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Intuitively it seems that to get a function back that was integrated, you would take the derivative. Instead, with the Fourier Transform we take an area under a curve of a modified function, and to get the function back we again take the area under a modified resulting curve.

I am trying to more intuitively understand the concept of the FT and inverse FT. I know there is a reason this double integration works, and it seems like it has to do with the fact that the resulting function, taken one way or the other, ends up being a function of another variable that you imposed on it with e^(+/-)jwt and the integral removes the original independent variable from the equation. Maybe I need to spend more time thinking about what it means to integrate a variable out.

I'm just curious if anyone has an explanation for how the FT and inverse FT fit together that makes the functions a little more intuitive. (Sorry if the question is a little open-ended seeming, but I think it's important in mathematics to intuitively understand equations where you can. At least for me, it's always helped me a lot with gaining confidence and using the equation faster.)

Thank you!

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2 Answers 2

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Integration and differentiation are inverse operations only in the case of either indefinite integrals or integrals with the independent variable as upper bound:

$$G(x)=\int_a^xg(t)dt\quad\Longleftrightarrow\quad G'(x)=g(x)\tag{1}$$

In the case of the Fourier transform we integrate from $-\infty$ to $\infty$, and we integrate over a function with two independent variables, namely $x(t)e^{-j2\pi ft}$, so $(1)$ doesn't apply.

The Fourier transform is an integral transform of the form

$$X(f)=\int_{-\infty}^{\infty}x(t)K(t,f)dt\tag{2}$$

where $K(t,f)$ is called the kernel of the transform. The integration kernel of the Fourier transform is

$$K(t,f)=e^{-j2\pi ft}\tag{3}$$

The inverse transform is achieved by using the inverse kernel which, in case of the Fourier transform, is simply $e^{j2\pi ft}$:

$$x(t)=\int_{-\infty}^{\infty}X(f)e^{j2\pi ft}df\tag{4}$$

I'm not sure if it helps your intuition, but the validity of $(4)$ given the transform $(2)$ with $K(t,f)=e^{-j2\pi ft}$ can be shown (in a not very rigorous way) using the following representation of the Dirac delta function:

$$\delta(t)=\int_{-\infty}^{\infty}e^{j2\pi ft}df\tag{5}$$

With $(5)$ we can express $x(t)$ as

\begin{align*} x(t) &= \int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau \\ &= \int_{-\infty}^{\infty}x(\tau)\int_{-\infty}^{\infty}e^{j2\pi f(t-\tau)}dfd\tau \\ &= \int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty}x(\tau)e^{-j2\pi f\tau}d\tau\right] e^{j2\pi ft}df \\ &= \int_{-\infty}^{\infty}X(f)e^{j2\pi ft}df\tag{6} \end{align*}

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  • $\begingroup$ I was thinking of answering this with the very same non-rigorous trick with $\delta(t-\tau)$. I would put it differently than you, Matt. But essentially the same point. May I be a little lazy and just insert into your answer what I think you're missing? It's about how to get to $$\delta(t)=\int\limits_{-\infty}^{\infty}e^{j2\pi ft} \, \mathrm{d}f$$ $\endgroup$ Nov 30, 2023 at 19:08
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    $\begingroup$ @robertbristow-johnson: You may, but don't use the inverse Fourier transform! :) $\endgroup$
    – Matt L.
    Nov 30, 2023 at 20:09
  • $\begingroup$ Agreed. That would defeat the whole point. $\endgroup$ Nov 30, 2023 at 21:46
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Intuition is always a very personal thing: what's intuitive to me might be confusing to you and vice versa, so take this with a grain of salt.

I like to think about the Fourier Transform as an orthogonal basis. We'll be using the DFT with sums first and start backwards with the inverse DFT which is given as

$$x[n] = \frac{1}{N}\cdot \sum_{k=0}^{N-1}X[k]\cdot e^{j2\pi\frac{k}{N}n}$$

The interpretation here is

Every sequence of length N can be represented as a weighted sum of N complex exponentials $W_k[n] = e^{j2\pi\frac{k}{N}n}$

We call the parameter $k$ the "frequency". We can consider the complex exponentials the "building blocks" from which we can assemble any sequence we want by choosing the proper "amount" of each basis function. That amount are the weighting factors $X[k]$.

There are other choices for bases functions we could use, but the complex exponentials just have some nice physical and mathematical properties.

One of these properties is that they are orthogonal with respect to $-k$, i.e.

$$\sum_{n=0}^{N-1} W_k[n] \cdot W_m[n] = N \cdot \delta[m+k] = \begin{cases} N & k = -m \\ 0 & k \ne -m \\ \end{cases} $$

where $\delta[n]$ is the Kronecker delta (or unit impulse). So if we want to figure out how much of a specific frequency is in a signal, we just need to multiply with the "orthogonal" basis function $W_{-k}$. Let's try that:

$$\sum_{n=0}^{N-1}x[n] W_{-k}[n] = \sum_{n=0}^{N-1} \left[\frac{1}{N}\cdot \sum_{m=0}^{N-1}X[m]\cdot W_m[n]\right]W_{-k}[n] =$$

$$ \frac{1}{N}\cdot \sum_{m=0}^{N-1}X[m] \left[\sum_{n=0}^{N-1}\cdot W_m[n]\cdot W_{-k}[n]\right] = \frac{1}{N}\cdot \sum_{m=0}^{N-1}X[m] \left[ N \cdot \delta[m-k] \right]= X[k]$$

So we have just derived the forward DFT just using the definition of the inverse DFT and the orthogonality property.

Both forward and inverse transform uses the same operation that's used in the definition for orthogonality: sums in this case.

The same argument can be done for the CFT as well. Here the orthogonality property uses an integral

$$\int e^{\omega_1 t} \cdot e^{\omega_2 t} dt = \delta(\omega_1+\omega_2) $$

so both forward and inverse transform use the integral as well.

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