A speech signal is being encoded using 8 bits per sample and then sampled at a rate of 26
kHz. The resulting binary data are then transmitted through 8-level Pulse Amplitude
Modulation over an AWGN baseband channel. The bandwidth required for transmission
is.
Answers
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13
Answer:
A speech signal of `3 kHz` is used to modulate a carrier signal of frequency `1 MHz`, using amplitude modulation. The frequencies of the side bands will be
Answered by
8
Given:
8 level PAM
n = 8
fs = 26kHz
To find:
The bandwidth required for transmission.
Solution:
Bandwidth = (n × fs / 2) / log₂M
Bandwidth = (8 × 26k / 2) / log₂8
Bandwidth = (8 × 26k) / (2 × 3)
Bandwidth = 34.67 kHz
Therefore, the required bandwidth for the transmission is equal to 34.67 kHz
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