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?

  • 3
    $\begingroup$ So, what do you think is the bandwidth of something that changes 6.46 million times a second? $\endgroup$ Commented Mar 14, 2022 at 16:33
  • 2
    $\begingroup$ You should read the first few chapters of any digital communications textbook (I recommend Lapidoth -- it's free). $\endgroup$
    – MBaz
    Commented Mar 14, 2022 at 17:48
  • $\begingroup$ @MarcusMüller I'd hazard a guess at ~6MHz, but I didn't want to assume that, since I don't know if a conversion from bits-per-second to cycles-per-second is valid (kinda like how liters and grams aren't compatible). $\endgroup$ Commented Mar 14, 2022 at 23:05
  • $\begingroup$ So you're proposing a means of transmission that has a bandwidth more than 6MHz because you want a bandwidth less than 6MHz? $\endgroup$
    – TimWescott
    Commented Mar 15, 2022 at 5:43
  • $\begingroup$ My proposal is obviously impossible, I'm just curious as to why. Bandwidth and frequency don't seem related at a casual glance - one is a single point on the frequency line, the other is a span on the frequency line. I'm just trying to resolve that confusion. $\endgroup$ Commented Mar 15, 2022 at 14:03

1 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.

Pulse Shaping


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