《DSP using MATLAB》Problem 5.22

代码:

%% ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
%%            Output Info about this m-file
fprintf('\n***********************************************************\n');
fprintf('        <DSP using MATLAB> Problem 5.22 \n\n');

banner();
%% ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

% -------------------------------------------------------------------
%          512-point DFT of a real-valued sequence x(n)     
%                                
% -------------------------------------------------------------------
 k = [0:511];
Xk_DFT = [zeros(1, 512)];
 
Xk_DFT(0+1) = 20;    Xk_DFT(5+1) = 20+j*30;    Xk_DFT(32+1) = -10+j*15;   Xk_DFT(152+1) = 17+j*23;
Xk_DFT(256+1) = 30;  Xk_DFT(360+1) = 17-j*23;  Xk_DFT(480+1) = -10-j*15;  Xk_DFT(507+1) = 20-j*30;

 N = length(Xk_DFT);


    magXk_DFT = abs( [ Xk_DFT ] );                                    % DFT magnitude
    angXk_DFT = angle( [Xk_DFT] )/pi;                                 % DFT angle
   realXk_DFT = real(Xk_DFT); 
   imagXk_DFT = imag(Xk_DFT);

EXk = sum(abs(Xk_DFT).^2)/N

 x = real(idft(Xk_DFT, N));
 n = [0:N-1];


figure('NumberTitle', 'off', 'Name', 'P5.22 X(k), DFT of x(n)')
set(gcf,'Color','white'); 
subplot(2,2,1); stem(k, magXk_DFT); 
xlabel('k'); ylabel('magnitude(k)');
title('magnitude DFT of x(n), N=512');  grid on;
subplot(2,2,3); stem(k, angXk_DFT);  
%axis([-N/2, N/2, -0.5, 50.5]);
xlabel('k'); ylabel('angle(k)');
title('angle DFT of x(n), N=512');  grid on;
subplot(2,2,2); stem(k, realXk_DFT); 
xlabel('k'); ylabel('real (k)');
title('real DFT of x(n), N=512');  grid on;
subplot(2,2,4); stem(k, imagXk_DFT);  
%axis([-N/2, N/2, -0.5, 50.5]);
xlabel('k'); ylabel('imag (k)');
title('imag DFT of x(n), N=512');  grid on;   


figure('NumberTitle', 'off', 'Name', 'P5.22 x(n)')
set(gcf,'Color','white'); 
%subplot(3,1,1); 
stem(n, x); 
xlabel('n'); ylabel('x(n)');
title('x(n), N=512');  grid on;

  运行结果:

       把alpha beta, k1、k2、k3求出来后,得知512点DFT有8个点不为零,如下图

        进行IDFT,得到时间域序列x(n)

        时间信号的能量

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转载自www.cnblogs.com/ky027wh-sx/p/9440211.html