1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '17 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 '17 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 '17 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 '18 at 15:48
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '17 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 '17 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 '17 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.