# Ramp function as derivative in frequency domain?

It is said that to get Laplacian of Gaussian in frequency domain, we may multiply the Fourier transform of Gaussian with two differentiating ramp function (1 ramp gives 1 order of derivative).

The description from the material that I was following: And the file can be found here

So how could we possibly get derivatives by multiplying ramp functions in the frequency domain? How does the math work here?

You may be familiar with the classical properties of the Fourier transform. One of them is the one regarding differentiation in the time domain:

$$x(t)\xrightarrow{\mathscr{F}}X(j\omega) \implies \frac{d}{dt}x(t)\xrightarrow{\mathscr{F}}j\omega X(j\omega)$$

And there appears the ramp function, $j\omega$.

The demonstration of why this is true is straightforward. If we have a signal $x(t)$ with Fourier transform $X(j\omega)$:

$$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j\omega)e^{j\omega t} \ \mathrm{d}\omega$$

Then by differentiating both sides:

$$\frac{d}{dt}x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}j\omega X(j\omega)e^{j\omega t} \ \mathrm{d}\omega$$

And that's the (pretty simple) proof.

Suppose:

$$f(t) = C_0 e^{i0t} + C_1 e^{it} + C_2 e^{i2t} + C_3 e^{i3t} + C_4 e^{i4t} + ...$$

Then

$$f'(t) = 0i \cdot C_0 e^{i0t} + i \cdot C_1 e^{it} + 2i \cdot C_2 e^{i2t} + 3i \cdot C_3 e^{i3t} + 4i \cdot C_4 e^{i4t} + ...$$

Which is the ramp function times the DFT coefficients as DFT coefficients.

Hope this helps.

Ced

Edit: Oops, I forgot the i's.