Aufgaben:Exercise 4.3: Operational Attenuation

• Return to book

Considered line model

If a communication link of length  $l$  is not terminated at both ends with its wave impedance  $Z_{\rm W}$,  reflections will always occur.

Instead of the wave attenuation  $a_{\rm W} = \alpha \cdot l$  one has to consider in this case the  operating attenuation  $a_{\rm B}$  (German:  "Betriebsdämpfung"   ⇒   subscipt  "B"),  which is given here without frequency dependence.  That means:  We always only consider a single frequency $f_0$:

$${ a}_{\rm B} = { a}_{\rm B}(f_0)= { a}_{\rm W}+ {\rm ln}\hspace{0.1cm} |q_1|+{\rm ln}\hspace{0.1cm} |q_2|+{ a}_{\rm IA}\hspace{0.05cm}.$$

The four parts  – all with the pseudo unit "Neper"  (Np) –  describe the following facts:

• The first summand  $a_{\rm W} = \alpha \cdot l$  models the  wave attenuation  of the wave propagating along the line.  Note that attenuations are denoted by  "$a$",  while the attenuation function per unit length is denoted by  "$\alpha$"   ⇒   read: "alpha".
• The second summand gives the  transmitter-side reflection loss.  This term takes into account the power loss due to reflections at the  "transmitter → line"  transition:

$${\rm ln}\hspace{0.1cm} |q_1|= {\rm ln}\hspace{0.1cm}\frac {R_1 + Z_{\rm W}}{2 \cdot \sqrt{R_1 \cdot Z_{\rm W}}} \hspace{0.05cm}.$$

• In an analogous way applies to the  receiver-side reflection loss  at the end of the line   ⇒   transition  "line → receiver":

$${\rm ln}\hspace{0.1cm} |q_2|= {\rm ln}\hspace{0.1cm}\frac {R_2 + Z_{\rm W}}{2 \cdot \sqrt{R_2 \cdot Z_{\rm W}}} \hspace{0.05cm}.$$

• The  interaction attenuation  (IA)  describes the signal attenuation due to the effect of a doubly reflected wave,  which can be constructively or destructively superimposed on the useful signal.  For this last part holds:

$${ a}_{\rm IA} = {\rm ln}\hspace{0.1cm} |1- r_1 \cdot r_2 \cdot {\rm e}^{- 2\hspace{0.05cm}\cdot\hspace{0.05cm} \gamma \hspace{0.05cm}\cdot\hspace{0.05cm} l}|$$

we use the following equations and nomenclature in this exercise:

$${a}_{\rm IA} = {\rm ln}\hspace{0.1cm}A, \hspace{0.3cm}A = |1- r_\alpha \cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2 \hspace{0.05cm}\cdot\hspace{0.05cm} \beta \hspace{0.05cm}\cdot\hspace{0.05cm} l}| \hspace{0.05cm},\hspace{0.3cm}r_\alpha = r_1 \cdot r_2\cdot {\rm e}^{-2 \hspace{0.05cm}\cdot\hspace{0.05cm} \alpha \hspace{0.05cm}\cdot\hspace{0.05cm} l},\hspace{0.3cm}r_1= \frac {R_1 - Z_{\rm W}}{R_1 + Z_{\rm W}}, \hspace{0.3cm}r_2= \frac {R_2 - Z_{\rm W}}{R_2 + Z_{\rm W}} \hspace{0.05cm}.$$

Notes:

• The exercise belongs to the chapter  Some Results from Line Transmission Theory.
• You can find essential information on the page  Influence of reflections - operational attenuation

• Use the following numerical values for numerical calculations:

$$Z_{\rm W} = 100\,{\rm \Omega}\hspace{0.05cm},\hspace{0.3cm}R_1 = 200\,{\rm \Omega}\hspace{0.05cm},\hspace{0.3cm} R_2 = 1\,{\rm k\Omega}\hspace{0.05cm},\hspace{0.3cm}l = 2\,{\rm km}\hspace{0.05cm},\hspace{0.3cm} \alpha = 0.1\,{\rm Np/km} \hspace{0.05cm}.$$

Questions

1

What would be the value of the operating attenuation  with matching,  i.e. if there were no reflections?

$a_{\rm B} \ = \ $ $\ \rm Np$

2

Calculate the two parts of the reflection loss for  $Z_{\rm W} = 100\,{\rm \Omega}$  as well as  $R_1 = 200\,{\rm \Omega}$  and  $ R_2 = 1\,{\rm k\Omega}$.

$\ln\ |q_1| \ = \ $ $\ \rm Np$

$\ln\ |q_2| \ = \ $ $\ \rm Np$

3

Calculate the reflection factors  $r_1$,  $r_2$ and  $r_\alpha$.

$r_1 \ = \ $

$r_2\ = \ $

$r_\alpha\ = \ $

4

Which condition must the auxiliary  $A = \vert 1- r_\alpha \cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2 \hspace{0.05cm}\cdot\hspace{0.05cm} \beta \hspace{0.05cm}\cdot\hspace{0.05cm} l}\vert $  satisfy,  so that constructive or destructive superposition occurs with respect to the interaction attenuation?

	$A$  is minimal   ⇒   constructive superposition,
	$A$  is maximal   ⇒   destructive superposition.

5

Specify the smallest possible value  $\beta_\text{min}$  for the phase function (per unit length)  $\beta(f_0)$ ,
so that constructive superposition occurs.

$\beta_\text{min} \ = \ $ $\ \rm rad/km$

6

What is the maximum size of the interaction attenuation component?  What are the requirements for this?

${\rm Max}\ \big [a_\text{IA}\big ] \ = \ $ $\ \rm Np$

Solution

Solution

(1)  With resistance matching  $(R_1 = R_2 =Z_{\rm W})$  only the first of the four summands remains:

$${a}_{\rm B} = {a}_{\rm W} =\alpha \cdot l = 0.1\,{\rm Np/km} \cdot 2\,{\rm km} \hspace{0.15cm}\underline{= 0.2\,{\rm Np}}\hspace{0.05cm}.$$

(2)  According to the given equations we obtain:

$$q_1 = \frac {R_1 + Z_{\rm W}}{2 \cdot \sqrt{R_1 \cdot Z_{\rm W}}}= \frac {200 + 100}{2 \cdot \sqrt{200 \cdot 100}}= 1.061 \hspace{0.2cm}\Rightarrow \hspace{0.2cm} {\rm ln}\hspace{0.1cm}|q_1| \hspace{0.15cm}\underline{= 0.059 \,{\rm Np}} \hspace{0.05cm},$$

$$q_2 = \frac {R_2 + Z_{\rm W}}{2 \cdot \sqrt{R_2 \cdot Z_{\rm W}}}= \frac {1000 + 100}{2 \cdot \sqrt{1000 \cdot 100}}= 1.739 \hspace{0.2cm}\Rightarrow \hspace{0.2cm} {\rm ln}\hspace{0.1cm}|q_2| \hspace{0.15cm}\underline{= 0.553 \,{\rm Np}}\hspace{0.05cm}.$$

(3)  With the given circuit resistors we obtain

$$r_1= \frac {200\,{\rm \Omega} - 100\,{\rm \Omega}}{200\,{\rm \Omega} + 100\,{\rm \Omega}} \hspace{0.15cm}\underline{= 0.333}\hspace{0.05cm},$$

$$r_2=\frac {1000\,{\rm \Omega} - 100\,{\rm \Omega}}{1000\,{\rm \Omega} + 100\,{\rm \Omega}}\hspace{0.15cm}\underline{ = 0.818}$$

$$\Rightarrow \hspace{0.3cm} r_\alpha = r_1 \cdot r_2\cdot {\rm e}^{-2 \hspace{0.05cm}\cdot\hspace{0.05cm} \alpha \hspace{0.05cm}\cdot\hspace{0.05cm} l}= 0.333 \cdot 0.818\cdot {\rm e}^{-4}\hspace{0.15cm}\underline{= 0.183} \hspace{0.05cm}.$$

(4)  Both statements are correct:

• With constructive superposition is   ⇒   ${a}_{\rm IA} = {\rm ln}\hspace{0.1cm}A < 0 \; \Rightarrow \; A < 1$  and minimal.
• In contrast, the maximum value of  $A $  (for which $A > 1$)   causes positive interaction attenuation, i.e., additional attenuation of the useful signal due to destructive superposition of outgoing wave and returning wave.

(5)  In the last subtask,  it was shown that constructive superposition is equivalent to the minimization of

$$A = |1- r_\alpha \cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2\hspace{0.05cm}\cdot\hspace{0.05cm} \beta \hspace{0.05cm}\cdot\hspace{0.05cm} l}|= |1- r_\alpha \cdot \cos(2 \beta l)+ {\rm j} \cdot \sin(2 \beta l)| = \sqrt {1- 2 \cdot r_\alpha \cdot \cos(2 \beta l)+ {r_\alpha}^2 \cdot \cos^2(2 \beta l)+ {r_\alpha}^2 \cdot \sin^2(2 \beta l)}$$

$$ \Rightarrow \hspace{0.3cm }A = \sqrt {1+ {r_\alpha}^2- 2 \cdot r_\alpha \cdot \cos(2 \beta l)} \hspace{0.05cm}.$$

• The minimum results for

$$\cos(2 \beta l) = +1 \hspace{0.2cm}\Rightarrow \hspace{0.2cm}2 \beta l= \pi,\ 2\pi,\ 3\pi, ... \hspace{0.2cm}\Rightarrow \hspace{0.2cm}\beta_{\rm min} = {\pi}({2l})\hspace{0.15cm}\underline{= 0.785\,{\rm rad/km}} \hspace{0.05cm}.$$

• In contrast, destructive superposition occurs if the phase function (per unit length) satisfies the following condition:

$$\cos(2 \beta l) = -1 \hspace{0.2cm}\Rightarrow \hspace{0.2cm}2 \beta l= {\pi}/{2},\ {3\pi}/{2},\ {5\pi}/{2},\text{ ...}$$

(6)  The argument  $A = \sqrt {1+ {r_\alpha}^2- 2 \cdot r_\alpha \cdot \cos(2 \beta l)}$  can become maximum  $A = 2$    ⇒   ${\rm Max}\ [a_\text{IA}] \; \underline {= 0.693 \ \rm Np}$.

The following requirements must be met for this:

• not terminated line  $(r_1 = r_2 = 1)$,
• short cable length,  so the term  $\alpha \cdot l$  is not effective  $(r_\alpha = 1)$,
• phase progression according to the subtask  (5).