--- title: "Measurement parameters and high VSWR measurement" subject: "RF and Microwave Engineering" chapter: 8 key: "semester_7/rf_and_microwave_engineering:vswr-greater-than-10-measurement" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/vswr-greater-than-10-measurement site: BCT Engineering --- # Measurement parameters and high VSWR measurement RF and Microwave Engineering · Chapter 8: RF and Microwave Measurements - Asked in: 2079 Bhadra, 2+6 marks - Video: [VSWR Measurment | Low & High | Microwave Engineering | Lec-121](https://www.youtube.com/watch?v=EParFhq0Pds) — Education 4u - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/vswr-greater-than-10-measurement - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-8 ## Question List out the major RF/MW measurement parameters. How the VSWR of any microwave transmitter (In case of VSWR > 10) can be measured? Explain. ## Short answer Major parameters: power, frequency, VSWR/Γ, impedance, attenuation, S‑parameters, Q, noise figure. For VSWR > 10, the double‑minimum method: measure guide wavelength λg from successive minima; locate the voltage minimum, find positions d1 and d2 where power doubles (3 dB up), then S ≈ λg/[π(d2−d1)]. ## Answer RF/MW measurement is mainly based on **power and wave patterns** because voltage and current cannot be measured directly at these frequencies. #### Major RF/MW measurement parameters 1. Power (low, medium, high; average and peak) 2. Frequency and guide wavelength (λg) 3. VSWR and reflection coefficient (Γ) 4. Impedance (load impedance, using Smith chart) 5. Attenuation and insertion loss 6. Scattering parameters (S11, S21, …) 7. Q factor of cavities 8. Noise figure, dielectric constant, phase shift #### Measurement of high VSWR (S > 10): double minimum method For high VSWR the minimum is very sharp and E_min is very small, so the direct method (E_max/E_min) is inaccurate: the crystal leaves its square-law region at E_max and the reading near the minimum is noisy. Instead the **width of the minimum** is measured. Set-up: ```text Source -> Isolator -> Atten. -> Freq. -> Slotted -> DUT (1 kHz AM) meter line | probe+crystal | VSWR meter ``` Procedure: 1. Find λg: λg = 2 × distance between two successive minima. 2. Move the probe to a voltage **minimum**; adjust gain so the meter reads a convenient value (e.g. 3 dB on the dB scale). 3. Move the probe to one side until the reading is **twice the minimum power** (3 dB higher, i.e. E = √2 E_min). Note position d1. 4. Move to the other side of the minimum and find the same level; note d2. 5. Calculate: ```text S = √[ 1 + 1 / sin²(π(d2 − d1)/λg) ] For large S: S ≈ λg / [π (d2 − d1)] ``` ```text |E| |\ / | \ / | \ √2Emin / |---\-----------/--- | \_________/ | d1 Emin d2 +--------------------> position ``` Example: λg = 4 cm, d2 − d1 = 0.1 cm S ≈ 4 / (π × 0.1) = 12.73; exact formula gives 12.78. Other method for high VSWR: the **calibrated attenuator method** – read the attenuator change (in dB) needed to give the same meter reading at E_max and E_min; S = 10^(dB/20). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is E_max/E_min inaccurate for S > 10? A. The slotted line is too short B. The minimum is very sharp and small, and the crystal leaves square-law at E_max C. The frequency is too low D. The source is not modulated Answer: B. Tiny E_min is noisy and the crystal is not square law at E_max, so the width of the minimum is measured instead. (Key point: Why direct method fails) 2. In the double minimum method, d1 and d2 are the positions where: A. The field is zero B. The field is maximum on both sides C. The power is twice the minimum (3 dB up) on either side of the minimum D. The reading is half the maximum Answer: C. At those points E = sqrt(2) E_min; the separation d2 - d1 gives S. (Key point: Double minimum procedure) 3. With lambda_g = 4 cm and d2 - d1 = 0.1 cm, what is S approximately? A. About 4 B. About 40 C. About 1.3 D. About 12.7 Answer: D. S ≈ lambda_g / [pi (d2 - d1)] = 4 / (pi x 0.1) = 12.73. (Key point: High-S formula) 4. How is lambda_g found on the slotted line? A. Twice the distance between two successive minima B. Equal to the distance between two successive minima C. Half the distance between a maximum and a minimum D. From the attenuator setting Answer: A. Successive minima are lambda_g/2 apart, so lambda_g = 2 x that distance. 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