Calculating cutoff frequency for Butterworth filter I have a problem while calculating cutoff frequency, suppose we have these specs. 


Firstly, I calculated the order of the filter and got $N=5.8858$ and round it up to get $N=6$.

Now I'm supposed to get $\Omega_c$. Using these equations:

\begin{cases}1+\left(\frac{0.2\pi}{\Omega_c}\right)^{2N} = \left(\frac{1}{0.89125}\right)^2 \quad&amp;\quad (1)\\
1+\left(\frac{0.3\pi}{\Omega_c}\right)^{2N} = \left(\frac{1}{0.17783}\right)^2\quad&amp;\quad (2)
\end{cases}

Now with $N=6$ and $T=1$, substituting in $(1)$

\begin{align}
\left(\frac{0.2\pi}{\Omega_c}\right)^{12} &amp;=\left(\frac{1}{0.89125}\right)^2 - 1 =0.25893\\
\implies (\Omega_c)^{12} &amp;= \frac{{(0.2\pi)}^{12}}{0.25893}\\
 &amp;= 40.29
\end{align}
But in the textbook it says $\Omega_c = 0.7032$, what I did wrong? Any help would be appreciated.
Everythng is alright. You made a mistake in the final calculation.

$$\Omega_c^{12} = \frac{{(0.2\pi)}^{12}}{0.25893} \implies \Omega_c = \sqrt[12]{\frac{{(0.2\pi)}^{12}}{0.25893}} =\frac{0.2\pi}{\sqrt[12]{0.25893}} = 0.7032$$

which is the desired result.