Aufgaben:Exercise 1.3: System Comparison at AWGN Channel

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System Comparison at the AWGN Channel

For the comparison of different modulation and demodulation methods with regard to noise sensitivity, we usually start from the so-called AWGN Channel  and present the following double logarithmic diagram:

• The y-axis indicates the sinkt-to-noise ratio (logarithmic SNR)  $10 · \lg ρ_v$  in dB.
•  $10 · \lg ξ$  is plotted on the x-axis; the normalized power parameter ("performance parameter") is characterized by:

$$ \xi = \frac{P_{\rm S} \cdot \alpha_{\rm K}^2 }{{N_0} \cdot B_{\rm NF}}\hspace{0.05cm}.$$

• Thus, the transmission  $P_{\rm S}$, the channel attenuation factor $α_{\rm K}$, the noise power density  $N_0$  and the bandwidth  $B_{\rm NF}$  of the message signal are suitably summarised together in  $ξ$ .
• Unless explicitly stated otherwise, the following values shall be assumed in the exercise:

$$P_{\rm S}= 5 \;{\rm kW}\hspace{0.05cm}, \hspace{0.2cm} \alpha_{\rm K} = 0.001\hspace{0.05cm}, \hspace{0.2cm} {N_0} = 10^{-10}\;{\rm W}/{\rm Hz}\hspace{0.05cm}, \hspace{0.2cm} B_{\rm NF}= 5\; {\rm kHz}\hspace{0.05cm}.$$

Two systems are plotted in the graph and their   $(x, y)$-curve can be described as follows:

• $\text{System A}$  is characterized by the following equation:

$$y = x+1.$$

• Accordingly,  $\text{System B}$ is characterized by:

$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right)\hspace{0.05cm}.$$

The additional axis labels drawn in green have the following meaning:

$$ x = \frac{10 \cdot {\rm lg} \hspace{0.1cm}\xi} {10 \,{\rm dB}}\hspace{0.05cm}, \hspace{0.3cm}y = \frac{10 \cdot {\rm lg} \hspace{0.1cm}\rho_v} {10 \,{\rm dB}}\hspace{0.05cm}.$$

Thus  $x = 4$  represents  $10 · \lg ξ = 40\text{ dB}$  or  $ξ = 10^4$  and  $y = 5$  represents  $10 · \lg ρ_v= 50\text{ dB}$ , i.e.,  $ρ_v = 10^5$.

Hints:

• This exercise belongs to the chapter   Quality Criteria.
• Particular reference is made to the page   Investigating at the AWGN Channel.
• By specifying the powers in watts, they are independent of the reference resistance, 𝑅.  $R$.

Questions

1

What is the sink-to-noise ratio (in dB) for  $\text{System A}$  with  $P_{\rm S}= 5 \;{\rm kW}$,   $\alpha_{\rm K} = 0.001$,   $N_0 = 10^{-10}\;{\rm W}/{\rm Hz}$,   $B_{\rm NF}= 5\; {\rm kHz}$?

$10 · \lg \hspace{0.05cm}ρ_v \ = \ $ $\ \text{dB}$

2

Now  $10 · \lg \hspace{0.05cm} ρ_v ≥ 60\text{ dB}$  is required.  Which independent measures can be taken to achieve this?

	Increasing the transmission power from  $P_{\rm S}= 5\text{ kW}$  to $10\text{ kW}$ .
	Increasing the channel transmission factor from nbsp;$α_{\rm K} = 0.001$  to  $0.004$.
	Reducing the noise power density to  $N_0=10^{–11 }\text{ W/Hz}$.
	Increasing the AF bandwidth from  $B_{\rm NF}= 5\text{ kHz}$  to  $\text{ kHz}$.

3

What is the signal-to-noise ratio for  $\text{System B}$  with  $10 · \lg ξ = 40\text{ dB}$?

$10 · \lg \hspace{0.05cm}ρ_v \ = \ $ $\ \text{dB}$

4

If the required signal-to-noise ratio is  $10 · \lg ρ_v = 50\text{ dB}$, what transmission power  $P_{\rm S}$ is sufficient to achieve this for  $\text{System B}$?

$P_{\rm S} \ = \ $ $\ \text{ kW }$

5

What value of  $10 · \lg ξ$  gives the greatest improvement of  $\text{System B}$  relative to  $\text{System A}$ ?

$10 · \lg \hspace{0.05cm} ξ \ = \ $ $\ \text{dB}$

Solution

Solution

(1)  The normalized performance parameter is calculated using these values as follows

$$\xi = \frac{5 \cdot 10^3\,{\rm W}\cdot 10^{-6} }{10^{-10}\,{\rm W}/{\rm Hz} \cdot 5 \cdot 10^3\,{\rm Hz}} = 10^4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 40\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x=4 \hspace{0.05cm}.$$

• This gives the auxiliary coordinate value   $y = 5$, which leads to a sink-to-noise ratio of   $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 50 \ \rm dB}$ .

(2) Answers 2 and 3 are correct:

This requirement corresponds to a $10$  dB, increase in the signal-to-noise ratio compared to the previous system, so  $10 · \lg \hspace{0.05cm}ξ$  must also be increased by $10$  dB:

$$10 \cdot {\rm lg} \hspace{0.1cm}\xi = 50\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \xi=10^5 \hspace{0.05cm}.$$

A tenfold larger   $ξ$ value is achieved – provided all other parameters are held constant in each case –

• by a transmission power of  $P_{\rm S} = 50$  kW  instead of   $5$  kW,
• by a channel transmission factor of   $α_{\rm K} = 0.00316$  instead of  $0.001$,
• by a noise power density of   $N_0 = 10^{ –11 }$  W/Hz  instead of  $10^{ –10 }$  W/Hz,
• by a bandwidth of  $B_{\rm NF} = 0.5$  kHz  instead of   $5$  kHz.

(3)  For  $10 · \lg \hspace{0.05cm} ξ = 40$  dB, the auxiliary value is   $x = 4$.  This gives the auxiliary y-value:

$$y= 6 \cdot \left(1 - {\rm e}^{-3} \right)\approx 5.7 \hspace{0.05cm}.$$

• This corresponds to a sink-to-noise ratio of   $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 57 \ \rm dB}$, a $7$  dB improvement over  $\text{System A}$ .

(4)  This problem is described by the following equation:

$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right) = 5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm e}^{-x+1} ={1}/{6}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \approx 2.79 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 27.9\,{\rm dB}\hspace{0.05cm}.$$

• For  $\text{System A}$  this required   $10 · \lg \hspace{0.05cm} \xi = 40$  dB , which was achieved with   $P_{\rm S} = 5$  kW and the other numerical values given. 
• Now the transmission power can be reduced by about   $12.1$  dB:

$$ 10 \cdot {\rm lg} \hspace{0.1cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}}= -12.1\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}} = 10^{-1.21}\approx 0.06\hspace{0.05cm}.$$

• This means that in  $\text{System B}$  the same system quality is achieved with only   $6\%$  of the transmission power of  $\text{System A}$  – i.e., with only   $P_{\rm S} \hspace{0.15cm}\underline{ = 0.3 \ \rm kW}$.

(5)  The larger sink-to-noise ratio of  $\text{System B}$  compared to  $\text{System A}$  we will denote with   $V$  (from German "Verbesserung" $\Rightarrow$ Improvement):

$$V = 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;B)} - 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;A)} = \left[6 \cdot \left(1 - {\rm e}^{-x+1} \right) -x -1 \right] \cdot 10\,{\rm dB}\hspace{0.05cm}.$$

• Durch Nullsetzen der Ableitung ergibt sich derjenige  $x$–Wert, der zur maximalen Verbesserung führt:

$$ \frac{{\rm d}V}{{\rm d}x} = 6 \cdot {\rm e}^{-x+1} -1\Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = \hspace{0.15cm}\underline {27.9\,{\rm dB}}\hspace{0.05cm}.$$

• Es ergibt sich also genau der in der Teilaufgabe  (4)  behandelte Fall mit  $10 · \lg ρ_υ = 50$  dB, während der Störabstand bei  $\text{System A}$  nur  $37.9$  dB beträgt. 
• Die Verbesserung ist demnach  $12.1$  dB.