## Limitation due to finite bandwidth (cont.)

#### Consider what happens if the channel has only sufficient bandwidth to pass the fundamental of the square wave. #### Providing that we are careful to maintain accurate levels throughout the system, and we sample the received signal at the correct time within each symbol period, then it is possible to recover the symbols and hence the data with a channel that has a passband of only 0.5 x 1/Ts Hz.
#### Because the square wave signal chosen represents the maximum and most extreme rate and range of signal change for the 8-ary example, we can simply infer that any other symbol pattern will require less bandwidth for transmission and hence 0.5 x 1/Ts is indeed sufficient bandwidth for all cases. It is also the minimum bandwidth needed as any less would result in the fundamental of the square wave being suppressed and no signal getting through the channel.

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