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Given a signal $x(t)$ and the $X(\omega)$ obtained from $x(t)$ using a FFT with a sampling time $Ts$, I get a subset of $X(\omega)$: $Y(\omega)$ obtained from $X(\omega)$ taking it between $\omega_0$ and $\omega_1$, the question is: if I make the IFFT of $Y(\omega)$, does the new signal $y(t)$ have the same sampling time $Ts$? Thanks in advance

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Starting from $x(t)$, you sample it with a sampling time $T_s$, then obtain a discrete time signal $x(n T_s)$ or $x(n)$, on which you apply FFT, to obtain the discrete spectrum $X(m)$ (the notation $X(\omega)$ usually denotes the Discrete Time Fourier Transform, which is different from the Discrete Fourier Transform, a fast version of the latter is FFT). Now, if you truncate $X(m)$ by keeping its values from $m_0$ up to $m_1$ and discarding the rest, without zero padding, then you are changing the FFT resolution, which is proportional to $1/N$, where $N$ is the number of your samples. The FFT resolution is the frequency distance between adjacent FFT points and corresponds to time distance between adjacent signal points (in the time domain). So yes, you are changing the sampling rate of your original signal. My advice is to use zero padding to avoid that.

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when I get $Y(\omega)$, I don't multiply with a window. In that case I would obtain a 'zero padding' of the whole signal $X(\omega)$. In my case, I get $only$ the signal between $\omega_0$ and $\omega_1$ – Riccardo.Alestra Oct 18 '12 at 14:01
I see. I will modify my answer for that case. – Manos Oct 18 '12 at 14:59

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