Question

Television channels are 6 MHz wide. How many bits/sec can be sent if four-level digital signals
are used? (Assume it is a noiseless channel.)

&

We have a low-pass channel with a bandwidth of 200 kHz. What is the maximum bit rate of this
channel

Answer 1

a)

Nyquist theorem developed for calculating the maximum data rate for a noiseless channel.

Using the Nyquist theorem,

Maximum data rate = 2 * H * log2V bits/sec.

= 2 *6MHz *log24 bits/sec

= 2 *6MHz * log222 bits/sec

= 2 * 2 * 6 MHz*bits/sec

=24 Mbps

We can sample 12 million times/sec. Four–level signals provide 2 bits per sample, so total data rate will be 24 Mbps.

b)

Baseband transmission of a digital signal that preserves the shape of the digital signal is possible only if we have a low-pass channel with an infinite or pass channel with an infinite or very wide bandwidth. In baseband transmission, the required bandwidth is proportional to the bit rate; bandwidth if we need to send bits faster, we need more bandwidth. is proportional to the bit rate; if we need to send bits faster, we need more bandwidth more bandwidth.

So, the maximum bit rate of this channel is 400kbps