Could you fix this code? I am trying to generate the Gaussian noise that is changing over time. X(t)=5+10*​cos(2*pi*t​+pi/6)+G(s​igma,t)

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Byungho
Byungho 2024-4-4
评论： Byungho 2024-4-4
Ran in:
deltat=0.1;
nsamples=2^10;
fs=1/deltat;
time=[0 : deltat : deltat*(nsamples-1)]
time = 1x1024
0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000 1.1000 1.2000 1.3000 1.4000 1.5000 1.6000 1.7000 1.8000 1.9000 2.0000 2.1000 2.2000 2.3000 2.4000 2.5000 2.6000 2.7000 2.8000 2.9000
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psdtot(1:nsamples/2+1) = zeros(1,nsamples/2+1);
nblock = 10;
for k=1:nblock
for i=1: nsamples
ydata(i) =5+10*cos(2*pi*time+pi/6)+15* randn(1);
end
[pxx,f] = periodogram(ydata,rectwin(nsamples),nsamples,fs);
psdtot = psdtot + pxx/nblock;
end
Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.
figure(1);
plot(time,ydata);
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Byungho
Byungho 2024-4-4
编辑：Byungho 2024-4-4
I'm trying to generate Gaussian noise from x(t)=5+10*cos(2*pi*t+pi/6)+G(sigma,t)
G(sigma,t) is Gaussian noise with a standard deviation of 15 (sigma G=15), detat=0.1(s), and nsample=2^10.
It is the start of calculating Power Spectral Density with FFT.
The overall variance of the signal is (sigma g)^2+(A)^2/2 as expected.
The generated random signal should be sinusoidal because X(t) has the cosine factor.
Byungho
Byungho 2024-4-4
Ran in:
deltat=0.1;
nsamples=2^10;
fs=1/deltat;
time=[0 : deltat : deltat*(nsamples-1)]
time = 1x1024
0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000 1.1000 1.2000 1.3000 1.4000 1.5000 1.6000 1.7000 1.8000 1.9000 2.0000 2.1000 2.2000 2.3000 2.4000 2.5000 2.6000 2.7000 2.8000 2.9000
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psdtot(1:nsamples/2+1) = zeros(1,nsamples/2+1);
nblock = 10;
for k=1:nblock
for i=1: nsamples
ydata(i) =randn(1);
end
[pxx,f] = periodogram(ydata,rectwin(nsamples),nsamples,fs);
psdtot = psdtot + pxx / nblock;
end
figure(1);
plot(time,ydata);
The above codes are the Gaussian noise without the signal bias and the sinusoidal signal.

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采纳的回答

Ayush Anand
Ayush Anand 2024-4-4
Hi Byungho,
The error you are getting is because you are trying to assign a vector 5+10*cos(2*pi*time+pi/6)+15*randn(1) to a single array index y(i). Since time is a vector of shape 1 x 1024, the expression 5+10*cos(2*pi*tim+pi/6) is also a vector with the same dimensions, and can't be assigned to y(i), a single array element. You might want to use cells or multidimensional arrays for the same. For that, you would need to predefine an nsamples x 1024 dimension array beforehand and then assign the values using the colon operator like y(i,:).
You can read more about the usage of the colon operator in MATLAB here:
https://www.mathworks.com/help/matlab/ref/colon.html
Hope this helps!
  1 个评论
Byungho
Byungho 2024-4-4
Ran in:
deltat=0.1;
nsamples=2^10;
fs=1/deltat;
time=[0 : deltat : deltat*(nsamples-1)]
time = 1x1024
0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000 1.1000 1.2000 1.3000 1.4000 1.5000 1.6000 1.7000 1.8000 1.9000 2.0000 2.1000 2.2000 2.3000 2.4000 2.5000 2.6000 2.7000 2.8000 2.9000
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
psdtot(1:nsamples/2+1) = zeros(1,nsamples/2+1);
nblock = 10;
for k=1:nblock
for i=1: nsamples
ydata(i,:) =5+10*cos(2*pi*time+pi/6)+15* randn(1);
end
[pxx,f] = periodogram(ydata,rectwin(nsamples),nsamples,fs);
psdtot = psdtot + pxx/nblock;
end
Arrays have incompatible sizes for this operation.
figure(1);
plot(time,ydata);
Could you let me know how to predefine an nsamples x 1024 dimension array?
Thank you for your input.

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