Exercise 1.3: System Comparison at AWGN Channel

System comparison at AWGN channel

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

The y-axis indicates the "sink-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 power $P_{\rm S}$, 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:

$$y = x+1.$$

$\text{System B}$ is instead 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 $R$.

Questions

$10 · \lg \hspace{0.05cm}ρ_v \ = \ $

$\ \text{dB}$

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

$10 · \lg \hspace{0.05cm}ρ_v \ = \ $

$\ \text{dB}$

$P_{\rm S} \ = \ $

$\ \text{ kW }$

$10 · \lg \hspace{0.05cm} ξ \ = \ $

$\ \text{dB}$

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 SNR 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 sink SNR 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 signal 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 SNR of $10 · \lg \hspace{0.05cm} ρ_v\hspace{0.15cm}\underline{ = 57 \ \rm dB}$ ⇒ $7$ 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 $\text{System A}$ $10 · \lg \hspace{0.05cm} \xi = 40$ dB is required, 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 SNR of $\text{System B}$ compared to $\text{System A}$ we will denote with $V$ (from German "Verbesserung" ⇒ "improvement"):

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

Setting the derivative to zero yields the $x$–value that leads to the maximum improvement:

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

This results in exactly the case discussed in subtask (4) with $10 · \lg ρ_υ = 50$ dB, while the sink SNR for $\text{System A}$ is only $37.9$ dB.
The improvement is therefore $12.1$ dB.