# Catalogue of series articles

Digital signal processing (DSP) is a very important research direction in the field of electronic communication. The blogger summarized the classic case analysis commonly used in digital signal processing (DSP), and analyzed the practical application of digital signal processing in detail based on algorithm analysis, MATLAB program implementation and signal image display.The first part is the MATLAB generation and graphic display of common discrete signals

MATLAB generation and graphic display of common discrete signals

Part II influence of zero pole distribution on system frequency response

Influence of zero pole distribution on system frequency response

# 1, Research purpose

1. Deepen the understanding of frequency response analysis of discrete systems and the concept of zero pole distribution.

2. Learn to use the geometric method of zero pole distribution to analyze and study the frequency response of the system.

# 2, Principle analysis

If the system function of the system is known, its zero pole distribution can be obtained. From the zero pole distribution, it is convenient to qualitatively analyze the frequency response of the system.

According to formulas (2.6.8) and (2.6.9) in the textbook, the amplitude characteristic of the system is divided by the product of the zero vector length divided by the product of the pole vector length. When the frequency changes from, observe the changes of the zero vector length and the pole vector length, focusing on those with short vector length. In addition, the analysis shows that the pole mainly affects the peak value of frequency response. The closer the pole is to the unit circle, the sharper the peak value is; The zero point mainly affects the valley value of frequency response. The closer the zero point is to the unit circle, the deeper the valley value is. If the zero point is on the unit circle, the frequency characteristic of the frequency point is zero. According to these laws, the amplitude characteristics of frequency response can be qualitatively drawn.

The peak frequency and valley frequency can be approximately expressed by the phase angle between the pole and zero of the response. For example, the peak frequency is approximately, the closer the pole is to the unit circle, the more accurate the estimation result is.

In this experiment, the frequency response of the system is analyzed by computer. The purpose is to master the geometric method of zero pole distribution to analyze and study the frequency response of the system. In the experiment, it is necessary to substitute it into the system function, select several points at equal intervals, and calculate its frequency response.

# 3, Experimental content

## 1. Situation I

It is assumed that the system is described by the following difference equation:

Assuming a = 0.7, a = 0.8 and a = 0.9, analyze the frequency characteristics of the system in three cases respectively, and print the amplitude characteristic curve.

## 2. Situation II

It is assumed that the system is described by the following difference equation:

Assuming a = 0.7, a = 0.8 and a = 0.9, analyze the frequency characteristics of the system in three cases respectively, and print the amplitude characteristic curve.

## 3. Case III

It is assumed that the system is described by the following difference equation:

Try to analyze its frequency characteristics, print its amplitude characteristic curve, calculate the peak frequency and valley frequency, and compare it with the phase angle of zero and pole.

# 4, Experimental code (matlab)

## 1. Situation I

%%3(1) B=[1]; a=0.7; A=[1,-a]; [H,w]=freqz(B,A,512,'whole'); figure; subplot(1,3,1); zplane(B,A); xlabel ('Re(z)'); ylabel('Im(z)');title('Pole-zero plot '); subplot(1,3,2); plot(w,abs(H)); xlabel ('w'); ylabel('|H(jw)|'); title('Amplitude frequency characteristic curve'); subplot(1,3,3); plot(w,angle(H)); xlabel ('w'); ylabel ('φ(jw)'); title('Phase frequency characteristic curve'); B=[1]; a=0.8; A=[1,-a]; [H,w]=freqz(B,A,512,'whole'); figure(2); subplot(1,3,1); zplane(B,A); xlabel ('Re(z)'); ylabel('Im(z)');title('Pole-zero plot '); subplot(1,3,2); plot(w,abs(H)); xlabel ('w'); ylabel('|H(jw)|'); title('Amplitude frequency characteristic curve'); subplot(1,3,3); plot(w,angle(H)); xlabel ('w'); ylabel ('φ(jw)'); title('Phase frequency characteristic curve'); B=[1]; a=0.9; A=[1,-a]; [H,w]=freqz(B,A,512,'whole'); figure(3); subplot(1,3,1); zplane(B,A); xlabel ('Re(z)'); ylabel('Im(z)');title('Pole-zero plot '); subplot(1,3,2); plot(w,abs(H)); xlabel ('w'); ylabel('|H(jw)|'); title('Amplitude frequency characteristic curve'); subplot(1,3,3); plot(w,angle(H)); xlabel ('w'); ylabel ('φ(jw)'); title('Phase frequency characteristic curve');

## 2. Situation II

%%3(2) a=0.7; B=[1,a]; A=[1]; [H,w]=freqz(B,A,512,'whole'); figure; subplot(1,3,1); zplane(B,A); xlabel ('Re(z)'); ylabel('Im(z)');title('Pole-zero plot '); subplot(1,3,2); plot(w,abs(H)); xlabel ('w'); ylabel('|H(jw)|'); title('Amplitude frequency characteristic curve'); subplot(1,3,3); plot(w,angle(H)); xlabel ('w'); ylabel ('φ(jw)'); title('Phase frequency characteristic curve'); a=0.8; B=[1,a]; A=[1]; [H,w]=freqz(B,A,512,'whole'); figure(2); subplot(1,3,1); zplane(B,A); xlabel ('Re(z)'); ylabel('Im(z)');title('Pole-zero plot '); subplot(1,3,2); plot(w,abs(H)); xlabel ('w'); ylabel('|H(jw)|'); title('Amplitude frequency characteristic curve'); subplot(1,3,3); plot(w,angle(H)); xlabel ('w'); ylabel ('φ(jw)'); title('Phase frequency characteristic curve'); a=0.9; B=[1,a]; A=[1]; [H,w]=freqz(B,A,512,'whole'); figure(3); subplot(1,3,1); zplane(B,A); xlabel ('Re(z)'); ylabel('Im(z)');title('Pole-zero plot '); subplot(1,3,2); plot(w,abs(H)); xlabel ('w'); ylabel('|H(jw)|'); title('Amplitude frequency characteristic curve'); subplot(1,3,3); plot(w,angle(H)); xlabel ('w'); ylabel ('φ(jw)'); title('Phase frequency characteristic curve');

## 3. Case III

%%3(3) B=[1,1]; A=[1,-1.273,0.81]; [H,w]=freqz(B,A,512,'whole'); figure; subplot(1,3,1); zplane(B,A); xlabel ('Re(z)'); ylabel('Im(z)');title('Pole-zero plot '); subplot(1,3,2); y=abs(H); plot(w,y); xlabel ('w'); ylabel('|H(jw)|'); title('Amplitude frequency characteristic curve'); subplot(1,3,3); plot(w,angle(H)); xlabel ('w'); ylabel ('φ(jw)'); title('Phase frequency characteristic curve'); f=find(y==max(y)); w(f)%Get the frequency point corresponding to the maximum amplitude (i.e. peak frequency) f=find(y==min(y)); w(f)%Get the frequency point corresponding to the minimum amplitude (i.e. Valley frequency)

# 5, Result image

## 1. Situation I

a=0.7

a=0.8

a=0.9

## 2. Situation II

a=0.7

a=0.8

a=0.9