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Posted 06 November 2011 - 10:30 AM
Posted 06 November 2011 - 11:22 AM
e^-i = e^(-1)i = cos(-1) + isin(-1) = cos(1) - isin(1)1320597004[/url]' post='4881088']
I've been implementing a (Discrete) Fourier transform (and it's inverse) in C++ and was wondering about two things in particular,
1. How does e^-i (negative) differ from e^i when representing this using Euler's formula?
I know that anything to the power of a negative value is it's reciprocal (e.g. 2^-1 = 1/2) but I'm a bit confused whether this changes the Euler's formula (cos(w) + isin(w)) or not.
1320597004[/url]' post='4881088']and
2. Why does the inverse Fourier transform get normalized at the end (i.e sum/numsamples) whereas the normal Fourier transform does not?
Posted 06 November 2011 - 12:15 PM
I know that anything to the power of a negative value is it's reciprocal (e.g. 2^-1 = 1/2) but I'm a bit confused whether this changes the Euler's formula (cos(w) + isin(w)) or not.
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Posted 10 November 2011 - 07:12 AM
Posted 10 November 2011 - 07:54 AM
That should be e^(-iw). And yes, it does mean that, but you can deduce it yourself from Euler's identity and the fact that sin(-x) = -sin(x). You can answer most questions in Math yourself, and you should try to get in the habit of doing that.Thanks for your replies.
So if e^-i is cos(w) - isin(w) does this mean you just negate the imaginary part?
Posted 11 November 2011 - 07:03 AM
Posted 11 November 2011 - 07:56 AM
Posted 11 November 2011 - 08:18 AM
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