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I have a problem while calculating cutoff frequency, suppose we have these specs. enter image description here

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&\quad (1)\\ 1+\left(\frac{0.3\pi}{\Omega_c}\right)^{2N} = \left(\frac{1}{0.17783}\right)^2\quad&\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} &=\left(\frac{1}{0.89125}\right)^2 - 1 =0.25893\\ \implies (\Omega_c)^{12} &= \frac{{(0.2\pi)}^{12}}{0.25893}\\ &= 40.29 \end{align} But in the textbook it says $\Omega_c = 0.7032$, what I did wrong? Any help would be appreciated.

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  • $\begingroup$ Note that the frequency response is for a continuous time filter, whereas the filter specifications are for a discrete-time filter. The piece you are missing is application of the impulse invariance method to go from the continuous time filter to a discrete time filter. $\endgroup$
    – Atul Ingle
    Jan 6, 2017 at 0:56
  • $\begingroup$ Yes I know that, those specs are for designing a discrete filter, but it supposes to be mapped into specs for a continuous filter to use one of the "continuous filter designs" like Butterworth in my case. Hence the impulse invariance method job is to do the mapping from a continuous filter into a discrete filter. Correct me if I did a mistake again. Thanks in advance $\endgroup$
    – Hossam Houssien
    Jan 6, 2017 at 1:47
  • $\begingroup$ you might like the bilinear transform mapping from $s$ to $z$ better than impulse invariant. but to each his own. $\endgroup$
    – robert bristow-johnson
    Jan 6, 2017 at 4:54
  • $\begingroup$ can you explain how did you calculate order and from where those equations come from to calculate Wc? $\endgroup$
    – Vivek
    Jun 21, 2018 at 15:48

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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.

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  • $\begingroup$ Thanks, I knew my mistake, I was assuming that $\pi = 180$ but I tried with $\pi=3.1415...$ Can I just ask you why did we use it like that and not $180 degrees$ $\endgroup$
    – Hossam Houssien
    Jan 6, 2017 at 18:39
  • $\begingroup$ Because $\pi$ is $\pi$ and not $180º$. Don't get confused with these two quantities, they are not the same thing, even though sometimes they are "equivalent" in a sense. However, in this case you can see that $\Omega_c$ must be a scalar, and if you don't use $\pi=3.14...$ then you would have units of degrees, which would be nonsense. $\endgroup$
    – Tendero
    Jan 6, 2017 at 19:04
  • $\begingroup$ @Tendro Thanks again, now I got it. Especially for the part "$\Omega_c$ must be a scalar" $\endgroup$
    – Hossam Houssien
    Jan 6, 2017 at 21:30

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