实验名称：数字滤波器的频域分析和实现
实验目的和任务：
（1）求滤波器的幅度响应和相位响应，观察对称性，判断滤波器类型。
（2）用Matlab函数实现系统的级联型和并联型结构，并对滤波器进行结构仿真。
实验内容：
系统传递函数的级联和并联实现：Q6.1, Q6.3, Q6.5, Q8.3,Q8.5, 滤波器的幅频特性分析：Q8.1, Q8.9,Q8.10,Q8.14
实验过程与结果分析：
Q6.1 使用程序P6.1，生成如下有限冲激响应传输函数的一个级联实现：
H1(z)=2+10z
?1+23z
?2+34z
?3+31z
?4+16z
?5
编写的MATLAB程序：
% Program P6_1
num = input('Numerator coefficient vector = '); den = input('Denominator coefficient vector = '); [b,a] = eqtflength(num,den); % make lengths equal [z,p,k] = tf2zp(b,a); sos = zp2sos(z,p,k)
Numerator coefficient vector = [2 10 23 34 31 16 4] Denominator coefficient vector = [1] sos =
2.0000 6.0000 4.0000 1.0000 0 0 1.0000 1.0000 2.0000 1.0000 0 0 1.0000 1.0000 0.5000 1.0000 0 0 画出级联实现的框图：
H1(z)不是一个线性相位传输函数。
3+8z-1 +12z-2 +7z-3 +2z-4 -2z-5 Q6.3 使用程序P6.1，H1(z)= -1-2-3-4-516+24z +24z +14z +5z +1z
Numerator coefficient vector = [3 8 12 7 2 –2] Denominator coefficient vector = [16 24 24 14 5 1] sos =
0.1875 -0.0625 0 1.0000 0.5000 0 1.0000 2.0000 2.0000 1.0000 0.5000 0.2500 1.0000 1.0000 1.0000 1.0000 0.5000 0.5000 画出级联实现的框图：
3+8z-1 +12z-2 +7z-3 +2z-4 -2z-5 Q6.5 使用程序P6.2，式H1(z)=所示因果无
16+24z-1 +24z-2 +14z-3 +5z-4 +1z-5 限冲激响应传输函数的两种不同并联形式实现。画出两种实现的框图： Parallel Form I Residues are
-0.4219 + 0.6201i -0.4219 - 0.6201i 2.3437 0.3437 - 2.5079i 0.3437 + 2.5079i
Poles are at
-0.2500 + 0.6614i -0.2500 - 0.6614i -0.5000 -0.2500 + 0.4330i -0.2500 - 0.4330i
Constant value
-2
Parallel Form II Residues are
-0.3047 - 0.4341i -0.3047 + 0.4341i -1.1719 1.0000 + 0.7758i
1.0000 - 0.7758i
Poles are at
-0.2500 + 0.6614i -0.2500 - 0.6614i -0.5000 -0.2500 + 0.4330i -0.2500 - 0.4330i
Constant value 0.1875
Q8.1 程序P8.1设计的滤波器类型为 IIR带阻滤波器 ，其指标为
?p=[0.4? 0.5?]
?S=[0.1? 0.8?] , RP=1dB，RS=30dB，滤波器的阶数为 4 。为了验证仿真需计算的
冲激响应样本为 5个。仿真是正确的吗？
Q8.3 生成Q8.1中产生的传输函数的一个级联实现，并编写一个程序来仿真它。其中每个单独的部分用一个直接Ⅱ型实现。验证仿真。
format short
num = input('Numerator coefficients = '); den = input('Denominator coefficients = '); Numfactors = factorize(num);
Denfactors = factorize(den);
disp('Numerator Factors'),disp(Numfactors) disp('Denominator Factors'),disp(Denfactors)
Numerator coefficients = [0.0571 0 -0.1143 0 0.0571] Denominator coefficients = [1.0000 -0.5099 1.2862 -0.3350 0.4479] Numerator Factors Columns 1 through 3
1.000000000000000 1.021143238995107 0 1.000000000000000 0.979294541463240 0 1.000000000000000 -1.021143238995070 0 1.000000000000000 -0.979294541463277 0 Denominator Factors Columns 1 through 3
1.000000000000000 -0.597633166080704 0.678485370310279 1.000000000000000 0.087733166080704 0.660146879504817
Q8.5 生成Q8.1中传输函数的一个并联Ⅰ型实现，并编写一个程序来仿真它。每一个单独的部分用直接Ⅱ型实现。验证仿真。
num = input('Numerator coefficient vector = '); den = input('Denominator coefficient vector = '); [r1,p1,k1] = residuez(num,den); [r2,p2,k2] = residue(num,den); disp('Parallel Form I') disp('Residues are');disp(r1); disp('Poles are at');disp(p1); disp('Constant value');disp(k1); disp('Parallel Form II') disp('Residues are');disp(r2); disp('Poles are at');disp(p2);
disp('Constant value');disp(k2); [b1,a1]=residuez(R1,P1,0); R1=[r1(1) r1(2) r1(3) r1(4)]; P1=[p1(1) p1(2) p1(3) p1(4)]; disp('b1=');disp(b1); disp('a1=');disp(a1);
Numerator coefficient vector = [0.0571 0 -0.1143 Denominator coefficient vector = [1.0000 -0.5099 1.2862 ]
Parallel Form I Residues are
-0.023475623417519 + 0.197736845648730i -0.023475623417519 - 0.197736845648730i -0.011716283258078 - 0.213037612139943i -0.011716283258078 + 0.213037612139943i Poles are at
0.298816583040352 + 0.767589747202480i 0.298816583040352 - 0.767589747202480i -0.043866583040352 + 0.811309190381311i -0.043866583040352 - 0.811309190381311i
Constant value
0.127483813351194
Parallel Form II Residues are
-0.158795680938490 + 0.041067400713457i -0.158795680938490 - 0.041067400713457i 0.173353325938490 - 0.000160296180734i 0.173353325938490 + 0.000160296180734i
Poles are at
0.298816583040352 + 0.767589747202480i 0.298816583040352 - 0.767589747202480i -0.043866583040352 + 0.811309190381311i -0.043866583040352 - 0.811309190381311i
0 0.0571] -0.3350 0.4479