Assuming that the channel is noiseless, if TV channels are 8 kHz wide with the bits/sample = 3Hz and signalling rate = 16 × 106 samples/second, then what would be the value of data rate? A. 16 Mbps B. 24 Mbps C. 48 Mbps D. 64 Mbps

[amp_mcq option1=”16 Mbps” option2=”24 Mbps” option3=”48 Mbps” option4=”64 Mbps” correct=”option3″]

The correct answer is C. 48 Mbps.

The data rate is the number of bits per second that are transmitted over a communication channel. It is calculated by multiplying the signalling rate by the bits/sample. In this case, the signalling rate is 16 × 106 samples/second and the bits/sample is 3 Hz. Therefore, the data rate is 16 × 106 × 3 = 48 Mbps.

Option A is incorrect because it is the signalling rate, not the data rate. Option B is incorrect because it is the product of the signalling rate and the bits/sample, but the bits/sample is not 2 Hz. Option D is incorrect because it is the square of the signalling rate, but the signalling rate is not 8 kHz.