# Tag Info

## New answers tagged lowpass-filter

0

In the first formulation, $\Omega_c$ is the $3$ $\textrm{dB}$ cut-off frequency, because for $\Omega=\Omega_c$ you obtain $|H(\Omega_c)|^2=\frac12$. In the other formulation, the attenuation at the pass band edge $\Omega_p$ is determined by the constant $\epsilon$. For $\Omega=\Omega_p$ you have $|H(\Omega_p)|=\frac{1}{1+\epsilon^2}$. I.e., by choosing the ...

0

From a slightly less "dsp-like" point of view, slightly more geometric / time series, but this also works: The relation between the sinusoid (of amplitude 1) and the unit circle is well known. Instead of thinking of a moving average as a geometric mean on a window that slides from left to right over the time series, you could also define it as the ...

0

Yes, you will need to make a 50Hz notch filter. I would suggest you to look at this post. If you tell what type of program you use (Python, R, etc) I could try to provide you some code aswell.

2

Below is the analytic result for both the actual max value of $0.901243$ and the maximum value found by the OP of $0.898464$ The reason you are not getting the predicted maximum is your samples of the sine wave are not located exactly at the peak. This is clear if you zoom in on the plot and compare the two peak locations for the number of samples given (as ...

1

Puzzle solved, thanks to Cedron Dawg and Dan Boschen! First, I ran a simple N point moving average of a sinewave, using the simulation model below: I used the OP's values: N = 10, P = 40, sinewave amplitude = 1 and a simulation step size, $\Delta t$, equal to unity. The results, shown in the next figure, are the same as those of the OP: The maximum ...

2

Okay, this takes a bit of algebra, Euler's formula, and the geometric series summation formula, and some plugging and chugging, but here is how you can calculate it directly: \begin{aligned} x[m] &= \frac{1}{n}\sum_{k=0}^{n-1} A \cos \left( (m-k) \frac{2\pi}{p} + \phi \right) \\ &= \frac{1}{n}\sum_{k=0}^{n-1} A \left[ \frac{e^{i\left( (m-k) \... 1 The amplitude reduction is simply given as the magnitude of the transfer function of moving average filter. A moving average filter has a rectangular impulse response so the transfer function will be a sinc() function. You need to sample the sinc() function at the frequency or your sign wave 1 Sound absorption is an example of filtering. You can build an enclosure (room) to lessen sound coming from your clothes washer. Adding insulation inside the wall helps further, as does offsetting studs on each side to minimize vibration transfer. This comes at a cost, but at some point you’ve made the problem unnoticeable—you don’t need 100% eradication. ... 2 Yes that is correct. The width of the transition band is inversely proportional to the length (or memory) of the filter which means to say that you would need an infinite amount of time to achieve your perfect filter. Therefore for practical reasons you decide how much aliasing would be tolerable (similar in many ways to deciding how many decimal places you ... 1 Scanned below is the fred harris' "Rule of Thumb" which included the filter taps as well as his other "rules". This is from DSP World ICSPAT Class Notes DSP World Workshops, Orlando, Florida, November 1-4, 1999. That course along with other similar presentations and courses by fred harris have significantly influenced my views and thinking of signal ... 3 I assume you are working with discrete-time, since continuous-time white noise has infinite power (\sigma^2). First, remember that the power of a stationary process is always equal to the autocorrelation at 0 (P_x = R_x[0]); and the variance is the autocovariance at 0 (\sigma^2_x = \rm{Cov}_x[0]). These 2 expressions are equal for processes with 0 ... 0 I think the question as is, is unanswerable, but here I am making an attempt of correct it with these comments as answer. Should I upsample before or after rectifying? For the algorithm, if you are finally assessing against a threshold in magnitude, you should not care about down|upsampling. Do not add that. Does this process look correct? Strictly ... 1 HINT: It looks like they shifted the frequency response by half the sampling frequency, i.e.,H_{HP}\left(e^{j\omega}\right)=H_{LP}\big(e^{j(\omega-\pi)}\big)\tag{1} Frequency shifting corresponds to modulation (multiplication) in the time domain. Now you just have to figure out the modulation sequence that achieves the correct frequency shift.

0

The solution was to take Fourier transform 3D for each slice, then to chose only the 2nd component of the Transform to transform it back to the spatial space, and that's it. The benefit of this is to detect if something is moving along the third axis(time in my case). for sl in range(img.shape[2]): #-----Fourier--H1---------------------------------------...

2

sampled at 25.6 Hz, 25.6 kHz is much more likely. Your signal contains a strong fundamental (probably the turbine itself) plus some "fuzzy" stuff on top which is the vibration signal the author is apparently after. You see this clearly in the spectrum: a strong peak at low frequency and a bit of fuzz between 4k and 8k Apply a high pass filter to get rid ...

2

A high-pass filtered signal has lost its low frequencies, so you can refocus the spectrum on a higher frequency range. When you rectify it, the first phenomenon is that the signal becomes positive. Hence, it is not zero-mean anymore, and thus recovers low-frequencies that were attenuated in the high-pass filtered signal. A second is the presence of higher ...

Top 50 recent answers are included