找回密码
 注册
关于网站域名变更的通知
查看: 588|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 |正序浏览 |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)
3 t; `& }" u. o$ z5 `$ R  K5 d) r+ \8 f1 m/ m  q/ d: T

, {2 F; D0 u/ ~% M* N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)
+ P" K, g5 i: N! e9 K/ H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。
: Z) U% L* d3 T9 _* I8 F; j% o+ e: N, }8 G+ {+ W; m9 o7 W8 ?
这篇就是为此而写。0 x& @4 _2 p% ~9 {
0 T4 q8 p, ]  x( Y; U# K0 c* L
案例:
+ I. W$ I6 ?: ]5 W) l: y  C7 X, `
4 K) u& C" z+ f" O3 V   E5 N4 C, B: }8 t- G9 a! i
- N. u% U9 C; V
想要基于有限样本数来确定他的频谱。
- {! }5 N: J: V& n
# P& a* Q* D! H+ @5 d; s下面我们分如下几种情况来分别讨论:/ N0 }: Y3 y3 b7 T% y8 ~. h& O

8 u+ z3 q" g6 H: U% }! Ka. 求出并画出   ,N = 10 的DFT以及DTFT;( l2 S$ {/ n% e. ]  q  d* b

2 b( c- N' X; ^' v. L5 ]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;! V, S2 w: _( ]* ~* ~0 T2 |

! B7 A8 a! K) A+ W' Mc.求出并画出   ,N = 100 的DFT以及DTFT;
6 U2 l% [: O  z: r0 \8 y* p5 Q
5 p& b$ j4 Y$ |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;3 q: h3 C$ X2 b/ Z# q
5 a* C$ `& k- r; f. I+ T: e
e. 比较c和d这两个序列的序列的DFT以及DTFT的异同。+ E. @: s5 h1 Q( i$ t

7 i" R- M& f: V' N) @+ R那就干呗!; n/ ^5 o4 W, M+ P* e
( F6 }7 \/ ^" k
题解:
- R- s& J7 E0 t# l* e: I  n4 ?; X- w* z4 R
a.
- {8 {* n: _# h6 S8 ?. N8 W
, L1 _0 R9 h6 _4 F- w
  • 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' J0 L2 x+ Q$ o  f  f0 [
      ; y. b7 O: V1 h
6 F4 H* `- A5 M' F

+ n' D4 w3 q7 b3 Ob.6 b$ n! P+ N" \/ ~
9 ]3 f# w) V2 D, W
  • 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
  • " N8 N; _8 H5 F' C. k4 v9 E
      # T8 r0 A$ b! g  k/ E+ _6 V  P; R

5 Z3 b" X# M' q
: \! U+ C7 V7 d8 r( h& S
! A. F/ X: k4 s& Rc.
1 K. Z7 |7 F5 k2 W( G( v9 N  I0 Y$ t! e& n1 u
  • 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
  • 2 \' |. e$ S/ K0 d
      : o: c, M* y" W" C& g8 {! k7 S( u

# d% K# w0 b3 |  S0 ^  V+ ]& j* i+ v' B( C9 T
9 a) S, c! q$ m0 @/ [
太小了,放大看:* f  P. V# @! o, [) t) `
- p: J5 x. X: z' M+ ~# d
/ C! C4 O# ~& H

( f6 X. F& g8 Q6 S. A   v, u/ q4 S# D2 B

- w0 X6 d% ~7 {2 V3 |d.
5 o4 K$ L/ e6 E4 Y* ?/ i# J% \- T- B5 q8 u
  • 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

  • 4 U5 }) {! Z( F; C
      
8 v1 _8 K0 e- e  Z) l( v# C# Q6 m
5 F! \; [: S7 K" K% F/ d 8 |3 ^9 |6 g( f0 X% n- x3 f6 G

) X6 c/ U3 f# n( N$ K) L; Pe., l+ q  j  X7 d6 A7 b

) w+ |) n) j5 k$ y, r& H
  • 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
    6 v& ?' c* r& C7 s- P
     5 U; u: M/ m0 e2 Z

) U& P& ~2 u+ Y5 S, x9 G/ R. O7 w( Y
( |4 B8 @; ?+ s. E* ^8 R+ E
局部放大看:2 _! A" k. _+ A4 L

0 n8 k1 W+ B5 t: d
! y$ C/ L6 P( i) F, G% Z& Q0 Y' i* `7 m. K
7 g7 n5 F) R) l( D8 w. w" T6 ^  t
  T! O5 Q5 X5 @7 l: h$ H
$ y8 R' {, Z7 w" b* i4 p
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, 2025-11-24 02:42 , Processed in 0.156250 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