New answers tagged filters
1
The filter is a discrete FIR filter, so the transfer function would be given as:
$$H(z) = .0287 + 0.1430z^{-1} + 0.3283z^{-2} + 0.3283z^{-3}+0.1430z^{-4}+0.0287z^{-5}$$
With the frequency response given by using the unit circle for the complex variable $z$, as in $z = e^{j\omega}$ for $\omega \in [0, 2\pi)$, with the sampling rate normalized to $\omega = 2\...
2
GMSK doesn't just use a Gaussian filter instead of a raised-cosine filter. It uses a Gaussian filter on the phase, before applying it to the modulator. This makes it a nonlinear operation.
When a raised-cosine filter is applied to some modulated signal, it is applied after modulation, as a linear operation.
So there's a whole lot of convenient rules about ...
2
(Assuming the context is FSK signals, where the pulse shape is applied to the phase, instead of the amplitude as in linear modulation).
The main reason is that the sideband and out-of-band emissions of GMSK are lower than those produced with a raised cosine filter. See for example this plot from Wikipedia: https://en.wikipedia.org/wiki/File:GMSK_PSD.png
This ...
7
The logical implications are the following:
"non-recursive" $\Longrightarrow$ FIR
IIR $\Longrightarrow$ "recursive"
But the opposites are not necessarily true because a FIR system can be implemented recursively (transfer function poles can be cancelled by zeros).
Of course, when referring to "recursive" or "non-recursive" we always talk ...
1
is it correct to say that FIR system is always non-recursive?
You answer that yourself:
We could express an finite accumulator up to past N inputs (FIR system) in both non-recursive and recursive forms.
exactly. A common example of a recursive filter that's in fact an FIR is the CIC filter.
Also is it correct to say that a non-recursive system is always ...
0
The reason the two tone method works to measure group delay as given is because group delay is determined by the derivative of phase with respect to frequency. Since the system is linear phase (constant delay at all frequencies) then we can over any two frequencies determine the delay from the difference in phase between any two frequencies at the input and ...
1
I'm aware I can subtract the mean of the data
That's actually a high-pass filter!
You can see in the time domain output that the filter takes some time to settle.
That is true for any causal system, i.e. any system that can't look into the future; you'll want to read up on "group delay"
Can anyone help with me with what parameters to tweek to ...
3
The filter you desire is often called a “DC removal” or a “DC cancelation” filter. Unfortunately there’s no way to design a useful IIR DC removal filter that has no “settling time”. Your designed filter’s “transition region” is the lower passband frequency (50 Hz) minus the upper stopband frequency (10 Hz). So your designed filter’s transition region is 40 ...
1
Note that the squared magnitude of the frequency response is given by
$$\big|H(j\omega)\big|^2=\frac{1}{1+\epsilon^2T^2_N\left(\frac{\omega}{\omega_c}\right)}\tag{1}$$
In the $s$-domain we have
$$H(s)H(-s)=\frac{1}{1+\epsilon^2T^2_N\left(\frac{s}{j\omega_c}\right)}\tag{2}$$
Computing the zeros of $(2)$ does not only result in the zeros of the filter's ...
1
In an RC high-pass filter, the output is taken across the resistor instead of the capacitor. Otherwise, if the output was taken across the capacitor, the RC high-pass filter would start to work as an RC low-pass filter.
An example RC high-pass filter circuit is given below:
The basic idea of using an RC high-pass filter is to filter out the input signal’s ...
0
An edge remains a concept that is a bit complicated to define, as it may involve a certain level of interpretation. For a pixel-wise point of view, I consider that a potential edge breaks down into three main features: it is singular (non-continuous, non-differentiable) across one direction, and more regular (smooth) in the other direction, at a certain ...
3
The purpose of pulse shaping filters is not to overcome ISI as is implied in the OP's question. The only reason for using a pulse shaping filter is spectral efficiency, and in the process ISI can be introduced if not done properly. In order to limit bandwidth, pulse shaping must extend the time domain response for each pulse beyond a symbol boundary, but can ...
0
OFDM itself doesn't use a pulse shaping filter. The spectral shape is inherent.
There are systems which apply a pulse shaping filter to the OFDM signal, but I'd call this "OFDM plus a pulse shaping filter", not just "OFDM".
5
If you remove (for the time being) that leading factor $A$ as a constant gain factor:
$$H(s)=\frac{s^2+\left(\frac{\sqrt{A}}{Q}\right)s + A}{As^2 + \left(\frac{\sqrt{A}}{Q}\right)s + 1}$$
what you get then is a symmetric, but otherwise general shelf that could be equally described as "LowShelf" or "HighShelf". In dB, the gain at the low ...
3
If you use a parameterization with a (pole or zero) frequency and a Q-factor for numerator and denominator of a biquadratic function you get the following general second-order transfer function
$$H(s)=G_{\infty}\frac{s^2+\frac{\omega_z}{Q_z}s+\omega_z^2}{s^2+\frac{\omega_p}{Q_p}s+\omega_p^2}\tag{1}$$
For a low shelving filter we want $H(\infty)=1$, i.e., $G_{...
0
While the exact confined Gaussian window mention above is hard to compute, it has an almost perfect approximation:
the "approximate confined gaussian window" which is simple to compute, see
https://www.researchgate.net/publication/261717241_Discrete-time_windows_with_minimal_RMS_bandwidth_for_given_RMS_temporal_width
0
do I need to change the number to “2” so that it is a second-order
Yes
and should I also double the cut-off frequency i.e change “W” to 0.3 from 0.15.
No
Strictly speaking, doubling up the filter will make the gain at the cutoff frequency -6dB instead of -3dB. So you have created a Linkwitz-Riley and not a Butterworth filter.
If you want a gain of exactly ...
2
I think the basic premise of your question is incorrect. Interpolating coefficients doesn't lead to unstable filters in biquads; moreover, it is very easy to prove. Just plot a1 vs a2 and you'll see that the stability region for the triplet of coefficients in the denominator is a convex spaces. Because of that, interpolating coeffs (as long as the same ...
0
Based on what you explained, the output signal $y(t)$ is given by:
$$
y(t) = \frac{dx(t)}{dt}
$$
where x(t) is the input signal.
By transforming this problem in the Laplace domain, we can re-express the above equation as:
$$
Y(s) = sX(s) - x(0^{-})
$$
If we know that $x(0^{-})=0$, then this can be simplified as $Y(s) = sX(s)$. Using this expression we can ...
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