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From: carlierm on 3 Jan 2010 21:10
>Tim Wescott wrote:
>> On Sun, 03 Jan 2010 08:29:00 -0600, carlierm wrote:
>>
>>> Hi
>>>
>>> I have a signal in the time domain, and I need to perform some
changes
>>> in the spectrum of magnitude of such signal, and then I want to
>>> reconstruct the signal and take it to the time domain again. The
problem
>>> I have is that i cannot get the signal in the time domain after the
>>> modifications. When I take ifft, I get a vector of complex numbers,
and
>>> I expect a vector or real numbers. The code I have is as follows:
>>>
>>> Fs=1024;
>>> t=0:1/Fs:1;
>>> x=sin(2*pi*t*200);
>>> y=fft(x);
>>> mx=abs(y);
>>> ma=angle(y);
>>> [C,I]=max(mx);
>>> mx(I)=100;
>>> for i=1:length(y)
>>> y4(i)=mx(i)*cos(ma(i))+ j*mx(i)*sin(ma(i));
>>> end
>>> y6=ifft(y4);
>>>
>>> I will appreciate any help.
>>>
>>> Monica
>>
>
>Consider this:
>
>If you fft a real time sequence then the real part is even and the
>imaginary part is odd.
>
>When you fft any sequence then the resulting sequence starts with an
>index of zero and ends with an index of N-1 - relative frequency - if
>the original sequence is a time sequence. And, you may assume that the
>value at N would be the same as the value at zero - as you may consider
>that it starts to repeat at that point. You may visualize this if you
>put the data on a circular axis (i.e. the axis is a circle) instead of a

>linear one.
>
>Because the real part is even, it mirrors around Fs/2 or around N/2. If

>the number of samples is even then there is a sample at that point and
>it is the N/2+1th sample and it must be purely real because the
>imaginary part is odd thus zero at that point. If the number of samples

>is odd then there is no sample at Fs/2.
>Because of the mirroring, the real part of the samples show up in equal
>pairs around zero and the imaginary parts of the samples show up in
>equal (but opposite sign) around zero.
>
>The magnitude is always even. Because of this, the magnitude is
>symmetrical about zero AND about Fs/2.
>
>I note that you changed the magnitude at a single sample point.
>Because of the above, what about the magnitude at the mirror image
>point? By changing a single sample you have perforce changed the
>evenness of the magnitude and I should say the real part. The resulting

>ifft can therefore *not* be real.
>
>Without solving the entire problem for you, I suggest you first make the

>change at both I and at length(y)-I ... check my math here to keep the
>magnitude an even function. Whether this guarantees that the real part
>remains even and the imaginary part remains odd I leave as an important
>exercise.
>
>Also, how can you be sure that the conversion from magnitude and phase
>back to real and imaginary is working OK. Have you considered phase
>wrap first? i.e. what happens when the phase goes beyond 2pi?
>
>P.S. Changing a single pair of mirrored samples (real and imaginary
>now...) in frequency (which is equivalent to multiplying by a pair of
>samples of selected value) is the same as convolving in time by a
>sinusoid. Is that what you're trying to do?
>
>Fred
>

Thank you very much Fred. You were right; I did the change at both points
as you suggested and it worked. Thanks again.

Monica
From: carlierm on 3 Jan 2010 21:12
>On Sun, 03 Jan 2010 08:29:00 -0600, carlierm wrote:
>
>> Hi
>>
>> I have a signal in the time domain, and I need to perform some changes
>> in the spectrum of magnitude of such signal, and then I want to
>> reconstruct the signal and take it to the time domain again. The
problem
>> I have is that i cannot get the signal in the time domain after the
>> modifications. When I take ifft, I get a vector of complex numbers,
and
>> I expect a vector or real numbers. The code I have is as follows:
>>
>> Fs=1024;
>> t=0:1/Fs:1;
>> x=sin(2*pi*t*200);
>> y=fft(x);
>> mx=abs(y);
>> ma=angle(y);
>> [C,I]=max(mx);
>> mx(I)=100;
>> for i=1:length(y)
>> y4(i)=mx(i)*cos(ma(i))+ j*mx(i)*sin(ma(i));
>> end
>> y6=ifft(y4);
>>
>> I will appreciate any help.
>>
>> Monica
>
>You don't say what tool you're doing this in; I assume it's Matlab, but
>it could as easily be Octave, Scilab, or something else yet. Letting us

>know would help.
>
>Assuming that it's one of the three that I mentioned, here's several
>things to check:
>
>First, check the magnitude of the complex part of your answer. Scilab,
>at least, nearly always decides that the return from an ifft is complex,

>but the complex part is often scraping the limit of computational
>accuracy. When you've got a real part with a maximum magnitude of 100,
>and an imaginary part with a maximum magnitude of 10^-15, you can often
>safely assume that you've got a vector of reals, with some round-off
>error that happens to be complex.
>
>Then, just to make sure that the world has not gone insane, see what you

>get when you take ifft(fft(x)).
>
>I _think_ that what you're doing by separating phase and magnitude is OK,

>but you should probably also try commenting out the mx(I) = ... line, and

>see what result you get.
>
>I can't think of anything else. Good luck, let us know what you find.
>
>--
>www.wescottdesign.com
>
I am using matlab. Thanks for your comments.
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