I have a problem in my DSP class that I could use some advice with.

An analog signal of length 2.5s is sampled with a Nyquist frequency and the sampled sequence has 5001 values. What is the highest frequency contained in the analog signal?

What I've gathered that the sampling frequency is $5001/2.5 = 2000.4 \text{ Hz}$ and thus $T=1/2000.4$

My textbook says that $\Omega_T \geq 2\Omega_m$, where $\Omega_T = 2\pi/T$ and $\Omega_m$ is the Nyquist frequency.

So I've found that $\Omega_T = 2\pi 2000.4$ so I get $2\pi 2000.4 \geq 2\Omega_m$ which leads to the highest frequency being $\Omega_m = 2000.4\pi$ or $f_m = \Omega_m/2\pi = 1000.2 \text{ Hz}$

Am I approaching this correctly?

  • $\begingroup$ for a signal w/ sample rate = 2000.4Hz, the highest frequency contained is 1000.2 Hz! $\endgroup$
    – ederwander
    Sep 17, 2014 at 16:37
  • $\begingroup$ ok that'w what I got, but is my assumption of the sampling rate right? $\endgroup$
    – dingari
    Sep 17, 2014 at 16:38

1 Answer 1


You are doing it correctly. One minor change that I would make is to divide 2.5 into 5000, not 5001, to get the sample rate. Two reasons- one, it appears that they purposely set up the problem to have a nice, neat number. Two, because they are implicitly saying that there was a sample at time 0 and a sample at time 2.5, so there is an "extra sample" because there is a sample at both ends, so you should subtract one.

Anyway, that is a triviality. You got the important concepts, and that is what matters.


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