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Threshold effect in FM; error probability of coherent PSK

Short answer

Threshold effect in FM occurs when CNR drops below ~10 dB, causing noise phasor to exceed carrier and produce 2π phase jumps (clicks), rapidly degrading output SNR making FM worse than AM. Coherent BPSK error probability is P_e = 1/2 erfc(√(E_b/N_0)) = Q(√(2E_b/N_0)), derived from correlator output with Gaussian noise and zero threshold.

Explain threshold effect in detection of FM signal. Derive the expression of error probability for coherent detection of Phase Shift Keying (PSK).

Threshold effect in FM detection

Threshold effect in FM. The FM improvement SNRo∝β2\text{SNR}_o\propto\beta^2 holds only when the carrier is much stronger than the noise at the discriminator input. When the carrier-to-noise ratio (CNR) falls below a threshold, about 10 dB (typically 10–13 dB), the noise phasor sometimes becomes larger than the carrier and the resultant phasor encircles the origin, causing a sudden 2π2\pi phase jump.

  high CNR: noise wobbles     low CNR: resultant can swing
  the tip slightly            around origin -> 2pi jump
        .-.                         .---.
  0 -->( * )                   0 --( *   )
        '-'                         '---'

Each 2π2\pi jump produces an impulse (spike) in the discriminator output, heard as clicks. The click rate rises rapidly as CNR drops, so the output SNR falls much faster than linearly and the signal is mutilated. Above threshold, FM is excellent; below it, FM is worse than AM. Wider deviation (larger β\beta) means larger BTB_T, more noise and a higher threshold.

 SNRo(dB)            . FM above threshold
     |             .   (slope 1, offset 3/2 beta^2)
     |           .
     |         .
     |       . <- threshold knee (CNR ~ 10 dB)
     |     .:
     |   . :     AM / baseband
     |  .  :
     +------------------> SNRi (dB)

The threshold can be lowered (threshold extension) by using FM feedback (FMFB) or PLL demodulators. Pre-emphasis/de-emphasis improves the output SNR above threshold but does not lower the threshold.

Error probability of coherent PSK

BPSK signals over 0≤t≤Tb0\le t\le T_b (antipodal):

s1(t)=Acos⁡ωct  (1),s2(t)=−Acos⁡ωct  (0),Eb=A2Tb2s_1(t)=A\cos\omega_ct\ \ (1),\qquad s_2(t)=-A\cos\omega_ct\ \ (0),\qquad E_b=\frac{A^2T_b}{2}

Coherent receiver: multiply by the locally generated cos⁡ωct\cos\omega_ct, integrate over TbT_b, sample, compare with threshold λ=0\lambda=0.

 r(t) -->(x)--> integrate --> sample --> y > 0 ? 1 : 0
          ^       0..Tb       t=Tb
          |
       cos(wc t)  (from carrier recovery)

Correlator output y=ai+ny=a_i+n with

a1=+ATb2,a2=−ATb2,σ2=N0Tb4a_1=+\frac{AT_b}{2},\qquad a_2=-\frac{AT_b}{2},\qquad \sigma^2=\frac{N_0T_b}{4}

The pdfs of yy are Gaussian centred at ±ATb/2\pm AT_b/2. Error when 0 is sent (y>0y>0):

Pe=∫0∞12πσe−(y+ATb/2)2/2σ2dy=12erfc⁡(ATb/22 σ)=12erfc⁡(ATb/2N0Tb/2)=12erfc⁡(A2Tb2N0)\begin{aligned} P_e &= \int_0^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-(y+AT_b/2)^2/2\sigma^2}dy=\frac12\operatorname{erfc}\left(\frac{AT_b/2}{\sqrt2\,\sigma}\right) \\ &= \frac12\operatorname{erfc}\left(\frac{AT_b/2}{\sqrt{N_0T_b/2}}\right)=\frac12\operatorname{erfc}\left(\sqrt{\frac{A^2T_b}{2N_0}}\right) \end{aligned}

By symmetry the error for a sent 1 is the same, so

Pe(BPSK)=12erfc⁡(EbN0)=Q(2EbN0)P_e(\text{BPSK})=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_b}{N_0}}\right)=Q\left(\sqrt{\frac{2E_b}{N_0}}\right)

BPSK is the best binary scheme: it needs 3 dB less Eb/N0E_b/N_0 than coherent ASK or FSK for the same PeP_e. Example: Eb/N0=10E_b/N_0=10 dB gives Pe=Q(20)≈3.9×10−6P_e=Q(\sqrt{20})\approx3.9\times10^{-6}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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