Why is the bandwidth in PCM half of the transmission rate?

[janu203]

When we use pulse code modulation , it is said that the min bandwidth requirement is half the transmission rate. Transmission rate given in Bits/sec. But why?

 

[Windadct]

The short answer is Nyquist theorem...

In a binary signal ( 0/1) the highest frequency that can be represented (inverse of the period), requires 2 samples.

Nyquist theorem is fundamental, and well worth the effort to study specifically, and IMO - be able to relate it to 3 real examples to get a handle on it. It is not a difficult concept.

 

[janu203]

The short answer is Nyquist theorem...

In a binary signal ( 0/1) the highest frequency that can be represented (inverse of the period), requires 2 samples.

Nyquist theorem is fundamental, and well worth the effort to study specifically, and IMO - be able to relate it to 3 real examples to get a handle on it. It is not a difficult concept.

So that means we can represent binary '1s and 0s' with two distinct pulses which combine to form one period. and taking inverse of that period '1/2T' gives us minimum bandwidth.

Is that right?

 

[Windadct]

Actually - I should have been more clear, it requires 2 periods!... - in reality, a sample has 2 parts of information: 1 period (T) , 2 data (0/1 in this case)

Also - breaking this down to "2 samples" actually makes it confusing! Better to think in terms of a continuous signal - or a large number of samples.

Think of the Maximum Frequency you can represent in this signal: 0 1 0 1 0 1 0 1, you can see we need 2 samples AND 2 periods... not one period. 2 periods = 1/2 the frequency.

-- Next I want to clarify what you are saying in your OP - "min bandwidth requirement is half the transmission rate", I think "requirement" is not the right word, the bandwidth LIMIT (of the signal) is 1/2 the transmission rate.

 

[sophiecentaur]

This needs to be tidied up a bit, I think. Nyquist doesn't talk about bit rate. He talks in terms of Sample Rate. The theory says that a signal can be reconstructed perfectly from a series of samples that are more than twice the frequency of the maximum frequency in the signal. By how much more than, depends on the form of the filtering that's used in the frequency limiting ('Nyquist') filter.
There is another thing which should be considered and that is Shannon's Theorem which tells us that the only limit to the rate of data that can be sent in a given bandwidth is the Signal to Noise Ratio in the channel. For example, Low Pass filtering a 'boxcar' waveform of zeros and ones will introduce inter symbol interference as each symbol is stretched in time by the filter. As long as the channel noise is less than the inter symbol interference and, if you take long enough and look over a long enough time window, you can recover the original data with any chosen error rate.