# Design of butterworth filter

I was trying to design a butterworth filter of arbitrary order. Then i came across this presentation:

Lecture 9: Poles, Zeros & Filters, by Peter Cheung, Imperial College London

I understand most of it, except for 2 things,

• Now, we use the fact that $-1=e^{j\pi(2k-1)}$ and $j=e^{\frac{j\pi}{2}}$.
I have no idea what's he saying here.
• And does he take a normalized cut-off frequency? If yes, how do i change it to arbitrary cut-off frequency.

Now, we use the fact that $-1=e^{j\pi(2k-1)}$ and $j=e^{\frac{j\pi}{2}}$.
I have no idea what's he saying here.

Well, he's simply stating mathematical truths:

• $f(x)=e^{jx}$ is periodic with $2\pi$, and $e^{j\pi(0-1)}=e^{-j\pi}=-1$. If that's not clear, draw a unit cycle in the complex plane and remember Euler's formula.
• $j=e^{\frac{j\pi}{2}}$: same business. $\frac\pi2$ is a quarter cycle, and thus you end up at the top point of the unit circle.

You really need to be familiar with the properties of the complex exponential function, or else you will have a very hard time understanding pole/zero diagrams in general.

And does he take a normalized cut-off frequency?

Yes, he does (very explicit on slide 9).

If yes, how do i change it to arbitrary cut-off frequency.

Um, same slide, explicitly:

We can design filters for any other cut-off frequency by substituting $s$ by $\frac s{\omega_c}$.

• I dunno how i missed the second question. But first one, i genuinely did not know. Thanks! – Aashu10 Nov 19 '16 at 13:55
• I don't understand one more thing. How does he go from $s_{k}=cos\frac{\pi}{2n}(2k+n-1)+j\,sin\frac{\pi}{2n}(2k+n-1),\quad k=1,2,3,...,n$ to $H(s)=\frac{1}{(s-s_{1})(s-s_{2}).....(s-s_{n})}.$. I am clearly missing something – Aashu10 Nov 19 '16 at 14:01
• Don't ask new questions in the comments, please. $s_k$ is the formula for the $k$. pole. And if your look at the fraction, whenever $s$ takes the value of any $s_k$, you get a pole of $H(s)$. You really need to brush up your understanding of what a pole is! – Marcus Müller Nov 19 '16 at 14:04