Why this message appears?
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clear all;
f1=200;
f2=250;
a1=1;
a2=1.5;
fe=10e3;
temps=0:1/fe:0.02-1/fe;
m1=a1*sin(2*pi*f1*temps);
m2=a2*sin(2*pi*f2*temps);
mt=m1*m2;
Error using *
Inner matrix dimensions must agree.
采纳的回答
Star Strider
2019-5-1
It appears because you are multiplying two row vector using matrix mulktiplication.
You need to use element-wise array multiplication (using .* instead of *):
mt = m1 .* m2;
10 个评论
taher zaouali
2019-5-1
Star Strider
2019-5-1
You have a ‘.n’ that makes no sense in the context of the rest of your code.
Try this instead:
spec_mt=fft(abs(mt));
however for best results, do this:
spec_mt=fft(abs(mt)) / numel(temps);
There is no ‘n’ in the code that you posted. If you defined ‘n’ but you did not post it and it is supposed to be the length of the fft, you need a comma, not a period, between ‘mt’ and ‘n’:
spec_mt=fft(abs(mt,n)) / numel(temps);
Walter Roberson
2019-5-1
Probably you want mt.*n but we would need to see more code to be more certain.
taher zaouali
2019-5-2
taher zaouali
2019-5-2
Star Strider
2019-5-2
Try this:
spec_yt=fft(abs(yt))/n;
taher zaouali
2019-5-2
Star Strider
2019-5-2
This takes the absolute values of the fft results, so the warning will not appear:
f1=200;
f2=250;
a1=1;
a2=1.5;
fe=10e3;
temps=0:1/fe:0.02-1/fe;
m1=a1*sin(2*pi*f1*temps);
m2=a2*sin(2*pi*f2*temps);
mt=m1.*m2;
n=length(temps);
freqs=0:fe/n:fe-1/n;
spec_mt=fft(abs(mt))/n;
figure(1)
subplot(211),plot(temps,mt);grid;
subplot(212),plot(freqs,abs(spec_mt));grid;
fmod=3e3;
xt=mt.*cos(2*pi*fmod*temps);
spec_xt=fft(abs(xt))/n;
figure(2)
plot(freqs,abs(spec_xt));grid;
yt=xt.*cos(2*pi*fmod*temps);
spec_yt=fft(abs(yt))/n;
y2=abs(spec_yt);
figure(3)
plot(freqs,y2);grid;
y2(200:n)=zeros(1,n-200+1);
spec_y2=abs(y2);
x2=real(ifft(y2));
figure(5)
subplot(211),plot(temps,mt);grid;
subplot(212),plot(temps,x2);grid;
You should probably shift the frequency axes so that ‘0’ is in the centre, or plot a one-sided fft. .
taher zaouali
2019-5-4
Star Strider
2019-5-4
As always, my pleasure.
I took the ‘n’ out of the fft call entirely. I was not certain where you wanted to put the ‘n’, or what you wanted to do with it. If ‘n’ is the same as the size as the time-domain vector, using it as the size of the fft does not change anything, because the length of the time-domain vector is the default length of the fft for that vector.
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