Question

A digital signaling system is required to operate at 9600 bps. a. If a signal element encodes a 4-bit word, what is the minimum required bandwidth of the channel

Answer 1

Answer:

1200 Hz

Explanation:

Given that the signal encodes 4-bit words, then

log2M = 4,

Again, we're told that the system operates at 9600 bps, so

C = 9600

If we bring back our earlier relation, we have

9600 = 2B * 4

9600 = 8B

B = 9600 / 8

B = 1200 Hz

Therefore, we can conclude that the minimum required bandwidth of the channel is 1200 Hz

Answer 2

The minimum required bandwidth of the channel is 1200 Hz

In a signal processing system, bandwidth(measured in Hertz) refers to the variation between a higher frequency and a lower frequency in a continuous range.

It can be calculated by using the formula:

$\mathbf{B_ a = \dfrac{f_s}{2} }$

where;

• The sampling rate $\mathbf{f_s = \dfrac{R}{n}}$

and;

• The operation rate of the digital signaling system R = 9600 bps
• The number of bits (n) = 4

∴

The sampling rate is:

$\mathbf{f_s = \dfrac{9600 \ bps}{4 \ bits}}$

= 2400 Hz

The minimum required bandwith $\mathbf{B_ a = \dfrac{f_s}{2} }$ is:

$\mathbf{B_a = \dfrac{2400 \ Hz}{2}}$

= 1200 Hz

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