Question

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.​

Answer 1

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

Answer 2

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