Why the sums of cos(x) over 2*pi range not zero?
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when evaluating the following codes:
t = linspace(-pi,pi,128); s = sin(t); c = cos(t); sum(s), sum(c)
ans =
6.811558403281303e-15
ans =
-0.999999999999978
Q: should not both be zero?
and try
sum(c(2:end))
ans =
2.5313e-14
also quad(@cos,-pi,pi)
ans =
4.0143e-09
Q: Why the discrepancy?
Thank you for your input.
Thanks J
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采纳的回答
Greg Heath
2012-7-11
T = fundamental period
N = number of samples
dt = T/N sampling interval
f0 = 1/T fundamental frequency
(n-1)*f0 harmonics 1<=n <= N
Orthogonality interval
t = t0:dt:t0+T-dt;
t = t0+[0:dt:T-dt];
t = t0+dt*[0:N-1];
help fft
doc fft
Hope this helps
Greg
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更多回答(3 个)
Luffy
2012-7-11
In matlab,
sin(pi) = 1.2246e-16
The expression sin(pi) is not exactly zero because pi is not exactly π
0 个评论
Wayne King
2012-7-11
编辑:Wayne King
2012-7-11
If you're trying to establish some equivalence between the integral of cos(t) from -pi and pi and the sum of cos(t), you're forgetting a very important part and that is the dt
t = linspace(-pi,pi,1000);
dt = (pi-(-pi))/length(t);
sum(cos(t))*dt
Using other integration routines in MATLAB is more robust than what I've done, but you see it gets you much closer to zero. Think about the formula for a Riemann sum.
Wayne King
2012-7-11
I don't think you can say that simply summing cos(t) on an arbitrary grid should be zero. You have to be careful how the grid is constructed. For example
k = 1;
N = 100;
t = 0:99;
sum(cos(2*pi*k/N*t))
sum(sin(2*pi*k/N*t))
are both zero, because I used a Fourier frequency and a specific discrete-time vector. This has to do with the orthogonality of the N-th roots of unity.
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