Computer-Networks
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Question 326
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A digital signaling system is required to operate at 9600 bps. If a signal element encodes a 4-bit
word, what is the minimum required bandwidth of the channel?
word, what is the minimum required bandwidth of the channel?
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1200Hz
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4800Hz
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19200Hz
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1900Hz
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Question 326 Explanation:
Using Nyquist’s equation: C = 2B log2M
We have C = 9600 bps a. log2M = 4, because a signal element encodes a 4-bit word Therefore,
C = 9600 = 2B x 4, and B = 1200 Hz
We have C = 9600 bps a. log2M = 4, because a signal element encodes a 4-bit word Therefore,
C = 9600 = 2B x 4, and B = 1200 Hz
Correct Answer: A
Question 326 Explanation:
Using Nyquist’s equation: C = 2B log2M
We have C = 9600 bps a. log2M = 4, because a signal element encodes a 4-bit word Therefore,
C = 9600 = 2B x 4, and B = 1200 Hz
We have C = 9600 bps a. log2M = 4, because a signal element encodes a 4-bit word Therefore,
C = 9600 = 2B x 4, and B = 1200 Hz