Question E7E06:
Why is de-emphasis used in FM communications receivers?  

Choices:  
A: For compatibility with transmitters using phase modulation  
B: To reduce impulse noise reception  
C: For higher efficiency  
D: To remove third-order distortion products    

Correct Answer is  A: For compatibility with transmitters using phase modulation  

Correct answer: A — For compatibility with transmitters using phase modulation

Why A is correct:
- Many “FM” transmitters in communications gear actually generate FM by starting with phase modulation (PM).
- In PM, the instantaneous frequency deviation is proportional to the derivative of the modulating signal:
  $$\phi_i(t) = \omega_c t + k_p m(t), \quad f_i(t) = \frac{1}{2\pi}\frac{d\phi_i}{dt} = f_c + \frac{k_p}{2\pi}\frac{dm(t)}{dt}$$
  This differentiating action boosts high audio frequencies (an effective pre-emphasis of about +6 dB/octave).
- FM receivers apply de-emphasis (about −6 dB/octave) so that the overall transmit–receive audio response is flat and compatible with PM-based transmitters.

Why the other choices are incorrect:
- B: To reduce impulse noise reception — FM limiting and the capture effect primarily reduce AM/impulse noise. De-emphasis is an audio-frequency equalizer; while it can reduce high-frequency hiss, that’s not its primary purpose.
- C: For higher efficiency — De-emphasis is a baseband (audio) filter; it does not affect RF power efficiency.
- D: To remove third-order distortion products — Intermodulation distortion is nonlinear; de-emphasis is a linear filter and does not remove such products.

Useful formulas and rules of thumb:
- PM “acts like” pre-emphasis because $\,\Delta f \propto \frac{dm}{dt}\,$, so higher audio frequencies get more deviation.
- Standard FM systems use complementary pre-emphasis and de-emphasis with approximately ±6 dB per octave so the end-to-end response is flat:
  $$H_{\text{total}}(f) = H_{\text{pre}}(f)\,H_{\text{de}}(f) \approx 1$$
- A common de-emphasis network is a single-pole RC low-pass:
  $$|H_{\text{de}}(f)| = \frac{1}{\sqrt{1+(2\pi f \tau)^2}}$$
  providing about −6 dB per octave above the break frequency $f_b \approx \frac{1}{2\pi\tau}$.