1
$\begingroup$

When a discrete-time signal is obtained by sampling a continuous-time signal, why we are removing the information about sampling interval and only deal it as a sequence of numbers?

$\endgroup$
1
  • $\begingroup$ Uniformly sampled. $\endgroup$ Commented Nov 11, 2020 at 21:06

2 Answers 2

3
$\begingroup$

Because they're defined like that – there's no specific reason, it's just what that word "discrete time signal" means: We consider a signal that's a sequence of numbers.

Especially note how the sample rate / interval has no effect on any of the DSP you do with the signal. Your digital FIR filter doesn't care at all. It multiplies and adds numbers. So, that info simply has no relation to the way we consider the signal.

At some point, you might want to interpret the signal w.r.t. to real-world frequencies again. Then you need to incorporate knowledge on the sampling rate again, but not at any earlier point – frequencies only matter in "fractions of the sample rate", not in Hz.

This ties in nicely with your previous question; for example: In digital domain, nothing cares whether the 3rd bin of your DFT corresponds to some real-world frequency. It's the 3rd bin, that what matters.

The trick really is trying to stop assigning real-world times and frequencies to your digital signal unless you're actively trying to specifically estimate something that has a real-world meaning. Always work in normalized frequencies – your digital 3-5 Hz bandpass in a 20 Hz-sampling rate system is identical (and not just similar!) to a 30 to 50 MHz bandpass in a 200 MHz-sampling rate system.

When you are designing such a filter, the first thing the algorithm does is divide 3 by 20 , and 30·10⁶ by 200·10⁶, and then calculates with the "3/20" normalized frequency.

$\endgroup$
3
  • $\begingroup$ That was fast, I shouldn't try to do three different things in the same time interval $\endgroup$ Commented Nov 11, 2020 at 13:49
  • $\begingroup$ "the sample rate / interval has no effect on any of the DSP you do with the signal" — well, the coefficients may depend on the sample rate. $\endgroup$ Commented Nov 11, 2020 at 21:09
  • 1
    $\begingroup$ no, my whole point, @RodrigodeAzevedo: they don't. They depend on the relationship of sample rate to some real-world thing, but never on the sample are itself. Once you're in discrete time, a filter is no longer something with a cut-off at 3 Hz, but something with a cut-off at 3/20. $\endgroup$ Commented Nov 11, 2020 at 21:11
5
$\begingroup$

Samples of (regularly sampled) discrete time signals, like pixels of a digital images, are sequences: they are unitless on the ordinal axis (albeit time or space), and unitless in amplitude. A sequence is

an enumerated collection of objects in which repetitions are allowed and order matters

This is close to what people call raw data.

However, to make a genuine "signal" or "image" object (meaningful in the real world or for devices) that can be better interpreted or manipulated, it is important to keep ancillary information, like: sampling rate, quantization, time-space coordinates, size, channels, amplitude units, calibration parameters, etc. Those additional parameters are found for instance in the header of most digital sound or image raster formats, and then followed by the unitless digital sequence.

Below is an example from the WAVE PCM soundfile format (converted in png for display). The Discrete signal is typically in the bottom "data" sub-chunk, yet requires the above information to be read, listened to, analyzed.

The canonical WAVE sound file format

$\endgroup$
1
  • 2
    $\begingroup$ very nice illustration! $\endgroup$ Commented Nov 11, 2020 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.