The magnitude spectrum of a sharpening filter 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.
$$h(m,n) = \begin{cases}
\frac{1}{25}, &amp; \text{for $|m| \leq 2, |n| \leq 2$} \\
 0, &amp; \text{otherwise}
\end{cases}$$
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 &amp;= \big[(1+\lambda)-\lambda H(u,v)\big] \big[(1+\lambda)-\lambda H^*(u,v)\big] \\
&amp; = (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.

But I get this.

Maybe there is a way to simplify this expression or maybe there is something that I'm doing wrong.
Thank your for your help!
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 &amp;= \big[(1+\lambda)-\lambda H(u,v)\big] \big[(1+\lambda)-\lambda H^*(u,v)\big] \\
&amp; = (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.