This question has already been asked and answered, but the motivation behind the use of normalized frequency units still evades me.

The Discrete Time Fourier Transform $$X(\tilde{ \omega }) = \sum_{n=-\infty}^{\infty} x_n e^{-i\tilde{\omega}n} $$ in my text is given in terms of the normalized (dimensionless) angular frequency $\tilde{\omega} = \omega \Delta t$ where $\omega$ and $\Delta t$ are the physical frequency and time interval between measurements, respectively.

I want to understand the merits of using normalized frequency units.


1 Answer 1


Consider a continuous signal $$ x(t) = \sin ( 2\pi f t) \ $$

of which we have the following measurent points: $$x_n = x(n T_s) = \sin \bigg(2\pi \frac{f}{f_s} n \bigg) \ . $$

We note that $\frac{f}{f_s} \equiv f_n$ arises naturally as the frequency of our discrete set of measurements, which is called the normalized frequency.

Its units are $$ \frac{[f]}{[f_s]} = \frac{\text{cycles/second}}{\text{samples/second}} = \text{cycles/sample} \ .$$

Hence, if we consider the (inverse) Discrete Time Fourier Transform, it is natural to use the normalized frequency $\tilde{\omega} = \omega T_s = 2\pi f_n$.


Note that the Nyquist sampling theorem requires $f_d < \frac{1}{2}$ to avoid aliasing. This puts a lower bound on our sampling rate $f_s$.


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