Sine and cosine transforms

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[edit] Fourier sine transform

In mathematics, the Fourier sine transform is a special case of the continuous Fourier transform, arising naturally when attempting to transform an odd function. Consider the general Fourier transform:

F(\omega) = \mathcal{F}(f)(\omega) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(t) e^{-i\omega t}\,dt.

We may expand the integrand by means of Euler's formula:

F(\omega)=\frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(t)(\cos\,{\omega t} - i\,\sin{ \,\omega t})\,dt,

or, written as the sum of two integrals:

F(\omega)=\frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(t)\cos\,{\omega t} \,dt - \frac{i}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(t)\sin\,{\omega t}\,dt.

Now notice that if we assume f(t) is an odd function, the product f(t)cosωt is also odd whilst the product f(t)sinωt is an even function. Since we are integrating over an interval symmetric about the origin (i.e. -∞ to +∞), the first integral must vanish to zero, and the second may be simplified to give:

F(\omega)= -i\,\sqrt{\frac{2}{\pi}} \int\limits_{0}^\infty f(t)\sin\,{\omega t} \,dt,

which is the Fourier sine transform for odd f(t). It is clear that the transformed function F(ω) is also an odd function, and a similar analysis of the general Inverse Fourier transform yields a second sine transform, namely:

f(t)= i\,\sqrt{\frac{2}{\pi}} \int\limits_{0}^\infty F(\omega)\sin\,{\omega t} \,d\omega.

Note that the numerical factors in the transforms are defined uniquely only by their product, as discussed for general continuous Fourier transforms. For this reason the imaginary units i and -i can be omitted, with the more commonly seen forms of the Fourier sine transforms being:

F(\omega)= \sqrt{\frac{2}{\pi}} \int\limits_{0}^\infty f(t)\sin\,{\omega t} \,dt,

and

f(t)= \sqrt{\frac{2}{\pi}} \int\limits_{0}^\infty F(\omega)\sin\,{\omega t} \,d\omega.

[edit] Fourier cosine transform

In mathematics, the Fourier cosine transform is a special case of the continuous Fourier transform, arising naturally when attempting to transform an even function. Consider the general Fourier transform:

F(\omega) = \mathcal{F}(f)(t) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(t) e^{-i\omega t}\,dt.

We may expand the integrand by means of Euler's formula:

F(\omega)=\frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(t)(\cos\,{\omega t} - i\,\sin{ \,\omega t})\,dt,

or, written as the sum of two integrals:

F(\omega)=\frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(t)\cos\,{\omega t} \,dt - \frac{i}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(t)\sin\,{\omega t}\,dt.

Now notice that if we assume f(t) is an even function, the product f(t)cosωt is also even whilst the product f(t)sinωt is an odd function. Since we are integrating over an interval symmetric about the origin (i.e. -∞ to +∞), the second integral must vanish to zero, and the first may be simplified to give:

F(\omega)= \sqrt{\frac{2}{\pi}} \int\limits_{0}^\infty f(t)\cos\,{\omega t} \,dt,

which is the Fourier cosine transform for even f(t). It is clear that the transformed function F(ω) is also an even function, and a similar analysis of the general inverse Fourier transform yields a second cosine transform, namely:

f(t)= \sqrt{\frac{2}{\pi}} \int\limits_{0}^\infty F(\omega)\cos\,{\omega t} \,d\omega.

Note that the numerical factors in the transforms are defined uniquely only by their product, as discussed for general continuous Fourier transforms.


[edit] See also

[edit] References

  • Mary L. Boas, Mathematical Methods in the Physical Sciences, 2nd Ed, John Wiley & Sons Inc, 1983. ISBN 0-471-04409-1