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Consider an extremely noisy channel having a bandwidth of 1kHz. What could be the channel capacity?

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Consider an extremely noisy channel having a bandwidth of 1kHz. What could be the channel capacity?

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written 2.2 years ago by |

Shannonâ€™s theorem gives the capacity of a system in the presence of noise.

$$ C = B\ log_2\ (1 + \frac SR)$$

Where,

C = Channel Capcity (bits/sec)

B = Channel Bandwidth

$\frac SN$ = Signal to Noise Ratio

**Channel Capacity in Extremely Noisy Channel -**

- Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero.
- In other words, the noise is so strong that the signal is faint.
- For this channel, the capacity C is calculated as:

$$ C = B\ log_2\ (1 + \frac SR)$$

$$ C = B\ log_2\ (1 + 0)$$

$$ C = B\ log_2\ 1$$

$$ C = B \times 0$$

$$ C = 0$$

- This means that the
**capacity of this channel is zero**regardless of the bandwidth. - In other words, we can not receive any data through this channel.

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