Fourier transformation and test function -- Question in the derivation

[Small bugs]

How does it work? (The derivative rules of FT)
We look at $$F[x(t)]=\hat{x}(f)$$
$$\mathcal{l} \text{ is a distribution, with}\tilde{x}=tx(t)$$
$$\mathcal{F}[Dl(x)]=\mathcal{F}l'(x)=2\pi il(\mathcal{F}\tilde{x})=2\pi i \mathcal{F}l(\tilde{x})$$
Till here I fully understand. But next step:
$$=2\pi i fFl(x)$$
How do they drag out f and make x~=tx(t) to x?

**Well, of course this is just a proof of derivative rule of FT, which is not hard
$$\mathcal{F}x'=2\pi i f \hat{x}$$

 

[jvicens]

The first step in understanding this proof is to understand the definition of the Fourier Transform (FT) and its properties. The FT is a mathematical operation that decomposes a function into its constituent frequencies. It is defined as follows:

$$\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x)e^{-2\pi i\xi x} dx$$

where $f(x)$ is the function being transformed and $\xi$ is the frequency variable.

One of the properties of the FT is that it is linear, which means that it follows the rules of linear algebra. This means that if we have two functions $f(x)$ and $g(x)$, and we take the FT of their sum, it is equal to the sum of their individual FTs. Mathematically, this can be written as:

$$\mathcal{F}[f(x)+g(x)] = \hat{f}(\xi) + \hat{g}(\xi)$$

Now, let's look at the first step in the proof that you mentioned:

$$\mathcal{F}[Dl(x)]=\mathcal{F}l'(x)=2\pi il(\mathcal{F}\tilde{x})=2\pi i \mathcal{F}l(\tilde{x})$$

In this step, we are using the linearity property of the FT. We are taking the FT of the derivative of a function $l(x)$, which is denoted as $Dl(x)$. This is equivalent to taking the derivative of the FT of $l(x)$, which we have denoted as $l'(x)$. This is a standard rule in calculus.

Next, we are using the fact that $\tilde{x}=tx(t)$. This means that we are multiplying the original function $x(t)$ by the variable $t$. In other words, we are stretching or compressing the function in the time domain. This is a common technique in signal processing.

Now, let's look at the next step:

$$=2\pi i fFl(x)$$

In this step, we are simply using the definition of the FT and plugging in the value of $\tilde{x}$, which we know is equal to $tx(t)$. This gives us:

$$=2\pi i \int_{-\infty}^{\infty