--- title: "FSK modulator, demodulator and signal space" subject: "Communication Systems" chapter: 7 key: "semester_6/communication_systems:fsk-mod-demod-signal-space" source: https://www.bctengineer.com/q/semester_6/communication_systems/fsk-mod-demod-signal-space site: BCT Engineering --- # FSK modulator, demodulator and signal space Communication Systems · Chapter 7: Digital Modulation Techniques - Asked in: 2072 Chaitra (CS II), 6 marks; 2069 Chaitra (CS II), 8 marks (asked 2 times) - Video: [BFSK signal space diagram, transmitter and receiver@fewminuteslearning9815](https://www.youtube.com/watch?v=qEjP2TmxfLA) — Few Minutes Learning - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/fsk-mod-demod-signal-space - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-7 ## Question Explain the modulator, demodulator and signal space diagram for FSK Modulation. ## Short answer BFSK uses two orthogonal frequencies for bits 0 and 1. The modulator switches between two oscillators. The coherent demodulator correlates with both basis functions and compares outputs. Signal space has points at (√Eb,0) and (0,√Eb) with a 45° decision boundary. ## Answer **Binary FSK (BFSK)** sends a bit by switching the carrier between two frequencies: one frequency for symbol 1 and another for symbol 0, while the amplitude stays constant. $$ s_i(t)=\begin{cases}\sqrt{\dfrac{2E_b}{T_b}}\cos(2\pi f_i t), & 0\le t\le T_b\\[4pt] 0, & \text{elsewhere}\end{cases}\qquad i=1,2 $$ with $f_i=\dfrac{n_c+i}{T_b}$ ($n_c$ an integer), so that the two tones are **orthogonal** over one bit. Symbol 1 is sent by $s_1(t)$ (frequency $f_1$) and symbol 0 by $s_2(t)$ (frequency $f_2$). #### Modulator The binary data in on-off form is applied to one product modulator, and its inverse to the other. Exactly one oscillator output reaches the adder in each bit interval. ```text b(t) on-off |---------------->(X)<---- cos 2*pi*f1*t | | | v | (+)------> BFSK s(t) | ^ v | [Inverter]-------->(X)<---- cos 2*pi*f2*t ``` - Bit 1: the upper channel is ON and the lower OFF, so the output is tone $f_1$. - Bit 0: the inverter turns the lower channel ON, so the output is tone $f_2$. - If the two oscillators are phase-synchronised (or one VCO is used), the phase is continuous (CPFSK). #### Coherent demodulator The receiver uses two correlators, one matched to each basis function, and compares their outputs. ```text +->(X)->[int 0..Tb]--x1--+ | ^ phi1(t) |+ x(t) ---+ (S)--> l = x1 - x2 | v phi2(t) |- | +->(X)->[int 0..Tb]--x2--+ v [l > 0 -> 1, l < 0 -> 0] ``` $\phi_1(t)=\sqrt{2/T_b}\cos 2\pi f_1t$ and $\phi_2(t)=\sqrt{2/T_b}\cos 2\pi f_2t$. If $l=x_1-x_2>0$ the receiver decides 1, otherwise 0. A non-coherent version uses band-pass filters at $f_1$ and $f_2$ followed by envelope detectors and a comparator. #### Signal space diagram BFSK is a two-dimensional scheme ($N=2$, $M=2$). The signal vectors are $$ \mathbf{s}_1=\begin{bmatrix}\sqrt{E_b}\\0\end{bmatrix},\qquad \mathbf{s}_2=\begin{bmatrix}0\\\sqrt{E_b}\end{bmatrix} $$ ```text phi2 | / decision boundary | Z2 / (x1 = x2 line) s2 * / |(0,sqrt Eb) | / | / Z1 | / | / +-----------*---------> phi1 0 s1 (sqrt Eb, 0) ``` - Distance between points: $d_{12}=\sqrt{2E_b}$ (smaller than $2\sqrt{E_b}$ for BPSK). - The decision boundary is the line $x_1=x_2$ through the origin at 45 degrees (the perpendicular bisector of the line joining $\mathbf{s}_1$ and $\mathbf{s}_2$). - Bit error probability (coherent): $P_e=\tfrac12\,\text{erfc}\sqrt{E_b/2N_0}$, so BFSK needs 3 dB more $E_b/N_0$ than BPSK for the same $P_e$. - Bandwidth is about $\lvert f_1-f_2\rvert+2/T_b$, larger than BPSK. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is fi = (nc + i)/Tb chosen for the two BFSK frequencies? A. So that the phase is always continuous B. So the two tones are orthogonal over one bit interval C. So that the bandwidth is minimum possible D. So that the amplitude of the two tones is equal Answer: B. With frequencies at integer multiples of 1/Tb the tones integrate to zero against each other over Tb, which allows independent correlators. (Key point: Orthogonal tones) 2. In the BFSK modulator, what does the inverter do? A. Inverts the phase of the carrier for bit 0 B. Doubles the frequency for bit 0 C. Turns on the f2 branch when the bit is 0, while the f1 branch is off D. Switches both oscillators on for bit 0 Answer: C. The on-off data gates f1; its inverse gates f2, so exactly one tone reaches the adder in each bit. (Key point: Modulator operation) 3. In the coherent BFSK demodulator, the receiver decides 1 when A. x1 + x2 > 0 B. x1 > 0 only C. The envelope exceeds half the peak D. l = x1 - x2 > 0 Answer: D. The difference of the two correlator outputs is compared with zero: larger f1 correlation means symbol 1. (Key point: Decision rule) 4. What are the BFSK signal points and decision boundary? A. s1 = (sqrt(Eb), 0), s2 = (0, sqrt(Eb)); boundary x1 = x2 B. s1 = (sqrt(Eb), 0), s2 = (-sqrt(Eb), 0); boundary x1 = 0 C. s1 = (sqrt(Eb), sqrt(Eb)), s2 = (-sqrt(Eb), -sqrt(Eb)); boundary x2 = 0 D. s1 = (0, 0), s2 = (sqrt(2Eb), 0); boundary x1 = sqrt(Eb/2) Answer: A. BFSK uses two orthogonal basis functions, so the points lie on separate axes and the bisector is the 45-degree line x1 = x2. (Key point: Signal space) 5. Why does coherent BFSK need 3 dB more Eb/N0 than BPSK for the same Pe? A. The distance between points is sqrt(2Eb), smaller than 2 sqrt(Eb) B. Its bandwidth is smaller than BPSK C. It requires non-coherent detection D. Its amplitude is lower than BPSK Answer: A. Distance sqrt(2Eb) versus 2sqrt(Eb) means the squared distance is half, so 3 dB more energy is needed. 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