--- title: "Doppler spread and Doppler shift derivation" subject: "Wireless Communication" chapter: 3 key: "semester_7/wireless_communication:doppler-spread-shift-derivation" source: https://www.bctengineer.com/q/semester_7/wireless_communication/doppler-spread-shift-derivation site: BCT Engineering --- # Doppler spread and Doppler shift derivation Wireless Communication · Chapter 3: Radio Wave Propagation in Mobile Networks - Asked in: 2081 Baisakh, 2+4 marks - Video: [Doppler Shift - Mobile Radio Propagation - Mobile Communication System](https://www.youtube.com/watch?v=nfk_9B3ZDFk) — Ekeeda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/wireless_communication/doppler-spread-shift-derivation - Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3 ## Question What is Doppler spread? Derive the expression for Doppler shift. ## Short answer Doppler spread is the spectral broadening of a received tone over fc-fm to fc+fm, with fm=v/lambda. For a mobile moving at speed v, path difference is v*t*cos(theta), giving phase change 2*pi*v*t*cos(theta)/lambda, so Doppler shift fd=(v/lambda)cos(theta). Maximum is v/lambda. ## Answer **Doppler spread (BD)** is the measure of spectral broadening of a received signal caused by the time rate of change of the mobile radio channel. It is the range of frequencies over which the received Doppler spectrum is non-zero. When a pure tone fc is sent, the received spectrum spreads between fc − fm and fc + fm, where fm = v/λ is the maximum Doppler shift; BD ≈ fm. #### Derivation of Doppler shift A mobile moves with constant speed *v* from point X to point Y in time Δt, receiving a signal from a remote source S. The direction of arrival makes angle θ with the direction of motion. ```text S (remote source) /| / | / | rays from S are / | almost parallel / θ | X ----------- Y ------> v |<---- l --->| path difference Δl = l cosθ ``` Distance travelled: ```text l = v·Δt ``` Since S is very far, the two rays (to X and to Y) are nearly parallel, so the difference in path length is: ```text Δl = l cosθ = v·Δt·cosθ ``` Phase change due to this path difference: ```text Δφ = 2π·Δl/λ = (2π·v·Δt·cosθ)/λ ``` The apparent change in frequency is the rate of change of phase divided by 2π: ```text fd = (1/2π)·(Δφ/Δt) fd = (v/λ)·cosθ ``` **Results:** - Moving towards the source (θ = 0): fd = +v/λ, received frequency increases. - Moving away (θ = 180°): fd = −v/λ, frequency decreases. - Moving perpendicular (θ = 90°): fd = 0. - Maximum Doppler shift fm = v/λ. **Example:** fc = 1850 MHz, v = 60 mph = 26.82 m/s, λ = 0.162 m → fm = 26.82/0.1622 = 165.4 Hz; moving towards the source, received frequency = 1850.000165 MHz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Doppler spread is: A. Delay spread B. Spectral broadening due to the channel's time variation C. Signal attenuation D. Path loss Answer: B. It is the range of frequencies where the Doppler spectrum is non-zero. (Key point: Doppler spread definition) 2. In the derivation, the path difference between X and Y is: A. lambda cos(theta) B. v/dt C. v dt cos(theta) D. v dt sin(theta) Answer: C. dl = l cos(theta) with l = v dt. (Key point: Path difference) 3. Doppler shift is the rate of change of phase over 2 pi, giving: A. 2 v / lambda B. v lambda cos(theta) C. (v/lambda) cos(theta) D. v/lambda sin Answer: C. fd = (1/2pi)(dphi/dt). (Key point: Doppler shift) 4. Moving directly away from the source (theta = 180 degrees) gives: A. fd = 2v/lambda B. fd = -v/lambda C. fd = +v/lambda D. fd = 0 Answer: B. cos 180 = -1, so the frequency decreases. (Key point: Moving away) 5. For fc = 1850 MHz and v = 26.82 m/s, fm is approximately: A. 1650 Hz B. 165 Hz C. 16.5 Hz D. 26.8 Hz Answer: B. lambda = 0.162 m, fm = 26.82/0.1622 = 165.4 Hz. (Key point: Numerical) ## Related in Wireless Communication - [Frequency dispersion and Doppler spread derivation](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Afrequency-dispersion-doppler-spread) — Wireless Communication, chapter 3 - [Fading factors and Doppler shift derivation](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Afading-factors-doppler-shift) — Wireless Communication, chapter 3 - [Fading types by delay spread and Doppler spread](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Afading-types-delay-doppler) — Wireless Communication, chapter 3, asked 2× - [Doppler spread, coherence bandwidth, fading classes](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Adoppler-coherence-fading-classification) — Wireless Communication, chapter 3 - [Short note: Doppler spread and coherence time](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Adoppler-spread-coherence-time-note) — Wireless Communication, chapter 3 - [Doppler spread fading; delay spread numerical](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Adoppler-fading-delay-spread-profile) — Wireless Communication, chapter 3 ## Related in other subjects - [Derivation of superelevation expression](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3Asuper-derive) — Transportation Engineering I, chapter 3, asked 7× - [Derive expression for tile drain spacing](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Atile-drain-spacing-derivation) — Irrigation and Drainage Engineering, chapter 9, asked 8× - [Derive development length expression](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aderive-development-length-expression) — Design of RCC Structure, chapter 6, asked 6× - [Derive PET expression by Penman's method](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aderive-penman-pet) — Engineering Hydrology, chapter 3