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:

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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?


2 Answers 2


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.



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


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

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

Hope this helps.


Edit: Oops, I forgot the i's.


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