为什么HHT时频谱图的高频区域会出现波浪线,低频区域却没有,只是散点? 5C

function HHT
clear;clc;clf;
N=2048;
%fft默认计算的信号是从0开始的
t=linspace(1,2,N);deta=t(2)-t(1);fs=1/deta;
x=5*sin(2*pi*10*t)+5*sin(2*pi*35*t);
z=x;
c=emd(z);

%计算每个IMF分量及最后一个剩余分量residual与原始信号的相关性
[m,n]=size(c);
for i=1:m;
a=corrcoef(c(i,:),z);
xg(i)=a(1,2);
end
xg;

for i=1:m-1
%--------------------------------------------------------------------
%计算各IMF的方差贡献率
%定义:方差为平方的均值减去均值的平方
%均值的平方
%imfp2=mean(c(i,:),2).^2
%平方的均值
%imf2p=mean(c(i,:).^2,2)
%各个IMF的方差
mse(i)=mean(c(i,:).^2,2)-mean(c(i,:),2).^2;
end;
mmse=sum(mse);
for i=1:m-1
mse(i)=mean(c(i,:).^2,2)-mean(c(i,:),2).^2;
%方差百分比,也就是方差贡献率
mseb(i)=mse(i)/mmse*100;
%显示各个IMF的方差和贡献率
end;
%画出每个IMF分量及最后一个剩余分量residual的图形
figure(1)
for i=1:m-1
disp(['imf',int2str(i)]) ;disp([mse(i) mseb(i)]);
end;
subplot(m+1,1,1)
plot(t,z)
set(gca,'fontname','times New Roman')
set(gca,'fontsize',14.0)
ylabel(['signal','Amplitude'])

for i=1:m-1
subplot(m+1,1,i+1);
set(gcf,'color','w')
plot(t,c(i,:),'k')
set(gca,'fontname','times New Roman')
set(gca,'fontsize',14.0)
ylabel(['imf',int2str(i)])
end
subplot(m+1,1,m+1);
set(gcf,'color','w')
plot(t,c(m,:),'k')
set(gca,'fontname','times New Roman')
set(gca,'fontsize',14.0)
ylabel(['r',int2str(m-1)])

%画出每个IMF分量及剩余分量residual的幅频曲线
figure(2)
subplot(m+1,1,1)
set(gcf,'color','w')
[f,z]=fftfenxi(t,z);
plot(f,z,'k')
set(gca,'fontname','times New Roman')
set(gca,'fontsize',14.0)
ylabel(['initial signal',int2str(m-1),'Amplitude'])

for i=1:m-1
subplot(m+1,1,i+1);
set(gcf,'color','w')
[f,z]=fftfenxi(t,c(i,:));
plot(f,z,'k')
set(gca,'fontname','times New Roman')
set(gca,'fontsize',14.0)
ylabel(['imf',int2str(i),'Amplitude'])
end
subplot(m+1,1,m+1);
set(gcf,'color','w')
[f,z]=fftfenxi(t,c(m,:));
plot(f,z,'k')
set(gca,'fontname','times New Roman')
set(gca,'fontsize',14.0)
ylabel(['r',int2str(m-1),'Amplitude'])

hx=hilbert(z);
xr=real(hx);xi=imag(hx);
%计算瞬时振幅
sz=sqrt(xr.^2+xi.^2);
%计算瞬时相位
sx=angle(hx);
%计算瞬时频率
dt=diff(t);
dx=diff(sx);
sp=dx./dt;
figure(6)
plot(t(1:N-1),sp)
title('瞬时频率')

%计算HHT时频谱和边际谱
[A,fa,tt]=hhspectrum(c);
[E,tt1]=toimage(A,fa,tt,length(tt));
figure(3)
disp_hhs(E,tt1) %二维图显示HHT时频谱,E是求得的HHT谱
pause
figure(4)
for i=1:size(c,1)
faa=fa(i,:);
[FA,TT1]=meshgrid(faa,tt1);%三维图显示HHT时频图
surf(FA,TT1,E)
title('HHT时频谱三维显示')
hold on
end
hold off
E=flipud(E);
for k=1:size(E,1)
bjp(k)=sum(E(k,:))*1/fs;
end
f=(1:N-2)/N*(fs/2);
figure(5)
plot(f,bjp);
xlabel('频率 / Hz');
ylabel('信号幅值');
title('信号边际谱')%要求边际谱必须先对信号进行EMD分解

function [A,f,tt] = hhspectrum(x,t,l,aff)

error(nargchk(1,4,nargin));

if nargin < 2

t=1:size(x,2);

end

if nargin < 3

l=1;

end

if nargin < 4

aff = 0;

end

if min(size(x)) == 1
if size(x,2) == 1
x = x';
if nargin < 2
t = 1:size(x,2);
end
end
Nmodes = 1;
else
Nmodes = size(x,1);
end

lt=length(t);

tt=t((l+1):(lt-l));

for i=1:Nmodes

an(i,:)=hilbert(x(i,:)')';
f(i,:)=instfreq(an(i,:)',tt,l)';
A=abs(an(:,l+1:end-l));

if aff
disprog(i,Nmodes,max(Nmodes,100))
end

end

function disp_hhs(im,t,inf)

% DISP_HHS(im,t,inf)
% displays in a new figure the spectrum contained in matrix "im"
% (amplitudes in log).
%
% inputs : - im : image matrix (e.g., output of "toimage")
% - t (optional) : time instants (e.g., output of "toimage")
% - inf (optional) : -dynamic range in dB (wrt max)
% default : inf = -20
%
% utilisation : disp_hhs(im) ; disp_hhs(im,t) ; disp_hhs(im,inf)
% disp_hhs(im,t,inf)

figure
colormap(bone)
colormap(1-colormap);

if nargin==1
inf=-20;
t = 1:size(im,2);

end

if nargin == 2
if length(t) == 1
inf = t;
t = 1:size(im,2);
else
inf = -20;
end
end

if inf >= 0
error('inf doit etre < 0')
end

M=max(max(im));

im = log10(im/M+1e-300);

inf=inf/10;

imagesc(t,fliplr((1:size(im,1))/(2*size(im,1))),im,[inf,0]);
set(gca,'YDir','normal')
xlabel(['time'])
ylabel(['normalized frequency'])
title('Hilbert-Huang spectrum')
function [f,z]=fftfenxi(t,y)
L=length(t);N=2^nextpow2(L);
%fft默认计算的信号是从0开始的
t=linspace(t(1),t(L),N);deta=t(2)-t(1);
m=0:N-1;
f=1./(N*deta)*m;
%下面计算的Y就是x(t)的傅里叶变换数值
%Y=exp(i*4*pi*f).*fft(y)%将计算出来的频谱乘以exp(i*4*pi*f)得到频移后[-2,2]之间的频谱值
Y=fft(y);
z=sqrt(Y.*conj(Y));![图片说明](https://img-ask.csdn.net/upload/201804/08/1523194608_477078.jpg)图片说明
为什么同样的程序我画出来的时频图,高频区域会出现波浪线,低频区域却没有,只是散点?求大神帮忙

1个回答

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索引超出矩阵维度,求改错

t=1:1:360; yuan1(t)=1000000/360+(1000000-1000000/360*(t-1))*0.049/12; yuan2(t)=1000000*0.049/12*(1+0.049/12)^360/((1+0.049/12 )^360-1)+t*0; plot(t,yuan1,'r',t,yuan2,'k') figure leiyuan(1)=yuan1(1); t=2:360; leiyuan(t)=leiyuan(t-1)+yuan1(t); plot(t,leiyuan) 索引超出矩阵维度。 出错 shuxueshiyanbaogao (line 110) leiyuan(t)=leiyuan(t-1)+yuan1(t);

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Hamlet's gambling

Description "There are a thousand Hamlets in a thousand people's eyes." --W.William Shakespeare Lotus is a big fan of Shakespeare. Her favorite tragedy written by Shakespeare is Hamlet, one of the most popular works in English language. However, since Lotus is really a merciful and kindhearted girl, she doesn't like the scene of the final match, in which Hamlet fenced against Laertes. Once she had a dream of Hamlet. In that dream, Hamlet fought with Laertes in another way: flipping coins. It sounds like gambling. In order to flip the coin in an absolutely fair way, Hamlet got a monkey to do this job instead of people. Before the monkey started to flip, Hamlet and Laertes respectively wrote down an arbitrary sequence of results (we call it a "pattern"). For example, Hamlet wrote down "Head, Tail, Head" while Laertes wrote down "Head, Head, Tail, Tail".It was guaranteed that the pattern of Hamlet's did not occur within the pattern of Laertes', nor did the pattern of Laertes' occur within the pattern of Hamlet's. Then the monkey began to flip the coin over and over and generated a sequence of results, like "Head, Head, Tail, Head, Tail ..." At any time, if the monkey obtained Hamlet's pattern, it stopped flipping and Hamlet won. Otherwise, if the monkey obtained Laertes's pattern, it also stopped flipping and Laertes won. One example of the gambling is like following: Hamlet's pattern was ""HHT" (H=Head, T=Tail) and his rival's pattern was "HTTT". The monkey flipped the coin and obtained "H, T, H, T, T, H, T, T, T". At that time, Laertes' pattern appeared at the end of the sequence, so the monkey stopped and the judge declared that Laertes won the game. Now your task is to decide that in Lotus' gambling, who has the higher probability to win the game. Pay attention that since the monkey has no bias, the probabilities to get head or tail in one flip are always equal. Input The input contains several test cases. The first line of each case contains two positive integers N and M (0 < N, M <= 100000). N is the length of Hamlet's pattern and M is the length of Laertes'. The second line contains a string of length N, representing the pattern of Hamlet's. The third line contains a string of length M, representing the pattern of Laertes'. All the strings only contain uppercase letters `H' and `T'. The input ends by a line of two zeros. Output For each test case, output a single line to show your answer. If Hamlet has a better chance to win, output the word 'Hamlet'. If Laertes has a better chance, output the word 'Laertes'. If they have the same probability to win the game, output the word 'Equal'. Sample Input 1 1 H T 1 2 T HH 0 0 Sample Output Equal Hamlet

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