New answers tagged filters
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Obviously, you might need to use the given frequency somehow to solve the problem. Guessing alone will usually not suffice. Since this is a homework type problem, and since you still need to learn a lot of basics, I'll give you a few hints to get you started.
First, the frequency domain behavior can be obtained from a (stable) transfer function by ...
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I don't know why but I couldn't get the equations to output any audio when using the Euler method given. Maybe I screwed it up somehow or this is not the best place for that type of approximation.
When I substituted $s=(2(z^{-1}−1))/(T(z^{-1}+1))$ it works perfectly. So problem solved either way. Thanks.
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First of all, it's important to understand that there is no single best way to transform a continuous-time system to a discrete-time system. The method you're using is called backward Euler method, and it is defined by the mapping
$$s\leftarrow\frac{1-z^{-1}}{T}\tag{1}$$
Note that in $(1)$ you scale by $1/T$, where $T$ is the sampling interval (i.e., $1/T$ ...
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It depends on what you mean by "meaningful". All outputs are meaningful in the sense that they combine the current input value with past input values. Initially, there are of course no past input samples. But if you agree that "no past samples" means "past samples with value zero" then that's exactly what happens if the delay elements are initialized with ...
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If it's a linear phase FIR filter, then the inputs signals will be shifted (delayed) by an amount of group delay at the output.
For a linear phase FIR filter of length $L = 2K+1$ the group delay will be $N = K$ samples. For even length $L = 2K$ FIR filters it will be at $N = (L-1)/2$ ; half-sample position.
For non-linear phase FIR filters, group delay ...
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The issue is that you're violating Nyquist which requires that the sampling rate be at least twice the rate of the highest frequency in the signal. In your case all of the sines that you're generating above 5000 Hz are folding back. For example, a 5100 Hz tone (5000+100) would fold back to 4900 Hz (5000-100) and so on.
The 1 kHz tone you highlighted is the ...
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The answer was there is no need to incorporate sample rate at all. Since I am using the simple case where Hb(s) is a constant, I just had to multiply that against the summed samples and subtract the result from both delay lines. It works irrespective of sample rate. I tried it with sampling at 11 kHz and 96 kHz and it sounds the same.
I don't really ...
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The answer was there is no need to incorporate sample rate at all. Since I am using the simple case where Hb(s) is a constant, I just had to multiply that against the summed samples and subtract the result from both delay lines. It works irrespective of sample rate. I tried it with sampling at 11 kHz and 96 kHz and it sounds the same.
I don't really ...
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The question is not very specific so this will also be just a general answer.
The stability of a composite linear-time-invariant (LTI) system composed of smaller LTI systems cannot be deduced from the stability of the component systems if the composition introduces feedback. Instead, you should test the stability of the full composite system. The system is ...
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I believe, mike, that the answer to your question is likely the Bilinear Transform. Make sure you identify the significant frequency (likely the resonant frequency of the LPF1 and LPF2 or $H_b(s)$ and apply prewarping to that frequency, so that the digital filter hits at the same place that the analog filter. Remember the transfer functions of the two ...
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Units of s represent the Laplace transform while units of z represent the z transform. It is often much easier to transform a time domain (units of t) signal to Laplace given that it translates integro-differential equations into simple algebra. The z transform does the same thing, but takes advantage of the repetition from sampling to make the transform ...
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Everything becomes very clear once you cinsder the following pair of signals and transforms:
If $g[n]$ has a Z-transform $G(z)$ then the expanded (zero stuffed) signal $h[n]$ has the Z-transform $H(z) = G(z^N)$, where
$$ h[n] = \begin{cases} {g[n/N] ~~~,~~~ n = m N \\ 0 ~~~,~~~ \text{otherwise} \tag{1}}\end{cases} $$
Looking at the given Z-transform :
$...
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@Ben I try to implement your idea and actually the amplitude of the signal is obtained starting from two components in quadrature.
But when i add an external signal, of the same amplitude, to my input the result is an oscillating signal
How can i reject the noise?
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For the cross correlation of the output of the filter with the input you should see the result of the two impulses in your filter convolved with the cross correlation properties of your actual waveform. If your waveform is random such that there is only a correlation at 0 lag then the result here would be a positive correlation peak at lag zero and then an ...
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I'd consider why you really want to do this - I personally can't think of a reason why I'd want to downsample to a specific sample number but I don't know your project
Floating an alternate idea, you could downsample until you're near around that level of decimation and then truncate? It won't be 100 samples exactly but it might be easier in the long run to ...
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First, all of these routines act on an input array. Your comment "values beyond the boundary of the signal are NOT zeros" implies that you want to process a continuous signal, or at least one that is longer than a single call and array. If you want to use these routines, you’ll need some buffer management of your signal. Second, for converting 611 to 100, ...
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This is my second answer. Technically it is introducing a high pass filter, but a very simple one with interesting properties for the OP's signal type.
Starting with your signal:
$$ x(t)=Ae^{j(at^2+bt+c)}+n(t)+i(t) $$
For a chosen value of $d$ (a lag time):
$$ x(t-d)=Ae^{j(a(t-d)^2+b(t-d)+c)}+n(t-d)+i(t-d) $$
Define the filtered signal as the ...
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This is what I would try:
Ignore the $n(t)$ and $i(t)$ since you know nothing about them. I am assuming that $x$ is complex and the rest of the variables are all real. Take the complex natural log of both sides:
$$ \ln \left( | x(t) | \right) + j \arg( x(t) ) = \ln(A) + j ( a t^2 + b t + c ) $$
Separate the real and imaginary parts. Using the imaginary ...
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Without understanding more detailed results on what i(t) is, one solution is to estimate either the phase or frequency first, and then pass that result through a filter.
An approach to do the frequency estimation digitally directly on the I (real) and Q (imaginary) samples of x(t) is the cross product frequency discriminator. After hard limiting x(t) to ...
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filtfilt() is a technique to achieve zero-phase filtering by applying the same filter twice to the data; with the output of the first stage reversed and filtered again in the second stage. Zero phase filtering is a desired property in image processing.
NaN means "not a number" and indicates those indeterminate conditions like $0/0$, $\infty/\infty$, $\infty ...
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I believe you are running into stability issues in an output power control loop design.
See below a diagram of similar power control loops that I have implemented, where for stability reasons any filtering in the loop is minimized and only done with the loop filter itself which is designed for stability. The noise you are trying to filter gets filtered by ...
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$$(A^TA)^{-1}A^T(y + noise) = \hat x$$
where $noise$ is vector of the same size as $y$ with all elements equal to unknown constant. That is all. You can get your estimated solution as a function of noise mean position in explicit form.
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When a 2D filter $h[n,m]$ is separable; i.e., $h[n,m] = f[n]g[m]$, then the 2D convolution of an image $I[n,m]$ with that filter can be decomposed into 1D convolutions between rows and columns of the image and the 1D filters $f[n]$ and $g[m]$ respectively.
Let me give you the MATLAB / OCTAVE code, I hope this is what you wanted to show ?
clc; clear all; ...
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If a 2D $K_2$ filter kernel is of rank $0$ or $1$, it can be written as a separable product of $2$ 1D kernels $K_1^r$ and $K_1^c$ on rows and columns. As such, it can implemented by 1D convolutions, as long as one properly reshape the 2D matrices into 1D ones, and take care about "out-of-range" values, to avoid wrap-around. For instance, you can pad in ...
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Attached Matlab sample code to apply an matched filter to a sine sweep for finding the playback position (the playback position estimate can be derived as the position of the peak in the matched filter plot):
% params
sampling_rate = 44000;
f_low = 10;
f_high = 100;
shift_percent = 0.0;
shift_samples = 30000;
% init
t = 0:1/sampling_rate:1;
N = length(t);
...
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Looking at Intel - An Investigation of Fast Real Time GPU Based Image Blur Algorithms By Filip Strugar it looks like the Kawase kernel is just a way of implementing a linear kernel quickly, but in a way that constrains the kernel somewhat.
This means that you could make such an algorithm. Either choose a set of spreads and adjust their weights (if that is ...
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What is the shape of your data? It's probably empty. Try print(data.shape) if it's a Numpy array.
Another remark, lfilter interprets the b and a as the coefficients of a discrete-time transfer function, so you cannot use it with your analog Butterworth filter coefficients.
You only need to worry about the axis parameter if your input data is ...
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See How many taps does an FIR filter need?
In your case you'd need more than 1000 taps depending on the allowable ripple, as your cut-off frequency is less than fs/500.
Alternatives :
use an IIR, a simple order-1 DC removal filter could work great
Average your signal and subtract the average in order to remove the DC
Rick Lyons proposes a clever ...
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i will agree with Hilmar that it can depend on the specific application.
if the application is to essentially losslessly store or transmit audio to later retrieve or receive that audio, including conversions of format (and this includes the A/D and D/A and SRC) then i would say that there is no good reason for a process to not be linear phase (which is ...
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Taking a step back, a standard image $I$ is composed of pixels. They are defined by a location (spatial coordinates, here denoted by $p$), and a "value", denoted by $I_p$. Smoothing or enhancing an image, in a large sense, consists in replacing each $I_p$ by $\hat{I}_p$, a value that would be: more probable, more consistent, more visually pleasant (choose ...
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I would start with the many resources on this site:
Is the Bilateral Filter a Solution of Some Variational Method?
How to Validate Bilateral Filter Implementation?
Comparison Between Guided Filter (Edge Preserving Filter) and Gaussian Filter.
What Is the Bilateral Filter Category: LPF, HPF, BPF or BSF?
Understanding the Parameters of the Bilateral Filter.
...
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http://ezcodesample.com/UAF/UAF.html
this is example with coding samples of nonlinear adaptive filtering.
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the group delay is D=(N-1)/2=20 samples
No. Group delay is a function of frequency. Assigning a single number to group delay is somewhat questionable. While the phase is indeed piece wise linear, at the "dips" of the combfilter, the phase jumps from $-\pi$ /2 to $\pi/2$. This is a real discontinuity, not a wrapping issue. At these frequencies the group ...
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Since this is an FIR, the group delay is D=(N-1)/2=20 samples.
No, since this is a linear phase (i.e. symmetric or anti-symmetric) filter, the group delay is half the length! (being a FIR isn't sufficient.)
The issue is that I get too peaks in the cross correlation, one at zero lag and another at 20 lag.
Write down the formula for auto-correlation at ...
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This is a common problem, and has an easy solution. The following assumes you are using a cascade of second-order direct-form-1 sections.
For a Butterworth highpass filter with a small ratio of Fcutoff to Fs, the numerator coefficients are close to [1 -2 1], and therefore the frequency response of the numerator by itself has a large attenuation at low ...
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An easy way to deal with this would be to actually let the sine run through your filter for a while, and then just at the right phase save the state of the IIR.
However, chances are: if the impulse response in this situation is a problem to you, your IIR isn't the right design.
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I would rephrase your terms as:
the "directional derivatives" are not so directional (although sometimes called similarly in lecture, they are only horizontal and vertical. Truer "directional derivatives" would allow angular refinement, cf. non-separable filters (Deformable Kernels for Early Vision, Perona)
the "directional derivatives" are not so ...
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Yeah, I answered my own question.
So tried the same filter adding zeros to a total of 32 taps.
Plays great now and spectrum is better in the highs.
So this is the filter:
1, -0.15, 0.026, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
(Total taps: 32)
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Probably not much advantage either way, possibly unless you've got a really high-order modulator (in which case someone else will have to answer!)
If the state-space representation is a single-input single-output system, and if it's linear, and if it's running in steady-state, and if data path widths aren't an issue, then there's no difference between it ...
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Well, look at your original picture: it's constant for all points but the edges, which means your derivative is zero for all points but these edges.
By applying a "rounding, smoothing" filter to it, you "smear" the edges enough to make the derivative be non-zero for multiple pixels, in every direction.
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That depends on how many of the accelerometer parameters (mostly drift and misalignment) you're trying to estimate.
If the IMU and the 'extra' accelerometer were in perfect alignment (and if their statistics are Gaussian), then the optimal combination of their outputs would be a simple weighted sum: $\vec {\hat a} = k_1 \vec a_1 + k_2 \vec a_2$ where (...
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The output variable holds the current output value given the current input value value. It is just the weighted sum of the current and ntaps-1 past input values. Note that this is not the most optimized version of an FIR filter routine.
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Most typical examples of the ideal filters are the classical brickwall frequency selective filters (lowpass, highpass, bandpass etc) which are defined as ideal because of their impossible to realize ideally selective frequency responses.
Their frequency responses include exactly flat passbands, exactly flat stopbands having absolute zero gain, and zero ...
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I have meet several definitions behind "being a realizable" filter. For instance, in Wikipedia Causal filter, there is:
Systems (including filters) that are realizable (i.e. that operate in
real time) must be causal
because:
In the context of physical systems, realizability is the property of
having some way of implementing a mathematically ...
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