Question

The figure of merit ratio of FM to PM for single tone modulating signal is

• A. 1
• B. 3
• C. 4
• D. \(\infty\)

Hint:

Figure of Merit (FOM) compares output SNR to some reference SNR.
For FM (single tone): \( (SNR)_o / (P_c/(N_0 B_m)) = \frac{3}{2} \beta^2 \), where \(\beta = \Delta f / f_m\).
For PM (single tone): \( (SNR)_o / (P_c/(N_0 B_m)) = \frac{1}{2} (\Delta\phi)^2 \), where \(\Delta\phi\) is peak phase deviation.
If modulation is set such that peak frequency deviation in FM is equal to peak phase deviation times \(f_m\) in PM (i.e., \(\Delta f = \Delta\phi \cdot f_m\)), then \(\beta_{FM} = \Delta\phi\). The ratio of FOMs becomes 3.

Answer 1

The Correct Option is B

The figure of merit (FOM) for an angle modulation system (like FM or PM) is a measure of its output Signal-to-Noise Ratio (SNR\(_o\)) improvement over the input Signal-to-Noise Ratio (SNR\(_i\)) or over a baseband system or an AM system. For wideband FM with single-tone modulation, the figure of merit (improvement in SNR over AM with \(m=1\)) is often given as: FOM\(_{FM} = \frac{3}{2} \beta^2 (\text{for input SNR defined as } P_S / (N_0 W))\) or \(3\beta^2 (\text{for input SNR defined as carrier power to noise power in message bandwidth})\) where \(\beta\) is the modulation index of FM (\(\beta = \Delta f / f_m\)). For PM with single-tone modulation \(m(t) = A_m \cos(\omega_m t)\), the phase deviation is \(\Delta\phi = k_p A_m\). The modulation index for PM is \(\beta_p = \Delta\phi\). The SNR improvement for PM is proportional to \(\beta_p^2\). Figure of Merit for FM: \( \text{FOM}_{FM} \propto (\frac{\Delta f}{W})^2 \), where \(W\) is message bandwidth. For single tone, \(W=f_m\), so \( \text{FOM}_{FM} \propto \beta^2 \). More precisely, \( \text{SNR}_{o,FM} / \text{SNR}_{o,AM} \approx \frac{3}{2} \beta^2 \). Figure of Merit for PM: \( \text{FOM}_{PM} \propto (\Delta \phi)^2 \). More precisely, \( \text{SNR}_{o,PM} / \text{SNR}_{o,AM} \approx \frac{1}{2} (\Delta \phi)^2 \). (These relations are with respect to output SNR of an envelope detected AM system with m=1). A different common comparison is for the SNR at the output of FM and PM demodulators compared to the SNR of the baseband signal if transmitted directly without modulation over the same channel noise power spectral density \(N_0/2\). \( (SNR)_o_ {FM} = \frac{3 A_c^2 k_f^2 P_m}{2 N_0 W^3} \) for a specific definition of \(P_m\), where \(W\) is related to highest freq. \( (SNR)_o_ {PM} = \frac{A_c^2 k_p^2 \omega_m^2 P_m'}{2 N_0 W} \) (PM output SNR depends on \(\omega_m^2\)). For single-tone modulation \(x(t) = A_m \cos(\omega_m t)\): For FM, the output SNR is \( \text{SNR}_{o,FM} = \frac{3 A_c^2 (\Delta f)^2}{4 N_0 f_m^3} \) if using \(f_m\) as bandwidth W. Or \( \frac{3}{2} \beta^2 \frac{P_c}{N_0 B_m} \). For PM, the output SNR is \( \text{SNR}_{o,PM} = \frac{A_c^2 (\Delta \phi)^2 f_m^2}{2 N_0 f_m^3} = \frac{1}{2} (\Delta \phi)^2 \frac{P_c}{N_0 B_m} \). The question asks for "figure of merit ratio of FM to PM". This is ambiguous as "figure of merit" can be defined differently. However, a well-known result, especially when considering pre-emphasis/de-emphasis, states that FM provides a 3 times (or 4.77 dB) better SNR than PM for certain types of message signals (like those with power spectral density that falls with frequency, typical for voice/audio). If the FOM for FM is taken as \(3\beta^2\) and for PM as \(\beta_p^2\) (where \(\beta_p = \Delta\phi_{peak}\)), and if we are comparing for "equivalent" conditions where peak frequency deviation in FM is made equal to peak phase deviation times modulating frequency for PM. This question is tricky without a precise definition of FOM being used. A common result cited is that for speech or music, FM is superior to PM by a factor of (W/f_avg)^2 where W is highest frequency and f_avg is average. However, option (b) "3" is a specific number often associated with comparisons. One specific comparison: Under conditions where both FM and PM produce the same transmission bandwidth and have the same carrier power, and for a message signal whose power spectrum is uniform, the SNR at the output of an FM system is three times the SNR at the output of a PM system, assuming optimal pre-emphasis and de-emphasis are used with the PM system to make it behave like FM for noise. Without pre-emphasis, the ratio can vary. If the question implicitly refers to the ratio of SNR improvement factors when \(\beta_{FM} = \beta_{PM} (\omega_{max}/\omega_m)\) or similar normalization. The factor '3' arises in some comparisons related to how noise power is distributed. For FM, noise power at output is proportional to \(f^2\), integrated over bandwidth. For PM, it's flat. The "figure of merit" often refers to \((SNR)_o / (SNR)_c\), where \( (SNR)_c = P_c / (N_0 B_m)\). Then for FM, \( FOM_{FM} = \frac{3}{2}\beta^2 \). For PM, \( FOM_{PM} = \frac{1}{2}(\Delta\phi)^2 \). If we set comparable conditions, for example, same peak deviation \(\Delta\omega_{peak}\). For FM, \(\Delta\omega_{peak} = \Delta f \cdot 2\pi\). For PM, \(\Delta\omega_{peak} = \Delta\phi \cdot \omega_m\). If \(\Delta f = \Delta\phi \cdot f_m\), then \(\beta_{FM} = \Delta f/f_m = \Delta\phi\). Then \(FOM_{FM} / FOM_{PM} = (\frac{3}{2}\beta^2) / (\frac{1}{2}(\Delta\phi)^2) = 3\). This holds if \(\beta\) in FM formula is peak freq deviation / message freq, and \(\Delta\phi\) in PM formula is peak phase deviation. And if we are comparing under condition that the modulation indices are effectively the same after considering the \(f_m\) scaling. Thus, the ratio is 3. \[ \boxed{3} \]