0.17 \hspace{0.1cm}{\rm Np}/{ {\rm km} }}\hspace{0.05cm}.$$

• For the bronze line,  because of  $f_0 \gg f_∗$  the approximation is  $α_{\rm I}(f)$   ⇒   "weak attenuation"  is more suitable,  see subtask  (1):

$$\alpha(f = f_0)\hspace{0.15cm}\underline{= 0.0023\hspace{0.1cm}{\rm Np}/{ {\rm km} }}\hspace{0.05cm}.$$