Suppose we have been given the CTFT (cont. time fourier transform) of a signal \(x\left ( t\right ) \), then this signal is sampled using sampling period \(T\), and we want now to determine the DTFT of the sequence \(x\left [ n\right ] \). We could offcourse apply the definition of DTFT on \(x\left [ n\right ] \), but we also obtain the DTFT spectrum directly from the CTFT. (I need to be careful here, we are here talking about magnitude spectrum only, double check on this).
Let \(F_{a}\left ( \Omega \right ) \) be the CTFT and let \(F\left ( \omega \right ) \) be the DTFT, then \(F\left ( \omega \right ) =\frac{1}{T}F\left ( T\Omega \right ) \) or \(F\left ( \omega \right ) =\frac{1}{T}F\left ( \frac{\Omega }{f_{s}}\right ) \) where \(f_{s}\) is the sampling frequency in samples per second (or Hz).
Hence if we want to find \(F\left ( \frac{\pi }{4}\right ) \), then this maps to \(\Omega =\frac{\omega }{T}=\frac{\pi }{4T}\), and so we read \(F\left ( \frac{\pi }{4T}\right ) \) from the CTFT, and then we multiply the result by \(\frac{1}{T}\), this gives \(F\left ( \frac{\pi }{4}\right ) \) l.583 — TeX4ht warning — “SaveEverypar’s: 2 at “begindocument and 3 “enddocument —