Complex inverse Fourier transform when using MATLAB

Question

I am using MATLAB and I am trying to find the ifft of a symmetric function, but I keep getting a complex result. I have tried using circshift, but I can't seem to get it figured out.

How can I fix it?

Here is the code:

t = 0:0.001:0.119;

for i = 1:120

    comp1(i) = 9.8*cos(2*pi*200*t(i));

    comp2(i) = 7.6*cos(2*pi*145*t(i) + 30/57.3);

    comp3(i) = 5.4*cos(2*pi*93*t(i) + 70/57.3);

    comp4(i) = 3.2*cos(2*pi*58*t(i) + 160/57.3);

    comp5(i) = cos(2*pi*35*t(i) + 320/57.3);

    YS = comp1 + comp2 + comp3 + comp4 + comp5;
end

Q = 1000/(2*60)*[-59:1:60];

Box = [zeros(1, 40), ones(1, 5), zeros(1, 30), ones(1, 5), zeros(1, 40)];

Box1 = circshift(Box, [0, 60]);

F = ifft(Box1);

Answer 1

(I am not sure what YS and Q in your code are for, so I just ignore them).

The transform X of a real signal x of length N has the property that

X(k) = X(N-k)*

for k=1,...,N-1 , where * is complex conjugate. For a real transform (as in your example) you can just ignore the * . Note that the indexing goes from 0 to N-1 , so for k=0 the above property is not defined, since the index of the last term is N-1 (there is no X(N) ).

In your case the signal Box does not have this property, nor does Box1 (check it!). Try for example:

Box = [zeros(1,41),ones(1,5),zeros(1,29),ones(1,5),zeros(1,40)];

This signal does have this property, and its ifft is indeed real.