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Nyquist Sampling Thm - Question 2

  1. Oct 1, 2006 #1

    cepheid

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    Hello,

    This is the same type of question as the one I just posted before. This time I have obtained an answer and would really appreciate it if someone would check my reasoning:

    Suppose a signal x(t) has a Nyquist sampling frequency [itex] \omega_s [/itex]. Compute the Nyquist sampling frequency for the following signal in terms of [itex] \omega_s [/itex]:

    dx/dt

    Again, my first thought was, how does the spectrum of the new signal compare to that of the original? Simple:

    [tex] \mathcal{F}\{\frac{dx}{dt}\} = j\omega X(j\omega) [/tex]

    So, I obtained an answer through reasoning rather than computation. Here was my reasoning: if the spectrum of the original signal x(t) has some maximum frequency [itex] \omega_M [/itex] i.e. it is band limited such that:

    |X(jw)| = 0 for |w| > w_M

    then we know that:

    [tex] \omega_s = 2\omega_M [/tex]

    Furthermore, |jwX(jw)| is zero whenever |X(jw)| is zero, therefore, the spectrum jwX(jw) of the differentiated signal is STILL zero outside of this frequency range i.e. it is band-limited in exactly the same way. Therefore, it's maximum frequency is also [itex] \omega_M [/itex], which means that its Nyquist sampling frequency is also given by [itex] 2\omega_M [/itex]. In other words, the Nyquist sampling frequency of the new signal is just the SAME as that of the old signal: [itex] \omega_s [/itex].

    Does this reasoning hold?
     
  2. jcsd
  3. Oct 1, 2006 #2
    Yes, it holds alright.
     
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