Error using + Matrix dimensions must agree.
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古いコメントを表示
by using white noise effect in the three level discrete F T:
clear all; close all; clc
fs=200; %sampling freq.
dt =1/fs;
N0=fs/50; %number of samples/cycle
m=3; %no. of cycles
t = dt*(0:200); %data window
fi=50; %Frequency test
ww=wgn(201,1,-40);
size(transpose(ww))
v=@(t) sin(2*pi*fi*t + 0.3)+transpose(ww);
tmax=1;
% v : as function of time
% fs : sampling frequency (Hz)
% tmax : time of final estimation
% to test: [t,f]=ZC(@(t)(220*sin(2*pi*50.1*t+pi/2)),50*512,1)
% to test: [t,f]=ZC(@(t)(220*sin(2*pi*50.1*t+pi/2)+randn(1)*.1),50*512,1)
n=N0-1:-1:0;
f0=50;
f=50.88;
Hc=2/N0*cos(2*pi*n/N0+pi/N0);
Hs=-2/N0*sin(2*pi*n/N0+pi/N0);
t_est=[];
f_est=[];
j_max=tmax*fs;
for j=1:j_max+1
x=v((200:j+N0-2).*dt);
c(j)=x*Hc';
s(j)=x*Hs';
if(j>N0)
Ac(j-N0)=sqrt(sum(c(end-N0+1:end).^2)/N0);
As(j-N0)=sqrt(sum(s(end-N0+1:end).^2)/N0);
cc(j-N0)=c(end-N0+1:end)*Hc';
ss(j-N0)=c(end-N0+1:end)*Hs';
if(j>2*N0)
Acc(j-2*N0)=sqrt(sum(cc(end-N0+1:end).^2)/N0);
Ass(j-2*N0)=sqrt(sum(ss(end-N0+1:end).^2)/N0);
ccc(j-2*N0)=cc(end-N0+1:end)*Hc';
ccs(j-2*N0)=cc(end-N0+1:end)*Hs';
ssc(j-2*N0)=ss(end-N0+1:end)*Hc';
sss(j-2*N0)=ss(end-N0+1:end)*Hs';
ff=f0*N0/pi*atan(tan(pi/N0)*((ccc(j-2*N0).^2+ccs(j-2*N0).^2)./(ssc(j-2*N0).^2+sss(j-2*N0).^2)).^.25);
t_est=[t_est;(j-1)*dt];
f_est=[f_est;ff];
end
end
end
t_est
f_est
plot(t_est,f_est,'red')
o=rms(fi)
c=rms(f_est)
RMSE = sqrt(mean(c - o).^2)
t_est;
f_est
plot(t_est, f_est,'red')
hold on
RMSE = sqrt(mean((f_est-fi).^2))
xlabel('time')
ylabel('frequency')
title('three LDFT white noise')
plot (t_est,fi*ones(size(t_est)))
hold off
the error is :
Error using +
Matrix dimensions must agree.
Error in @(t)sin(2*pi*fi*t+0.3)+transpose(ww)
Error in threeDFT (line 33)
x=v((200:j+N0-2).*dt);
2 件のコメント
Rik
2020 年 12 月 3 日
Please write your comments in the comments, not in flags. You can use links to the comment itself if you want an unambiguous reference.
採用された回答
VBBV
2020 年 12 月 3 日
%t = dt*(0:200); %data window COMMENT THIS LINE IN YOUR PROGRAM
fi=50; %Frequency test
ww=wgn(201,1,-40);
size(transpose(ww))
v=@(t) sin(2*pi*fi*t + 0.3)+transpose(ww); % using function handle
since you are using fuinction handle to define v in terms of t and calling inside the for loop with varying length of t
for j=1:j_max+1
x=v((200:j+N0-2).*dt); % ....
you dont have to define data window at beginning, comment that line
3 件のコメント
Walter Roberson
2020 年 12 月 3 日
There is no point in defining t as a symbolic variable. Any initilization you do to t is ignored when you create an anonymous function with named parameter t: inside the body of the anonymous function, any named parameter will be replaced by the value passed into the function handle.
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