I am learning DSP and finding difficulty understanding the term Normalized frequency often used with DFT & DTFT.

  1. What does normalized frequency mean in DSP and how it is different from analog frequency?

  2. What is the significance of normalized frequency in DSP?

  3. Why is the limit of normalized frequency $2\pi$?

  4. How does the FFT deal with normalized frequency?


3 Answers 3


Normalized frequency is frequency in units of cycles/sample or radians/sample commonly used as the frequency axis for the representation of digital signals.

When the units are cycles/sample, the sampling rate is 1 (1 cycle per sample) and the unique digital signal in the first Nyquist zone resides from a sampling rate of -0.5 to +0.5 cycles per sample. This is the frequency equivalent of representing the time axis in units of samples instead of an actual time interval such as seconds.

When the units are radians/sample, the sampling rate is $2\pi$ ($2\pi$ radians per sample) and the unique digital signal in the first Nyquist zone resides from a sampling rate of $-\pi$ to $+\pi$.

How this comes about can be seen from the following expressions:

For an analog signal given as $$x(t)=\sin(2\pi F t)$$ where F is the analog frequency units in Hz,

When sampled at a sampling frequency of $F_s$ Hz, the sampling interval is $T_s=1/F_s$ so the signal after being sampled is given as:

$$x(nT_s)=\sin(2\pi F nT_s) = \sin\left(\frac{2\pi F}{F_s}n\right)$$

Where the units of normalized frequency, either $\frac{F}{F_s}$ in cycles/sample or $\frac{2\pi F}{F_s}$ in radians/sample is clearly shown.

This is illustrated below using $\Omega = 2\pi F$

Update: As @Fat32 points out in the comments, the units for sampling rate $F_s$ in the figure below is "samples/sec" (which is "Hz") in order for the normalized frequency to become radians/sample.

Normalized Frequency

To visually see the concept of "radians/sample" (and most other DSP concepts dealing with frequency and time) it has helped me considerably to get away from viewing individual frequency tones as sines and/or cosines and instead view them as spinning phasors ($e^{j\omega t} = 1 \angle (\omega t)$) as depicted in the graphic below, which shows a complex phasor spinning at a rate of 2 Hz and it's associated cosine and sine (being the real and imaginary axis). Each point in a DFT is an individual frequency tone represented as a single rotating phasor in time. Such a tone in an analog system would continuously rotate (counter-clockwise if a positive frequency and clockwise if a negative frequency) at F rotations per second where F is the frequency in Hz, or cycles/second. Once sampled, the rotation will be at the same rate but will be in discrete samples where each sample is a constant angle in radians, and thus the frequency can be quantified as radians/sample representing the rate of rotation of the phasor.

Euler's Identitiy

  • 2
    $\begingroup$ I was only half way through my coffee! Complete mistake, fixed now $\endgroup$ Jun 7, 2017 at 14:11
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    $\begingroup$ The FFT typically uses yet another unit for the frequency axis referred to as the Frequency Index which goes from 0 to N-1 where N is the number of samples used in the FFT . This maps to Normalized Frequency by equating N to 1 cycle/sample. Thus if you divide the FFT Frequency by N you get Normalized Frequency in cycles/sample. For example if I have 10 samples in the FFT in a system sampled at 100 Hz, the frequency bins in the FFT result will go from 0, 10, 20 ....90 Hz. N-1= 9 and 100 Hz represents 1 sample per cycle. Hope that helped. $\endgroup$ Jun 7, 2017 at 15:21
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    $\begingroup$ Yes I see- the sampling rate should be given in units of "Samples per Second" to make that work, good point @Fat32! Consistent units are important. $\endgroup$ Jun 7, 2017 at 15:45
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    $\begingroup$ @user6363 The sampling rate is 1 cycle/sample means when you use Normalized frequency then whatever the sampling rate is becomes 1 (cycle per sample), for example if the sampling rate is 100 MHz, then 100 MHz maps to 1, and a tone at 25 MHz for example would map to 0.25 (cycles/sample). When the units are radians/sample, the 100 MHz sampling rate would map to $2\pi$ and the tone at 25 MHz would map to $0.5\pi$. In my example the waveform extends in bandwidth from +6π/20 on a radians/sample normalized scale. If the sampling rate was 100 MHz, then this would be +/-3/20 *100 MHz = +/-15MHz $\endgroup$ Jun 7, 2017 at 23:54
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    $\begingroup$ No I said bin 9 is 90 Hz and that is correct. It is also correct that bin[9] is -10 Hz. Look a the graphs of frequency that both i and Fat32 have posted and you can see that the frequencies from 0 to Fs are the same as 0 to Fs/2 and -Fs/2 to 0! So for a 10 pt FFT in a 100 Hz system, the frequencies are both 0, 10, 20, 30, 40, 50, 60, 70, 80, 90 AND equally 0, 10, 20, 30, 40, 50, -40, -30, -20, -10! Check out FFTSHIFT in Matlab as it does this translation. If this is still confusing, this would be a good different question to ask (or search if it has already been answered) $\endgroup$ Jun 8, 2017 at 3:31

The following figure also displays a simplified graphical view of the frequency normalization procedure as a result of the sampling of a continuous-time signal enter image description here


Short answer

What is normalized frequency?

Normalized frequency $f'$ appears in discrete-time signal equation, it corresponds to regular frequency $f$ seen in continuous-time signal equation. In their respective equations $f$ and $f'$ determine how quickly the periodic signal repeats.

  • Continuous-time: $x(t) = \cos (2 \pi \color{blue} {f t})$
  • Discrete-time: $x[n] = \cos (2 \pi \color{green} {f' n})$

Their values are different because the continuous-time signal is a function of time $\color{blue} {t}$ but the discrete-time signal is a function of sample index $\color{green} {n}$.

Note normalized frequency is $f'$ and the quantity $2 \pi f'$ is known as normalized radian frequency.

For the two functions to match at sampling times, $f'$ has to combine frequency and sampling rate: $f' = \frac {f} {f_s}$. For a 100Hz cosine sampled at 1600Hz:

  • $x(t) = \cos (2 \pi \color{blue} {100 t})$
  • $x[n] = \cos (2 \pi \color{green} {\frac {100} {1600} n})$

By definition of sampling interval $t = \frac {1} {1600} n$. Both functions deliver the same value for any sampling time:

enter image description here

More details and answers to your other questions are provided below.

Normalized vs analog frequency

What does normalized frequency mean in DSP and how it is different from analog frequency?

$f'$ is not a true frequency as it hasn't the $\frac {1} {s}$ dimension, however discrete-time product $\color {blue} {f' n}$ and continuous-time product $\color {green} {f t}$ are exactly comparable, they have no dimension, they are seen as angles in radians (angles have no dimension).

A sample sequence doesn't save the actual analog signal frequency. Time information is lost and only retrievable via the sampling rate, a separate quantity. This is why the sampling rate must be known when the analog signal is reconstructed from the samples.

On the other hand, many operations have a result depending on the waveform but not on the analog frequency. For these operations, the normalized frequency is sufficient. Discrete Fourier transform (DFT) is one of such operations. Fast Fourier transform (FFT), the algorithm used to efficiently compute a DFT, manipulates only normalized frequencies.

Normalized frequency in Discrete Fourier Transform

How does the FFT deal with normalized frequency?

The sampling rate is not passed to the DFT function, it works only on the samples. So how can the function know the actual frequency of the signal? It doesn't!

The sequence of $N$ samples passed to DCT must represent a signal period, this is a condition. Saying $N$ samples are present in a signal period is equivalent to saying the normalized frequency of the samples is $\frac {1} {N}$. So analog frequency is ignored, but normalized frequency is not.

DFT works assuming $\frac {1} {N}$ is the normalized frequency, and computes spectral coefficients $X_k$ associated to normalized frequencies defined relatively to $N$.

Corresponding analog frequencies can be retrieved at any time by multiplying each DFT bin normalized frequency by the separate sampling rate.

Normalized frequency use cases

What is the significance of normalized frequency in DSP?

Time-less samples and normalized frequency are sufficient, and allow generalization to any frequency for various transformations of the signal:

  • When signal waveform is the relevant factor, e.g. in spectral analysis.
  • When relative bandwidth is the relevant factor, e.g. when designing filters.

Wrapping of normalized frequency at 1

Why is the limit of normalized frequency 2π?

Normalized radian frequency wraps at $2 \pi$. The duration of a sample, expressed as a fraction of a signal cycle is the inverse of $f'$. After $k$ samples, the samples repeat. Therefore they repeat after $k f' = \frac {1} {f'} f' = 1$ cycle of $x$ and $f'$ has limits $[0, 1]$ or $[-\frac {1} {2}, \frac {1} {2}]$.

A consequence is normalized radian frequency ($\hat \omega = 2 \pi \frac {f} {f_s}$) has limits $[0, 2 \pi]$ or $[-\pi, \pi]$.

  • $\begingroup$ This is interesting reading. I wouldn't say that "time disappears when sampling" but I like this answer. $\endgroup$ Jan 3, 2023 at 16:29

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