why the plot doesn't work ,in the code i'm finding the estimation value of the phase using montecarlo simulation

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mhamad Yaacoub
mhamad Yaacoub 2022 年 1 月 21 日
コメント済み: mhamad Yaacoub 2022 年 1 月 21 日
%phase astimation using monte carlo
clear all
close all
clc
%signal model x(n)=Acos(2pif0n+thetha)+w(n)
phase=0.75;
A=1;
sigma=0.1;
f0=0.035;
SNR=A^2/2*(sigma^2)
SNR = 0.0050
for m=1:length(SNR)
M=5;
for i=1:M;
s=A*cos(2*pi*f0*i+phase);
w=randn(size(s));
x(i)=s+sigma*w
y1=sum(x*sin(2*pi*f0*i))
y2=sum(x*cos(2*pi*f0*i))
phasehat(i)=-atan(y1+y2)
%
% mat(i,:)=phasehat
end
%compute bias
b(m)=(1/M)*sum(mean(phasehat(i))-phase)
%mean square error
mse=mean((phasehat(i)-phase)^2)
var=1/M*SNR
end
x = 0.7006
y1 = 0.1528
y2 = 0.6837
phasehat = -0.6966
x = 1×2
0.7006 0.4581
y1 = 0.4933
y2 = 1.0484
phasehat = 1×2
-0.6966 -0.9954
x = 1×3
0.7006 0.4581 0.4058
y1 = 0.9588
y2 = 1.2361
phasehat = 1×3
-0.6966 -0.9954 -1.1433
x = 1×4
0.7006 0.4581 0.4058 0.0162
y1 = 1.2178
y2 = 1.0075
phasehat = 1×4
-0.6966 -0.9954 -1.1433 -1.1485
x = 1×5
0.7006 0.4581 0.4058 0.0162 -0.1706
y1 = 1.2563
y2 = 0.6401
phasehat = 1×5
-0.6966 -0.9954 -1.1433 -1.1485 -1.0855
b = -0.3671
mse = 3.3692
var = 1.0000e-03
figure(1)
plot(10*log(SNR),b)
figure(2)
plot(10*log(SNR),mse)
figure(3)
plot(10*log(SNR),10*log(var))

回答 (1 件)

Chris
Chris 2022 年 1 月 21 日
編集済み: Chris 2022 年 1 月 21 日
since length(SNR) is 1, you are running one loop iteration and generating one point. plot() doesn't show one point unless you specify a marker:
figure(1)
plot(10*log(SNR),b,'o')
figure(2)
plot(10*log(SNR),mse,'x')
figure(3)
plot(10*log(SNR),10*log(var),'^')
or equivalently, use scatter:
figure(1)
scatter(10*log(SNR),b)
% ...etc
  4 件のコメント
mhamad Yaacoub
mhamad Yaacoub 2022 年 1 月 21 日
Thank you a lot, and sorry for disturbing

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