New answers tagged lowpass-filter
0
The primary purpose of those filters is to pass the desired signal bandwith and reject the carrier feed-through at $\omega_h$ and the double frequency component at $2\omega$. Given that, I would recommend designing the filters with a passband and rejection requirement at those frequencies in mind with a goal of minimizing filter complexity (which means ...
2
Of course.
The difference is linear operation so you simply get
$$H(z) = H_1(z)-H_2(z)$$
The Z transform of a moving average filter of length N is simply $H_N(z) = \frac{1-z^{-N}}{1-z^{-1}}$ so in your case you get
$H(z) = \frac{1-z^{-10}}{1-z^{-1}}-\frac{1-z^{-21}}{1-z^{-1}}$
Pop in $z = e^{-j\omega}$ for your frequency of interest and solve for amplitude ...
2
The inverse Fourier transform of $H(f) = \operatorname{rect}\left(\frac{f}{2B}\right)$, the transfer function of the ideal lowpass filter of bandwidth $B$ Hz, is, as the OP says, $h(t) = 2B\operatorname{sinc}(2Bt)$. Now, the power transfer function of this ideal lowpass filter (ILPF) of bandwidth $B$ Hz is $|H(f)|^2$ to which we apply low-level math such as ...
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I have managed to get the whole thing to work.
Code to do so:
/**
*
* @author Edward Jenkins
* @version 1.2
*/
public abstract class SincSpecs implements ISimpleFilter{
// instance variables
private double cutoffFrequency;
private double sampleRate;
private double resonance;
private boolean highPass;
private double value;
...
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... that still satisfies the specifications?
First you need a complete set of specifications that can be quantitatively verified. That typically includes max allowable deviation from the target by frequency or by band either my RMS deviation, peak deviation, max absolute deviation, min attenuation for stopbands, etc. It may also include max latency, max ...
2
General Background: Representing a signal numerically in a computer requires the signal to be discrete. That means it's periodic in the other domain. When you use something like a DFT you need the signal to be discrete in both domains: That means the signals are ALSO periodic in both domains.
If I apply an ideal low pass filter, I should get an infinite ...
1
To eliminate gravity you need a high pass filter. From your source
This can be achieved by applying a high-pass filter. Conversely, a low-pass filter can be used to isolate the force of gravity.
You implemented a low pass filter. Instead try
$$G[n] = p \cdot G[n-1] + 0.5 \cdot (1-p) \cdot (a[n]-a[n-1])$$
where p is the location of your pole. I would start ...
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