# How does one compute the band width of a signal in the time domain $x(t)$?

Sorry if this is a super easy question, I am really not a signals guy (more of a statistician, computer scientist). But I was trying to reconstruct/learn a sinusoidal function $x(t)$ with linear regression $f(x) = \langle w , \phi(x) \rangle$ (I'm adding as many polynomial features as I may need) over a fixed interval $[-1,1]$. Then I realized that there is this thing called the Nyquist-Shannon sampling theorem that might help figure out if I have enough samples. The reason that I care about it is because the performance of my regression model is really bad on the test set and I wondering if I am just not getting enough samples from my ground truth signal (which for the moment I control synthetically). I know now that I need my sample frequency $f_s$ to be larger than twice the band width $B$ of my signal:

$$f_s > 2B$$

to my understanding bandwidth is just the range of the frequencies present in the signal $x(t)$ (I even asked to clarify that to me just in case here: What is the definition of the bandwidth of a signal?). Thus if that is correct then band width is:

$$B = f_{max} - f_{min}$$

is there a general way to get $f_{max}$,$f_{min}$ from a signal in the time domain?

If I understand this correctly then I just need to get the frequencies that define $x(t)$? Right?

If that is correct then the only way I know is if we know the analytic form of the $x(t)$, which I do for my example, then I just read of the frequencies from the Fourier series $f(x) = \frac{1}{2}a_0 + \sum^{\infty}_{n=1} a_n cos(nx) + \sum^{\infty}_{n=1} b_n sin(nx)$. Say $n_{max},n_{min}$ are the largest values modifying the inside of sin or cos terms then we read them off and do:

$$B = \frac{n_{max} }{2 \pi} - \frac{n_{min}}{2 \pi}$$

right? What happens if there is only 1 single term like it only a sin like $x(t) = sin(n t)$, how do we get the smaller frequency if there is no other frequency?

Now that I understand much better what the term frequency means I assume that the answer will point me that if I don't know the analytic form of my signal (as probably happens most in practice) we need to resort to some type of transform used. Probably the Fourier transform? But a quick google search to the frequency domain yielded a list of methods:

1. Fourier series – repetitive signals, oscillating systems
2. Fourier transform – nonrepetitive signals, transients
3. Laplace transform – electronic circuits and control systems
4. Z transform – discrete-time signals, digital signal processing Wavelet transform - image analysis, data compression

I don't know when to apply each but it seems the wikipedia article links to more articles for each one...

Is this correct or am I totally off? (which might be the case since I'm not very familiar with this field). I am assuming there are probably engineering details I am totally missing like flow pass filters or something else that might be important for real signals in practice. Though its just a random guess based on the definition of low-pass filter (filters high frequencies out). Though not sure.

Further reading What is the definition of the bandwidth of a signal? makes me suspect that the actual correct answer is as follows:

1. Consider a time domain signal $x(t)$.
2. Consider its frequency description $X(\omega)$

the the support of $X(\omega)$ is the bandwidth. In other words, "the summations" of the points in the domain (to be more precise, the integral of the domain). So if the domain is the whole interval $[-W,W]$ the the bandwidth would be:

$$B = \int^{W}_{-W}d \omega = 2W$$

so in general is it:

$$B = \int_{w \in supp(X(\omega))} d\omega$$

would be my guess.

• Your last paragraph is correct, except that by convention one only counts the positive frequencies; that is, the bandwidth would be $W$, not $2W$. Also: the bandwidth of a sine wave is zero; if you limit it to the range $[-1,1]$, however, you're multiplying it by a rectangular pulse and then its bandwidth becomes infinite.
– MBaz
Oct 20 '17 at 16:35
• @MBaz but the Nyquist-Shannon sampling theorem must apply to also signals (functions in time say) that have countable frequencies like things with fourier series, right? In the simplest example is considering $x(t) = sin(kt)$. I feel there must be a way to sample it optimally, no? Oct 20 '17 at 23:10
• Yes, the sampling theorem applies to periodic and aperiodic signals. So, a signal like a pure sine wave with frequency $f_0$ can be reconstructed from samples taken at a rate $f_s>2f_0$. There are other, simpler ways to specify that signal, though: for instance, by specifying its frequency, amplitude and phase, or by projecting it on other orthonormal bases. Is this useful? I'm not sure I understand what your question is, specifically.
– MBaz
Oct 20 '17 at 23:19
• @MBaz yes that was useful. I guess my question seems to be a bit more related to the definition of bandwidth or how Nyquist-Shannon theorem changes when applied to signals with countable frequencies. So $B=\int_{w\in supp(X(\omega)} d\omega$ or $B=f_{max} - f_{min}$, or Nyquist-Shannon theorem must be something different depending on the signal. I am not sure which one it is though. Oct 21 '17 at 3:29
• I get your question now. See chapters 6--9 of this free book: afidc.ethz.ch/A_Foundation_in_Digital_Communication/Home.html
– MBaz
Oct 21 '17 at 14:32

As far as I know there is no way to get $f_{max}$ and $f_{min}$ directly from time domain (zero crossing rate will not help if the signal has a non-zero DC offset for example) you need to look at the frequency domain.
If there is just one term, the smaller frequency is just $0$
Given a time signal and its sampling rate ($Fs$), you can take the discrete Fourier Transform to study its frequency spectrum. Unfortunately you cannot use this to decide your sampling rate if your initial sampling rate ($Fs$) is lesser than the 2$*$bandwidth (2$*$[highest frequency] if you are looking for perfect reconstruction) as you have already lost the data during initial sampling(look up aliasing for further reading).
• yes, a constant term has to be represented by a zero frequency (not true for piecewise constants, it has to be constant all throughout). A constant $c$ can be looked up as $c*cos(w_0 t)$, with $w_0 = 0$ Oct 21 '17 at 21:13