Difference between revisions of "Aufgaben:Exercise 3.10: Noise Power Calculation"
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− | [[File:P_ID1115__Mod_A_3_9.png|right|frame| | + | [[File:P_ID1115__Mod_A_3_9.png|right|frame|Noise power densities of PM and FM]] |
− | + | Consider the phase and frequency modulation of a cosine oscillation with frequency $f_{\rm N}$. . First, let the message frequency be $f_{\rm N} = f_5 = 5 \ \rm kHz$ and the modulation index (phase deviation) be $η = 5$. | |
− | + | In the presence of additive Gaussian noise with noise power density $N_0$ , the PM demodulator results in a constant noise power density ${\it \Phi}_{v {\rm , \hspace{0.08cm}PM} }(f) = {\it \Phi}_0$, which also depends on the modulation index $η$ : | |
:$${\it \Phi}_0 = \frac{N_0}{\eta^2} \hspace{0.05cm}.$$ | :$${\it \Phi}_0 = \frac{N_0}{\eta^2} \hspace{0.05cm}.$$ | ||
− | + | For the calculation of the noise power $P_{\rm R}$ , only the frequency range $±f_{\rm N}$ is relevant (see graph). | |
− | + | The noise power density after FM demodulation with the frequency deviation $Δf_{\rm A}$ is: | |
:$${\it \Phi}_{v {\rm , \hspace{0.08cm}FM} } (f) = N_0 \cdot \left(\frac{f}{\Delta f_{\rm A}}\right)^2 \hspace{0.05cm}.$$ | :$${\it \Phi}_{v {\rm , \hspace{0.08cm}FM} } (f) = N_0 \cdot \left(\frac{f}{\Delta f_{\rm A}}\right)^2 \hspace{0.05cm}.$$ | ||
− | * | + | *The signal-to-noise ratio $10 · \lg ρ_v = 50 \ \rm dB$ is given for phase modulation with $f_{\rm N} = 5 \ \rm kHz$. |
− | * | + | *In this task, we are looking for the S/N ratio of FM for the message frequency $f_{\rm N} = 5 \ \rm kHz$ as well as the resulting S/N ratios of PM and FM for message frequency $f_{\rm N} = f_{10} = 10 \ \rm kHz$. |
− | '' | + | |
− | * | + | |
− | * | + | |
− | + | ||
+ | |||
+ | |||
+ | |||
+ | ''Hints:'' | ||
+ | *This exercise belongs to the chapter [[Modulation_Methods/Influence_of_Noise_on_Systems_with_Angle_Modulation|Influence of Noise on Systems with Angle Modulation]]. | ||
+ | *Particular reference is made to the section [[Modulation_Methods/Influence_of_Noise_on_Systems_with_Angle_Modulation#System_comparison_of_AM.2C_PM_and_FM_with_respect_to_noise| System comparison of AM, PM and FM with respect to noise]]. | ||
+ | |||
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {What is the signal-to-noise ratio for <u>phase modulation</u> and $f_{\rm N} = 10 \ \rm kHz$? Interpret the result. |
|type="{}"} | |type="{}"} | ||
$10 · \lg ρ_v \ = \ $ { 46.99 3% } $\ \rm dB$ | $10 · \lg ρ_v \ = \ $ { 46.99 3% } $\ \rm dB$ | ||
− | { | + | {Calculate the signal-to-noise ratio for <u>frequency modulation</u> and $f_{\rm N} = 5 \ \rm kHz$. What is the modulation index for this configuration? |
|type="{}"} | |type="{}"} | ||
$10 · \lg ρ_v \ = \ $ { 54.77 3% } $\ \rm dB$ | $10 · \lg ρ_v \ = \ $ { 54.77 3% } $\ \rm dB$ | ||
− | { | + | {Calculate the signal-to-noise ratio for <u>frequency modulation</u> and $f_{\rm N} = 10 \ \rm kHz$. Interpret the result in comparison to your answers for '''(1)''' and '''(2)'''. |
|type="{}"} | |type="{}"} | ||
− | $10 · \lg ρ_v \ = \ $ { 45.74 3% } $\ \rm | + | $10 · \lg ρ_v \ = \ $ { 45.74 3% } $\ \rm dB$ |
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' The signal-to-noise power ratio (sink SNR) $ \rho_{v }$ is the quotient of useful power $P_{\rm S}$ and noise power $P_{\rm R}$. For phase modulation, it is: |
:$$ \rho_{v } = \frac{P_{\rm S}}{P_{\rm R}} = \frac{P_{\rm S}}{{\it \Phi}_0 \cdot 2 f_{\rm N} } =\frac{\eta^2}{2} \cdot \frac{P_{\rm S}}{N_0 \cdot f_{\rm N} }\hspace{0.05cm}.$$ | :$$ \rho_{v } = \frac{P_{\rm S}}{P_{\rm R}} = \frac{P_{\rm S}}{{\it \Phi}_0 \cdot 2 f_{\rm N} } =\frac{\eta^2}{2} \cdot \frac{P_{\rm S}}{N_0 \cdot f_{\rm N} }\hspace{0.05cm}.$$ | ||
− | * | + | *The measurement with $f_{\rm N} = f_5 = 5 \ \rm kHz$ gives the SNR $ \rho_{v } = 10^5$ $($corresponding to $10 · \lg ρ_v =50\ \rm dB)$ . |
− | * | + | *Doubling the message frequency results in half the SNR, since twice the noise power is now effective: |
:$$ \rho_{v }= 0.5 \cdot 10^5 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}\underline {\approx 46.99\,{\rm dB}}\hspace{0.05cm}.$$ | :$$ \rho_{v }= 0.5 \cdot 10^5 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}\underline {\approx 46.99\,{\rm dB}}\hspace{0.05cm}.$$ | ||
− | + | This result can also be derived using the relationship $ρ_v = η^2/2 · ξ$ . | |
− | * | + | *For phase modulation, $η$ is independent of the message frequency. |
− | * | + | *DThe SNR loss is due to the fact that now the performance parameter $ξ = P_{\rm S}/(N_0 · f_{\rm N})$ is halved. |
+ | |||
+ | |||
− | '''(2)''' | + | '''(2)''' For frequency modulation and a message frequency $f_{\rm N} = 5 \ \rm kHz$ the noise power is given by: |
:$$P_{\rm R} = \int_{-f_{\rm N}}^{ + f_{\rm N}} {\it \Phi}_{v {\rm , \hspace{0.08cm}FM} } (f)\hspace{0.1cm}{\rm d}f = \frac{2 \cdot N_0}{\Delta f_{\rm A}^{\hspace{0.1cm}2}} \cdot \int_{0}^{ f_{\rm N}} f^2\hspace{0.1cm}{\rm d}f = \frac{2 \cdot N_0 \cdot f_{\rm N}^{\hspace{0.1cm}3}}{3 \cdot \Delta f_{\rm A}^2} \hspace{0.05cm}.$$ | :$$P_{\rm R} = \int_{-f_{\rm N}}^{ + f_{\rm N}} {\it \Phi}_{v {\rm , \hspace{0.08cm}FM} } (f)\hspace{0.1cm}{\rm d}f = \frac{2 \cdot N_0}{\Delta f_{\rm A}^{\hspace{0.1cm}2}} \cdot \int_{0}^{ f_{\rm N}} f^2\hspace{0.1cm}{\rm d}f = \frac{2 \cdot N_0 \cdot f_{\rm N}^{\hspace{0.1cm}3}}{3 \cdot \Delta f_{\rm A}^2} \hspace{0.05cm}.$$ | ||
− | + | *Thus, considering the frequency deviation $Δf_{\rm A} = η · f_{\rm N}$ , we get: | |
:$$P_{\rm R} = \frac{2 \cdot N_0 \cdot f_{\rm N}}{3 \cdot \eta^2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v }= \frac{3 \cdot \eta^2 \cdot P_{\rm S}}{2 \cdot N_0 \cdot f_{\rm N}} = 3 \cdot \rho_{v {\rm , \hspace{0.08cm}PM}}\hspace{0.05cm}.$$ | :$$P_{\rm R} = \frac{2 \cdot N_0 \cdot f_{\rm N}}{3 \cdot \eta^2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v }= \frac{3 \cdot \eta^2 \cdot P_{\rm S}}{2 \cdot N_0 \cdot f_{\rm N}} = 3 \cdot \rho_{v {\rm , \hspace{0.08cm}PM}}\hspace{0.05cm}.$$ | ||
− | + | *This means: frequency modulation is better than phase modulation by a factor of $3$ $($or $4.77 \ \rm dB)$ : | |
:$$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v }= 50\,{\rm dB} + 10 \cdot {\rm lg} \hspace{0.15cm}{3}\hspace{0.15cm}\underline {\approx 54.77\,{\rm dB}}\hspace{0.05cm}.$$ | :$$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v }= 50\,{\rm dB} + 10 \cdot {\rm lg} \hspace{0.15cm}{3}\hspace{0.15cm}\underline {\approx 54.77\,{\rm dB}}\hspace{0.05cm}.$$ | ||
− | '''(3)''' | + | |
+ | |||
+ | '''(3)''' According to the answer to question '''(2)''' , together with $f_{10} = 10 \ \rm kHz$, we get: | ||
:$$P_{\rm R} = \frac{2 \cdot N_0 \cdot f_{\rm 10}}{3 \cdot \eta_{10}^{\hspace{0.1cm}2}} = \frac{ f_{\rm 10} \cdot \eta_{5}^{\hspace{0.1cm}2}}{ 3 \cdot f_{\rm 5} \cdot \eta_{10}^{\hspace{0.1cm}2}}\cdot \frac{2 \cdot N_0 \cdot f_{\rm 5}}{\eta_{5}^{\hspace{0.1cm}2}} \hspace{0.05cm}.$$ | :$$P_{\rm R} = \frac{2 \cdot N_0 \cdot f_{\rm 10}}{3 \cdot \eta_{10}^{\hspace{0.1cm}2}} = \frac{ f_{\rm 10} \cdot \eta_{5}^{\hspace{0.1cm}2}}{ 3 \cdot f_{\rm 5} \cdot \eta_{10}^{\hspace{0.1cm}2}}\cdot \frac{2 \cdot N_0 \cdot f_{\rm 5}}{\eta_{5}^{\hspace{0.1cm}2}} \hspace{0.05cm}.$$ | ||
− | + | *the second term gives the noise power of the comparison system $($PM, $f_{\rm N} = f_5)$ , which led to the result of $10 · \lg ρ_v = 50\ \rm dB$ . | |
− | + | *However, in frequency modulation, the modulation index $η$ is now inversely proportional to the message frequency, so the quotient is $η_5^2/η_{10}^2 = 4$ . | |
+ | *Thus, the pre-factor is $8/3$. Due to the larger noise power, the SNR is smaller: | ||
:$$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v }= 50\,{\rm dB} - 10 \cdot {\rm lg} \hspace{0.15cm}({8}/{3})\hspace{0.15cm}\underline {\approx 45.74\,{\rm dB}}\hspace{0.05cm}.$$ | :$$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v }= 50\,{\rm dB} - 10 \cdot {\rm lg} \hspace{0.15cm}({8}/{3})\hspace{0.15cm}\underline {\approx 45.74\,{\rm dB}}\hspace{0.05cm}.$$ | ||
− | |||
− | * | + | |
− | * | + | At the same message frequency $f_{\rm N} = 10 \ \rm kHz$ , FM is now $1.25 \ \rm dB$ worse than PM, since the halving of $η$ – a factor of $4$ after squaring– snow has a greater effect than the system-dependent factor of $3$ by which FM was superior to PM. |
+ | |||
+ | *The comparison of subtasks '''(2)''' and '''(3)''' shows a difference by a factor of $8$ and $9.03 \ \rm dB$, respectively. | ||
+ | *The less favorable value for the larger message frequency $f_{\rm N} = 10 \ \rm kHz$ results from the halved modulation index and the doubled large noise bandwidth. | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
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− | [[Category: | + | [[Category:Modulation Methods: Exercises|^3.3 Noise Influence with PM and FM^]] |
Latest revision as of 18:00, 17 March 2022
Consider the phase and frequency modulation of a cosine oscillation with frequency $f_{\rm N}$. . First, let the message frequency be $f_{\rm N} = f_5 = 5 \ \rm kHz$ and the modulation index (phase deviation) be $η = 5$.
In the presence of additive Gaussian noise with noise power density $N_0$ , the PM demodulator results in a constant noise power density ${\it \Phi}_{v {\rm , \hspace{0.08cm}PM} }(f) = {\it \Phi}_0$, which also depends on the modulation index $η$ :
- $${\it \Phi}_0 = \frac{N_0}{\eta^2} \hspace{0.05cm}.$$
For the calculation of the noise power $P_{\rm R}$ , only the frequency range $±f_{\rm N}$ is relevant (see graph).
The noise power density after FM demodulation with the frequency deviation $Δf_{\rm A}$ is:
- $${\it \Phi}_{v {\rm , \hspace{0.08cm}FM} } (f) = N_0 \cdot \left(\frac{f}{\Delta f_{\rm A}}\right)^2 \hspace{0.05cm}.$$
- The signal-to-noise ratio $10 · \lg ρ_v = 50 \ \rm dB$ is given for phase modulation with $f_{\rm N} = 5 \ \rm kHz$.
- In this task, we are looking for the S/N ratio of FM for the message frequency $f_{\rm N} = 5 \ \rm kHz$ as well as the resulting S/N ratios of PM and FM for message frequency $f_{\rm N} = f_{10} = 10 \ \rm kHz$.
Hints:
- This exercise belongs to the chapter Influence of Noise on Systems with Angle Modulation.
- Particular reference is made to the section System comparison of AM, PM and FM with respect to noise.
Questions
Solution
- $$ \rho_{v } = \frac{P_{\rm S}}{P_{\rm R}} = \frac{P_{\rm S}}{{\it \Phi}_0 \cdot 2 f_{\rm N} } =\frac{\eta^2}{2} \cdot \frac{P_{\rm S}}{N_0 \cdot f_{\rm N} }\hspace{0.05cm}.$$
- The measurement with $f_{\rm N} = f_5 = 5 \ \rm kHz$ gives the SNR $ \rho_{v } = 10^5$ $($corresponding to $10 · \lg ρ_v =50\ \rm dB)$ .
- Doubling the message frequency results in half the SNR, since twice the noise power is now effective:
- $$ \rho_{v }= 0.5 \cdot 10^5 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v } \hspace{0.15cm}\underline {\approx 46.99\,{\rm dB}}\hspace{0.05cm}.$$
This result can also be derived using the relationship $ρ_v = η^2/2 · ξ$ .
- For phase modulation, $η$ is independent of the message frequency.
- DThe SNR loss is due to the fact that now the performance parameter $ξ = P_{\rm S}/(N_0 · f_{\rm N})$ is halved.
(2) For frequency modulation and a message frequency $f_{\rm N} = 5 \ \rm kHz$ the noise power is given by:
- $$P_{\rm R} = \int_{-f_{\rm N}}^{ + f_{\rm N}} {\it \Phi}_{v {\rm , \hspace{0.08cm}FM} } (f)\hspace{0.1cm}{\rm d}f = \frac{2 \cdot N_0}{\Delta f_{\rm A}^{\hspace{0.1cm}2}} \cdot \int_{0}^{ f_{\rm N}} f^2\hspace{0.1cm}{\rm d}f = \frac{2 \cdot N_0 \cdot f_{\rm N}^{\hspace{0.1cm}3}}{3 \cdot \Delta f_{\rm A}^2} \hspace{0.05cm}.$$
- Thus, considering the frequency deviation $Δf_{\rm A} = η · f_{\rm N}$ , we get:
- $$P_{\rm R} = \frac{2 \cdot N_0 \cdot f_{\rm N}}{3 \cdot \eta^2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v }= \frac{3 \cdot \eta^2 \cdot P_{\rm S}}{2 \cdot N_0 \cdot f_{\rm N}} = 3 \cdot \rho_{v {\rm , \hspace{0.08cm}PM}}\hspace{0.05cm}.$$
- This means: frequency modulation is better than phase modulation by a factor of $3$ $($or $4.77 \ \rm dB)$ :
- $$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v }= 50\,{\rm dB} + 10 \cdot {\rm lg} \hspace{0.15cm}{3}\hspace{0.15cm}\underline {\approx 54.77\,{\rm dB}}\hspace{0.05cm}.$$
(3) According to the answer to question (2) , together with $f_{10} = 10 \ \rm kHz$, we get:
- $$P_{\rm R} = \frac{2 \cdot N_0 \cdot f_{\rm 10}}{3 \cdot \eta_{10}^{\hspace{0.1cm}2}} = \frac{ f_{\rm 10} \cdot \eta_{5}^{\hspace{0.1cm}2}}{ 3 \cdot f_{\rm 5} \cdot \eta_{10}^{\hspace{0.1cm}2}}\cdot \frac{2 \cdot N_0 \cdot f_{\rm 5}}{\eta_{5}^{\hspace{0.1cm}2}} \hspace{0.05cm}.$$
- the second term gives the noise power of the comparison system $($PM, $f_{\rm N} = f_5)$ , which led to the result of $10 · \lg ρ_v = 50\ \rm dB$ .
- However, in frequency modulation, the modulation index $η$ is now inversely proportional to the message frequency, so the quotient is $η_5^2/η_{10}^2 = 4$ .
- Thus, the pre-factor is $8/3$. Due to the larger noise power, the SNR is smaller:
- $$10 \cdot {\rm lg} \hspace{0.15cm}\rho_{v }= 50\,{\rm dB} - 10 \cdot {\rm lg} \hspace{0.15cm}({8}/{3})\hspace{0.15cm}\underline {\approx 45.74\,{\rm dB}}\hspace{0.05cm}.$$
At the same message frequency $f_{\rm N} = 10 \ \rm kHz$ , FM is now $1.25 \ \rm dB$ worse than PM, since the halving of $η$ – a factor of $4$ after squaring– snow has a greater effect than the system-dependent factor of $3$ by which FM was superior to PM.
- The comparison of subtasks (2) and (3) shows a difference by a factor of $8$ and $9.03 \ \rm dB$, respectively.
- The less favorable value for the larger message frequency $f_{\rm N} = 10 \ \rm kHz$ results from the halved modulation index and the doubled large noise bandwidth.