# Nyquist Sampling Thm - Question 2

1. Oct 1, 2006

### cepheid

Staff Emeritus
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?

2. Oct 1, 2006

### doodle

Yes, it holds alright.

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