How to use csdn in C for the Z transformed filter

The Z-transformed filter can be implemented in C in the form of a digital filter. The specific implementation method is as follows:1

Firstly, the z-transform is used to convert the difference equation in the time domain into a transfer function in the frequency domain. For example, the difference equation of a second-order low-pass filter is converted into a transfer function in the z-domain: h(z) = (b0 + b1*z -1 + b2*z -2) (1 + a1*z -1 + a2*z -2)2

The transfer function is then converted into the form of a digital filter. 3.In C, a digital filter can be implemented in the form of a difference equation.

For example, for the second-order low-pass filter described above: float xn; input signal samplefloat yn; output signal sample float xn 1 = 0; sample float xn 2 = 0 at the previous moment of the input signal; Sample float yn 1 = 0 at the last time of the input signal; sample float yn 2 = 0 at the previous moment of the output signal; Sample float b0 = ; Filter coefficient float b1 = ; float b2 = ;float a1 = ;float a2 = ;for(int i = 0; i \u003c n;i++) where n is the number of sampling points of the signal, xn, yn, xn 1, xn 2, yn 1, yn 2 all represent the signal samples and output signal samples at different times, and b0, b1, b2, a1, and a2 are the coefficients of the filter. By converting the difference equation into the form of a digital filter, a Z-transformed filter can be implemented in C.

h(z)=bz 1-az can be judged from a, b before bai feed and after du feed coefficient, of course, zhi can be drawn as straight dao2 type can also be drawn as cascade type, if you want to know the filter high pass and low pass, look at the amplitude frequency response.

Find the frequency properties of the transfer function, which is the Bode diagram. The main thing is the amplitude and frequency characteristics, and the phase frequency characteristics are generally not required. Look at the change in gain and cutoff frequency in the frequency domain, and then divide it into low-pass and high-pass these.

Looking at the numerator in the second order, first replace the formula with the jw form of the time domain, the denominator must be a quadratic formula, and then look at the numerator, the numerator is in the square form of w, that is, it is high-pass, because the higher the w, the larger the modulus of the fraction, the numerator is w, the first formula is bandpass, and the numerator without w is low-pass, because the larger the w, the modulus of the denominator must increase, and the value of the whole fraction decreases.

h(z)=bz 1-az can be used to judge the feedforward and backward coefficients from a and b, of course, it can be drawn as a straight 2 type or as a cascade type, is this what you mean by the type?

If you want to know the filter high-pass and low-pass, just look at the amplitude-frequency response.

In the z-transform, the position of the zero copy point represents the system's

Valley", the position of the pole represents the "peak" of the system, we see the place where there is a peak as the place where the signal can pass through, and the place where there is a valley as the place where the signal is cut off. And if we choose the unit circle as a period in the frequency domain, then we can conclude that if there is no zero point, the pole is high pass on the left half of the imaginary axis, and the pole is low pass on the right side of the imaginary axis; If there is no pole, and the zero point on the left side of the imaginary axis is low-pass, and on the right side of the imaginary axis is high-pass; If there are both zeros and poles, and the modulus of the vector pointing to the unit circle with the zero pointing to the unit circle divided by the modulus pointing to the unit circle, for a first-order system, the closer the pole and the zero point are, the greater the bandwidth.

The above system has a zero point of 1 and a pole, so when starting from r=1, the modulus from the zero point to the unit circle of 0 radians is zero, so it is high pass, and the -10 hour width of this problem.

The zero pole distribution form in the filter determines the amplitude and frequency characteristics of the system, the zero point controls the trough of the amplitude-frequency curve, the pole controls the peak of the amplitude-frequency curve, and the response curve of the transverse axis frequency normalization and non-normalization in the amplitude-frequency response refers to the digital filter, which has no effect, just to better correspond to the frequency.

This professional can find the answer for you, I don't know.

The second problem is that it can be converted, and you look at the textbook of Xi'an Electronics, which is more detailed.

Determine the filter type: use the Fourier transform to find h(f), and draw the amplitude-frequency characteristic curve to see if the high-frequency part is "through".

There are 0 and 1 times of s in the denominator, and the numerator is 1 order of s, so it is the higher one, referred to as "higher order".

H(s)=A (BS+C) has a "low order" on the molecule, so it is a low pass.

H(s)=AS2 (BS2+CS+D) has a "higher order" on the molecule, so it is Qualcomm.

H(S)=A (BS2+CS+D) has a "low order" on the molecule, so it is lowpass.

H(s)=AS (BS2+CS+D) has an "intermediate order" on the molecule, so it is bandpass.

Main Categories:

According to the signal processed, it is divided into two types: analog filter and digital filter.

Low-pass filter: It allows the passage of low-frequency or DC components in the signal, suppressing high-frequency components or interference and noise;

High-pass filter: it allows the passage of high-frequency components in the signal, suppressing low-frequency or DC components;

Bandpass filter: It allows signals in a certain frequency band to pass through, suppressing signals, interference, and noise below or above that frequency band;

Let's go white, your tracing is unknown...

Quickly determine the filter type based on the system function:

1. The dead method is to use the Fourier transform to find h(f), and draw the amplitude-frequency characteristic curve to see if the high-frequency part is "through".

2. Use the Rass transform to find h(s), and then remember the sentence: what is on the molecule is what is passed.

The unit impulse response h(n) can be used to represent a linear time-invariant discrete system.

In this case, y(n)=x(n)*h(n) takes the z transformation on both sides: y(z)=x(z)h(z) is defined as a system function. It is the z-transform of the unit impulse response.

The system function z=e on the unit circle is the frequency response of the system. Therefore, the z-transform of the unit impulse response can be used to describe the linear time-invariant discrete system.

For example, h(s)=as (bs+c) has a "higher order" on the molecule, so it is high-pass. The "higher order" here means this: there are 0 and 1 times of s in the denominator, and the numerator is 1 order of s, so it is the higher one, referred to as "higher order".

H(s)=A (BS+C) has a "low order" on the molecule, so it is a low pass.

H(s)=AS2 (BS2+CS+D) has a "higher order" on the molecule, so it is Qualcomm.

H(S)=A (BS2+CS+D) has a "low order" on the molecule, so it is lowpass.

H(s)=AS (BS2+CS+D) has an "intermediate order" on the molecule, so it is bandpass.

Isn't this question too broad ...

The pole is related to the high and low pass, you can think of it, the high pass has a large response to the amplitude of the high frequency, that is, the pole is close to the imaginary axis (because the pole is on the denominator of the filter z). In the same way, if the pole is close to the real axis, it is a low pass.

The poles of the z-transform are within the unit circle. The relationship with high-pass and low-pass can be converted to Rastalk transform first, and then bring in a few special points, such as special points on the real axis and the imaginary axis, observe the function value of the filter function at this time, and judge the high-pass, low-pass, or band-pass, band-pass, and band resistance, I don't know if it can help you.

It cannot be said that there is an obvious relationship between the zero pole and the high and low pass, but generally we derive the frequency response of the system in an appropriate way, and then according to the requirements of the stability of the system (this is related to the zero pole), and finally deduce the high and low pass characteristics of the system.

And the normalized frequency:

For a low-pass filter with a cut-off frequency of a certain Wc, let s wc replace S in the normalized prototype filter system, i.e., .

s-->s/wc）

For the high-pass filter, the band transformation method can be used, and the normalized prototype filter is obtained by frequency band transformation.