Given Information:
Quantizing level = 128
bandwidth = 12 kHz
Required Information:
a) data rate = ?
b) sampling rate = ?
Answer:
a) data rate = 24,000 bits/s
b) sampling rate = 3 kHz
Explanation:
a. Find the rate (b/s) at which information is transmitted through the channel.
The data rate is given by
data rate = 2*Bandwidth
data rate = 2*12,000
data rate = 24,000 bits/s
b. Find the rate at which the analog signal is sampled. What is the maximum possible value for the highest frequency component of the analog signal?
Quantizing level = 128
The corresponding number of bits transmitted per sample is given by
Log₂(128) = 7 bits
We are given that a synchronizing pulse is added at the end of each code word representing a sample of the analog signal, therefore, total number of bits are
7 bits + 1 bit = 8 bits
Therefore, the sample rate is
sampling rate = data rate/number of bits
sampling rate = 24,000/8
sampling rate = 3,000 Hz
sampling rate = 3 kHz
According to the Nyquist criteria the sampling rate must be at least twice of the highest frequency component of the input signal, so
sampling rate = 2*f
3 kHz = 2*f
f = 3 kHz/2
f = 1.5 kHz