--- title: "Armstrong FM transmitter numerical" subject: "Communication Systems" chapter: 3 key: "semester_6/communication_systems:armstrong-numerical-200khz-25hz" source: https://www.bctengineer.com/q/semester_6/communication_systems/armstrong-numerical-200khz-25hz site: BCT Engineering --- # Armstrong FM transmitter numerical Communication Systems · Chapter 3: Angle Modulation - Asked in: 2077 Chaitra (CS I), 8 marks - Video: [(a)Design (the block diagram of) an Armstrong indirect FM modulator to generate an FM carrier wit…](https://www.youtube.com/watch?v=gCoiXTR6_nk) — Engineer Thileban Explains - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/armstrong-numerical-200khz-25hz - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3 ## Question For a given Armstrong FM transmitter, the NBFM output has a frequency of 200 kHz and frequency deviation of 25 Hz. This signal is then frequency multiplied by 65 and passed through mixer of oscillation frequency 10.8 MHz. The resulting signal is then fed to a frequency multiplier with n = 50. Calculate the maximum frequency deviation and the valid carrier frequency of the WBFM for commercial use. Draw the necessary diagram. ## Short answer After ×65: 13 MHz, deviation 1.625 kHz. Mixing with 10.8 MHz yields 2.2 MHz. After ×50: maximum deviation = 81.25 kHz, carrier frequency = 110 MHz. ## Answer In the Armstrong method, frequency multipliers multiply **both** carrier frequency and deviation; the mixer shifts the carrier frequency but leaves the deviation unchanged. #### Diagram ```text NBFM x65 Mixer x50 200 kHz -->[n1]-->13 MHz-->[X]-->2.2M-->[n2]--> WBFM df=25 Hz df=1625 Hz ^ df=1625 110 MHz | df=81.25k [Osc 10.8 MHz] ``` #### Step 1: after first multiplier (n₁ = 65) $$ \begin{aligned} f_1 &= 65 \times 200\ \text{kHz} = 13\ \text{MHz} \\ \Delta f_1 &= 65 \times 25 = 1625\ \text{Hz} \end{aligned} $$ #### Step 2: mixer with 10.8 MHz The mixer gives $13 \pm 10.8$ MHz, i.e. 2.2 MHz (difference) or 23.8 MHz (sum). Deviation stays 1625 Hz. #### Step 3: second multiplier (n₂ = 50) $$ \begin{aligned} \Delta f &= 50 \times 1625 = 81250\ \text{Hz} = 81.25\ \text{kHz} \\ f_c &= 50 \times 2.2 = 110\ \text{MHz} \quad (\text{difference}) \\ f_c &= 50 \times 23.8 = 1190\ \text{MHz} \quad (\text{sum}) \end{aligned} $$ #### Valid carrier frequency The sum term gives 1190 MHz (UHF), which is far outside the VHF FM broadcast range and is rejected by filtering after the mixer. The difference term is the usable one, giving **Answer:** maximum frequency deviation **Δf = 81.25 kHz**; valid WBFM carrier **f_c = 110 MHz** (from the 2.2 MHz difference output). Note: 110 MHz lies just above the 88–108 MHz broadcast band and 81.25 kHz is slightly above the 75 kHz standard, so in practice the oscillator frequency or multiplier would be adjusted slightly; with the given data, 110 MHz is the only practical (VHF) choice. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does a frequency multiplier do to carrier and deviation? A. Multiplies only the carrier B. Multiplies only the deviation C. Multiplies both by n D. Leaves both unchanged but filters harmonics Answer: C. The instantaneous frequency is multiplied, so both carrier and deviation scale by n. (Key point: Effect of multiplier) 2. What does the mixer do to the deviation? A. It multiplies it by the oscillator frequency B. It divides it by 10.8 C. It adds 10.8 MHz to it D. Nothing; it only shifts the carrier frequency Answer: D. Mixing translates the carrier but the deviation is unchanged. (Key point: Effect of mixer) 3. What is the final maximum deviation? A. 25 x 65 x 50 = 81.25 kHz B. 25 x 65 = 1.625 kHz C. 25 x 50 = 1.25 kHz D. 81.25 Hz Answer: A. Only the two multipliers (65 and 50) change deviation: 25 x 3250. (Key point: Maximum deviation) 4. Which carrier is chosen and why? A. 1190 MHz, because it is higher B. 110 MHz from the 2.2 MHz difference, since the 1190 MHz sum is outside the VHF FM range C. 110 MHz from the sum term 23.8 MHz D. 13 MHz, as the mixer output is not multiplied Answer: B. (13-10.8) = 2.2 MHz, x50 = 110 MHz; the sum gives 1190 MHz which is unsuitable. 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