Analog to Digital (conversion time)

I have this question, and my work in the lab seems to conflict my calcultations


Given an ADC(analog to digital conversion) time of 25 micro seconds, how many data points can be taken if the frequency is 50 Hz? 200Hz? 1 KHz? 1 MHz?

what I did is I too 1/50= .02 secs per oscillation

then .02 = 20,000 micro seconds.

then I did 20,000 micro sec/ 25 micro sec
to give me 800 data points taken.

then as I did the calculations for the 200 Hz and so on, the number of data points taken decreases.

does this make sense?

how does, please explain it in a physical way if you can, in terms of how we would see a general sign waves(if that relationship works at all).

much appreciated,
johnny5

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Given an ADC(analog to digital conversion) time of 25 micro seconds,

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ADC time ? you mean sample frequency?

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how many data points can be taken if the frequency is 50 Hz? 200Hz? 1 KHz? 1 MHz?

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the frequency of what...? the input signal ?


/btq :)

I think they mean "sample frequency", I don't know why they word it the way they do.

And yes, the frequency of the "input signal". wouldn't they call that the "analog wave" too?

the number of points "taken" of an inputsignal depends only on the
sample frequency. If you sample it at a frequency of 44100 you get
a point every 1/44100 second.
is this an answer to the question or am I missing somethin ? :)

/btq

>Given an ADC(analog to digital conversion) time of 25 micro
>seconds, how many data points can be taken if the frequency is
>50 Hz? 200Hz? 1 KHz? 1 MHz?

I assume that the ADC time is the sample period of the ADC, which means that every 25 microseconds a sample is taken of the input signal. It does not matter what the frequency of this input signal is.

Now the physical way. Take a piece of paper and draw in the same graph a couple of sines with different frequencies. Now decide you take samples every n seconds for every signal. If you paint a point at each sample of the signal, you see the result of the sampling.

Note that some, or perhaps all, sampled signals don't look like the original signal, this is caused by your sampling frequency. The sampling theorem says that if you sample with a sampling frequency of N Hz, then you can only sample signals of N/2 Hz without loss of information.


Sampling every 25 microseconds is the same as a sampling frequency of 40 Hz. According to the sampling theorem you can only sample signals of 20 Hz without signal loss.

Sounds like you understand mathmatically the inverse relationship of frequency and period, but that you don't quite have it intuitively yet. Your results are right.

Look at it this way. Suppose that your sample frequency is 1 hz. Then every sample takes 1 sec. Now sample a 1/1000 hz signal. You have 1000 sec and take 1000 1 sec samples. Raise the frequency to 1 hz. You get 1 sec and get 1 sample.

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Sampling every 25 microseconds is the same as a sampling frequency of 40 Hz. According to the sampling theorem you can only sample signals of 20 Hz without signal loss.

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Don't you mean 40 KHz and 20 KHz?