Exercise 3.10: Noise Power Calculation

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

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.