Editing Fourier transform (section)

== Example ==
The following figures provide a visual illustration of how the Fourier transform's integral measures whether a frequency is present in a particular function. The first image depicts the function <math>f(t) = \cos(2\pi\ 3 t) \ e^{-\pi t^2},</math> which is a 3&nbsp;[[hertz|Hz]] cosine wave (the first term) shaped by a [[Gaussian function|Gaussian]] [[Envelope (waves)|envelope function]] (the second term) that smoothly turns the wave on and off. The next 2 images show the product <math>f(t) e^{-i 2\pi 3 t},</math> which must be integrated to calculate the Fourier transform at +3&nbsp;Hz. The real part of the integrand has a non-negative average value, because the alternating signs of <math>f(t)</math> and <math>\operatorname{Re}(e^{-i 2\pi 3 t})</math> oscillate at the same rate and in phase, whereas <math>f(t)</math> and <math>\operatorname{Im} (e^{-i 2\pi 3 t})</math> oscillate at the same rate but with orthogonal phase. The absolute value of the Fourier transform at +3&nbsp;Hz is 0.5, which is relatively large. When added to the Fourier transform at -3&nbsp;Hz (which is identical because we started with a real signal), we find that the amplitude of the 3&nbsp;Hz frequency component is 1.

[[File:Onfreq.png|class=skin-invert-image|left|thumb|695x695px|Original function, which has a strong 3&nbsp;Hz component. Real and imaginary parts of the integrand of its Fourier transform at +3&nbsp;Hz.]]
{{clear}}However, when you try to measure a frequency that is not present, both the real and imaginary component of the integral vary rapidly between positive and negative values. For instance, the red curve is looking for 5&nbsp;Hz. The absolute value of its integral is nearly zero, indicating that almost no 5&nbsp;Hz component was in the signal. The general situation is usually more complicated than this, but heuristically this is how the Fourier transform measures how much of an individual frequency is present in a function <math> f(t).</math><gallery widths="360px" heights="360px" class="skin-invert-image">
File:Offfreq i2p.svg| Real and imaginary parts of the integrand for its Fourier transform at +5&nbsp;Hz.
File:Fourier transform of oscillating function.svg| Magnitude of its Fourier transform, with +3 and +5&nbsp;Hz labeled.
</gallery>

To re-enforce an earlier point, the reason for the response at &nbsp;<math>\xi=-3</math> Hz&nbsp; is because &nbsp;<math>\cos(2\pi 3t)</math>&nbsp; and &nbsp;<math>\cos(2\pi(-3)t)</math>&nbsp; are indistinguishable.  The transform of &nbsp;<math>e^{i2\pi 3t}\cdot e^{-\pi t^2}</math>&nbsp; would have just one response, whose amplitude is the integral of the smooth envelope: <math>e^{-\pi t^2},</math>&nbsp; whereas &nbsp;<math>\operatorname{Re}(f(t)\cdot e^{-i2\pi 3t})</math> is &nbsp;<math>e^{-\pi t^2} (1 + \cos(2\pi 6t))/2.</math>