Difference between revisions of "Aufgaben:Exercise 1.3: System Comparison at AWGN Channel"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Modulation_Methods/Quality_Criteria |
}} | }} | ||
− | [[File:P_ID960__Mod_A_1_3.png|right|frame| | + | [[File:P_ID960__Mod_A_1_3.png|right|frame|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 [[Modulation_Methods/Quality_Criteria#Some_remarks_on_the_AWGN_channel_model|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: |
:$$ \xi = \frac{P_{\rm S} \cdot \alpha_{\rm K}^2 }{{N_0} \cdot B_{\rm NF}}\hspace{0.05cm}.$$ | :$$ \xi = \frac{P_{\rm S} \cdot \alpha_{\rm K}^2 }{{N_0} \cdot B_{\rm NF}}\hspace{0.05cm}.$$ | ||
− | * | + | *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: |
:$$P_{\rm S}= 5 \;{\rm kW}\hspace{0.05cm}, \hspace{0.2cm} \alpha_{\rm | :$$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} = | K} = 0.001\hspace{0.05cm}, \hspace{0.2cm} {N_0} = | ||
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B_{\rm NF}= 5\; {\rm kHz}\hspace{0.05cm}.$$ | B_{\rm NF}= 5\; {\rm kHz}\hspace{0.05cm}.$$ | ||
− | + | 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: |
:$$y = x+1.$$ | :$$y = x+1.$$ | ||
− | * | + | * $\text{System B}$ is instead characterized by: |
:$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right)\hspace{0.05cm}.$$ | :$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right)\hspace{0.05cm}.$$ | ||
− | + | The additional axis labels drawn in green have the following meaning: | |
:$$ x = \frac{10 \cdot {\rm lg} \hspace{0.1cm}\xi} {10 \,{\rm dB}}\hspace{0.05cm}, \hspace{0.3cm}y = \frac{10 \cdot {\rm lg} \hspace{0.1cm}\rho_v} {10 \,{\rm dB}}\hspace{0.05cm}.$$ | :$$ x = \frac{10 \cdot {\rm lg} \hspace{0.1cm}\xi} {10 \,{\rm dB}}\hspace{0.05cm}, \hspace{0.3cm}y = \frac{10 \cdot {\rm lg} \hspace{0.1cm}\rho_v} {10 \,{\rm dB}}\hspace{0.05cm}.$$ | ||
− | + | *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., $ρ_v = 10^5$. | ||
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− | '' | + | |
− | * | + | |
− | * | + | |
− | * | + | ''Hints:'' |
+ | *This exercise belongs to the chapter [[Modulation_Methods/Quality_Criteria|Quality Criteria]]. | ||
+ | *Particular reference is made to the page [[Modulation_Methods/Quality_Criteria#Investigating_at_the_AWGN_channel|Investigating at the AWGN channel]]. | ||
+ | *By specifying the powers in watts, they are independent of the reference resistance $R$. | ||
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {What is the sink signal-to-noise ratio (in dB) for $\text{System A}$ with $P_{\rm S}= 5 \;{\rm kW}$, $\alpha_{\rm |
− | K} = 0.001$, $N_0 = 10^{-10}\;{\rm W}/{\rm Hz}$ | + | K} = 0.001$, $N_0 = 10^{-10}\;{\rm W}/{\rm Hz}$, $B_{\rm NF}= 5\; {\rm kHz}$? |
|type="{}"} | |type="{}"} | ||
$10 · \lg \hspace{0.05cm}ρ_v \ = \ $ { 50 3% } $\ \text{dB}$ | $10 · \lg \hspace{0.05cm}ρ_v \ = \ $ { 50 3% } $\ \text{dB}$ | ||
− | { | + | {Now $10 · \lg \hspace{0.05cm} ρ_v ≥ 60\text{ dB}$ is required. Which independent measures can be taken to achieve this? |
|type="[]"} | |type="[]"} | ||
− | - | + | - 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 $10\text{ kHz}$. |
− | { | + | {What is the sink signal-to-noise ratio for $\text{System B}$ with $10 · \lg ξ = 40\text{ dB}$? |
|type="{}"} | |type="{}"} | ||
$10 · \lg \hspace{0.05cm}ρ_v \ = \ $ { 57 3% } $\ \text{dB}$ | $10 · \lg \hspace{0.05cm}ρ_v \ = \ $ { 57 3% } $\ \text{dB}$ | ||
− | { | + | {If the required sink signal-to-noise ratio is $10 · \lg ρ_v = 50\text{ dB}$, what transmission power $P_{\rm S}$ is sufficient to achieve this for $\text{System B}$? |
|type="{}"} | |type="{}"} | ||
$P_{\rm S} \ = \ $ { 0.3 3% } $\ \text{ kW }$ | $P_{\rm S} \ = \ $ { 0.3 3% } $\ \text{ kW }$ | ||
− | { | + | {What value of $10 · \lg ξ$ gives the greatest improvement for $\text{System B}$ relative to $\text{System A}$ ? |
|type="{}"} | |type="{}"} | ||
$10 · \lg \hspace{0.05cm} ξ \ = \ ${ 27.9 3% } $\ \text{dB}$ | $10 · \lg \hspace{0.05cm} ξ \ = \ ${ 27.9 3% } $\ \text{dB}$ | ||
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' The normalized performance parameter is calculated using these values as follows: |
:$$\xi = \frac{5 \cdot 10^3\,{\rm W}\cdot 10^{-6} }{10^{-10}\,{\rm W}/{\rm Hz} \cdot 5 \cdot 10^3\,{\rm Hz}} = 10^4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 40\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x=4 \hspace{0.05cm}.$$ | :$$\xi = \frac{5 \cdot 10^3\,{\rm W}\cdot 10^{-6} }{10^{-10}\,{\rm W}/{\rm Hz} \cdot 5 \cdot 10^3\,{\rm Hz}} = 10^4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 40\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x=4 \hspace{0.05cm}.$$ | ||
− | + | *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)''' | + | '''(2)'''<u> Answers 2 and 3</u> 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: | |
:$$10 \cdot {\rm lg} \hspace{0.1cm}\xi = 50\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \xi=10^5 \hspace{0.05cm}.$$ | :$$10 \cdot {\rm lg} \hspace{0.1cm}\xi = 50\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \xi=10^5 \hspace{0.05cm}.$$ | ||
− | + | 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)''' | + | '''(3)''' For $10 · \lg \hspace{0.05cm} ξ = 40$ dB, the auxiliary value is $x = 4$. This gives the auxiliary $y$–value: |
:$$y= 6 \cdot \left(1 - {\rm e}^{-3} \right)\approx 5.7 \hspace{0.05cm}.$$ | :$$y= 6 \cdot \left(1 - {\rm e}^{-3} \right)\approx 5.7 \hspace{0.05cm}.$$ | ||
− | + | *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)''' | + | |
+ | '''(4)''' This problem is described by the following equation: | ||
:$$ y= 6 \cdot \left(1 - {\rm e}^{-x+1} \right) = 5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm e}^{-x+1} ={1}/{6}\hspace{0.3cm} | :$$ 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}.$$ | \Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \approx 2.79 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 27.9\,{\rm dB}\hspace{0.05cm}.$$ | ||
− | + | *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: | ||
:$$ 10 \cdot {\rm lg} \hspace{0.1cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}}= -12.1\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}} = 10^{-1.21}\approx 0.06\hspace{0.05cm}.$$ | :$$ 10 \cdot {\rm lg} \hspace{0.1cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}}= -12.1\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac{P_{\rm S}}{ 5 \;{\rm kW}} = 10^{-1.21}\approx 0.06\hspace{0.05cm}.$$ | ||
− | + | *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)''' | + | '''(5)''' The larger sink SNR of $\text{System B}$ compared to $\text{System A}$ we will denote with $V$ (from German "Verbesserung" ⇒ "improvement"): |
:$$V = 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;B)} - 10 \cdot {\rm lg} \hspace{0.1cm}\rho_v \hspace{0.1cm}{\rm (System\;A)} | :$$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}.$$ | = \left[6 \cdot \left(1 - {\rm e}^{-x+1} \right) -x -1 \right] \cdot 10\,{\rm dB}\hspace{0.05cm}.$$ | ||
− | + | *Setting the derivative to zero yields the $x$–value that leads to the maximum improvement: | |
:$$ \frac{{\rm d}V}{{\rm d}x} = 6 \cdot {\rm e}^{-x+1} -1\Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = \hspace{0.15cm}\underline {27.9\,{\rm dB}}\hspace{0.05cm}.$$ | :$$ \frac{{\rm d}V}{{\rm d}x} = 6 \cdot {\rm e}^{-x+1} -1\Rightarrow \hspace{0.3cm} x = 1+ {\rm ln} \hspace{0.1cm}6 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = \hspace{0.15cm}\underline {27.9\,{\rm dB}}\hspace{0.05cm}.$$ | ||
− | + | *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. | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
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− | [[Category: | + | [[Category:Modulation Methods: Exercises|^1.2 Quality Criteria^]] |
Latest revision as of 17:54, 23 March 2022
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
Solution
- $$\xi = \frac{5 \cdot 10^3\,{\rm W}\cdot 10^{-6} }{10^{-10}\,{\rm W}/{\rm Hz} \cdot 5 \cdot 10^3\,{\rm Hz}} = 10^4 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.1cm}\xi = 40\,{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x=4 \hspace{0.05cm}.$$
- 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.