Various frequency relations in digital signal processing

4 frequencies and their number relationships

The actual physical frequency represents the frequency of the ad acquisition of physical signals, FS is the sampling frequency, by the Nyquist sampling theorem can be known, FS must ≥ the signal maximum frequency of twice times before the signal aliasing, so FS can sample the highest frequency of the signal is FS/2.

The angular frequency is 2*pi times the physical frequency, which is also called the analog frequency.

Normalized frequency is the result of the physical frequency by FS normalization, the highest signal frequency of FS/2 corresponds to the normalized frequency 0.5, which is why in the MATLAB fdtool tool normalized frequency why the maximum of only 0.5 reasons.

The circumference frequency is 2*pi times the normalized frequency, which is also called the digital frequency.

The analysis of FFT frequency and actual physical frequency

Do a n-point FFT, indicating that the original signal in the time domain took n points to do spectral analysis, N-point FFT transformation results are still n points.

In other words, the 2PI digital frequency w is divided into n parts, and the range of the entire digital frequency w covers the analog frequency range from the 0-2PI*FS. The FS here is the sampling frequency . And we usually only care about the spectrum in 0-pi, because according to Naikost law, only the signal within the F=FS/2 range is the valid signal that is sampled. So, in the range of W, the resulting spectrum is definitely about N/2 symmetry.

For example, if you do a 16-point FFT analysis, the maximum frequency of your original analog signal f=32khz, the sampling frequency is 64khz,n range is 0,1,2...15. At this point, the 64kHz analog frequency is divided into 16 points, each part is 4kHz, this is called frequency resolution . So in the horizontal axis, the corresponding F is 4kHz n=1, n=2 corresponds to 8kHz, n=15 corresponds to 60kHz, your spectrum is about n=8 symmetry. You only need to care about n=0 to 7 of the spectrum is enough, because the original signal is the highest analog frequency is 32kHz.

There are two conclusions here.

• First, we must know the original signal sampling frequency FS is how much, we can know the actual frequency of each n corresponds to the actual frequency of the K point is calculated as F (k) =k* (fs/n)
• Second, you 64kHz do 16 points after the FFT, because the frequency resolution is 4kHz, if the original signal in 5kHz or 63kHz has a component, you are invisible in the spectrum, this means that the more you want the spectrum to be realistic, you have to take more points to do fft,n is bigger, You have to take a longer signal sample to do the analysis in the time domain. However, due to the principle of discrete sampling, you can not completely accurately draw the original continuous time signal of the real spectrum, only infinitely close (that is, n infinity), this is called the frequency leakage. In the case of sampling frequency FS, the frequency leakage can be improved by taking more points, or by adding windows to the pre-FFT, which is another topic.

Periodic discussion on Fourier transform of discrete signal

To analyze this, we start with the relationship between the Laplace transformation and the Z-transform.

The graph of Z-plane and S-plane

The relationship in the diagram has the following points:

• The imaginary axis of the s plane is mapped to the unit circle on the Z-plane
• The negative half axis of the s plane is mapped to the unit circle in the Z-plane
• The positive half axis of the s plane is mapped to the unit circle of the Z-plane

The Laplace transform is a transform used for continuous signals, and the corresponding Z-transform is a transformation applied to the Z-plane. So from another angle, the angular frequency (analog frequency) mentioned above corresponds to the S-plane, and the circumferential frequency corresponds to the Z-plane (and the reason why it is called the circumferential frequency).

Now let's take a look at the transformation of the analog frequency on the S-plane virtual axis, which will result in how the Z-Plane unit circle changes:

• When the analog frequency changes from 0 to FS on the imaginary axis of the s plane, the digital frequency changes from 0 to 2 pi on the z-plane unit circle.
• When the analog frequency changes from 2FS to 4fs on the imaginary axis of the s plane, the digital frequency still changes from 0 to 2 pi on the z-plane unit circle.
• 。。。。。。 The z-plane repeats like this.

We know that the Fourier transform of the discrete signal corresponds to the z-transform on the unit circle , so the above conclusion verifies why the Fourier transform of the discrete signal is cyclical: the root cause is the periodicity on the unit circle.

Considering that we can choose a cycle in the actual application, this can also explain: because the actual signal frequency is always below FS/2, this corresponds to the Z Plane unit circle on the 0~pi, in a period of time can be the signal analysis.

In the spectrum analysis of the continuous signal, its spectrum function of the independent variable with the analog frequency (some book with symbols), and the discrete signal spectrum analysis, its spectrum function variables with the digital frequency (some book with symbols), here the analog frequency and digital frequency is and (some books and), in addition, here TS is the sampling interval of the signal , and its relationship to the sampling frequency is:. Since the spectrum of the sampled signal is a periodic function, its period is the sampling frequency (or the analog angular frequency), so the period of the digital frequency used in the spectrum analysis of the discrete signal is.

Li Zeguang
Links: https://www.zhihu.com/question/29863278/answer/86161994
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Various frequency relations in digital signal processing