Aufgaben:Exercise 2.6: Free Space Attenuation

• Return to book

Photo of a transmitter

A shortwave transmitter operated according to the modulation method "DSB-AM with carrier" works with carrier frequency  $f_{\rm T} = 20 \ \rm MHz$  and transmit power $P_{\rm S} = 100\ \rm kW$.  It is designed for a bandwidth of  $B_{\rm NF} = 8 \ \rm kHz$ .

For test operation, a mobile receiver is used, which operates with a synchronous demodulator. If this is located at distance   $d$  from the transmitter, the attenuation function of the transmission channel can be approximated as follows:

$$\frac{a_{\rm K}(d, f)}{\rm dB} = 34 + 20 \cdot {\rm lg }\hspace{0.2cm}\frac{d}{\rm km} + 20 \cdot {\rm lg }\hspace{0.2cm}\frac{f}{\rm MHz} \hspace{0.05cm}.$$

This equation describes so-called  free space attenuation, which also depends on the (carrier) frequency.

It can be assumed that the entire DSB-AM spectrum is attenuated like the carrier frequency. This means that

• the slightly larger attenuation of the upper sideband (USB), and
• the slightly smaller attenuation of the lower sideband (LSB)

are compensated for by a corresponding pre-distortion at the transmitter.

Let the effective noise power density at the receiver be  $N_0 = 10^{–14} \ \rm W/Hz.$

For the first two subtasks, it is assumed that the transmitter transmits only the carrier, which is equivalent to the modulation depth being  $m = 0$ .

Hints:

• This exercise belongs to the chapter   Synchronous Demodulation.
• Particular reference is made to the page  Sink SNR and the performance parameter.

Questions

1

What power is received at a distance  $d = 10 \ \rm km$  from the transmitter when only the carrier is transmitted  $(m = 0)$?

$P_{\rm E} \ = \ $ $\ \rm mW$

2

At what distance  $d$  from the transmitter is the receiver located when the received power is  $P_{\rm E} = 100 \ \rm µ W$??

$d \ = \ $ $\ \rm km$

3

Which sink SNR results from the distance  $d$ calculated in subtask  (2)  when the modulation depth is  $m = 0.5$ ?

$10 · \lg ρ_v \ = \ $ $\ \text{dB}$

4

What is the minimum modulation depth  $m$  that can be chosen for a resulting sink-to-noise ratio of  $60 \ \rm dB$ ?

$m_{\min} \ = \ $

5

Which of the following statements are true?

	DSB–AM with carrier does not make sense for energy reasons if a synchronous demodulator is used.
	DSB–AM without carrier does not make sense for energy reasons if a synchronous demodulator is used.
	A small carrier component can be helpful for the required frequency and phase synchronization.

Solution

Solution

(1)  According to the equation for free space attenuation, when   $d = 10\ \rm km$  and  $f_{\rm T} = 20 \ \rm MHz$, then:

$$\frac{a_{\rm K}(d, f_{\rm T})}{\rm dB} = 34 + 20 \cdot {\rm lg }\hspace{0.1cm}\frac{d}{\rm km} + 20 \cdot {\rm lg }\hspace{0.1cm}\frac{f_{\rm T}}{\rm MHz}= 34 + 20 \cdot {\rm lg }\hspace{0.1cm}(10) + 20 \cdot {\rm lg }\hspace{0.1cm}(20)\approx 80\hspace{0.1cm}{\rm dB} \hspace{0.05cm}.$$

• This corresponds to a power reduction by a factor of   $10^{8}$:

$$P_{\rm E}= 10^{-8} \cdot P_{\rm S}= 10^{-8} \cdot 100\,{\rm kW}\hspace{0.15cm}\underline {= 1\, {\rm mW} \hspace{0.05cm}}.$$

(2)  From  $P_{\rm S} = 10^5 \ \rm W$, $P_{\rm E} = 10{^–4}\ \rm W$  follows a free space attenuation of  $90 \ \rm dB$.  From this, we further obtain:

$$20 \cdot {\rm lg }\hspace{0.1cm}\frac{d}{\rm km} = ( 90-34 - 26)\hspace{0.1cm}{\rm dB}= 30\,{\rm dB}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} d = 10^{1.5}\,{\rm km}\hspace{0.15cm}\underline { = 31.6\,{\rm km}\hspace{0.05cm}}.$$

(3)  For DSB–AM without a carrier, that is, for a modulation depth  $m → ∞$, the following would hold:

$$ \rho_{v } = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{{N_0} \cdot B_{\rm NF}} = \frac{ P_{\rm E}}{{N_0} \cdot B_{\rm NF}}= \frac{10^{-4}\,{\rm W}}{10^{-14}\,{\rm W/Hz}\cdot 8 \cdot 10^{3}\,{\rm Hz} } = 1.25 \cdot 10^6\hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg }\hspace{0.1cm}\rho_{v } \approx 61\,{\rm dB}\hspace{0.05cm}.$$

• With modulation depth  $m = 0.5$  the sink SNR becomes smaller by a factor of  $[1 +{2}/{m^2}]^{-1} = {1}/{9}$ .  Thus, the signal-to-noise ratio at the sink is also smaller:

$$ 10 \cdot {\rm lg }\hspace{0.1cm}\rho_{v } = 61\,{\rm dB}- 10 \cdot {\rm lg }\hspace{0.1cm}(9) \hspace{0.15cm}\underline {\approx 51.5\,{\rm dB}\hspace{0.05cm}}.$$

(4)  According to the calculations in subtask  (3) , the following condition must be satisfied:

$$ 10 \cdot {\rm lg }\hspace{0.1cm}\left({1 + {2}/{m^2}}\right) < 1\,{\rm dB}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} 1 +{2}/{m^2} < 10^{0.1}=1.259 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{2}/{m^2} < 0.259 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} m > \sqrt{8}\approx 2.83 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} m_{\rm min} \hspace{0.15cm}\underline {= 2.83} \hspace{0.05cm}.$$

(5)  Answers 1 and 3 are correct:

• When using a synchronous demodulator, the addition of the carrier makes no sense unless the former is useful for the required carrier recovery.
• Since the carrier cannot be used for demodulation, only a fraction of the transmit power is available for demodulation   $($one third for   $m = 1$, one ninth for  $m = 0.5)$.