Ideal Channel (Noise less): According to Nyquist Theorem, the maximum numbers of bits (I.e., Theoretical Limit) that can be transmitted over an ideal communication link cannot exceed to

C: maximum number of bits can be transmitted per sec (bps)

B: Channel Bandwidth (Hz)

M: # of discreet levels in a signal (depends on which modulation scheme we are using)

2B: Baud Rate or Modulation Rate (# of samples generated per seconds)

log₂(M) will provide how many bits transmitted per sample

Note:

ANSWER THE FOLLOWING USING SAID METHOD:

Question 1: Television channels are 6 MHz wide. How many bits/sec can be sent if fourlevel digital signals are used? Assume a noiseless channel.

Question 2: Television channels are 12 MHz wide. How many bits/sec can be sent if 8level digital signals are used? Assume a noiseless channel.

Question 3: A signal is transmitted digitally over a 4kHz noiseless channel with the modulation rate of 8000 samples per second. How many bits per second are actually sent for each of these methods?

1. Delta Modulation (2 discrete levels)
2. 6 level signals

Question 4: How many levels would be required to communicate at 9600 bps over a channel with the bandwidth of 4KHz.

Question 5: What is the minimum bandwidth required to transmit the data at 40 Kbps with 4-levels signaling scheme.

Question 6: Find the Baud Rate of a channel that is capable of transmitting the data at rate of 12Kbps using 10 levels signaling scheme?

Question 7: How many number of bits transmitted per sample if bandwidth is 4 KHz with maximum data rate of 512 bps.