Aufgaben:Exercise 4.4: Coaxial Cable - Frequency Response

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Various coaxial cable types

A so-called normal coaxial cable of length  $l$  with

• core diameter  $\text{2.6 mm}$,  and
• outer diameter  $\text{9.5 mm}$

has the following frequency response:

$$H_{\rm K}(f) = {\rm e}^{- \alpha_0 \hspace{0.05cm} \cdot \hspace{0.05cm} l} \cdot {\rm e}^{- \alpha_1 \hspace{0.05cm}\cdot \hspace{0.05cm}l \hspace{0.05cm}\cdot \hspace{0.05cm}f} \cdot {\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot \sqrt{f}} \cdot {\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1 \hspace{0.05cm}\cdot \hspace{0.05cm} l \hspace{0.05cm}\cdot \hspace{0.05cm}f} \cdot {\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot \sqrt{f}} \hspace{0.05cm}.$$

The attenuation parameters  $\alpha_0$,  $\alpha_1$ and  $\alpha_2$ are to be used in  "Neper per kilometer"  (Np/km)  and the phase parameters  $\beta_1$ and  $\beta_2$ in  "Radian per kilometer"  (rad/km).  The following numerical values apply:

$$\alpha_0 = 0.00162 \hspace{0.15cm}{\rm Np}/{\rm km} \hspace{0.05cm},$$

$$\alpha_1 = 0.000435 \hspace{0.15cm} {\rm Np}/{{\rm km} \cdot {\rm MHz}} \hspace{0.05cm},$$

$$\alpha_2 = 0.2722 \hspace{0.15cm}{\rm Np}/{{\rm km} \cdot \sqrt{\rm MHz}} \hspace{0.05cm}.$$

For the system-theoretical description of a coaxial cable  (German:  "Koaxialkabel"   ⇒   subscipt  "K"),  one uses

• the attenuation function  (in Np or dB):

$${ a}_{\rm K}(f) = - {\rm ln} \hspace{0.10cm}|H_{\rm K}(f)|= - 20 \cdot {\rm lg} \hspace{0.10cm}|H_{\rm K}(f)| \hspace{0.05cm},$$

• the phase function  (in rad or degree):

$$b_{\rm K}(f) = - {\rm arc} \hspace{0.10cm}H_{\rm K}(f) \hspace{0.05cm}.$$

In practice,  one often uses the approximation

$$H_{\rm K}(f) = {\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot \sqrt{f}} \cdot {\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot \sqrt{f}}$$

$$\Rightarrow \hspace{0.3cm} a_{\rm K}(f) = \alpha_2 \cdot l \cdot \sqrt{f}, \hspace{0.8cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot {\rm rad}/{\rm Np}\hspace{0.05cm}.$$

This is allowed because  $\alpha_2$  and  $\beta_2$  have exactly the same numerical value and differ only by different pseudo units.

With the definition of the characteristic cable attenuation  (in Neper or decibel)

$${a}_{\rm \star(Np)} = {a}_{\rm K}(f = {R}/{2}) = 0.1151 \cdot {a}_{\rm \star(dB)}$$

digital systems with different bit rate  $R$  and cable length  $l$  can be treated uniformly.

Notes:

• The exercise belongs to the chapter  Properties of Coaxial Cables.

• You can use the interactive  "HTML 5/JS" applet  Attenuation of Copper Cables  to check your results.

Questions

1

Which terms of  $H_{\rm K}(f)$  do not lead to distortions? The

	$\alpha_0$–term,
	$\alpha_1$–term,
	$\alpha_2$–term,
	$\beta_1$–term,
	$\beta_2$–term.

2

What length  $l_{\rm max}$  could such a cable have so that a DC signal is attenuated by no more than  $1\%$ ?

$l_\text{max} \ = \ $ $\ \rm km$

3

What is the attenuation  (in Np)  at frequency  $f = 70 \ \rm MHz$  when the cable length is  $\underline{l = 2 \ \rm km}$?

$a_{\rm K}(f = 70\ \rm MHz) \ = \ $ $\ \rm Np$

4

Assuming all other things are equal,  what is the attenuation when only the  $\alpha_2$–term is considered?

$a_{\rm K}(f = 70\ \rm MHz) \ = \ $ $\ \rm Np$

5

What is the formula for the conversion between  $\rm Np$  and  $\rm dB$?  What is the $\rm dB$ value that results for the attenuation calculated in  (4)?

$a_{\rm K}(f = 70\ \rm MHz) \ = \ $ $\ \rm dB$

6

Which statements are true if we restrict ourselves to the  $\alpha_2$–value with respect to the attenuation function?

	One can also do without the phase term with  $\beta_1$.
	One can also do without the phase term with  $\beta_2$.
	$a_\star \approx 40 \ \rm dB$  holds for a system with  $R = 70 \ \rm Mbit/s$  and  $l = 2 \ \rm km$.
	$a_\star \approx 40 \ \rm dB$  holds for a system with  $R = 140 \ \rm Mbit/s$  and  $l = 2 \ \rm km$.
	$a_\star \approx 40 \ \rm dB$  holds for a system with  $R = 560 \ \rm Mbit/s$  and  $l = 1 \ \rm km$.

Solution

Solution

(1)  Solutions 1 and 4  are correct:

• The  $\alpha_0$–term causes only a frequency-independent attenuation.
• The  $\beta_1$–term  (linear phase)  results in a frequency-independent delay.
• All other terms contribute to the  (linear)  distortions.

(2)  With  ${\rm a}_0 = \alpha_0 \cdot l$  the following equation must be satisfied:

$${\rm e}^{- {\rm a}_0 } \ge 0.99 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm a}_0 < {\rm ln} \hspace{0.10cm}\frac{1}{0.99}\approx 0.01\,\,{\rm (Np)} \hspace{0.05cm}.$$

• This gives the maximum cable length:

$$l_{\rm max} = \frac{{\rm a}_0 }{\alpha_0 } = \frac{0.01\,\,{\rm Np}}{0.00162\,\,{\rm Np/km}}\hspace{0.15cm}\underline{\approx 6.173\,\,{\rm km}} \hspace{0.05cm}.$$

(3)  The following applies to the attenuation curve  when all terms are taken into account:

$${a}_{\rm K}(f) = [\alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt{f}\hspace{0.05cm}] \cdot l = \big[0.00162 + 0.000435 \cdot 70 + 0.2722 \cdot \sqrt{70}\hspace{0.05cm}\big]\, \frac{\rm Np}{\rm km} \cdot 2\,{\rm km}$$

$$\Rightarrow \hspace{0.3cm}{a}_{\rm K}(f = 70\ \rm MHz) = \big[0.003 + 0.061 + 4.555 \hspace{0.05cm}\big]\, {\rm Np}\hspace{0.15cm}\underline{= 4.619\, {\rm Np}}\hspace{0.05cm}.$$

(4)  According to the calculation in subtask  (3),  the attenuation value  ${a}_{\rm K}(f = 70\ \rm MHz)\hspace{0.15cm}\underline{=4.555 \ \rm Np}$ is obtained here.

(5)  For any positive quantity  $x$  the following holds:

$$x_{\rm Np} = {\rm ln} \hspace{0.10cm} x = \frac{{\rm lg} \hspace{0.10cm} x}{{\rm lg} \hspace{0.10cm} {\rm e}} = \frac{1}{{20 \cdot \rm lg} \hspace{0.10cm} {\rm e}} \cdot (20 \cdot {\rm lg} \hspace{0.10cm} x) = 0.1151 \cdot x_{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{\rm dB} = 8.6859 \cdot x_{\rm Np}\hspace{0.05cm}.$$

The attenuation value  $4.555 \ {\rm Np}$  is thus identical to  ${a}_{\rm K}(f = 70\ \rm MHz)\hspace{0.15cm}\underline{=39.56 \ \rm dB}$.

(6)  Solutions 1, 4 and 5 are correct.  Explanation:

• With the restriction to the attenuation term with  $\alpha_2$,  the following applies to the frequency response:

$$H_{\rm K}(f) = {\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot \sqrt{f}} \cdot {\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1 \hspace{0.05cm}\cdot \hspace{0.05cm} l \hspace{0.05cm}\cdot f} \cdot {\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot \sqrt{f}} \hspace{0.05cm}.$$

• If the  $\beta_1$–phase term is omitted,  nothing changes with respect to the distortions.   Only the phase and group delay would be  (both equal)  smaller by the value  $\tau_1 = (\beta_1 \cdot l)/(2\pi)$.

• If,  on the other hand,  we omit the  $\beta_2$–term,  we obtain completely different conditions:

(a) The frequency response  $H_{\rm K}(f)$  no longer fulfills the requirement of a causal system;  in such a case,  $H_{\rm K}(f)$  would have to be in minimum phase.

(b) The impulse response  $h_{\rm K}(t)$  is symmetrical at  $t = 0$  with real frequency response,  which does not correspond to the conditions.

• Therefore,  as an approximation for the coaxial cable frequency response,  the following is allowed:

$${a}_{\rm K}(f) = \alpha_2 \cdot l \cdot \sqrt{f},$$

$$b_{\rm K}(f) = a_{\rm K}(f) \cdot {\rm rad}/{\rm Np}\hspace{0.05cm}.$$

• That means:  ${a}_{\rm K}(f)$  and  ${b}_{\rm K}(f)$  of a coaxial cable are in first approximation identical in shape and differ only in their units.

• For a digital system with bit rate  $R = 140 \ \rm Mbit/s$   ⇒   $R/2 = 70 \ \rm Mbit/s$  and cable length  $l = 2 \ \rm km$ ,   $a_\star \approx 40 \ \rm dB$  holds (see solution to the last sub-task).
• A system with four times the bit rate  $R/2 = 280 \ \rm Mbit/s$  and half the length  $(l = 1 \ \rm km)$  results in the same characteristic cable attenuation.
• In contrast,  the following holds for a system with  $R/2 = 35 \ \rm Mbit/s$  and  $l = 2 \ \rm km$:

$${a}_\star = 0.2722 \hspace{0.15cm}\frac {\rm Np}{\rm km \cdot \sqrt{MHz}} \cdot {\rm 2\,km}\cdot\sqrt{\rm 35\,MHz} \cdot 8.6859 \,\frac {\rm dB}{\rm Np} \approx 28\,{\rm dB} \hspace{0.05cm}.$$