--- title: "Bit error probability of coherent BFSK" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:bfsk-coherent-ber" source: https://www.bctengineer.com/q/semester_6/communication_systems/bfsk-coherent-ber site: BCT Engineering --- # Bit error probability of coherent BFSK Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2081 Baisakh (CS II), 6 marks - Video: [8 Probability Error Of BFSK Explained Module 2 | 5th Sem Digital Communication ECE VTU](https://www.youtube.com/watch?v=GyiCXOiq44o) — VTU Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/bfsk-coherent-ber - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question Derive the bit error probability for the coherent binary FSK system. ## Short answer Coherent BFSK uses orthogonal tones with energy E_b per bit. The decision variable is Gaussian with mean ±AT_b/2 and variance N_0T_b/2. Bit error probability is P_e = ½erfc(√(E_b/2N_0)) = Q(√(E_b/N_0)). This is 3 dB worse than BPSK because orthogonal signals are √(2E_b) apart versus 2√E_b for antipodal signals. ## Answer In **binary FSK**, bit 1 and bit 0 are sent as two different carrier frequencies $f_1$ and $f_2$ with the same amplitude. **Coherent BFSK signals** over $0\le t\le T_b$: $$ s_1(t)=A\cos\omega_1t\ \ (1),\qquad s_2(t)=A\cos\omega_2t\ \ (0),\qquad E_b=\frac{A^2T_b}{2} $$ The tones are chosen **orthogonal** over $T_b$: $\int_0^{T_b}\cos\omega_1t\cos\omega_2t\,dt=0$ (e.g. $\lvert f_1-f_2\rvert=n/2T_b$). **Coherent receiver:** two correlators, one per tone; their outputs are subtracted and compared with zero. ```text +->(x)--> integ 0..Tb --> y1 --+ | ^ cos w1t |(+) r(t) --+ (sum)--> l = y1 - y2 | v cos w2t |(-) l > 0 ? 1 : 0 +->(x)--> integ 0..Tb --> y2 --+ ``` **Signal part of $l$.** - 1 sent: $y_1=\frac{AT_b}{2}$, $y_2=0$ (orthogonality) so $l=+\frac{AT_b}{2}$. - 0 sent: $y_1=0$, $y_2=\frac{AT_b}{2}$ so $l=-\frac{AT_b}{2}$. **Noise part.** $n_1=\int w\cos\omega_1t\,dt$ and $n_2=\int w\cos\omega_2t\,dt$ each have variance $\frac{N_0T_b}{4}$ and are uncorrelated (orthogonal tones), so $n=n_1-n_2$ has $$ \sigma^2=\frac{N_0T_b}{4}+\frac{N_0T_b}{4}=\frac{N_0T_b}{2} $$ **Error probability.** $l$ is Gaussian with mean $\pm AT_b/2$ and variance $\sigma^2$; threshold 0. For 0 sent, error if $l>0$: $$ \begin{aligned} P_e &= \frac12\operatorname{erfc}\left(\frac{AT_b/2}{\sqrt2\,\sigma}\right)=\frac12\operatorname{erfc}\left(\frac{AT_b/2}{\sqrt{N_0T_b}}\right) \\ &= \frac12\operatorname{erfc}\left(\sqrt{\frac{A^2T_b}{4N_0}}\right) \end{aligned} $$ With $E_b=A^2T_b/2$: $$ P_e(\text{coherent BFSK})=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_b}{2N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right) $$ Check with the general formula: $E_d=\int(s_1-s_2)^2dt=E_b+E_b=2E_b$, so $\frac12\operatorname{erfc}\sqrt{E_d/4N_0}=\frac12\operatorname{erfc}\sqrt{E_b/2N_0}$. **Comparison:** BFSK has the same $P_e$ as ASK (for equal average energy) and needs **3 dB more $E_b/N_0$ than BPSK**, because orthogonal signals are $\sqrt{2E_b}$ apart while antipodal ones are $2\sqrt{E_b}$ apart. Non-coherent FSK gives $P_e=\frac12e^{-E_b/2N_0}$, which is slightly worse but needs no carrier phase. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are the two BFSK tones chosen orthogonal over T_b? A. So the cross-correlator output is zero and each correlator responds only to its own tone B. So the bandwidth is minimised to zero C. So the noise variance becomes zero D. So that the signal energy doubles Answer: A. Orthogonality gives y2 = 0 when 1 is sent (and y1 = 0 when 0 is sent), so l = +/- A T_b/2. (Key point: Orthogonal tones) 2. What is the variance of n = n1 - n2 in the coherent BFSK receiver? A. N0 Tb / 4 B. N0 Tb / 2 C. N0 Tb D. 2 N0 Tb Answer: B. n1 and n2 are uncorrelated, each with variance N0Tb/4, so the variances add to N0Tb/2. (Key point: Noise variance of decision variable) 3. What is Pe of coherent BFSK? A. (1/2) erfc( sqrt(Eb/N0) ) B. Q( sqrt(2 Eb/N0) ) C. (1/2) erfc( sqrt(Eb/2N0) ) = Q( sqrt(Eb/N0) ) D. (1/2) exp(-Eb/2N0) Answer: C. Substituting Eb = A^2 Tb/2 gives Pe = (1/2) erfc(sqrt(Eb/2N0)). The last option is non-coherent FSK. (Key point: BFSK error probability) 4. Why does coherent BFSK need 3 dB more Eb/N0 than BPSK? A. Its noise is twice as strong B. Orthogonal signals are sqrt(2Eb) apart while antipodal ones are 2 sqrt(Eb) apart C. It uses twice the bandwidth only D. It needs non-coherent detection Answer: B. Smaller signal distance means Ed = 2Eb instead of 4Eb, so 3 dB more energy is needed. 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