MATLAB ------- 用不同样本数的同一个有限长序列作 DTFT 比对

ulppknot · 发表于 2019-12-9 11:16

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上篇我们讨论了：MATLAB ------- 用 MATLAB 得到高密度谱和高分辨率谱的方式比对（附MATLAB脚本）

可是还是觉得不过瘾，还有下面的情况需要比对。于是就有了这篇。

案例：

想要基于有限样本数来确定他的频谱。

下面我们分如下几种情况来分别讨论：

a. 求出并画出

的DTFT；

b. 求出并画出

的DTFT；

clc;clear;close all;
n = 0:99;
x = cos(0.48*pi*n) + cos(0.52*pi*n);
n1 = 0:9;
y1 = x(1:10);
subplot(2,2,1)
stem(n1,y1);
title('signal x(n), 0 <= n <= 9');
xlabel('n');ylabel('x(n) over n in [0,9]');
Y1 = dft(y1,10);
magY1 = abs(Y1);
k1 = 0:1:9;
N = 10;
w1 = (2*pi/N)*k1;
subplot(2,2,2);
% stem(w1/pi,magY1);
% title('DFT of x(n) in [0,9]');
% xlabel('frequency in pi units');
%In order to clearly see the relationship between DTFT and DFT, we draw DTFT on the same picture.
%Discrete-time Fourier Transform
K = 500;
k = 0:1:K;
w = 2*pi*k/K; %plot DTFT in [0,2pi];
X = y1*exp(-j*n1'*w);
magX = abs(X);
% hold on
plot(w/pi,magX);
% hold off
subplot(2,2,3)
stem(n,x);
title('signal x(n), 0 <= n <= 99');
xlabel('n');ylabel('x(n) over n in [0,99]');
Xk = dft(x,100);
magXk = abs(Xk);
k1 = 0:1:99;
N = 100;
w1 = (2*pi/N)*k1;
subplot(2,2,4);
% stem(w1/pi,magXk);
% title('DFT of x(n) in [0,99]');
% xlabel('frequency in pi units');
%In order to clearly see the relationship between DTFT and DFT, we draw DTFT on the same picture.
%Discrete-time Fourier Transform
K = 500;
k = 0:1:K;
w = 2*pi*k/K; %plot DTFT in [0,2pi];
X = x*exp(-j*n'*w);
magX = abs(X);
hold on
plot(w/pi,magX);
hold off
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