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A 22 KHz baseband channel is used by a digital transmission system. Suppose ideal pulses are sent at the Nyquist rate, and the pulses can take 1024 levels. There is no noise in the system. What is the bit rate of this system

Answer :

Answer:

Bit rate = 440 kBits/sec

Step-by-step explanation:

Band width = W= 22 kHz

Number of levels = L = 1024 levels

Bit per sample:

 [tex]m=log_2 L\\\\m =log_2(1024)\\\\m=10 bits/sample[/tex]

Ideal pulses are sent at the Nyquist rate then bit rate = 2 x W x m

[tex]bit\,\,rate= 2\times 22\times 10^3\times 10\\\\bit\,\,rate= 440\times 10^3 bits\,sec^{-1}[/tex]

bit rate = 440 kBits/sec

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