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I'm trying to derive an expression for the magnitude spectrum of the following sharpening filter.

$$ g(m,n) = \delta(m,n)+\lambda (\delta(m,n) - h(m,n)) $$

where $\lambda$ is some positive constant and $h(m,n)$ is a low pass filter.

$$h(m,n) = \begin{cases} \frac{1}{25}, & \text{for $|m| \leq 2, |n| \leq 2$} \\ 0, & \text{otherwise} \end{cases}$$

Using the linearity of Fourier transform, I get

$$ G(u,v) = 1+\lambda(1-H(u,v))$$

where $$ H(u,v) = \frac{1}{25} \cdot \frac{\sin\frac{5\mu}{2}\sin\frac{5\nu}{2}}{\sin\frac{\mu}{2}\sin\frac{\nu}{2}} \cdot e^{j\cdot 6(\mu+\nu)} $$

So, to get the magnitude spectrum, I make the following steps \begin{align*} |G(u,v)|^2 &= \big[(1+\lambda)-\lambda H(u,v)\big] \big[(1+\lambda)-\lambda H^*(u,v)\big] \\ & = (1+\lambda)^2-(1+\lambda)\lambda\cdot 2\Re \big\{H(u,v)\big\}+\lambda^2|H(u,v)|^2 \end{align*}

where $$ \Re\big\{H(u,v)\big\} = \frac{1}{25} \cdot \frac{\sin\frac{5\mu}{2}\sin\frac{5\nu}{2}}{\sin\frac{\mu}{2}\sin\frac{\nu}{2}} \cdot \cos(6(\mu+\nu)) $$

Therefore, to obtain the magnitude spectrum I should take the square root of the expression above.

However, the picture I get when I'm trying to plot it does not coincide with something that I'm expecting to see. I mean it's a high pass filter, hence there should be something like that.

enter image description here

But I get this.

enter image description here

Maybe there is a way to simplify this expression or maybe there is something that I'm doing wrong.

Thank your for your help!

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UPD: I've figured it out myself. I was inattentive with the exponentials while deriving the expression for the low-pass filter $H(u,v)$. There shouldn't be $\cos(6(\mu+\nu))$ term, then the result is fine and $$\Re \big\{H(u,v)\big\}=\frac{1}{25} \cdot \frac{\sin\frac{5\mu}{2}\sin\frac{5\nu}{2}}{\sin\frac{\mu}{2}\sin\frac{\nu}{2}}$$

Therefore, \begin{align*} |G(u,v)|^2 &= \big[(1+\lambda)-\lambda H(u,v)\big] \big[(1+\lambda)-\lambda H^*(u,v)\big] \\ & = (1+\lambda)^2-(1+\lambda)\lambda\cdot 2\Re \big\{H(u,v)\big\}+\lambda^2|H(u,v)|^2 = [1+\lambda - \lambda H(u,v)]^2 \end{align*}

And the result is exactly what I was expecting to get.

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