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Suppose I have 4 of 300bps synchronous digital input and 1 of analogue input with a bandwidth of 500Hz. The analogue input is sampled at 4bit using PCM. All the channels are multiplexed with TDM.

As one of the input is of analogue, I need to convert that to digital input first before TDM can be carried out. Since the analogue signal will be sampled at 4bits, my initial idea was to use the Nyquist sampling theorem to have 500Hz × 2 = 1000 samplings per second. But on a second thought, this is merely the bandwidth and I cannot do this unless it is said that the highest frequency is 500Hz.

However, then how should I find out the input bit rate of the analogue input channel for the TDM multiplexing?

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  • $\begingroup$ You're not necessarily on the wrong track; under the right conditions, you can do something called bandpass sampling to sample at ~2x the bandwidth of the signal irrespective of its center frequency. $\endgroup$ – Jason R Sep 23 '11 at 14:01
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Theoretically speaking you need 2 samples per second for 1 Hz of bandwidth (regardless whether the signal is centered around DC or not), so in this case a sample rate of 1 kHz would be enough. Practically speaking, you always need more since the theoretical requirements is based on the assumption of an ideal low pass (or bandpass) filter. The correct choice of sample rate depends on how the band limiting is achieved and what bandwidth you are really interested in and what noise and sampling accuracy you need. Most audio systems use 44.1 kHz or 48 kHz to achieve 20 kHz usable bandwidth. Another consideration is that you already have some digital inputs. These are derived from some master clock anyway. If you can use the same clock to sample your input, you can make the subsequent processing a lot easier.

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