# Exercise 3.1Z: Frequency Response of the Coaxial Cable

Some coaxial cable types

A so-called  "standard coaxial cable"

• with core diameter  $2.6 \ \rm mm$,
• outer diameter  $9.5 \ \rm mm$  and
• length  $l$

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 entered in  "Neper"  $(\rm Np)$,  the phase parameters  $\beta_1$  and  $\beta_2$  in  "radian"  $(\rm rad)$.  The following numerical values apply:

$$\alpha_0 = 0.00162 \hspace{0.15cm}\frac{\rm Np}{\rm km} \hspace{0.05cm},\hspace{0.2cm} \alpha_1 = 0.000435 \hspace{0.15cm}\frac{\rm Np}{\rm km\cdot{\rm MHz}} \hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km\cdot\sqrt{\rm MHz}} \hspace{0.05cm}.$$

Often,  to describe a  "linear time-invariant system"  $\rm (LTI)$  in terms of system theory,  one uses

• the attenuation function  $($in  $\rm Np$  or  $\rm dB)$:
$$a_{\rm K}(f) = - {\rm ln} \hspace{0.10cm}|H_{\rm K}(f)|= - 20 \cdot {\rm lg} \hspace{0.10cm}|H_{\rm K}(f)| \hspace{0.05cm},$$
• the phase function  $($in  $\rm rad$  or  $\rm degrees)$:
$$b_{\rm K}(f) = - {\rm arc} \hspace{0.10cm}H_{\rm K}(f) \hspace{0.05cm}.$$

In practice one often uses the approximation

$$H_{\rm K}(f) = {\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot \sqrt{f}} \cdot {\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot \sqrt{f}}$$
$$\Rightarrow \hspace{0.3cm} a_{\rm K}(f) = \alpha_2 \cdot l \cdot \sqrt{f}, \hspace{0.2cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot \frac{\rm rad}{\rm Np}\hspace{0.05cm}.$$

This is allowed because  $\alpha_2$  and  $\beta_2$  have exactly the same numerical value – just different pseudo units.

Using the definition of the  characteristic cable attenuation  (in Neper or decibels)

$$a_{\rm * (Np)} = a_{\rm K}(f = {R_{\rm B}}/{2}) = 0.1151 \cdot a_{\rm * (dB)}$$

digital systems with different bit rate  $R_{\rm B}$  and cable length  $l$  can be treated uniformly.

Notes:

### 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 to attenuate a DC signal by no more than  $1\%$?

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

3

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

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

4

All other things being equal,  what attenuation results 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 for the attenuation calculated in subtask  (4)?

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

6

Which of the statements are true provided that one restricts oneself to the  $\alpha_2$–value with respect to the attenuation function?

 One can also do without the phase term  $\beta_1$. One can also do without the phase term  $\beta_2$. $a_* ≈ 40\,{\rm dB}$  is valid for a system with  $R_{\rm B} = 70\,{\rm Mbit/s}$  and  $l = 2\,{\rm km}$. $a_* ≈ 40\,{\rm dB}$  is valid for a system with  $R_{\rm B} = 140\,{\rm Mbit/s}$  and  $l = 2\,{\rm km}$. $a_* ≈ 40\,{\rm dB}$  is valid for a system with  $R_{\rm B} = 560\,{\rm Mbit/s}$  and  $l = 1\,{\rm km}$.

### Solution

#### Solution

(1)  Solutions 1 and 4  are correct:

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

(2)  With  $a_0 = \alpha_0 \cdot l$,  the following equation must be satisfied:

$${\rm e}^{- a_0 } \ge 0.99 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}a_0 < {\rm ln} \hspace{0.10cm}\frac{1}{0.99}\approx 0.01\,\,{\rm (Np)} \hspace{0.05cm}.$$
• This gives the following for the maximum cable length:
$$l_{\rm max} = \frac{a_0 }{\alpha_0 } = \frac{0.01\,\,{\rm Np}}{0.00162\,\,{\rm Np/km}}\hspace{0.15cm}\underline {\approx 6.173\,\,{\rm km}} \hspace{0.05cm}.$$

(3)  For the attenuation curve,  considering all terms:

$$a_{\rm K}(f) \ = \ \big[\alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt{f}\hspace{0.05cm}\big] \cdot l = \big [0.00162 + 0.000435 \cdot 70 + 0.2722 \cdot \sqrt{70}\hspace{0.05cm}\big]\, \frac{\rm Np}{\rm km} \cdot 2\,{\rm km} = \hspace{0.15cm}\underline {= 4.619\, {\rm Np}}\hspace{0.05cm}.$$

(4)  According to the calculation at point  (3),  the attenuation value  $\underline {4.555\,{\rm Np}}$  is obtained here.

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

$$x_{\rm Np} = {\rm ln} \hspace{0.10cm} x = \frac{{\rm lg} \hspace{0.10cm} x}{{\rm lg} \hspace{0.10cm} {\rm e}} = \frac{1}{{20 \cdot \rm lg} \hspace{0.10cm} {\rm e}} \cdot (20 \cdot {\rm lg} \hspace{0.10cm} x) = 0.1151 \cdot x_{\rm dB}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{\rm dB} = 8.686 \cdot x_{\rm Np}\hspace{0.05cm}.$$
• Thus,  the attenuation value  $4.555\,{\rm Np}$  is identical to  $\underline{39.57\,{\rm dB} }$.

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

• With the restriction to the attenuation term with  $\alpha_2$,  the following holds for 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 we omit the  $\beta_1$  phase term,  nothing changes with respect to the distortions.  Only the phase and group delay would become smaller  (both equal)  by the value  $\tau_1 = (\beta_1 \cdot l)/2\pi$.
• On the other hand,  if the  $\beta_2$–term is omitted,  completely different ratios result:
1. The frequency response  $H_{\rm K}(f)$  now no longer satisfies the requirement of a causal system;  for such a system  $H_{\rm K}(f)$  must be minimum-phase.
2. The impulse response  $h_{\rm K}(t)$  is symmetrical about  $t = 0$  with real frequency response,  which does not correspond to the conditions.
• Therefore as an approximation for the coaxial cable frequency response is allowed:
$$a_{\rm K}(f) = \alpha_2 \cdot l \cdot \sqrt{f}, \hspace{0.2cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot \frac{\rm rad}{\rm Np}\hspace{0.05cm}.$$
• That means:  $a_{\rm K}(f)$  and  $b_{\rm K}(f)$  of a coaxial cable are identical in form and differ only in their units.
• For a digital system with bit rate  $R_{\rm B} = 140\,{\rm Mbit/s}$   ⇒   $R_{\rm B}/2 = 70\,{\rm Mbit/s}$  and cable length  $l = 2\,{\rm km}$:
$a_* ≈ 40\,{\rm dB}$ is indeed valid – see solution to subtask  (5).
• A system with four times the bit rate  $(R_{\rm B}/2 = 280\,{\rm Mbit/s})$  and half the length  $(l = 1\,{\rm km})$  results in the same characteristic cable attenuation.
• In contrast,  for a system with  $R_{\rm B}/2 = 35\,{\rm Mbit/s}$  and  $l = 2\,{\rm km}$  holds:
$$a_{\rm dB} = 0.2722 \frac{\rm Np}{\rm km\cdot \sqrt{\rm MHz}} \cdot 2 \ \rm km \cdot \sqrt{35 \ \rm MHz} \cdot 8.6859 \frac{\rm dB}{\rm Np} ≈ 28 \ \rm dB.$$