系统辨识理论及应用(课后题答案第三章3.2、3.3)国防工业出版社

1、系统辨识——连续系统传递函数——脉冲传递函数function h=Continuous_system_transferFcn(N,G,dt)

% N——系统阶数

% G——采样数据（个数大于等于2N+1）

% G为一维行向量

% dt——采样间隔

if nargin<3

errordlg('not enough input varibles','error hint');

else

g_NN=zeros(N,N);

for i=1:N

g_NN(i,:)=G(i+1:i+1+N-1);

end

g_N=-G(1:N)';

a=inv(g_NN)*g_N;

%% x的求解

syms x

for i=1:N

X(i)=x^i;

end

f=X*a+1;

x=double(solve(f));

%%极点的求解

p=log(x)/dt;

c_NN=zeros(N,N);

for i=1:N

c_NN(i,:)=x.^(i-1);

end

c_N=G(1:N)';

%%增益求解

k=inv(c_NN)*c_N;

p

k

z=zeros(1,N);

p=p';

k=k';

Continuous_TransferFcn=0;

for i=1:N

Continuous_TransferFcn=Continuous_TransferFcn+zpk(z(i),p(i),k(i)); end

Continuous_TransferFcn

end

end

例题 3.1(P32)

>>G=[0 0.1924 0.2122 0.1762];

>> N=2;

>> dt=1;

>> Continuous_system_transferFcn(N,G,dt) p =

-0.4934

-0.7085

k =

1.6280

-1.6280

Continuous_TransferFcn =

0.35024 s

---------------------

(s+0.4934) (s+0.7085)

Continuous-time zero/pole/gain model.

习题3.2(P34)

>> G=[0 0.196 0.443 0.624 0.748 0.831]; >> N=3;

>> dt=0.2;

>> Continuous_system_transferFcn(N,G,dt) p =

-0.0633

-1.7846

-11.1860

k =

1.1249

-1.3399

0.2150

Continuous_TransferFcn =

-0.08507 s (s-253.1)

-------------------------------

(s+0.06329) (s+1.785) (s+11.19) Continuous-time zero/pole/gain model.

2 系统辨识——离散系统传递函数——脉冲传递函数

function h=Discrete_system_transferFcn(N,G,dt)

% N——系统阶数

% G——采样数据（个数大于等于2N+1）

% G为一维行向量

% dt——采样间隔

if nargin<3

errordlg('not enough input varibles','error hint');

else

g_NN=zeros(N,N);

for i=1:N

g_NN(i,:)=G(i+1:i+1+N-1);

end

g_N=-G(N+2:2*N+1)';

a1=inv(g_NN)*g_N;

a=zeros(N,1);

for j=1:N

a(j,1)=a1(N+1-j,1);

end

B=zeros(N+1,N+1);

B=diag(linspace(1,1,N+1));

for i=1:N+1

for j=1:N+1

if (i==j)&(i

B(i+1:N+1,j)=a(1:N+1-i,1);

end

end

end

g__N=G(1:N+1)';

b=B*g__N;

a

b

num=b';

den=[1 a'];

Discrete_TransferFcn=tf(num,den,dt);

Discrete_TransferFcn

end

end

例题 3.2(P33)

>> G=[0 7.157039 9.491077 8.563839 5.930506 2.845972 0.144611]; >> N=3;

>> dt=0.05;

>> Discrete_system_transferFcn(N,G,dt)

a =

-2.2300

1.7606

-0.4950

b =

7.1570

-6.4691

-0.0009

Discrete_TransferFcn =

7.157 z^2 - 6.469 z - 0.0008933

--------------------------------

z^3 - 2.23 z^2 + 1.761 z - 0.495

Sample time: 0.05 seconds

Discrete-time transfer function.

习题3.3(P34)

>> G=[10 6.989 4.711 3.136 2.137 1.559 1.252 1.096 0.938 0.860]; >> N=3;

>> dt=0.1;

>> Discrete_system_transferFcn(N,G,dt)

a =

-2.1919

1.7166

-0.4794

b =

10.0000

-14.9295

6.5581

0.0139

Discrete_TransferFcn =

10 z^3 - 14.93 z^2 + 6.558 z + 0.01389

--------------------------------------

z^3 - 2.192 z^2 + 1.717 z - 0.4794

Sample time: 0.1 seconds

Discrete-time transfer function.