Aufgaben:Exercise 3.10: Noise Power Calculation

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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  $η$ :

$${\it \Phi}_0 = \frac{N_0}{\eta^2} \hspace{0.05cm}.$$

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

1

What is the signal-to-noise ratio for  phase modulation  and  $f_{\rm N} = 10 \ \rm kHz$? Interpret the result.

$10 · \lg ρ_v \ = \ $ $\ \rm dB$

2

Calculate the signal-to-noise ratio for frequency modulation and  $f_{\rm N} = 5 \ \rm kHz$. What is the modulation index for this configuration?

$10 · \lg ρ_v \ = \ $ $\ \rm dB$

3

Calculate the signal-to-noise ratio for frequency modulation and  $f_{\rm N} = 10 \ \rm kHz$.  Interpret the result in comparison to your answers for   (1)  and  (2).

$10 · \lg ρ_v \ = \ $ $\ \rm dB$

Solution

Solution

(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:

$$ \rho_{v } = \frac{P_{\rm S}}{P_{\rm R}} = \frac{P_{\rm S}}{{\it \Phi}_0 \cdot 2 f_{\rm N} } =\frac{\eta^2}{2} \cdot \frac{P_{\rm S}}{N_0 \cdot f_{\rm N} }\hspace{0.05cm}.$$

• 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.