Question 7116 – Computer-Networks
May 7, 2024Question 11871 – Computer-Networks
May 7, 2024Computer-Networks
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Question 454
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Suppose a digitized voice channel is made by digitizing 8 kHz bandwidth analog voice signal. It is required to sample the signal at twice the highest frequency (two samples per hertz). What is the bit rate required, if it is assumed that each sample requires 8 bits?
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32 kbps
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64 kbps
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128 kbps
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256 kbps
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Question 454 Explanation:
t is required to sample the signal at twice the highest frequency (two samples per hertz).
Bit rate = current bandwidth * 2 * sample size.
Bit rate = 8 KHz * 2 * 8 bit.
= 8 K / sec * 2 * 8 bit.
= 128 kb / sec.
= 128 Kbps.
Bit rate = current bandwidth * 2 * sample size.
Bit rate = 8 KHz * 2 * 8 bit.
= 8 K / sec * 2 * 8 bit.
= 128 kb / sec.
= 128 Kbps.
Correct Answer: C
Question 454 Explanation:
t is required to sample the signal at twice the highest frequency (two samples per hertz).
Bit rate = current bandwidth * 2 * sample size.
Bit rate = 8 KHz * 2 * 8 bit.
= 8 K / sec * 2 * 8 bit.
= 128 kb / sec.
= 128 Kbps.
Bit rate = current bandwidth * 2 * sample size.
Bit rate = 8 KHz * 2 * 8 bit.
= 8 K / sec * 2 * 8 bit.
= 128 kb / sec.
= 128 Kbps.