- #1

DuncanM

- 98

- 2

[tex]g(x) = \int_{0}^{\infty} A(\omega )cos(2\pi \omega x)d\omega - \int_{0}^{\infty} B(\omega )sin(2\pi \omega x)d\omega[/tex]

where

[tex]A(\omega ) = 2 \int_{-\infty}^{\infty} g(x)cos(2\pi \omega x)dx[/tex]

and

[tex]B(\omega ) = 2 \int_{-\infty}^{\infty} g(x)sin(2\pi \omega x)dx[/tex]

For an even function, B(ω) = 0

For an odd function A(ω) = 0

(2) For discrete points, the function is declared as

[tex]g_{j} = a_{0} + \sum_{k = 1}^{[N/2]} a_{k} cos(kj\frac{2\pi}{N}) + \sum_{k = 1}^{[N/2]} b_{k} sin(kj\frac{2\pi}{N})[/tex]

where

[tex]a_{k} = \frac{2}{N} \sum_{j = 0}^{N-1} g_{j} cos(jk\frac{2\pi}{N})[/tex]

and

[tex]b_{k} = \frac{2}{N} \sum_{j = 0}^{N-1} g_{j} sin(jk\frac{2\pi}{N})[/tex]

The text I am reading states that the discrete coefficients approximate the exact Fourier coefficients of a periodic function.

Now, taking a baby step, say I consider the function g(x) = sin(x), the plain, old sine function.

Since the sine function is odd, the Fourier integral transform indicates that A(ω) = 0.

Since the exact Fourier coefficients of a periodic function approximate discrete coefficients, I expected the [itex]a_{k}[/itex] values of a DFT to be 0. But they are not. In fact, substituting the results of a DFT back into the original equation do not work without

*both*the [itex]a_{k}[/itex] and [itex]b_{k}[/itex] values.

Here is what I did:

I assumed a period of 1 time unit, over 1 period, and divided the motion into uniform intervals. Calculating the sine values at each interval, this data was then fed into a DFT. The resultant transform includes both sine and cosine coefficients.

Shouldn't it include only coefficients for the sine terms?

What is going on? Have I made an incorrect leap in logic?