代码:
%% ---------------------------------------------------------------------------- %% Output Info about this m-file fprintf(‘\n***********************************************************\n‘); fprintf(‘ <DSP using MATLAB> Problem 4.14 \n\n‘); banner(); %% ---------------------------------------------------------------------------- %format rat; %% -------------------------------------------------------------------- %% using the method on Page 110 to get the inverse-z transform %% -------------------------------------------------------------------- b = [-2, 5.65, -2.88]; a = [1, -0.1, 0.09, 0.648]; [R, p, C] = residuez(b, a) Mp = (abs(p))‘ % pole magnitudes Ap = (angle(p))‘/pi % pole angle in pi units [delta, n] = impseq(0, 0, 7); x_chk = filter(b, a, delta) figure(‘NumberTitle‘, ‘off‘, ‘Name‘, ‘Problem 4.14 X(z) pole-zero‘) set(gcf,‘Color‘,‘white‘); zplane(b, a); title(‘pole-zero plot‘); grid on; %% ----------------------------------------------------------------- %% Now, we use the invCCPP function to get x(n) sequence %% ----------------------------------------------------------------- [b11, a11] = residuez(R(1:2), p(1:2), C) b0 = b11(1); b1 = b11(2); % numerator coefficient a1 = a11(2); a2 = a11(3); % denumerator [As, Ac, r, v0] = invCCPP(b0, b1, a1, a2); %% ------------------------------------------------------------------------ %% x1(n)=Ac*(r^n)*cos(pi*v0*n)*u(n) + As*(r^n)*sin(pi*v0*n)*u(n) %% ------------------------------------------------------------------------ n_start = 0; n_end = 7; n = [n_start : n_end]; x1 = Ac * (r.^n) .* cos( pi * v0 .* n) .* stepseq(0, n_start, n_end) + As * (r.^n) .* sin(pi * v0 .* n ) .* stepseq(0, n_start, n_end); x_ori = x1 + R(3)*(p(3).^n) .* stepseq(0, n_start, n_end)
运行结果:
可见有一对复共轭极点和一个实数极点(-0.8),从零极点图也可看出:
计算一对复共轭极点相对应的有理分式的分子分母的系数b11和a11;这里只显示原始时间序列的前8个元素:
最后写出z变换表达式如下:
原文地址:https://www.cnblogs.com/ky027wh-sx/p/8463180.html
时间: 2024-11-04 08:40:45