--- title: "Armstrong FM modulator: find n1 and n2" subject: "Communication Systems" chapter: 3 key: "semester_6/communication_systems:armstrong-n1-n2-numerical" source: https://www.bctengineer.com/q/semester_6/communication_systems/armstrong-n1-n2-numerical site: BCT Engineering --- # Armstrong FM modulator: find n1 and n2 Communication Systems · Chapter 3: Angle Modulation - Asked in: 2074 Bhadra (CS I), 9 marks - Video: [ARMSTRONG METHOD || GENERATION OF FM WAVE || ADC || Electronics and Communication Engineering](https://www.youtube.com/watch?v=pGzV1bsXoMk) — Friend's Explanation - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/armstrong-n1-n2-numerical - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3 ## Question An Armstrong FM modulator is required in order to transmit an audio signal of bandwidth 50 Hz to 15 kHz. The Narrow Band (NB) phase modulator used utilizes an oscillator providing carrier frequency fc1 = 0.2 MHz. The output of the NB phase modulator is multiplied by n₁ by multiplier and then passed to mixer with a local oscillator frequency fc2 = 10.925 MHz. The desired FM wave at the transmitter output has a carrier frequency fc = 90 MHz and a frequency deviation Δf = 75 kHz, which is obtained by multiplying the mixer output frequency with n₂ by using another multiplier. Find n₁ and n₂. Assume that the NBFM signal at the output of NB phase modulator has modulation index, β = 0.5. ## Short answer With worst-case f_m = 50 Hz, Δf_1 = β_1 f_m = 25 Hz, so n_1 n_2 = 75k/25 = 3000. Solving 90 = n_2(0.2 n_1 – 10.925) with integer factors gives n_1 = 64, n_2 = 48. ## Answer **Given:** audio 50 Hz – 15 kHz; $f_{c1} = 0.2$ MHz; $\beta_1 = 0.5$; $f_{c2} = 10.925$ MHz; final $f_c = 90$ MHz, $\Delta f = 75$ kHz. ```text NB PM x n1 Mixer x n2 0.2 MHz->[n1]->0.2n1 MHz->[X]->f2->[n2]-> 90 MHz df1=25 Hz ^ df=75 kHz [10.925 MHz] ``` #### Step 1: deviation at the NB modulator The NB phase modulator (with integrator) must keep $\beta_1 \le 0.5$ for every audio frequency. Since $\beta_1 = \Delta f_1/f_m$, the worst case is the **lowest** audio frequency, 50 Hz: $$ \Delta f_1 = \beta_1f_{m,min} = 0.5 \times 50 = 25\ \text{Hz} $$ #### Step 2: total multiplication needed The mixer does not change the deviation, so $$ n_1n_2 = \frac{\Delta f}{\Delta f_1} = \frac{75 \times 10^3}{25} = 3000 $$ #### Step 3: carrier frequency condition The mixer output is the difference frequency (taking $n_1f_{c1} > f_{c2}$): $$ f_c = n_2(n_1f_{c1} - f_{c2}) \;\Rightarrow\; 90 = n_2(0.2n_1 - 10.925) $$ Using $n_1n_2 = 3000$: $0.2n_1n_2 = 600$, so $$ \begin{aligned} 90 &= 600 - 10.925n_2 \\ n_2 &= \frac{510}{10.925} = 46.68, \qquad n_1 = \frac{3000}{46.68} = 64.27 \end{aligned} $$ Multipliers must be integers. Choosing $n_1 = 64$ and solving the frequency equation exactly: $$ \begin{aligned} 0.2 \times 64 &= 12.8\ \text{MHz} \\ f_2 &= 12.8 - 10.925 = 1.875\ \text{MHz} \\ n_2 &= \frac{90}{1.875} = 48 \end{aligned} $$ #### Check $$ \begin{aligned} f_c &= 48 \times 1.875 = 90\ \text{MHz} \\ \Delta f &= n_1n_2\Delta f_1 = 64 \times 48 \times 25 = 76800\ \text{Hz} = 76.8\ \text{kHz} \end{aligned} $$ The carrier is exactly 90 MHz and the deviation is within 2.4% of 75 kHz (exactly 75 kHz is reached by reducing $\beta_1$ slightly to 0.488). Both values are practical: $64 = 2^6$ (six doublers) and $48 = 2^4 \times 3$ (four doublers and a tripler). **Answer:** $n_1 = 64$, $n_2 = 48$ (ideal product 3000; continuous solution $n_1 \approx 64.3$, $n_2 \approx 46.7$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is Delta f1 = 25 Hz? A. beta1 must stay 0.5 even at the lowest audio frequency 50 Hz, so Delta f1 = 0.5 x 50 B. Because beta1 = 0.5 at the highest frequency 15 kHz C. Because Delta f1 = fc1 / 8000 D. Because the mixer reduces it to 25 Hz Answer: A. beta = Delta f/fm is largest at the lowest fm, which limits Delta f1. (Key point: Deviation at NB modulator) 2. What is n1 n2? A. 90 MHz / 0.2 MHz = 450 B. 75 kHz x 25 Hz C. 15 kHz / 50 Hz = 300 D. 75 kHz / 25 Hz = 3000 Answer: D. Only multipliers raise deviation (mixer does not). (Key point: Product n1 n2) 3. Which carrier equation is used? A. fc = n1 n2 (fc1 + fc2) B. fc = n2(n1 fc1 - fc2) C. fc = n1(n2 fc1 - fc2) D. fc = n1 n2 fc1 - fc2 Answer: B. The mixer takes the difference after the first multiplier, and n2 then multiplies it. (Key point: Carrier equation) 4. Which integer pair is chosen? A. n1 = 48, n2 = 64 B. n1 = 50, n2 = 60 C. n1 = 64, n2 = 48 D. n1 = 100, n2 = 30 Answer: C. 64 x 0.2 = 12.8 MHz; minus 10.925 = 1.875 MHz; x48 = 90 MHz. 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