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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 \omega_s. Compute the Nyquist sampling frequency for the following signal in terms of \omega_s:
dx/dt
Again, my first thought was, how does the spectrum of the new signal compare to that of the original? Simple:
\mathcal{F}\{\frac{dx}{dt}\} = j\omega X(j\omega)
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 \omega_M i.e. it is band limited such that:
|X(jw)| = 0 for |w| > w_M
then we know that:
\omega_s = 2\omega_M
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 \omega_M, which means that its Nyquist sampling frequency is also given by 2\omega_M. In other words, the Nyquist sampling frequency of the new signal is just the SAME as that of the old signal: \omega_s.
Does this reasoning hold?
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 \omega_s. Compute the Nyquist sampling frequency for the following signal in terms of \omega_s:
dx/dt
Again, my first thought was, how does the spectrum of the new signal compare to that of the original? Simple:
\mathcal{F}\{\frac{dx}{dt}\} = j\omega X(j\omega)
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 \omega_M i.e. it is band limited such that:
|X(jw)| = 0 for |w| > w_M
then we know that:
\omega_s = 2\omega_M
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 \omega_M, which means that its Nyquist sampling frequency is also given by 2\omega_M. In other words, the Nyquist sampling frequency of the new signal is just the SAME as that of the old signal: \omega_s.
Does this reasoning hold?
