New answers tagged filtering
1
I see that the difference falls down as 20dB/dec.
Because you high pass also only has a 20 dB/octave slope. For first order filters the upwards slope of the high pass is symmetrical to the downwards slope of (1 - highpass) other than bilinear distortion. Your difference cannot fall faster than the high pass rises.
How can I make it fall faster?
You need ...
3
The reason is the decorrelation due to the varying phase shift between the input and output. See the plot below showing the phase of the OP's high pass filter, and how closely it follows the cancellation when plotted on a log log scale (the higher frequency components of the two signals are better aligned in time and therefore have higher correlation leading ...
1
Those kind of algorithms are called Non Local algorithms.
The most known algorithms of this family is the - Non Local Means which is a decent Noise Reduction (Denoising) algorithm.
Until the Deep Learning boom, this approach has been extended and usually means working in the Patch Space of image - Patch Based Models and Algorithms for Image Denoising: A ...
1
If you're familiar with LaPlace transforms, you can see the Z transform by analogy.
The unit circle is equivalent to the jw axis, with zero frequency at 1+j0 and the Nyquist rate at -1+j0.
1
BIBO stability of LTI systems implies that their impulse response is absolutely summable, that is,
\begin{equation}
\sum_{n=-\infty}^{+\infty}|h(n)| < +\infty
\end{equation}
That exact same relationship is a sufficient condition for the Fourier Transform of the impulse response - the so-called Frequency Response - to converge.
Convergence of the ...
0
We get the Fourier Transform of a signal at the unit circle. If the ROC does not include the unit circle, that means that the Fourier transform does not converge which means that the system is unstable. Also, please read bores signal processing basics website and Alan Oppenheim. Its explained really well there.
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