--- title: "MSK modulation scheme" subject: "Communication Systems" chapter: 7 key: "semester_6/communication_systems:msk-modulation" source: https://www.bctengineer.com/q/semester_6/communication_systems/msk-modulation site: BCT Engineering --- # MSK modulation scheme Communication Systems · Chapter 7: Digital Modulation Techniques - Asked in: 2080 Chaitra, 4 marks - Video: [MSK / Minimum Shift Keying /in Digital Communication](https://www.youtube.com/watch?v=cDwwTGSK9pM) — Engineering Pathshala - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/msk-modulation - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-7 ## Question Explain the modulation scheme for MSK with appropriate example. ## Short answer MSK is continuous-phase FSK with frequency deviation 1/(4Tb). It is equivalent to OQPSK with half-sinusoid pulse shaping, giving constant envelope and low sidelobes. The phase changes by ±π/2 per bit. Bit error probability equals BPSK. ## Answer **Minimum Shift Keying (MSK)** is continuous-phase binary FSK in which the two tones are separated by the minimum spacing that still keeps them orthogonal, $\Delta f=f_1-f_2=\dfrac{1}{2T_b}$ (deviation ratio $h=0.5$). #### Signal $$ s(t)=\sqrt{\frac{2E_b}{T_b}}\cos\left[2\pi f_c t+\theta(t)\right],\qquad \theta(t)=\theta(0)\pm\frac{\pi}{2T_b}t $$ - Bit 1: the phase increases by $\pi/2$ over the bit ($f_1=f_c+\tfrac{1}{4T_b}$). - Bit 0: the phase decreases by $\pi/2$ over the bit ($f_2=f_c-\tfrac{1}{4T_b}$). - The phase is **continuous** at bit boundaries, so there are no sudden jumps. Expanding in quadrature form, MSK is equivalent to offset QPSK with half-sinusoid pulse shaping: $$ s(t)=\sqrt{\tfrac{2E_b}{T_b}}\cos\theta(t)\cos 2\pi f_ct-\sqrt{\tfrac{2E_b}{T_b}}\sin\theta(t)\sin 2\pi f_ct $$ #### Modulator ```text data -> [S/P: odd/even bits, each 2Tb long] | I bits | Q bits (offset Tb) v v (X) cos(pi t/2Tb) (X) sin(pi t/2Tb) | | (X) cos 2pi fc t (X) sin 2pi fc t | | +-------->(+)<--------+ | MSK out ``` #### Example Data 1 1 0 1, starting phase $\theta(0)=0$: | Bit | Phase change | Phase at end of bit | |---|---|---| | 1 | $+\pi/2$ | $\pi/2$ | | 1 | $+\pi/2$ | $\pi$ | | 0 | $-\pi/2$ | $\pi/2$ | | 1 | $+\pi/2$ | $\pi$ | The phase moves along straight lines (the phase trellis) and never jumps. For $T_b=1\ \mu\text{s}$ the tones are $f_c\pm 250$ kHz, a spacing of 500 kHz $=1/(2T_b)$. #### Features - Constant envelope, so efficient non-linear amplifiers can be used. - The main lobe is 1.5 times wider than that of QPSK, but the sidelobes are much lower, falling as $1/f^4$ (less adjacent-channel interference). - Same $P_e$ as BPSK/QPSK with coherent detection: $\tfrac12\text{erfc}\sqrt{E_b/N_0}$. - Gaussian filtering of the data before MSK gives GMSK (used in GSM). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the tone spacing of MSK? A. 1/Tb, so the tones are orthogonal B. 1/(2Tb), the minimum spacing that keeps the tones orthogonal C. 1/(4Tb), the minimum for continuous phase D. 2/Tb, to reduce the sidelobes Answer: B. MSK is CPFSK with deviation ratio h = 0.5, i.e. frequency difference 1/(2Tb). (Key point: Frequency spacing) 2. In MSK, what happens to the phase over a bit of 1 or 0? A. It jumps by pi at every bit boundary B. It changes by +pi for 1 and -pi for 0 C. It changes linearly by +pi/2 for 1 and -pi/2 for 0, continuous at boundaries D. It stays fixed and only the amplitude changes Answer: C. Phase moves along straight lines in the trellis, never jumping. (Key point: Phase behaviour) 3. For data 1 1 0 1 starting at phase 0, what is the phase at the end of the third bit? A. pi B. 0 C. 3pi/2 D. pi/2 Answer: D. Phase goes 0 to pi/2, pi, then the 0 bit subtracts pi/2 giving pi/2. (Key point: Example) 4. Which is a feature of MSK? A. Constant envelope with low sidelobes falling as 1/f^4 B. Variable envelope needing linear amplifiers C. Narrower main lobe than QPSK D. Worse error probability than BPSK Answer: A. Constant envelope allows efficient non-linear amplifiers, and the smooth phase gives much lower sidelobes though the main lobe is 1.5 times wider than QPSK. (Key point: MSK features) ## Related in Communication Systems - [Why SSB preferred; AM power at 100% modulation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Assb-preferred-am-power-1000w) — Communication Systems, chapter 2 - [QAM modulation, demodulation and constellation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aqam-mod-demod-constellation) — Communication Systems, chapter 7 ## Related in other subjects - [MSK modulation and DSSS](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Amsk-dsss) — Wireless Communication, chapter 4 - [MSK as CPFSK/OQPSK; compare OQPSK, MSK, GMSK](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Amsk-oqpsk-gmsk-comparison) — Wireless Communication, chapter 4 - [MSK, GMSK and advantages of spread spectrum](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Amsk-gmsk-spread-spectrum-advantages) — Wireless Communication, chapter 4 - [MSK, GMSK and OFDM modulator/demodulator](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Amsk-gmsk-ofdm-modem) — Wireless Communication, chapter 4