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How tf is this Lowpass?!

Anonymous Fri 11 Oct 04:41:51 2019 No.11048768 View Reply Original Report

Quoted By: >>11049423 >>11049993

$H(z)=-\frac{1}{8}z^{-2}+\frac{1}{4}z^{-1}+\frac{3/4}+\frac{1}{4}z-\frac{1}{8}z^{2}$

Anonymous

Anonymous Fri 11 Oct 2019 04:45:20 No.11048778 Report

Quoted By:

$$H(z)=-\frac{1}{8}z^{-2}+\frac{1}{4}z^{-1}+\frac{3/4}+\frac{1}{4}z-\frac{1}{8}z^{2}$[\math]$

Anonymous

Anonymous Fri 11 Oct 2019 04:46:25 No.11048783 Report

Quoted By:

$H(z)=-\frac{1}{8}z^{-2}+\frac{1}{4}z^{-1}+\frac{3/4}+\frac{1}{4}z-\frac{1}{8}z^{2}$

Anonymous

Anonymous Fri 11 Oct 2019 04:47:29 No.11048784 Report

Quoted By: >>11049399

$H(z)=-\frac{1}{8}z^{-2}+\frac{1}{4}z^{-1}+\frac{3/4}+\frac{1}{4}z-\frac{1}{8}z^{2}$

Anonymous

Anonymous Fri 11 Oct 2019 04:48:40 No.11048789 Report

Quoted By:

$$H(z)=-\frac{1}{8}z^{-2}+\frac{1}{4}z^{-1}+\frac{3}{4}+\frac{1}{4}z-\frac{1}{8}z^{2}$[\math]$

Anonymous

Anonymous Fri 11 Oct 2019 04:49:42 No.11048792 Report

Quoted By:

$H(z)=-\frac{1}{8}z^{-2}+\frac{1}{4}z^{-1}+\frac{3}{4}+\frac{1}{4}z-\frac{1}{8}z^{2}$

Anonymous

Anonymous Fri 11 Oct 2019 04:51:07 No.11048797 Report

Quoted By: >>11049399

$H(z)=-\frac{1}{8}z^{-2}+\frac{1}{4}z^{-1}+\frac{3}{4}+\frac{1}{4}z-\frac{1}{8}z^{2}$

Finally.

Anonymous

Anonymous Fri 11 Oct 2019 09:33:49 No.11049381 Report

Quoted By:

$H(z)=-\frac{1}{8}z^{-2}+\frac{1}{4}z^{-1}+\frac{3}{4}+\frac{1}{4}z-\frac{1}{8}z^{2} [\math]$

Anonymous

Anonymous Fri 11 Oct 2019 09:34:52 No.11049383 Report

Quoted By:

How did you do it?

Anonymous

Anonymous Fri 11 Oct 2019 09:45:14 No.11049399 Report

Quoted By:

>>11048784
>>11048797
>gets it right on third try
>keeps fucking it up for another two tries

Anonymous

Anonymous Fri 11 Oct 2019 09:47:33 No.11049404 Report

Quoted By:

\begin{equation}
H(z)=-\frac{1}{8}z^{-2}+\frac{1}{4}z^{-1}+\frac{3}{4}+\frac{1}{4}z-\frac{1}{8}z^{2}
\end{equation}

Anonymous

Anonymous Fri 11 Oct 2019 09:59:02 No.11049423 Report

Quoted By: >>11049471

>>11048768
Doesn't seem like a low-pass filter, it massively amplifies large and very small z:s.
H goes off to negative infinity at large z:s and z=0, and has local maxima at z=1,H=1 and z=-1,H=0.

by the way, for those confused about $\textrm{/sci/}\TeX$ , type
$H(z)=-\frac{1}{8}z^{-2}+\frac{1}{4}z^{-1}+\frac{3}{4}+\frac{1}{4}z-\frac{1}{8}z^2$
without the space in " $" to get <script type="math/tex">H(z)=-\frac{1}{8}z^{-2}+\frac{1}{4}z^{-1}+\frac{3}{4}+\frac{1}{4}z-\frac{1}{8}z^2$ .
Getchoself 4chan-x so you can get TeX previews before you post

Anonymous

Anonymous Fri 11 Oct 2019 10:42:22 No.11049471 Report

Quoted By:

>>11049423
Thank you,
${{{{N_i}_g}_g}_e}_r$

${{{{N_i}_g}_g}_e}_r$

${{{{N_i}_g}_g}_e}_r$

${{{{N_i}_g}_g}_e}_r$

Anonymous

Anonymous Fri 11 Oct 2019 16:21:22 No.11049990 Report

Quoted By:

$f(x)=xcos$\theta$^x [\math]$

Anonymous

Anonymous Fri 11 Oct 2019 16:23:45 No.11049993 Report

Quoted By:

>>11048768
Draw the Bode plot and find out why. Imagine fucking up TeX this badly.

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