找回密码
 注册
关于网站域名变更的通知
查看: 554|回复: 1
打印 上一主题 下一主题

MATLAB ------- 用 MATLAB 得到高密度谱和高分辨率谱的方式比对(附MATLAB脚本)

[复制链接]

该用户从未签到

跳转到指定楼层
1#
发表于 2019-12-9 11:11 | 只看该作者 |只看大图 回帖奖励 |倒序浏览 |阅读模式

EDA365欢迎您登录!

您需要 登录 才可以下载或查看,没有帐号?注册

x
MATLAB ------- 使用 MATLAB 得到高密度谱(补零得到DFT)和高分辨率谱(获得更多的数据得到DFT)的方式对比(附MATLAB脚本)/ [# K$ `& J, R! A& m& d# v, y

! D7 |& Z6 `" |- T. R
4 d- U5 {) ^3 R上篇分析了同一有限长序列在不同的N下的DFT之间的不同: MATLAB ------- 用 MATLAB 作图讨论有限长序列的 N 点 DFT(含MATLAB脚本)
; f) e' s4 q6 j* M那篇中,我们通过补零的方式来增加N,这样最后的结论是随着N的不断增大,我们只会得到DTFT上的更多的采样点,也就是说频率采样率增加了。通过补零,得到高密度谱(DFT),但不能得到高分辨率谱,因为补零并没有任何新的信息附加到这个信号上,要想得到高分辨率谱,我们就得通过获得更多的数据来进行求解DFT。
) |  O: N. U! D! D/ H7 ^6 m( u" R% D. O
这篇就是为此而写。1 K2 y% z" u$ b1 b
% C: u3 f3 d: B% Y! ?& m, P" O$ g
案例:
% v8 x' J; g8 q( f, t* V' q1 z$ g9 O  a1 E: `
9 e( {- d6 _' v9 S" j! h( m8 y
* G  m7 p( A) N. @1 l& T
想要基于有限样本数来确定他的频谱。2 y  g, E* Z0 X: u3 S: N

$ Y! J! `: {  J5 [  {下面我们分如下几种情况来分别讨论:
) [% V! y6 p  T# O" ~, H
3 v; U. ~% P6 j  g& c7 ba. 求出并画出   ,N = 10 的DFT以及DTFT;
$ @$ m6 z9 a+ h; o/ ~3 W7 @7 {6 `1 Z4 G
b. 对上一问的x(n)通过补零的方式获得区间[0,99]上的x(n),画出 N = 100点的DFT,并画出DTFT作为对比;% h% E) ^/ d* k& a9 G% q# @

# \, \* _8 J4 [2 s1 p3 G. jc.求出并画出   ,N = 100 的DFT以及DTFT;
+ m! j8 H) C7 C% X' p: K* Q+ s7 V. f1 N' R% T1 b
d.对c问中的x(n)补零到N = 500,画出 N = 500点的DFT,并画出DTFT作为对比;6 F! z  v+ l) s, S! M& b  N
8 |2 {1 H5 |& p+ ?) G+ ]* E- _
e. 比较c和d这两个序列的序列的DFT以及DTFT的异同。, O# i/ h) c8 f

' [, S& V6 Y, `8 F- K0 c& h6 J那就干呗!
* B, t" c( @0 a1 i. ~
" l. A- U: b9 R7 b+ z" m8 x" f题解:) M* X3 H  p6 D$ R8 M- g
" W. P- w( ~- }$ J
a.1 z" z0 _; c# ?4 X% j2 l$ F# P. Z
  s9 R# E8 x5 M7 Z% {
  • 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(2,1,1)
  • 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(2,1,2);
  • 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 off1 X3 v) u" o+ L' f! n2 D
      2 j% S! H0 x' |: n* {9 C
4 [+ f% D/ k) ?: S& j

. W  O; u6 M' Q% m( b& s' ]& ?% Kb.
0 _8 M* b2 @% U6 i0 n, N- x5 \
7 T, I# Z: c9 ~" z# z" }6 e+ Q
  • clc;clear;close all;
  • n = 0:99;
  • x = cos(0.48*pi*n) + cos(0.52*pi*n);
  • % zero padding into N = 100
  • n1 = 0:99;
  • y1 = [x(1:10),zeros(1,90)];
  • subplot(2,1,1)
  • stem(n1,y1);
  • title('signal x(n), 0 <= n <= 99');
  • xlabel('n');ylabel('x(n) over n in [0,99]');
  • Y1 = dft(y1,100);
  • magY1 = abs(Y1);
  • k1 = 0:1:99;
  • N = 100;
  • w1 = (2*pi/N)*k1;
  • subplot(2,1,2);
  • 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);
  • % w = [-fliplr(w),w(2:K+1)];   %plot DTFT in [-pi,pi]
  • % X = [fliplr(X),X(2:K+1)];    %plot DTFT in [-pi,pi]
  • magX = abs(X);
  • % angX = angle(X)*180/pi;
  • % figure
  • % subplot(2,1,1);
  • hold on
  • plot(w/pi,magX);
  • % title('Discrete-time Fourier Transform in Magnitude Part');
  • % xlabel('w in pi units');ylabel('Magnitude of X');
  • % subplot(2,1,2);
  • % plot(w/pi,angX);
  • % title('Discrete-time Fourier Transform in Phase Part');
  • % xlabel('w in pi units');ylabel('Phase of X ');
  • hold off

  • 7 l3 L0 D# F! R' w+ e! c0 M
      
, q& P. J) w  o8 x+ w 1 W; F3 T- O$ `; E

6 c# X$ k& d( k( ?0 S1 Z% l2 p
c.
0 G$ R8 c6 Z, t& y6 Z: O( Q3 x5 m  i# [5 a6 b3 i
  • clc;clear;close all;
  • n = 0:99;
  • x = cos(0.48*pi*n) + cos(0.52*pi*n);
  • % n1 = 0:99;
  • % y1 = [x(1:10),zeros(1,90)];
  • subplot(2,1,1)
  • 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(2,1,2);
  • 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);
  • % w = [-fliplr(w),w(2:K+1)];   %plot DTFT in [-pi,pi]
  • % X = [fliplr(X),X(2:K+1)];    %plot DTFT in [-pi,pi]
  • magX = abs(X);
  • % angX = angle(X)*180/pi;
  • % figure
  • % subplot(2,1,1);
  • hold on
  • plot(w/pi,magX);
  • % title('Discrete-time Fourier Transform in Magnitude Part');
  • % xlabel('w in pi units');ylabel('Magnitude of X');
  • % subplot(2,1,2);
  • % plot(w/pi,angX);
  • % title('Discrete-time Fourier Transform in Phase Part');
  • % xlabel('w in pi units');ylabel('Phase of X ');
  • hold off
  • ' j. i9 o6 c2 w. K0 o
      : A( o2 p$ a0 t( a( v0 u

0 c$ Z) X7 y3 o7 K% E1 J
8 A( a% C( g; _: W, Y$ P
4 }6 [$ S8 {3 O太小了,放大看:) L, a( K) U3 S: h3 }4 U

6 H1 S* ?" _; p; }* V4 M
, H" o* @3 x5 K# p! R% N$ ~: Y! z  J0 m/ r0 B& l$ U& D  Z7 v

! K# n/ C6 u+ c+ }1 @1 T
: P6 f5 l# l9 i  B; Z. dd." `2 C' v! u7 M& _% P
' g# Y  K5 l2 y9 c+ p, a
  • clc;clear;close all;
  • n = 0:99;
  • x = cos(0.48*pi*n) + cos(0.52*pi*n);
  • % n1 = 0:99;
  • % y1 = [x(1:10),zeros(1,90)];
  • %zero padding into N = 500
  • n1 = 0:499;
  • x1 = [x,zeros(1,400)];
  • subplot(2,1,1)
  • stem(n1,x1);
  • title('signal x(n), 0 <= n <= 499');
  • xlabel('n');ylabel('x(n) over n in [0,499]');
  • Xk = dft(x1,500);
  • magXk = abs(Xk);
  • k1 = 0:1:499;
  • N = 500;
  • w1 = (2*pi/N)*k1;
  • subplot(2,1,2);
  • % stem(w1/pi,magXk);
  • stem(w1/pi,magXk);
  • title('DFT of x(n) in [0,499]');
  • 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 = x1*exp(-j*n1'*w);
  • % w = [-fliplr(w),w(2:K+1)];   %plot DTFT in [-pi,pi]
  • % X = [fliplr(X),X(2:K+1)];    %plot DTFT in [-pi,pi]
  • magX = abs(X);
  • % angX = angle(X)*180/pi;
  • % figure
  • % subplot(2,1,1);
  • hold on
  • plot(w/pi,magX,'r');
  • % title('Discrete-time Fourier Transform in Magnitude Part');
  • % xlabel('w in pi units');ylabel('Magnitude of X');
  • % subplot(2,1,2);
  • % plot(w/pi,angX);
  • % title('Discrete-time Fourier Transform in Phase Part');
  • % xlabel('w in pi units');ylabel('Phase of X ');
  • hold off

  • . b$ ?" ?: O8 m3 T
       ( M& s* `6 W* k0 W9 H) `* l

# I, E- A$ ^+ y: f& d, U! G, _0 d" J
$ Y7 l/ d4 o. D8 ^1 N: e7 L
, I8 I4 r6 s  K* @e.9 h/ K  j1 y4 |/ y

2 v1 F) `) f: N' m! u
  • clc;clear;close all;
  • n = 0:99;
  • x = cos(0.48*pi*n) + cos(0.52*pi*n);
  • subplot(2,1,1)
  • % 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;
  • 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);
  • % w = [-fliplr(w),w(2:K+1)];   %plot DTFT in [-pi,pi]
  • % X = [fliplr(X),X(2:K+1)];    %plot DTFT in [-pi,pi]
  • magX = abs(X);
  • % angX = angle(X)*180/pi;
  • % figure
  • % subplot(2,1,1);
  • hold on
  • plot(w/pi,magX);
  • % title('Discrete-time Fourier Transform in Magnitude Part');
  • % xlabel('w in pi units');ylabel('Magnitude of X');
  • % subplot(2,1,2);
  • % plot(w/pi,angX);
  • % title('Discrete-time Fourier Transform in Phase Part');
  • % xlabel('w in pi units');ylabel('Phase of X ');
  • hold off
  • % clc;clear;close all;
  • %
  • % n = 0:99;
  • % x = cos(0.48*pi*n) + cos(0.52*pi*n);
  • % n1 = 0:99;
  • % y1 = [x(1:10),zeros(1,90)];
  • %zero padding into N = 500
  • n1 = 0:499;
  • x1 = [x,zeros(1,400)];
  • subplot(2,1,2);
  • % subplot(2,1,1)
  • % stem(n1,x1);
  • % title('signal x(n), 0 <= n <= 499');
  • % xlabel('n');ylabel('x(n) over n in [0,499]');
  • Xk = dft(x1,500);
  • magXk = abs(Xk);
  • k1 = 0:1:499;
  • N = 500;
  • w1 = (2*pi/N)*k1;
  • stem(w1/pi,magXk);
  • title('DFT of x(n) in [0,499]');
  • 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 = x1*exp(-j*n1'*w);
  • % w = [-fliplr(w),w(2:K+1)];   %plot DTFT in [-pi,pi]
  • % X = [fliplr(X),X(2:K+1)];    %plot DTFT in [-pi,pi]
  • magX = abs(X);
  • % angX = angle(X)*180/pi;
  • % figure
  • % subplot(2,1,1);
  • hold on
  • plot(w/pi,magX,'r');
  • % title('Discrete-time Fourier Transform in Magnitude Part');
  • % xlabel('w in pi units');ylabel('Magnitude of X');
  • % subplot(2,1,2);
  • % plot(w/pi,angX);
  • % title('Discrete-time Fourier Transform in Phase Part');
  • % xlabel('w in pi units');ylabel('Phase of X ');
  • hold off- n! I8 r" }2 u2 I
     : O# b- F- d$ O
) U! [: Y5 O1 P% P5 y9 [! ]

- N9 A1 d; b7 F% U* }" T0 _9 a! W5 n* d7 N% U: n2 T
局部放大看:3 I7 I, z$ i+ j' a- w3 D

( Z. e$ E8 u4 K+ V+ z! f6 [7 _+ b 3 S( b: u/ Q7 ^7 P& t6 }

* j4 r' G8 h9 p! H4 ] ( n/ S) f- c! K+ }- X2 Z% C

9 T. \' L2 D  x" j
4 W6 r1 V, d2 e: h5 L6 E8 m
您需要登录后才可以回帖 登录 | 注册

本版积分规则

关闭

推荐内容上一条 /1 下一条

EDA365公众号

关于我们|手机版|EDA365电子论坛网 ( 粤ICP备18020198号-1 )

GMT+8, 2025-8-11 21:08 , Processed in 0.125000 second(s), 26 queries , Gzip On.

深圳市墨知创新科技有限公司

地址:深圳市南山区科技生态园2栋A座805 电话:19926409050

快速回复 返回顶部 返回列表