# Calculating maximum signal frequency knowing its sampling rate?

I was curious as to how I could calculate the maximum frequency of a signal knowing only its sampling rate (for example 125 nanoseconds). Does the sampling theorem have anything to do with it (sampling rate is at least two times the maximum frequency)? So is the max frequency half of the sampling rate?

• There is a unit consistency issue in your question. "Rate" can be ambiguous. First, you have a sampling rate of 125 nanoseconds (time), then you compare the sampling rate with two times the sampling frequency (Hertz). – Laurent Duval Oct 18 '16 at 21:32
• SE.DSP wishes you a happy new year 2017, with a kind reminder that your question and its answers may require some action (votes, acceptance, etc.) – Laurent Duval Dec 31 '16 at 17:07

The maximum possible frequency in the sampled signal is half the sampling rate, assuming that you are not bandpass sampling and/or there is an anti-aliasing filter. The question is whether the signal has content up to that rate or whether it stops at a lower frequency.

One way to determine this would be to take your sampled signal $x[n]$ and transform it to the frequency domain via FFT to get $X[k]$.

Then find the largest value of $k$ for which $$\left |X[k] \right| > t$$ for $0 \le k \le \frac{K}{2}$ where $K$ is the length of $X$ and $t$ is some "threshold" below which you decide the signal level is effectively zero.

• Not true for undersampled sufficiently narrow-band signals (with low enough sample jitter, etc.) The samples could represent a much higher frequency frequency band, as long as the bandwidth was narrow enough to prevent aliasing or folding across multiples of half the sample rate. – hotpaw2 Oct 18 '16 at 22:58
• @hotpaw2 Agreed! I was assuming the band pass sampling didn't come into it. Most questions of this sort don't want to consider it. And most sampling systems have anti-aliasing filters to prevent it. YMMV – Peter K. Oct 18 '16 at 23:32

Once a continuous signal is sampled at frequency $f_s=1/T_s$ (here $T_s=$ 125 nanoseconds), the maximum "observable" frequency of the sampled signal, without further hypotheses, is indeed half the sampling frequency. [EDIT following @hotpaw2 answer] There is a risk that some frequency components of the sampled signal, and especially a maximum frequency (e.g. computed after @Peter K.), are caused by aliasing, and not so related to a "maximum frequency" in the continuous signal.

The sampling theorem has something to do with it. But the sampling rate is in seconds, you have to convert it to a sampling frequency (in Hertz) by computing the inverse.

You can't know from just the sample rate. It's unknown. The signal could have been baseband, bandlimited to under half the sample rate, or a much higher frequency narrow-band signal that was undersampled, or some aliased combination or mix of the above.

You have to know something about the signal before it was sampled (for instance, was it properly band-pass, low-pass, and/or anti-alias filtered, etc.) to know anything about the frequencies represented by the samples.

• In fact, given that there is no such thing as a perfect zero-level stop-band anti-alias filter, the max frequency will be of some tiny portion of the aliased noise floor representing a near infinite frequency (say, the reciprocal of Planck time?). – hotpaw2 Oct 18 '16 at 23:04