An FM signal with modulation index M_f passes through an ideal frequency doubler. What is the modulation index at the output of the doubler (assume the modulating frequency remains the same)?

Difficulty: Easy

Correct Answer: 2 M_f

Explanation:


Introduction / Context:
Frequency multipliers are common in FM transmitters and test setups. They multiply the carrier frequency by an integer n and also scale the peak frequency deviation. Knowing how modulation index transforms is crucial for meeting deviation and spectral mask limits after multiplication stages.


Given Data / Assumptions:

  • Input FM has modulation index Mf = Δf / fm.
  • A frequency doubler multiplies instantaneous frequency deviation and carrier frequency by 2.
  • Modulating frequency fm is unchanged by the multiplier.


Concept / Approach:

For an n-times frequency multiplier, Δfout = n * Δfin, and fm,out = fm,in. Therefore, Mf,out = Δfout / fm,out = n * Δfin / fm,in = n * Mf,in. With n = 2, the output modulation index doubles.


Step-by-Step Solution:

Write Mf = Δf / fm.For a doubler, Δf → 2Δf; fm unchanged.Hence Mf,out = 2 * Mf,in.


Verification / Alternative check:

Spectral lines (Bessel components) shift and scale consistently with the multiplier, reflecting increased deviation (wider spectrum) proportional to n.


Why Other Options Are Wrong:

  • Mf or Mf/2: contradicts the standard transformation.
  • Mf^2: not supported by FM theory.
  • “Unchanged with limiter”: a limiter clips amplitude, not frequency deviation.


Common Pitfalls:

Assuming the modulating frequency doubles—only the carrier and deviation are multiplied by the frequency multiplier.


Final Answer:

2 M_f

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion