Answer:
a. Nyquist rate = 30 kHz
b. n = 16 bits
c. Bit rate = 480,000 bits/s
d. Bit rate = 705,600 bits/s
Step-by-step 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.