Aufgaben:Exercise 4.5Z: Impulse Response once again

• Return to book

Impulse response of a coaxial cable (representation with or without running time)

As in  Exercise 4.5 , we consider a binary transmission system with bit rate  $R$  ⇒   symbol duration  $T= 1/R$.

A  "standard coaxial cable"  $\text{(2.6 mm}$  core diameter,  $\text{9.5 mm}$  outer diameter$)$  of length  $l = 1 \ \rm km$  with the following frequency response is used as transmission medium:

$$H_{\rm K}(f) = {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1 \hspace{0.05cm}\cdot \hspace{0.05cm} f \hspace{0.05cm}\cdot \hspace{0.05cm}l} \cdot {\rm e}^{- \alpha_2 \hspace{0.01cm} \sqrt{f} \hspace{0.05cm}l} \cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm} \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm}l} = H_1(f) \cdot H_2(f) \cdot H_3(f)$$

The partial frequency responses  $H_1(f)$,  $H_2(f)$  and  $H_3(f)$  are used here only as abbreviations.  The line parameters are:

$$\beta_1 = 21.78\, \frac{\rm rad}{\rm km \cdot MHz}\hspace{0.05cm}, $$

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

$$ \beta_2 = 0.2722\, \frac{\rm rad}{\rm km \cdot \sqrt{MHz}} \hspace{0.05cm}.$$

The graph shows the resulting impulse response  $h_{\rm K}(t\hspace{0.05cm}')$, where  $t\hspace{0.05cm}' = t/T$  represents the normalized time. Without considering the (normalized) phase running time  $\tau\hspace{0.05cm}' = \tau/T$ ,  $h_{\rm K}(t\hspace{0.05cm}')$  can be written as follows:

$$h_{\rm K}(t\hspace{0.05cm}') = \frac {1}{T} \cdot \frac {a_\rm \star/\pi}{ \sqrt{2 \hspace{0.05cm}t\hspace{0.05cm}'^3}}\cdot {\rm e}^{ -{a_\rm \star^2}/( {2\pi \hspace{0.05cm}t\hspace{0.05cm}')} } \hspace{0.05cm}, \hspace{0.2cm} \hspace{0.15cm} {\rm with}\hspace{0.15cm}{a}_{\rm \star}\hspace{0.15cm} {\rm in}\hspace{0.15cm} {\rm Neper}\hspace{0.05cm}.$$

• This equation gives the Fourier retransform of the product  $H_2(f) \cdot H_3(f)$ .
• The characteristic cable attenuation  ${a}_{\rm \star} = \alpha_2 \cdot \sqrt {R/2} \cdot l \hspace{0.05cm}$ is used here.

Notes:

• The exercise belongs to the chapter  Properties of Coaxial Cables.

• You can use the  (German language)  interactive SWF applet  "Zeitverhalten von Kupferkabeln"   ⇒   "Time behavior of copper cables"  to check your results.
• In  Exercise 4.5  the maximum value of the normalized impulse response was calculated as follows:

$${\rm Max}\, \big[T \cdot h_{\rm K}(t)\big ] = \frac {\sqrt{13.5 \pi} \cdot {\rm e}^{-1.5} }{{a}_{\rm \star}^2} \approx \frac {1.453 }{{a}_{\rm \star}^2} \hspace{0.05cm}, \hspace{0.2cm} \hspace{0.15cm} {\rm with}\hspace{0.15cm}{a}_{\rm \star}\hspace{0.15cm} {\rm in}\hspace{0.15cm} {\rm Neper}\hspace{0.05cm}.$$

Questions

1

Which partial frequency response is responsible for the phase running time  $\tau$ ?

	$H_1(f)$,
	$H_2(f)$,
	$H_3(f)$.

2

Determine the bit rate of the binary system when  $\tau\hspace{0.05cm}' = \tau/T = 694$ .

$R \ = \ $ $\ \rm Mbit/s$

3

Give the characteristic cable attenuation  ${a}_{\rm \star}$  for the combined description of the frequency responses  $H_2(f)$  and  $H_3(f)$ .

${a}_{\rm \star} \ = \ $ $\ \rm Np$

4

Determine the (normalized) maximum value of the impulse response.

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

5

Which of the following statements are true?

	Distortions are reproduced correctly without  $H_1(f)$ .
	Distortions are reproduced correctly without  $H_2(f)$ .
	Distortions are reproduced correctly without  $H_3(f)$ .

Solution

Solution

(1)  Only solution 1 is correct:

• The spectral representation of a running time term is  ${\rm e}^{-{\rm j} \hspace{0.05cm}\cdot\hspace{0.05cm} 2 \pi \hspace{0.05cm}\cdot\hspace{0.05cm} f \hspace{0.05cm}\cdot\hspace{0.05cm}\tau}$.
• A comparison with the information provided shows that  $H_1(f)$  exactly satisfies this approach.

(2)  According to the information provided, the following applies:

$$2\pi \cdot f \cdot \tau = \beta_1 \cdot f \cdot l \Rightarrow \hspace{0.3cm}\tau= \frac {\beta_1 \cdot l}{2\pi} = \frac {21.78\, {\rm rad}/{({\rm km \cdot MHz})}\cdot 10\,{\rm km}}{2\pi} = 34.7\,{\rm µ s}$$

$$\Rightarrow \hspace{0.3cm}\tau '= {\tau}/{T} = 694 \Rightarrow \hspace{0.3cm} T = \frac {34.7\,{\rm µ s}}{700} \approx 0.05\,{\rm µ s}\hspace{0.05cm}.$$

• The bit rate is equal to the reciprocal of the symbol duration:

$$\underline{R = 20 \ \rm Mbit/s}.$$

(3)  For the characteristic cable attenuation one obtains:

$${a}_{\rm \star} = \alpha_2 \cdot \sqrt {R/2} \cdot l = 0.2722\, \frac{\rm Np}{\rm km \cdot \sqrt{MHz}} \cdot \sqrt {10\,{\rm MHz}} \cdot 10\,{\rm km} \hspace{0.15cm}\underline{\approx 8.6\,{\rm Np}}\hspace{0.05cm}.$$

• The corresponding dB value is  ${a}_{\rm \star} = 75 \ \rm dB$.

(4)  Using the given equation and the result of subtask  (3) , we obtain:

$${\rm Max}\, \big[T \cdot h_{\rm K}(t)\big] \approx \frac {1.453 }{{a}_{\rm \star}^2} = \frac {1.453 }{8.6^2} \hspace{0.15cm}\underline{ \approx 0.02}\hspace{0.05cm}.$$

(5)  Only solution 1 is correct:   $H_1(f)$  describes the frequency-independent running time which does not result in any distortion.

On the other hand,  $H_2(f)$  or  $H_3(f)$  should never be omitted for the calculation of the impulse response, otherwise serious errors would occur:

• The impulse response  $h_2(t)$  as the Fourier retransform of  $H_2(f)$  is an even function with the maximum at  $t = 0$  and extends in both directions over hundreds of symbols.
• In contrast, the Fourier retransform of  $H_3(f)$  is an odd function with a point of discontinuity at  $t = 0$.
• For  $t > 0$ ,   $h_3(t)$  drops similarly – but not exactly – to an exponential function. For negative times  $t$ ,   $h_3(t) = - h_3(|t|)$ is valid.
• Only the convolution  $h_2(t) \star h_3(t)$  yields the causal impulse response, but without the phase running time  $\tau$, which is considered in this model by  $H_1(f)$ .