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:

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 $[-\pi, \pi]$.