Minimum recording time for a signal

I am familiar with both Nyquist frequency and Nyquist rate. What I cannot seem to find information on is the minimum time necessary to detect a signal.

In my use case I am using a multiplexer to measure signals from multiple analog sources, each source carrying information from 60 to 100 Hz. I would, ideally, like to take these samples as quickly as possible.

Lets assume that my lowest frequency is 10 Hz. to detect a 10 Hz signal, you need at least 200ms of data, right? You cant detect a signal that comes once every 100 ms without measuring for at least twice as long. Or can you? This is part one of my question.

Part two is, how does the signal appear if I tried to measure a 10 Hz signal using only a 100 ms bin? Obviously it appears as a random point, but does it mask itself as a higher frequency? Is it just combobulated noise? What does the aliasing look like?

Lets assume that my lowest frequency is 10 Hz. to detect a 10 Hz signal, you need at least 200ms of data, right? You cant detect a signal that comes once every 100 ms without measuring for at least twice as long. Or can you? This is part one of my question.

Consider the Discrete Fourier Transform (DFT) of that "subsampled" block:

$$X_k = \frac{1}{N} \cdot \sum_{n=0}^{N-1}x_n \cdot \left( \cos(\frac{-2\pi kn}{N}) + i \sin(\frac{-2 \pi kn}{N}) \right)$$

This formula has "no problem" in producing the $k^{th}$ sum, when $k \ll N$. So, yes, information about that harmonic is still available in the signal.

At the same time, notice that now, this sum is going to be scaled by the limited availability of data. So, if you "subsample" a 10Hz sinusoid by examining a block of data that contains half a period, you can still evaluate the sum for that low frequency, but the amplitude that you will "read" back will be half the true amplitude, simply because the sum did not have enough samples to "evolve" in.

In a noise free environment, that would not be too much of a problem but in practical conditions, this extra scaling would make reliable detection / extraction of that harmonic even more difficult in a noisy environment.

...how does the signal appear if I tried to measure a 10 Hz signal using only a 100 ms bin? Obviously it appears as a random point, but does it mask itself as a higher frequency? Is it just combobulated noise? What does the aliasing look like?

I am not sure how does the "...random point..." represent the signal in here (?), but the closest you can get to how that would look like is to take a "window" of the samples of a 10Hz sinusoid. Obviously, if you are sampling this at different (or random) times, you will be "catching" it at some random phase. So, the amplitude component will be more stable than its phase.

This "mental picture" probably takes us nicely into the discussion about aliasing:

Strictly speaking, there is no aliasing here because $F_s>f_{foi}$ (frequency of interest).

BUT!, if you are trying to estimate the amplitude and phase of those components via this kind of "blocked" recordings, then the frequency of sampling of the block does come into play.

It is not exactly clear form your question if the frequency by which you collect blocks of data is stable or even if, through the analog muxer, the signals are truly sampled at the same time.

So, to cut a long story short: If you are sampling in windows but do it regularly, there should be no aliasing. Your estimation of the amplitude will be scaled and the estimation of the phase will have a systematic error that is proportional to the rate at which the blocks are acquired and the initial phase that the first block happened to "catch" the sinusoid at the input. But, if you sample these blocks irregularly, with some jitter because of whatever reason, then this will creep in to your estimates. Again, the amplitude will probably be more stable than the phase but the latter will now vary in an irregular way that depends on the jitter.

Hope this helps.

The time needed for frequency estimation depends on the waveform, its frequency, and the S/N. In zero noise, one only needs 3 or 4 non-aliased sample points to reconstruct all the parameters of a pure unmodulated sinusoid. The more noise, the longer one needs to sample for a given accuracy. If the frequency of interest is near DC or Fs/2 and of unknown period with relation to the sampling aperture, one may need to sample longer to separate a signal from its conjugate image in the spectrum.

And obviously, to detect an intermittent signal, the probability is higher the higher the ratio of sampling to not sampling that source channel.