We know that integration gives area under the curve. And that FT is also an integration. How do we interpret FT in terms of the area under the curve especially because e^(jwt) is a complex term? In general, while dealing with complex terms in integration can we relate to some area?

  • $\begingroup$ Its best to interpret it as an infinite sum, rather than an area. $\endgroup$
    – Paul
    Commented May 1, 2015 at 14:27
  • $\begingroup$ Since the function is continuous it is actually similar to area under the curve rather than a sum. $\endgroup$ Commented May 1, 2015 at 14:50

2 Answers 2



$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt$$

you get for the real part of $X(\omega)$ (assuming that $x(t)$ is real)

$$X_R(\omega)=\int_{-\infty}^{\infty}x(t)\cos(\omega t)dt\tag{1}$$

and for the imaginary part

$$X_I(\omega)=-\int_{-\infty}^{\infty}x(t)\sin(\omega t)dt\tag{2}$$

So if you like, from (1) and (2) you can interpret the real part of the Fourier transform as the area under the curve $x(t)\cos(\omega t)$, and the imaginary part would then be the area under the curve $-x(t)\sin(\omega t)$.

Note that interpreting an integral as the area under a curve does not always help intuition. In the case of the Fourier transform it is more natural to interpret the integrals (1) and (2) as inner products.

  • $\begingroup$ @seethaRama I agree with Matt L. in the sense that the Fourier transform's most natural interpretation is as an inner product. If you want to think in terms of "area under the curve" I advice you to think about the inner product of two function rather than the Fourier transform. You could think about why there is a sum in the inner product of two vectors and how it is used to measure dissimilarity, extend these tought to the continuous case then to the fourier case $\endgroup$ Commented May 1, 2015 at 21:02

Actually the Fourier transform (and similar transforms like Laplace, and so on ) is not an area calculation, but a change of basis.

This is more intuitive if you see a (continous or discrete) signal $x$ as a vector expressed in some basis (the time-domain basis). i.e

$$x = \sum a^i x_i$$

Then by linear algebra a change of basis is :

$$y = \sum b^i y_i$$

where $a$ and $b$ basis coefficients are related through a linear transformation. Now go to the continous limit (so sums become integrals) and use the roots of unity (i.e $e^{2\pi\omega t}$) as basis components and you have the Fourier transform.

You see it is not related to an area under the signal curve, but it is a linear change a of basis to another (the frequency) domain. This domain encodes the signal in such a way that further information can be extracted more easily (change of signal representation)


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