1
$\begingroup$

I know that a continuous-time digital signal with sharp edges (e.g. jumping from one y-Value to another discontinuously at the same x-value) will have infinite bandwidth. But what about sampled signals, that have a limited sampling frequency?

I read that aperiodic time limited signals will have an infinite bandwidth, but I can reconstruct any signal by doing:

reconstructed_signal = ifft(fft(some_signal))
reconstructed_signal ≈ some_signal # true

Is taking fft(some_signal) already aliased?

What is the bandwidth of a random signal?

noise = randn(ComplexF64, 1000)

Will it have infinite bandwidth or is the bandwidth equal to the sampling rate?

$\endgroup$
3
  • $\begingroup$ By "digital signal" do you mean a two-valued signal in continuous time? I.e. $x(t) = 0\ \forall\ t < 0$ and $x(t) = 1\ \forall\ t \ge 0$? $\endgroup$
    – TimWescott
    Mar 4 at 0:28
  • $\begingroup$ Yes, that's what I actually meant. Thanks for clarifying. $\endgroup$
    – Soeren
    Mar 4 at 7:37
  • $\begingroup$ I didn't notice you were a newbie and so forgot to remind you to edit your question (Stackexchange is a bit unique in that it wants a crafted Q & A, rather than having answers and clarifications and whatnot buried in the discussion). So I went ahead and edited your question. $\endgroup$
    – TimWescott
    Mar 4 at 15:06

3 Answers 3

2
$\begingroup$

I know that a digital signal with sharp edges (going from one y-Value to another at the same x-Value) will have infinite bandwidth.

A digital signal never has "sharp edges"; by the virtue of being digital, it's only defined at discrete points in time. So, it's just a sequence of numbers; these numbers don't have "edges" in between; the space in between two samples is simply undefined, has no value.

It does not have infinite bandwidth, by no understanding of the word I'd subscribe to. Being digital, the "whole world of bandwidths it could have" is the Nyquist rate. So, no.

Only when we take a signal composed of a comb of dirac deltas in the time-continuous domain we see a periodic spectrum – indeed, infinite bandwidth. But that's not the original digital signal, that's a very specific conversion of it. For time-continuous signals, we actually have infinitely extending spectrum.

Usually, and more usefully, we assume that, if there is a time-continuous signal equivalent to the digital signal (that's not generally the case! It helps trying to remind oneself of that, once in a while), that it was bandwidth-limited to begin with – otherwise, no digital signal could represent it, anyways.

So, I think your premise is a bit confused here.

I read that aperiodic time limited signals will have an infinite bandwidth

In continuous-time, under the continuous Fourier transform,

reconstructed_signal = ifft(fft(some_signal))

You're still in discrete-time domain. So, the statement above has little to do with your statement here!

What is the bandwidth of a random signal?

In continous-time, it could be up to infinite.

In discrete-time, it could be up to Nyquist.

"up to", because "random" doesn't tell us much – you could have a random signal that is weak-sense stationary, correlated and has a limited spectrum in either case.

$\endgroup$
7
  • 1
    $\begingroup$ (+1) I think you can describe a digital waveform as having infinite bandwidth to the same extent you could allow a discrete time waveform to extend for infinite time. Mathematically we can describe periodic discrete waveforms in time which extend to infinity, but repeat. To the same extent we can mathematically describe a waveform in frequency which extends to infinity but repeats. This is what the spectrum for a discrete time waveform looks like- it is periodic in frequency extending to infinity and due to this periodicity we only need to show what is within Nyquist. $\endgroup$ Mar 4 at 12:14
  • 1
    $\begingroup$ This model does help with simplifying the understanding of multi-rate concepts and better unifies the continuous and discrete time concepts and is (mathematically) accurate. Nothing in the causal physical world can extend for infinite bandwidth in either continuous or discrete time. All math is a model anyway and wanted to highlight that there is utility in such a model with the frequency domain extending to infinity for a discrete time signal. $\endgroup$ Mar 4 at 12:15
  • 1
    $\begingroup$ @DanBoschen very well-put! It's a bit a matter of how you want to think about the frequency domain of discrete-time signals: if you want to think about it as defined over $\mathbb R$, it's right to think of it as periodically repeating (but you get into trouble e.g. Parseval if you don't explicitly state Nyquist, and then you're kind of admitting it's not really "just" periodic), or you can think of it as over $S_1$ instead (i.e. on a 1-dimensional sphere, a "ring", in the purely geometrical sense). Then you get into trouble with intuition. $\endgroup$ Mar 4 at 12:20
  • 1
    $\begingroup$ Fancy R: $\mathbb R$ :) $\endgroup$ Mar 4 at 12:29
  • 1
    $\begingroup$ the all-over-the-reals is also very necessary when you need to convert to and from time-continuous (and also, I guess, if you're doing non-rational resamplers) $\endgroup$ Mar 4 at 12:31
2
$\begingroup$

Maybe to extend the answer of Marcus. Here is a visualization:Visualization

Your implicit model, when doing the FFT is always that your signal is band-limited as Marcus pointed out. So we know the that our signal is composed out of several complex sinusoids (See, e.g. here). Now since we know our complex sinusoid model, we can artificailly increase the sampling frequency of our implicit model, what is shown in the dashed line.

$\endgroup$
1
$\begingroup$

I read that aperiodic time limited signals will have an infinite bandwidth, but I can reconstruct any signal by performing an inverse FFT on the FFT of the signal.

Any aperiodic time limited signal in continuous time will have an infinite bandwidth, because in order to be aperiodic then it must have discontinuities in some higher derivative. Those discontinuities translate into infinite bandwidth.

The FFT does not operate on continuous-time signals, or even signals of infinite extent. It only operates on discrete-time, finite span signals. The "sensible" spectrum of a discrete-time signal with even sample intervals is periodic, repeating every $2 \pi$ radians per sample interval. Or, depending on who you're talking to or the problem you're working on, a discrete-time signal has a spectrum that's only defined over an interval that's exactly $2 \pi$ radians long, and talking about frequencies outside that interval is meaningless.

Is taking fft(some_signal) already aliased?

In the context that you're asking the question, yes. It's more accurate to say that if you start with some continuous-time signal $x(t)$ and sample it: $x_k = x(k T_s)$, then at that point the signal is aliased (basically munging the entire possible spectrum of $x(t)$ onto the $2 \pi$ radians that are possible for $x_k$). Then if $x(t)$ is of infinite extent you have to truncate $x_k$, and then you can take an FFT.

What is the bandwidth of some random signal?

Complicated. As you defined it, we're talking about a discrete-time signal $x_k$ of finite extent, where each $x_k$ is independent of any other sample $x_n,\ n \ne k$. In that case, the expected value of magnitude of the spectrum of the signal will be flat across the output of the FFT.

That's as infinite as you can get with a sampled-time signal, but I think it's fair to say that it fits your "bandwidth equal to the sample rate".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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