Exercise 3.1: Impulse Response of the Coaxial Cable

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Impulse response of a coaxial cable

The frequency response of a coaxial cable of length  $l$  can be represented by the following formula:

$$H_{\rm K}(f) \ = \ {\rm e}^{- \alpha_0 \hspace{0.05cm} \cdot \hspace{0.05cm} l} \cdot {\rm e}^{- (\alpha_1 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1) \hspace{0.05cm}\cdot f \hspace{0.05cm}\cdot \hspace{0.05cm}l} \cdot \ {\rm e}^{- (\alpha_2 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2) \hspace{0.05cm}\cdot \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm}l} \hspace{0.05cm}.$$
  • The first term of this equation is due to the ohmic losses.
  • The second term is due to the transverse losses.
  • Dominant,  however,  is the skin effect,  which is expressed by the third term.

With the coefficients valid for a so-called  "standard coaxial cable"  $(2.6 \ \rm mm$  core diameter and  $9.5 \ \rm mm$  outer diameter$)$

$$\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km \cdot \sqrt{\rm MHz}} \hspace{0.05cm}, \hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac{\rm rad}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm},$$

the frequency response can also be represented as follows:

$$H_{\rm K}(f) \approx {\rm e}^{- 0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km} \hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}} } \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km} \hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}}} \hspace{0.05cm}.$$

That means: Attenuation curve  $a_{\rm K}(f)$  and phase curve  $b_{\rm K}(f)$  are identical except for the pseudo units  "$\rm Np$"  and  "$\rm rad$".

If one defines the characteristic cable attenuation  $a_*$  at half the bit rate  $(R_{\rm B}/2)$,  one can treat digital systems of different bit rate and length uniformly:

$$a_{\star} = a_{\rm K}(f = {R_{\rm B}}/{2}) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{- a_{\star} \cdot \sqrt{2f/R_{\rm B}}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} a_{\star} \cdot \sqrt{2f/R_{\rm B}}}\hspace{0.4cm}{\rm with}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np} \hspace{0.05cm}.$$

The corresponding  $\rm dB$  value is larger by a factor of  $8.688$.  For a binary system,  $R_{\rm B} = 1/T$  applies,  so that the characteristic cable attenuation is then related to the frequency  $f = 1/(2T)$. 

The Fourier transform of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$,  which can be given in closed-analytic form for a coaxial cable with the approximations described here.  For a binary system holds:

$$h_{\rm K}(t) = \frac{ a_{\star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot {\rm exp} \left[ - \frac{a_{\star}^2}{2 \pi \cdot t/T}\hspace{0.1cm}\right] \hspace{0.4cm}{\rm with}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np} \hspace{0.05cm}.$$

Subtask  (5)  is related to the basic receiver pulse  $g_r(t) = g_s(t) * h_K(t)$,  where  $g_s(t)$  should be assumed to be a rectangular pulse with height  $s_0$  and duration  $T$. 




What is the length  $l$  of a standard coaxial cable,  if for the bit rate  $R_{\rm B} = 140 \ \rm Mbit/s$  the characteristic cable attenuation is  $a_* = 60 \ \rm dB$? 

$l \ = \ $

$\ \rm km $


At what time  $t_{\rm max}$  does  $h_{\rm K}(t)$  have its maximum?  Let  $a_* = 60 \ \rm dB$ be further valid.

$t_{\rm max}/T= \ $


What is the maximum value of the impulse response?  Let  $a_* = 60 \ \rm dB$  continue to hold.

${\rm Max}\ \big [h_{\rm K}(t)\big ]= \ $

$\ \cdot 1/T $


At what time  $t_{\rm 5\%}$  is  $h_{\rm K}(t)$  less than  $5\%$  of the maximum?  Consider only the first term of the given formula as an approximation.

$t_{\rm 5\%}/T= \ $


Which statements are true for the basic receiver pulse  $g_r(t)$? 

$g_r(t)$  is twice as wide as  $h_{\rm K}(t)$.
It is approximately  $g_r(t) = h_{\rm K}(t) \cdot s_0 \cdot T$.
$g_r(t)$  can be approximated by a Gaussian pulse.


(1)  The characteristic cable attenuation  $a_* = 60 \ \rm dB$  corresponds to about  $6.9 \ \rm Np$.  Therefore,  it must hold:

$$\alpha_2 \cdot l \cdot {R_{\rm B}}/{2} = 6.9\,\,{\rm Np}$$
$$\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rm Np}}{0.2722\,\,\frac{\rm Np}{{\rm km} \cdot \sqrt{\rm MHz}} \cdot \sqrt{70\,\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 3\,\,{\rm km}} \hspace{0.05cm}.$$

(2)  With the substitutions

$$x = { t}/{ T}, \hspace{0.2cm} K_1 = \frac{a_*/T}{\sqrt{2\pi^2 }}, \hspace{0.2cm} K_2 = \frac{a_*^2}{2\pi}$$

the impulse response can be described as follows:

$$h_{\rm K}(x) = K_1 \cdot x^{-3/2}\cdot {\rm e}^{-K_2/x} \hspace{0.05cm}.$$
  • By setting the derivative to zero,  it follows:
$$- {3}/{2} \cdot K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\rm e}^{-K_2/x}\cdot (-K_2) \cdot (-x^{-2})= 0 \hspace{0.05cm}$$
$$\Rightarrow \hspace{0.3cm} {3}/{2} \cdot x^{-5/2} = K_2 \cdot x^{-7/2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} x_{\rm max} = {2}/{3} \cdot K_2 = { a_{\star}^2}/({3 \pi}) \hspace{0.05cm}.$$
  • This gives for  $60 \ \rm dB$  cable attenuation  $(a_* ≈ 6.9 \ \rm Np)$:
$$x_{\rm max} = { t_{\rm max}}/{ T} \hspace{-0.1cm}: \hspace{0.2cm} { t_{\rm max}}/{ T} = { 6.9^2}/({3\pi})\hspace{0.15cm}\underline {\approx 5} \hspace{0.2cm}.$$

(3)  Substituting the result of  (2)  into the given equation,  we obtain  (using  $a$  instead of  $a_*$):

$$h_{\rm K}(t_{\rm max}) \ = \ \frac{1}{T} \cdot \frac{ a}{ \sqrt{2 \pi^2 \cdot \frac{a^6}{(3\pi)^3}}}\hspace{0.1cm} \cdot {\rm exp} \left[ - \frac{a^2}{2\pi} \cdot \frac{3\pi}{a^2}\hspace{0.1cm}\right]= \frac{1}{T} \cdot \frac{1}{a^2}\cdot \sqrt{\frac{27 \pi }{2}} \cdot {\rm e}^{-3/2}\hspace{0.15cm}\approx \frac{1}{T} \cdot \frac{1.453}{a^2} \hspace{0.05cm}.$$
  • Thus,  with  $a = 6.9$,  we arrive at the final result:
$${\rm Max} \ [h_{\rm K}(t)] = \frac{1.453}{{6.9\,}^2} \cdot {1}/{T}\hspace{0.15cm}\underline {\approx 0.03 \cdot {1}/{T}} \hspace{0.05cm}.$$

(4)  Using the result of subtask  (3),  the determining equation is:

$$\frac{ a/T}{ \sqrt{2 \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 \cdot {1}/{T}= 0.0015 \cdot {1}/{T} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} (t_{5\%}/T)^{3/2} = \frac{a}{\sqrt{2} \cdot \pi \cdot 0.0015}\approx 1036 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}t_{5\%}/T \hspace{0.15cm}\underline {\approx 103.5}\hspace{0.05cm}.$$
  • This value is slightly too large because the second term  ${\rm e}^{\rm – 0.05} ≈ 0.95$  was neglected.
  • The exact calculation gives  $t_{\rm 5\%}/T ≈ 97$.

(5)  The second solution  is correct.  In general:

$$g_r(t) = g_s(t) \star h_{\rm K}(t) = s_0 \cdot \int_{t-T/2}^{t+T/2} h_{\rm K}(\tau) \,{\rm d} \tau .$$
  • Since the channel impulse response  $h_{\rm K}(t)$  changes only insignificantly within a symbol duration,  it can also be written for this purpose:
$$g_r(t) \approx h_{\rm K}(t) \cdot s_0 \cdot T .$$