Why do ATSC channels need 6MHz of bandwidth when they're digital?

Question

If I'm understanding it correctly, an ATSC channel is a digital channel that uses 6 megahertz of bandwidth to send 19.38Mbps using 8VSB modulation. For analog broadcasts, that makes sense, but I'm struggling to understand why digital data needs such a wide bandwidth and have assumptions that are probably wrong.

For instance, why couldn't we just have a single carrier wave that changes amplitude between one of eight levels at a rate of 6,460,000 times per second? The lowest ATSC channel, channel 2, has a carrier frequency of 54MHz, so each change in amplitude would persist for about 8.4 cycles of the carrier (up to 107 cycles for channel 51 at 692MHz), which seems like it should be enough for receivers to recognize the level and greatly reduce the bandwidth needed.

Clearly the above is an incorrect, unworkable idea, but I'm curious as to why it wouldn't work. Is it because changing the amplitude that quickly takes up about the same amount of bandwidth?

Answer 1

A key point is bandwidth is proportional to the symbol rate, or rate of change of the modulation. If the symbol is rectangular shaped, the spectrum is a Sinc function with the first null in Hz at the inverse of the symbol duration in seconds.

The diagram below demonstrates simply what happens if we had a single carrier that changed amplitudes between one of eight levels at a given symbol rate (each level corresponds to a different "symbol" so the rate of change between levels is the symbol rate). In this graphic I had the symbol rate was 1 MHz, and the blue frequency spectrum in the lower part of the plot corresponds to the blue trace which showing the time domain waveform, and what spectrum we would expect if the waveform were to instantly change from one level to the next. This requires the highest frequency content to change so rapidly (infinite theoretically), and with that we see only a portion of the very wide spectrum that would result. In practice the spectral occupancy is reduced through pulse shaping, which means to transition slowly from one level to the next, reducing the overall spectrum to be below reasonable levels. In this case, with proper pulse shaping, the spectrum required is only slightly higher than the symbol rate. I am not describing the specifics of 8VSB modulation but rather demonstrating how when we change the amplitude and phase of an arbitrary carrier with time (modulation) more spectrum overall is needed. For the users simple example of changing the level of a carrier, if done with proper pulse shaping the overall bandwidth required would be just over 6.46 MHz for the dominant portion of the spectrum sufficient to not interfere with adjacent channels. Practically implementations would range from 15 to 30% over this bandwidth.