j = square root of negative one ( sqrt(-1) )

Remember the DFT returns complex numbers.


--- [EMAIL PROTECTED] wrote:
> Ok, I have arrived at this point in trying to understand the DFT.
> 
> X(k)= x(n)[cos(2*PI*n*k/N) - j sin(2*PI*n*k/N)]
> 
> In this equation, N is the number of samples we are processing,
>                   n is the sample index, starting at 0,
>                   x(n) is the value of sample number n,
>                   k is the frequency index,
>                   j is ???? What is the value of j????
> 
> For any value of k, we get X(k), which is the content of the signal
> at the
> frequency
> kF/N, with F being the sampling frequency.
> Example:
> If we sample at 50Hz and take 50 samples. Each value of k would
> represent a
> frequency of: kF/N=k*50/50=k*1=k, so, for k=1, X(k)=X(1) is the
> content of
> the signal at 1Hz. For k=2, X(k)=X(2) is the content of the signal at
> 2Hz.,
> and so on. 
> 
> My 2 questions are: 
> 1. Using the above equation for X(k)=X(2), with 50 samples sampled at
> 50Hz,
> with n=2 and x(n)=9, what is the value of X(2)?
> 2. What is j in the equation and is it always the same, how can I
> determine
> it, how can I use it?
> 
> Thank you.
> Dan
> 
> 
> 
> 
> At 03:12 PM 9/19/00 -0500, you wrote:
> >"If you are not well founded in the calculus there is precious
> little
> >available in the literature."(Quote from the book UNDERSTANDING THE
> FFT by
> >Anders E. Zonst)
> >
> >I totallt agree. I bought this book because it states on the cover
> "A
> >tutorial on the algorithm & Software for Laymen, Students,
> Technicians &
> >Working Engineers"
> >
> >Well, I am a laymen and I still can't follow this book. I haven't
> even been
> >able to understand even one complete chapter.
> >
> >What ever happened to the teachers?  I have always believed that if
> a
> >person really knows a subject, he can teach it. He just goes step by
> step,
> >giving examples (in real number calculations, not greek and latin).
> >
> >Can anyone help me find a true tutorial, book, or guide to the FFT?
> I have
> >looked at about 15 different Web pages on the topic and the only
> book I
> >have seen, on the net, that might present FFT's in a way that a
> layman
> >could learn from is priced at about $80.00. And from what I have
> seen and
> >experienced I wouldn't bet on this book either.
> >I am a good student when I set my mind on a subject. 
> >
> >Thanks
> >Dan
> >
> >
> >
> >
> >                                              
> >
>
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> >
> >
> 
>                                               
> 
>
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