Aufgaben:Exercise 4.7: Copper Twin Wire 0.5 mm

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Impulse response of the copper twin wire

The time response of a copper twin wire with diameter  $d = 0.5 \ \rm mm$  is to be analyzed.

The frequency response with the line length  $l = 1.5 \ \rm km$  and the bit rate  $R = 10 \rm Mbit/s$:

$$H_{\rm K}(f) = {\rm e}^{-{\rm a}_0 } \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} 2 \pi \cdot f \hspace{0.05cm} \cdot \hspace{0.01cm}\tau_{\rm P}} \cdot {\rm e}^{-{\rm a}_1 \hspace{0.05cm}\cdot \hspace{0.02cm}2f/R}\cdot {\rm e}^{-{\rm a}_2 \hspace{0.05cm}\cdot \hspace{0.05cm}\sqrt{2f/R}}\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} b_2 \hspace{0.05cm}\cdot \hspace{0.05cm}\sqrt{2f/R}} \hspace{0.05cm}.$$

The following quantities are used, which can be derived from the damping and phase function per unit length:

$${a}_0 = \alpha_0 \cdot l\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}\alpha_0 = 0.5066\,\, \frac{\rm Np}{\rm km}\hspace{0.05cm},$$

$$ \tau_{\rm P} = \frac{\beta_1 \cdot l}{2 \pi} \hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}\beta_1 = 30.6\,\, \frac{\rm rad}{\rm km \cdot MHz}\hspace{0.05cm},$$

$$ {a}_1 = \alpha_1 \cdot l \cdot {{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm} \alpha_1 = 0.136\,\, \frac{\rm Np}{\rm km \cdot MHz}\hspace{0.05cm},$$

$$ {a}_2 = \alpha_2 \cdot l \cdot \sqrt{{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm} \alpha_2 = 1.1467\,\, \frac{\rm Np}{\rm km \cdot \sqrt{MHz}}\hspace{0.05cm},$$

$$ {b}_2 = \beta_2 \cdot l \cdot \sqrt{{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm} \beta_2 = 1.1467\,\, \frac{\rm rad}{\rm km \cdot \sqrt{MHz}}\hspace{0.05cm}.$$

The impulse response can thus be expressed in the form

$$h_{\rm K}(t ) = K \cdot \big [ \delta(t - \tau_{\rm P})\star h_{1}(t) \star h_{2}(t) \big ]$$

where

• the partial impulse response  $h_1(t)$  is due to the third term in the above equation, and
• $h_2(t)$  indicates the joint time-domain representation of the last two terms.

The graph shows as red curve the part  $h_2(t)$  of the impulse response and the convolution product  $h_1(t) \star h_2(t)$  ⇒   blue curve).
Here  $h_2(t)$  is equal to coaxial cable impulse response with the characteristic cable attenuation  ${a}_\star = {a}_2$.

Notes:

• The exercise belongs to the chapter  Properties of Balanced Copper Pairs.

• The parameters  $\alpha_0$,  $\alpha_1$ and  $\alpha_2$ were converted from the  $k$–parameters as shown in  Exercise 4.6  gezeigt.
• The phase function parameter  $\beta_2$  was set here numerically equal to the attenuation function parameter  $\alpha_2$. 
• Therefore, the attenuation component  ${a}_2$  and the phase component  ${b}_2$  differ only in units.
• On the page  Discussion of the approximate solution found,  it is explained why this measure is necessary.
• You can use the  (German language)  interactive SWF applet  "Zeitverhalten von Kupferkabeln"   ⇒   "Time behavior of copper cables"  to check your results.

Questions

1

Calculate the constant  $K$  of the impulse response  $h_{\rm K}(t )$.

$K \ = \ $

2

Calculate the phase delay  $\tau_P$, related to the symbol duration  $T$.

$\tau_P/T \ = \ $

3

What is the characteristic attenuation  $a_\star$  of the comparable coaxial cable?

$a_\star \ = \ $ $\ \rm dB$

4

What are the characteristics of the partial impulse response  $h_{\rm 1}(t )$ ?

	$h_{\rm 1}(t )$  is an even function.
	The maximum of  $h_{\rm 1}(t )$  is at  $t = 0$.
	The integral over  $h_{\rm 1}(t )$  gives the value  $2$.

5

Which properties can you recognize in the function  $h_1(t ) \star h_2(t )$?

	$h_1(t ) \star h_2(t )$  completely reproduces the distortions of  $h_{\rm K}(t )$.
	$h_1(t ) \star h_2(t )$  differs from  $h_{\rm K}(t )$  only by one factor.

Solution

Solution

(1)  With  $a_0 = \alpha_0 \cdot l \approx 0.76 \ \rm Np$,  for the constant  $K$, which indicates the influence of the coefficient  $ \alpha_0$  on the impulse response, we obtain:

$$K = {\rm e}^{-{a}_0 }= {\rm e}^{-0.76} \hspace{0.15cm}\underline{= 0.468} \hspace{0.05cm}.$$

(2)  For the phase delay, using the given equation:

$$\tau_{\rm P} = \frac{\beta_1 \cdot l}{2 \pi}= \frac{30.6 \cdot 1.5}{2 \pi}\, {\rm µ s}\approx 7.31\, {\rm µ s}\hspace{0.05cm},$$

and related to the symbol duration  $T = 0.1 \ µ \rm s$:   

$${\tau_{\rm P}}/{T} \hspace{0.15cm}\underline{ \approx 73}\hspace{0.05cm}.$$

(3)  The impulse response of a coaxial cable is approximately equal to  $h_2(t)$, if the cable has the following characteristic cable attenuation:

$${a}_\star ={a}_2 = \alpha_2 \cdot l \cdot \sqrt{{R}/{2}} = 1.1467\,\, \frac{\rm Np}{\rm km \cdot \sqrt{MHz}} \cdot 1.5\,{\rm km} \cdot \sqrt{\frac{10\,{\rm MHz}}{2}} = 2.93\,{\rm Np} = 2.93\,{\rm Np} \cdot8.686\,\frac {\rm dB}{\rm Np} \hspace{0.15cm}\underline{ =25.5\,{\rm dB}}\hspace{0.05cm}.$$

(4)  Statements 1 and 2 are correct:

• The Fourier transform  $H_1(f) = {\rm e}^{-A \hspace{0.05cm} \cdot \hspace{0.05cm} |f|}$ with $A = 2 \hspace{0.05cm}\cdot \hspace{0.05cm} {a}_1/R$  is real and even, so  $h_1(t)$  is also real and even.
• Due to the low–pass characteristic of  $H_1(f)$,  the maximum is at  $t = 0$.
• The last statement, on the other hand, is incorrect:   the integral over  $h_1(t)$  in the entire time domain  $ \pm \infty$  is equal to  $H_1(f=0) = 1$.

(5)  Only solution 1 is correct:

• The partial impulse response  $h_1(t ) \star h_2(t )$  takes into account the influence of  $\alpha_1$,  $\alpha_2$  and  $\beta_2$  and thus all terms leading to distortions.
• In contrast,   $\alpha_0$  leads only to a frequency-independent attenuation and  $\beta_1$  only to a constant running time for all frequencies.
• Solution 2, on the other hand, does not apply:   first  (for small  $t$–values)    $h_1(t ) \star h_2(t )$  is smaller than  $h_2(t )$. Then, for large  $t$–values, the blue curve lies above the red one.
• This means:   $\alpha_1$  and thus also  $h_1(t )$  actually cause additional distortions, even when they are not very significant.