找回密码
 注册
关于网站域名变更的通知
查看: 569|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)
' w  `9 h6 B* m4 M! {) L3 J/ `% `2 s; i5 Z
* U- a9 a$ }# b( M& ?7 Y6 l
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)
! a1 Y- M$ P6 b" j) ~$ i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。# ]  H4 K. i+ ]% N2 |7 D- i

7 l6 ?" s# H* C8 Z这篇就是为此而写。
# a4 ?: o, u4 ]! g% v9 K; f# E  v0 u) Q
案例:
6 y7 w* A3 U; Y# n; B/ i- ^& y% I
, l' j/ Q+ Z0 {
% g: ], d9 |* q2 M
4 @  R1 e8 g- }( r2 T想要基于有限样本数来确定他的频谱。0 X7 q. K7 V/ y% R0 ?
' q6 J  o& i& J- O- V) m
下面我们分如下几种情况来分别讨论:
7 \- K* q7 k& v$ c% Q0 S. {7 {" _$ L& v/ R5 k8 Y. w
a. 求出并画出   ,N = 10 的DFT以及DTFT;
7 s) U% A$ `9 _; v2 ~7 H4 J; K/ a% x
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;$ k/ m. t  V, f( Y  n1 @
* h$ g! }5 }8 f1 F
c.求出并画出   ,N = 100 的DFT以及DTFT;
0 q/ p8 \' |% a: T; o) e+ d8 A$ o. T' G. j, ^
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;
! V$ {+ q! e) @8 b9 Z3 q  |! W% c! m8 {0 ?/ ^3 f
e. 比较c和d这两个序列的序列的DFT以及DTFT的异同。; S% a4 ]8 x6 @" |9 Y  e" R* o& ?6 B7 [
7 K) h+ n2 |. y
那就干呗!
% X( s8 d; R2 p& b; r; Q$ M3 l/ |; K1 d$ n0 o4 ?* L; q
题解:# w; W& t6 h% W2 z  a
( v0 ^4 S- G5 m! ]" h5 z8 T" v
a.
% T* i% `/ @2 i
) y  P7 U7 r5 X; ]& b4 ?% i
  • 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 off
    * N( w4 l  Y8 \5 L7 [
      8 y9 |" T9 ~. B3 n; O

! V! L: R, t1 z$ J/ K9 z) k
7 l3 U( b0 W& t8 W* h* _( [b.
0 k, Z( J, |( X8 [* A) Z9 a: O+ M. O2 s+ V
  • 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

  • 4 F; B+ o' A- D3 I: r
      
+ v  m9 X; `8 A; }, m/ r, G/ c
0 D4 I; T5 e% H9 ~: V . M. Z6 T4 [8 K3 g  B9 L" P$ p

# B, I% K  _0 Kc.
" S( T9 e& j% V: u9 l1 ~# p/ L
0 h0 g& d5 ^, k4 d4 h0 A4 `
  • 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
  • 1 r2 \* M. K) v
      * b( S( g  z9 q4 r- ]
6 F/ G* n7 c6 p# T0 p& O( @( ?% e

8 r; K4 e7 r2 z1 O: p: @* I+ H, o' a% o  e
太小了,放大看:2 Z5 i9 o2 l4 B" G* V& ~. w$ t

; u1 Z7 U2 n7 n0 F# x
3 ~& ]* c: D+ |) ?* @4 E0 ^9 E! g: W
5 K9 R9 O- I' r1 p, S' @. [0 F; A; @
0 L# i! J% P1 _. K! n; B1 H
d.$ w4 H0 D, @$ ^' m
. M0 m5 O  C# Y  B7 E
  • 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
  • 7 v! j9 H# j5 N/ \
      
# E) T& s6 x2 ~7 B1 U ( |. O9 J2 t9 C, a

! I- v2 {* F" y0 O7 ~+ ^4 _/ N: S/ M7 @5 U& @
e.
5 e7 X! H' ^* f% e! L
" S/ `* Y, _& ], y: h. h' A
  • 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 off9 m& o( p' Z5 `/ N0 W3 V
     
; u/ J9 R" G$ L" o$ s# c: g ! o! a7 K5 g7 O6 h0 b; S
5 V1 V( ?" u) _, }) G7 v8 ]' }
! q% y- m0 ]. J5 {( C: ^2 i
局部放大看:9 S" u9 K6 d% w) q( q! B/ Q

3 M0 [" m9 Q) ]6 _: M + y) n+ S* |" R- t6 q" E1 A
! O4 m+ w1 K; L& V. y* L

1 E8 Q) [2 n1 R0 B& `
% k7 i: s4 F# Z/ S5 b( E! ?  X! j5 y3 r

! f; l/ A3 H5 u4 E) h
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, 2025-10-5 19:26 , Processed in 0.125000 second(s), 27 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