New answers tagged butterworth
2
It is well-known that the poles of a normalized continuous-time $n^{th}$-order Butterworth lowpass filter lie on a semi-circle with radius $1$, centered at $s=0$:
$$p_k= e^{j\pi(2k-1+n)/2n},\qquad k=1,\ldots,n\tag{1}$$
Note that for odd order $n$, there is a single pole at $s=-1$.
Combining the complex conjugate pole pairs, we can construct a polynomial
$$D(...
1
Okay, since
$$ H(j \Omega) = H(s) \Big|_{s=j\Omega} $$
then
$$\begin{align}
\Big| H(j\Omega) \Big|^2 &= \frac{1}{1 + \left(\frac{\Omega}{\Omega_0}\right)^{2N}} \\
\\
\Big| H(s) \Big|^2 &= \frac{1}{1 + \left(\frac{s}{j\Omega_0}\right)^{2N}} \\
\\
\end{align} $$
Poles, $p_n$, occur at values of $s$ where the denominator goes to zero.
$$\begin{align}
...
1
Because your passband is too narrow, leading to zeros and poles very close to the unit circle. Quantization error finally results in poles outside the unit circle and consequently an unstable filter.
First design a bandpass filter with $f_0 = 1/60$ Hz, $f_1 = 1$ Hz and $f_s = 10$ Hz.
from scipy.signal import butter
from scipy.signal import freqz
from scipy....
Top 50 recent answers are included
