Exercise 1.2: Coaxial Cable

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Various coaxial cables

The frequency response of a normal coaxial cable of length  $l$  (with a diameter of   $2.6 \ \text{mm}$  of the inner conductor and  an external diameter of $9.5 \ \text{mm}$)  is for frequencies  $f > 0$:

$$H(f) = {\rm e}^{-\alpha_{0\hspace{0.02cm}} \hspace{0.05cm} \cdot \hspace{0.05cm} l} \cdot {\rm e}^{-(\alpha_1 + {\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \beta_1)\hspace{0.05cm} \cdot \hspace{0.05cm} f \cdot \hspace{0.05cm} l}\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm e}^{-(\alpha_2 + {\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}\beta_2) \hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm} l}.$$

• The first term in this equation, arising from the ohmic losses, is described by the so-called kilometric damping  $α_0 = 0.00162\, \text{Np/km}$ .

• The frequency-proportional damping component   ⇒   $α_1 · f · l$  mit  $α_1 = 0.000435 \,\text{Np/(km · MHz)}$  is due to the lateral losses. This only becomes noticeable at very high frequencies and is neglected in the following.

• Also, the frequency-proportional phase  $β_1 · f · l$  with  $β_1 = 21.78 \,\text{rad/(km · MHz)}$  is left out of consideration because this only leads to an equal transit time for all frequencies.

Hence, for frequencies between  $200 \ \text{kHz}$  and  $400 \ \text{MHz}$  the frequency response of the coaxial cable is predominantly determined by the influence of

• the damping constant  $α_2 = 0.2722 \,\text{Np/(km · MHz}^{0.5})$, and
• the phase constant  $β_2 = 0.2722 \,\text{rad/(km · MHz}^{0.5})$,

which are due to the so-called skin effect. For positive frequencies the following applies:

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

Because of the same numerical values of  $α_2$  and  $β_2$  this can be expressed as follows:

$$H(f) = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} },$$

where the parameter  $f_0$  equally accounts for the two constants  $α_2$  and  $β_2$  and the cable length  $l$ .

Please note:

• This task belongs to the chapter  System Description in Frequency Domain.

Questions

1

What is the frequency response constant  $K$  for a cable length of  $l = 5\,\text{ km}$?

$K \ = \ $

2

What length  $l_{\rm max}$  could a cable have such that a direct signal is not damped by no more than  $3\%$ ?

$l_{\rm max} \ = \ $  $\text{ km}$

3

What is the characteristic frequency  $f_0$  for a cable length of  $l = 5\,\text{ km}$. Consider the relation  $\rm \sqrt{2j} = 1 + j$.

$f_0\ = \ $  $\text{ MHz}$

4

A cosine signal of frequency  $f_x = f_0$  is applied to the cable input with a power of  $P_x = \,\text{1 W}$. What is the output power  $P_y$?

$P_y \ = \ $  $\text{ mW}$

5

What output power is obtained with the signal frequency  $f_x = 10 \ \rm MHz$?

$P_y \ = \ $  $\text{ mW}$

Sample solution

Solution

(1)  Für den Gleichsignalübertragungsfaktor gilt:

$$K = H(f=0) = {\rm e}^{-\alpha_0 \hspace{0.05cm}\cdot \hspace{0.05cm} l} = {\rm e}^{-0.00162 \hspace{0.05cm} \cdot \hspace{0.05cm} 5}\hspace{0.15cm}\underline{ \approx 0.992}.$$

(2)  Mit  ${\rm a_0 } = α_0 · l$  müsste folgende Gleichung erfüllt sein:

$${\rm e}^{\rm -a_0 } \ge 0.97 \hspace{0.2cm} \Rightarrow \hspace{0.2cm} {\rm a_0 } < \ln \frac{1}{0.97 } \approx 0.0305\,{\rm Np}.$$

• Damit erhält man für die maximale Länge  $l_{\rm max} = 0.0305 \ \text{Np}/0.00162 \ \text{Np/km} \rm \underline{\: ≈ \: 18.8 \: km}$.

(3)  Wegen  $β_2 = α_2$  und der angegebenen Beziehung  $\rm 1 + j = \sqrt{2j}$  kann für den Frequenzgang auch geschrieben werden:

$$H(f) = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f \hspace{0.05cm} \cdot \hspace{0.05cm} {\alpha_2}^2 \hspace{0.05cm} \cdot \hspace{0.05cm} l^2} }= K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} }.$$

• Durch Koeffizientenvergleich mit der vorne angegebenen Gleichung erhält man:

$${1}/{f_0} = \alpha_2^2 \hspace{0.05cm} \cdot \hspace{0.05cm} l^2 = \big ( \frac { {\rm 0.272} }{\rm km \hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{MHz} }\big )^2 \cdot ({\rm 5 \hspace{0.05cm} km})^2 = \frac{1.855}{ {\rm MHz} }\hspace{0.2cm} \Rightarrow \hspace{0.2cm} f_0 \hspace{0.15cm}\rm \underline{= 0.540 \: MHz}.$$

(4)  Für den Frequenzgang gilt:

$$\begin{align*}H(f) & = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} } = K \cdot {\rm e}^{- \sqrt{ f/f_0} } \cdot {\rm e}^{- {\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}\sqrt{ f/f_0} } \hspace{0.05 cm} \Rightarrow \hspace{0.05 cm} |H(f)|^2 = K^2 \cdot {\rm e}^{- 2\hspace{0.05cm} \cdot \hspace{0.05cm}\sqrt{ f/f_0} }.\end{align*}$$

• Für  $f = f_0$  erhält man hierfür  $K^2 \cdot \rm e^{–2} ≈ 0.135$. Daraus folgt weiter:

$$P_y = P_x \cdot |H(f = f_0)|^2 \hspace{0.15cm}\underline{\approx135\hspace{0.05cm}{\rm mW}}.$$

(5)  Mit der höheren Frequenz  $f_x = 10\:\text{MHz}$  ist die Ausgangsleistung gegenüber  $f_x = 0.54\:\text{MHz}$  signifikant kleiner:

$$P_y = P_x \cdot {\rm e}^{- 2\hspace{0.05cm} \cdot \hspace{0.05cm}\sqrt{ 10/0.54} }\approx P_x \cdot {\rm e}^{- 8.6 } \hspace{0.15cm}\underline{\approx 0.184 \hspace{0.1cm}{\rm mW}}.$$