# What is the Center Frequency of Gabor Filter?

If the standard form of a Gabor function is as follows,

$$g_{\lambda, \theta,\varphi, \sigma,\gamma}(x, y)=\exp\left(-\frac{x'^2+\gamma^2y'^2}{2\sigma^2}\right)\cos\left(2\pi\frac{x'}{\lambda}+\varphi\right)$$

where $$x'=x\cos\theta+y\sin\theta\\ y'=-x\sin\theta+y\cos\theta$$ How can I find the Center Frequency from this equation?

• That is only the real part of a Gabor filter. Is that what you actually meant or did you forget the imaginary one? May 28, 2016 at 13:30
• @Tendero, I am actually working with crack detection algorithms. So, I am trying to implement a Gabor Filter Bank. I am seeing that this guy youtube.com/watch?v=-NZakhhB_Do is only real part. So, I understood that the real part is only what I need. What do you say?
– user18425
May 28, 2016 at 16:11
• I was just checking that you hadn't forgotten about the imaginary one, just that May 28, 2016 at 16:16
• @Tendero, this is related dsp.stackexchange.com/questions/31061/…
– user18425
May 28, 2016 at 16:44
• @Tendero this is also related dsp.stackexchange.com/questions/31046/…
– user18425
May 28, 2016 at 16:47

The center frequency of a Gabor function is given by the reciprocal of its wavelength, which in turn is determined by the value of the parameter $$\lambda$$ in the equation.

To find the value of $$\lambda$$, we can rewrite the cosine term as follows:

$$\cos\left(2\pi\frac{x'}{\lambda}+\varphi\right) = \cos\left(\frac{2\pi x'}{\lambda} \right)\cos(\varphi) - \sin\left(\frac{2\pi x'}{\lambda} \right)\sin(\varphi)$$

Substituting the expressions for $$x'$$ and $$y'$$ in terms of $$x$$ and $$y$$ yields:

$$\cos\left(2\pi\frac{x\cos\theta+y\sin\theta}{\lambda}+\varphi\right) = \cos\left(\frac{2\pi x\cos\theta+2\pi y\sin\theta}{\lambda} \right)\cos(\varphi) - \sin\left(\frac{2\pi x\cos\theta+2\pi y\sin\theta}{\lambda} \right)\sin(\varphi)$$

Then, we can use the trigonometric identity $$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$$ to write:

$$\cos\left(2\pi\frac{x\cos\theta+y\sin\theta}{\lambda}+\varphi\right) = \cos\left(\frac{2\pi x}{\lambda}\cos\theta+\frac{2\pi y}{\lambda}\sin\theta\right)\cos(\varphi) - \sin\left(\frac{2\pi x}{\lambda}\cos\theta+\frac{2\pi y}{\lambda}\sin\theta\right)\sin(\varphi)$$

We can now recognize the expression inside the cosine and sine functions as the dot product of the vector $$(x,y)$$ with the vector $$(\cos\theta,\sin\theta)$$, and write:

$$\cos\left(2\pi\frac{x\cos\theta+y\sin\theta}{\lambda}+\varphi\right) = \cos\left(\frac{2\pi}{\lambda}(x\cos\theta+y\sin\theta)\right)\cos(\varphi) - \sin\left(\frac{2\pi}{\lambda}(x\cos\theta+y\sin\theta)\right)\sin(\varphi)$$

Finally, we can use the fact that the wavelength $$\lambda$$ corresponds to a full period of the cosine term, which occurs when the argument of the cosine function increases by $$2\pi$$, to obtain:

$$\frac{2\pi}{\lambda} = 1$$

Therefore, the center frequency is given by:

$$f_c = \frac{1}{\lambda} = \frac{1}{2\pi}$$

Note that the center frequency is independent of the other parameters of the Gabor function, such as the orientation $$\theta$$, the phase $$\varphi$$, and the width parameters $$\sigma$$ and $$\gamma$$.

• You are saying that lambda is always 2pi? But it’s a parameter settable by the user! Aug 7, 2023 at 13:49
• @CrisLuengo, Please feel free to edit my poster answer. I made it a community wiki. Aug 7, 2023 at 13:52