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Explain fading effects due to Doppler spread.

Mumbai University > Electronics and telecommunication engineering> Sem 7 > mobile communication

Marks: 5M

Year: Dec 2015

1 Answer
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  • Depending on how rapidly the transmitted baseband signal changes as compared to the rate of change of the channel, a channel may be classified either as a fast fading or slow fading channel. That is the velocity of the mobile (or velocity of objects in the channel) and the baseband signaling determines whether a signal undergoes fast fading or slow fading.
  • Fast Fading: In a fast fading channel, the channel impulse response changes rapidly within the symbol duration. That is, the coherence time of the channel is smaller than the symbol period of the transmitted signal. This causes frequency dispersion due to Doppler spreading, which leads to signal distortion.
  • In frequency domain signal distortion due to fast fading increases with increasing Doppler spread relative to the bandwidth of the transmitted signal. Therefore, a signal undergoes fast fading if

    $T_S \gt T_C \hspace{0.2cm}and \hspace{0.2cm}B_S \lt B_D$

    Where: $T_C$ and $T_S$ = coherence time $B_S$ and $B_D$ = Doppler spread.

  • When a channel is specified as a fast or slow fading channel, it does not specify whether the channel is flat fading or frequency selective in nature. Fast fading only deals with the rate of change of the channel due to motion and fast fading only occurs for very low data rates.
  • Slow Fading: In a slow fading channel, the channel impulse response changes at a rate much slower than the transmitted baseband signal s(t).
  • In the frequency domain, this implies that the Doppler spread of the channel is much less than the bandwidth of the baseband signals. Therefore, a signal undergoes slow fading if

    $T_S \lt\lt T_D \hspace{0.2cm} and \hspace{0.2cm} B_S \gt\gt B_D$

    Where: $T_C$ and $T_S$ = coherence time $B_S$ and $B_D$ = Doppler spread.

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