Series

Sample rate — the series

2 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. A 9.0 kHz input sampled at 10 kHz arrives as 1.0 kHz. computed by solving, not by drawing. The dots are the samples. The input at 9.00 kHz is above half the 10 kHz rate, and every dot also lies on the 1.00 kHz curve drawn beside it — the two sample sequences differ by 9.3e-15, which is the arithmetic and not a small effect. Nothing is attenuated and nothing is distorted: the samples are the samples of a different signal, at full amplitude, and there is no measurement of them that could say which one was there.

    The frequency a sample rate invents

    Every other boundary on this site is a model getting gradually worse. This one has no gradient at all: below half the sample rate a set of samples has one sinusoid through it, above half the sample rate it has another, and the two sets of numbers are identical to three parts in ten thousand billion. Nothing is attenuated, nothing is distorted, and there is no measurement of the samples that could say which signal was there.

    part 1 · digital
  2. What the reconstruction returns, either side of 5.0 kHz. computed by solving, not by drawing. The Whittaker–Shannon sum is evaluated on the samples and compared with two things: the signal that was sampled, and the frequency the samples report. Below 5.00 kHz these are the same curve and the error is 4.88e-3 — the truncation of the sum at sixty-four samples either side, and nothing else. Above it they part: at 9.00 kHz the reconstruction is 1.59e-3 from the alias and 2.000 from the input. The small number is the interesting one. A reconstruction cannot be improved into the right answer, because it is already an exact answer to a different question.

    An exact answer to a different question

    The reconstruction that turns samples back into a signal is normally introduced as the thing that recovers what was there. Measured on both sides of half the sample rate it does something more interesting than failing: above the boundary it returns the alias to 1.6 parts in a thousand, which is the same accuracy it returns the input with below the boundary, and it is wrong about the input by twice the amplitude. Its error is not a degradation. It is exactness about something else.

    part 2 · digital

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