--- title: "BFSK generation, non-coherent detection and signal space" subject: "Communication Systems" chapter: 7 key: "semester_6/communication_systems:bfsk-noncoherent-signal-space" source: https://www.bctengineer.com/q/semester_6/communication_systems/bfsk-noncoherent-signal-space site: BCT Engineering --- # BFSK generation, non-coherent detection and signal space Communication Systems · Chapter 7: Digital Modulation Techniques - Asked in: 2076 Chaitra (CS II), 8 marks - Video: [DC#44 Binary Frequency Shift Keying (BFSK) signal Generation, coherent & non coherent detection ||EC](https://www.youtube.com/watch?v=-zJrQmiS9m0) — EC 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-noncoherent-signal-space - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-7 ## Question Explain the generation and non coherent detection for BFSK signal and also show the signal space diagram for a BFSK signal. ## Short answer BFSK transmits symbols 1 and 0 as tones f1 and f2 using gated oscillators. Non-coherent detection compares envelope outputs of bandpass filters centered at f1 and f2; the larger envelope decides the bit. Signal space points are (√E_b,0) and (0,√E_b), orthogonal, with distance √(2E_b) and decision boundary x1=x2. ## Answer **Binary FSK** transmits symbol 1 as a tone of frequency $f_1$ and symbol 0 as a tone of frequency $f_2$, each lasting $T_b$ and having constant amplitude: $$ s_i(t)=\sqrt{\frac{2E_b}{T_b}}\cos 2\pi f_it,\qquad f_i=\frac{n_c+i}{T_b},\ i=1,2,\ 0\le t\le T_b $$ The choice $f_i=(n_c+i)/T_b$ makes the tones orthogonal over a bit. #### Generation ```text data b(t) (on-off) ------->(X)<-- sqrt(2Eb/Tb) cos w1 t | | | (+)--------> BFSK | | +-->[Inverter]-------->(X)<-- sqrt(2Eb/Tb) cos w2 t ``` 1. The binary data is in on-off (unipolar) form: 1 is "on", 0 is "off". 2. For bit 1 the upper product modulator passes the $f_1$ oscillator; the inverter blocks the lower one. 3. For bit 0 the inverter output is 1, so the $f_2$ oscillator passes. 4. The adder combines both branches; only one tone appears in each bit. If the two oscillators are synchronised, phase is continuous at bit boundaries (CPFSK). A single VCO driven by polar data gives the same result. #### Non-coherent detection The phase of the received tone is not needed. Two filter-envelope branches compare energy at $f_1$ and $f_2$. ```text +->[BPF f1]->[Env. det]--l1--+ | | x(t) --+ [Comparator]--> data | | l1>l2: 1 +->[BPF f2]->[Env. det]--l2--+ l1l_2$, else 0. An equivalent form uses quadrature correlators (cos and sin) in each branch and squares and adds their outputs. Error probability: $P_e=\tfrac12 e^{-E_b/2N_0}$, slightly worse than coherent BFSK but with a much simpler receiver (no carrier recovery). #### Signal space diagram Orthonormal basis: $\phi_i(t)=\sqrt{2/T_b}\cos 2\pi f_it$, $i=1,2$. Then $$ \mathbf{s}_1=(\sqrt{E_b},\,0),\qquad \mathbf{s}_2=(0,\,\sqrt{E_b}) $$ ```text phi2 | * s2 (0, sqrt Eb) / boundary x1 = x2 | Z2 / | / | / Z1 | / +---------------*--------> phi1 0 s1 (sqrt Eb, 0) ``` - Two-dimensional, two orthogonal message points. - Euclidean distance $d=\sqrt{2E_b}$. - Decision boundary: the bisector $x_1=x_2$. - Coherent $P_e=\tfrac12\text{erfc}\sqrt{E_b/2N_0}$, i.e. 3 dB worse than BPSK because of the smaller distance. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the generator, what happens for bit 0? A. Both branches are off B. The inverter output turns on the f2 branch, so tone f2 appears C. The f1 tone is phase inverted D. Both tones are sent together Answer: B. Only one tone appears in each bit. (Key point: BFSK generation) 2. What does the non-coherent receiver compare? A. The correlator output with zero B. The phase of the received tone with a recovered carrier C. The envelope outputs l1 and l2 of filters at f1 and f2 D. The delayed and present bit Answer: C. Decide 1 if l1 > l2, otherwise 0, without needing phase. (Key point: Non-coherent detection) 3. Compared with coherent BFSK, non-coherent BFSK has A. Better Pe and more complexity B. The same Pe and the same complexity C. Better Pe and no filters D. Slightly worse Pe, (1/2)e^(-Eb/2N0), but a much simpler receiver Answer: D. Not needing carrier recovery costs a small error penalty. (Key point: Non-coherent Pe) 4. Which basis and points are used? A. phi_i = sqrt(2/Tb)cos(2 pi fi t); s1 = (sqrt(Eb), 0), s2 = (0, sqrt(Eb)); d = sqrt(2Eb) B. One basis; points at +-sqrt(Eb); d = 2 sqrt(Eb) C. Two bases; points at (+-sqrt(Eb/2), +-sqrt(Eb/2)); d = sqrt(Eb) D. One basis; points at 0 and sqrt(2Eb); d = sqrt(2Eb) Answer: A. Orthogonal tones give two axes and a boundary x1 = x2. (Key point: Signal space) ## Related in Communication Systems - [Threshold effect in non-coherent FM detection and its correction](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Athreshold-noncoherent-fm-correction) — Communication Systems, chapter 9 - [Coherent vs non-coherent detection; carrier recovery](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acoherent-noncoherent-carrier-recovery) — Communication Systems, chapter 2 - [BPSK modulation and signal space diagram](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Abpsk-signal-space) — Communication Systems, chapter 7 - [Bit error probability of coherent BFSK](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Abfsk-coherent-ber) — Communication Systems, chapter 9 - [Phase error in coherent DSB-SC detection; AM demodulation by PLL](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aphase-error-dsbsc-am-pll) — Communication Systems, chapter 2 - [Noise performance of coherent DSB-FC detection](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adsbfc-noise-coherent-100pct) — Communication Systems, chapter 9 ## Related in other subjects - [Four-segment instruction pipeline and space-time diagram](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Afour-segment-pipeline) — Computer Organization and Architecture, chapter 4, asked 5× - [Hydropower generation with block diagram](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ahydropower-generation-block-diagram) — Power System Analysis I, chapter 1 - [Signal flow graph of G1–G3, H1–H3 block diagram](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asfg-g1g2g3-h1h2h3) — Control Systems, chapter 3