Compact disc(CD) records audio signals digitally by means of a binary code. Assume an audio signal bandwidth of 15 kHz. a. What is the Nyquist rate? b. If the Nyquist samples are quantized into 65,536 levels (L = 65,536) and then binary coded, what number of binary digits is required to encode a sample. c. Determine the number of binary digits per second (bits/s) required to encode the audio signal. d. For practical reasons discussed in the text, signals are sampled at a rate well above the Nyquist rate. Practical CDs use 44,100 samples/s. If L = 65,536, determine the number of pulses per second required to encode the signal.

a. Nyquist rate = 30 kHz

b. n = 16 bits

c. Bit rate = 480,000 bits/s

d. Bit rate = 705,600 bits/s

Explanation:

compact disc(CD) records audio signals digitally by means of a binary code. Assume an audio signal bandwidth of 15 kHz.

a. What is the Nyquist rate?

According to the Nyquist criteria, the sampling frequency must be at least 2 times the bandwidth.

Nyquist rate = 2 x 15  kHz

Nyquist rate = 30 kHz

So that means we need to take at least 30,000 sample/s to meet Nyquist criteria.  

b. If the Nyquist samples are quantized into 65,536 levels (L = 65,536) and then binary coded, what number of binary digits is required to encode a sample.

We have L = 65,536

The required number of binary digits is given by

65,536 = 2ⁿ

2¹⁶ = 2ⁿ

n = 16

Which means that we need n = 16 bits to encode each sample.

c. Determine the number of binary digits per second (bits/s) required to encode the audio signal.

The corresponding bit rate is given by

Bit rate = 16 bits/sample x 30,000 sample/s

Bit rate = 480,000 bits/s

Therefore, we need 480,000 bits/s to encode the audio signal.

d. For practical reasons discussed in the text, signals are sampled at a rate well above the Nyquist rate. Practical CDs use 44,100 samples/s. If L = 65,536, determine the number of pulses per second required to encode the signal.

Now the sampling rate has been changed from 30,000 sample/s to 44,100 samples/s

Then the corresponding bit rate is

Bit rate = 16 bits/sample x 44,100 sample/s

Bit rate = 705,600 bits/s

Therefore, we need 705,600 bits per second to encode the signal.

answer:

yes

explanation:

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