The output of a video surveillance camera is an analog signal with a bandwidth that extends from 0 Hz to 1.5 MHz. The signal is sampled at 1.2 times its Nyquist rate.
The resulting PAM signal is converted to 6 bit digital words with an analog to digital converter and the words are transmitted serially over a twisted pair used as the cable for data transmission.
a. What is the sampling frequency?
b. What is bit rate of the transmission on the twisted pair?
c. If the digital outputs of three identical video cameras are combined onto a single cable using time division multiplexing, what is the bit rate on the cable?© BrainMass Inc. brainmass.com July 18, 2018, 10:46 am ad1c9bdddf
Given, bandwidth of analog signal = B = 1.5 MHz
So, Nyquist sampling rate for this signal = fN = 2B = 2*1.5 MHz = 3 MHz or 3M samples/second
a. Since the signal is sampled at 1.2 times its Nyquist rate, the sampling frequency = 1.2 fN = 1.2*3 ...
In part (c), solution assumes that the Time Division Multiplexer reads data from each video camera at same rate as the sampling frequency.