Lеt's say that i have a plot where i can see the spectrum of some discrete signal and let's say that it's x axis is normalized (by normalized i mean $\omega/\pi$ on the $x$ axis). Now, i know that in order to get to the real frequency at some particular point on the plot, i need to know the sampling frequency $f_s$ and then multiply that particular point with $f_s/2$ in order to get the actual frequency.

Now, i know this is maybe something trivial, but i am having struggle to understand why is this so, i mean, why multiplication by $f_s/2$ gives us the actual frequency in hertz?

Any help appreciated!


The maximum frequency of a discrete-time signal is half the sampling frequency. A signal with maximum frequency in discrete time is an alternating sequence, and since its period is two sample intervals, its frequency is $f_s/2$. All frequencies in the discrete domain are normalized by that maximum frequency.

This implies that a discrete-time system's behavior can be described independently of the sampling frequency. E.g., if you have a low pass filter, the cut-off frequency is fixed in terms of relative frequency. The actual cut-off frequency in Hertz depends on the sampling frequency.

Often you will encounter the normalized frequency in radians:


where $f_s$ is the sampling frequency, and $f$ is the frequency in Hertz. Note that the frequency response of a discrete-time system is $2\pi$-periodic with respect to $\omega$, i.e., with respect to $f$ it is periodic with period $f_s$.

  • $\begingroup$ I do not get the second paragraph about describing a discrete-time system independently of sampling frequency... $\endgroup$ – jomegaA Feb 2 '20 at 20:46
  • $\begingroup$ @jomegaA: What I meant is that the frequency response of a discrete-time LTI system is only determined in terms of relative frequency. $\endgroup$ – Matt L. Feb 3 '20 at 10:35

Discrete time sequence have frequency $$-\dfrac{1}{2}\leq f\leq \dfrac{1}{2}\tag{1}$$

Continuous time signal $x(t)=A\cdot\sin(2\pi Ft)\tag{2}$

The discrete sequence can be obtained from CT signal through periodic sampling.

$$t=n\cdot T= \dfrac{n}{F_s}\tag{3}$$

Plugging the above relation in $(2)$ we get


Again plugging the $f$ in $(1)$ we get

$$-\dfrac{F_s}{2}\leq F \leq \dfrac{F_s}{2}\tag{5}$$

The maximum frequency is $\dfrac{F_s}{2}$


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