# Tag Info

2

"is it ok to resample FIR filter coefficients?" Yes What are possible downsides of such approach? Down sampling always contains trade offs: you will need to apply a proper low pass filter to deal with aliasing. The choice of low pass filter will determine that trade off between band, residual aliasing, time domain distortion, preservation of ...

1

Generally yes. Think about it in frequency domain: there, convolution is simply multiplication, and if either the signal or the impulse response have no energy above some frequency, the resulting output will not have any either. The only issue is that impulse responses tend to be short, and resampling may use windowing and stretch them in time domain. Make ...

0

If somehow a filter could look into the future, then it could also react to future events. E.g.: y(t) = x(t-k/2) + x’(t+k/2) Where t is «now», k is a positive constant and x’ is a prediction of a future input would be a linear phase, lagless comb filter of sorts I guess. So perhaps «to the degree that you are able to predict the future signal», then a ...

2

Image Processing Context In classic Image Processing the filters used are known. Hence being separable is a property of a given filter which is suitable to the task. In this context, separability only means we can have a more efficient way to apply the filter computationally while the end result is the same. So, in Image Processing, if you have a filter ...

0

For any given problem definition, there's a filter that -- if you ignore execution time and hardware expense -- is "best"*. In general, that "best" filter isn't separable. Depending on the problem at hand, the degree to which the optimum degrades if you find the best separable filter will vary. So -- sometimes a non-separable filter ...

1

Here is a demonstration that confirms what @Hilmar said: We recall the initial relationship: $$s_{hp}[n]=(−1)^ns_{lp}[n]$$ Let us calculate its discrete Fourier transform: $$DTFT\{s_{hp}[n]\}=S_{hp}[k]=\sum_{n=0}^{N-1}\{(−1)^ns_{lp}[n]\}e^{-j2\pi \frac{nk}{N}}$$ But $(-1)^n$ is also equal to $e^{j\pi n}$...

0

That's a common technique in audio Audio modulation effects like Flanger, Phaser and specifically Chorus Pitch shifting, sample players, auto-tune, etc. In a sense "sample rate conversion" would qualify FM modulation synthesizers.

2

Of course it's valid for real coefficients because real number is a subset of complex number and the proof doesn't make any assumption whether the coefficient is complex or real. For a complex $d$  \begin{aligned} |A(z)|^2 &= A(z)A^*(z) = \frac{1-d^*z}{z-d} \frac{1-dz^*}{z^*-d^*}\\ &=\frac{1-(d^*z+dz^*)+|d|^2|z|^2}{|z|^2 - (d^*z+dz^*) + |d|^2} \end{...

4

Given a system with a known frequency response in the S-domain. Is there a way to find whether the system is linear and time invariant? If by "known frequency response in the s-domain" you mean a Laplace transfer function* as a ratio of polynomials in s -- yes. Laplace transform analysis on a system is not valid unless the system is linear and ...

0

@Hilmar, I solved this question. I am the OP. I forgot the password of my old account. Poles are not in complex conjugated pairs. The solution is simple. It uses this formula.

0

I want to convert those numbers into poles/zeros That is impossible. These are just two points out of the frequency response, and you need the full frequency response for that. There's almost certainly an analog filter with many-poles/zeros involved here as anti-aliasing filter. You can't even with a perfect model of that filter get all these coefficients ...

1

In the windowing design method for FIR filters, the impulse response for the desired frequency response filter is selected with a window. Typically the desired frequency response is a rectangular function in frequency (a brickwall filter, pass a group of frequencies and block the rest), in which case the ideal impulse response is a Sinc function. In this ...

0

Here is the trick-: I just found it out somewhere

2

We need to separate the concept of edge detection from the tools we use to apply the procedure. Edges are local property of the image. Being so local means we don't analyze the image in frequency domain but in spatial domain. Yet, a common step for edge detection is applying High Pass / Gradient Filter. Since those are Linear Shift Invariant operators we may ...

0

Let us assume that your desired spectrum can be expressed as analytical or numerically and it is denoted as $H(f)$. First, you should take the inverse Fourier Transform of $H(f)$ in order to obtain the impulse response $h(t)$ (You can do this numerically or analytically as well.). Then, apply convolution on your signal by utilizing $h(t)$. Recall that $h(t)$ ...

0

I feel like the other answers don't make it clear how, when and why and at what stage the matched filter is actually used in a step by step process, before you can actually understand the mathematical explanation. Also there is a confusion about how the reference signal is known. Instead, the reference symbol template is known. There is a matched filter ...

1

What is Passband Ripple? Passband ripple $\delta_1$ is typically specified as the 0 to peak difference in the passband gain in the magnitude response of a filter. For a filter with unity gain (1), the ripple will oscillate between $1-\delta_1$ and $1+\delta_1$. Ripple causes some frequencies in the passband to be amplified and others to be attenuated. In ...

2

Each column represents a wedge, like this: Your program needs to fill the wedge rather than just drawing a line. For each point on a column, you have to draw an arc rather than a single point. The width of the arc (in degrees) is given by the number of columns and the viewing angle of the device. You have to draw arcs along your line of the appropriate ...

0

A box filter, like many other filters, has a low-pass characteristic. The box filter is really bad at it though, the reason people use it is because it is so easy to implement and so cheap to compute. In the frequency domain, the box filter is a sinc. Higher frequencies are attenuated, but the filter goes to zero very, very slowly, thus not attenuating ...

0

Low-pass filters globally show up in the frequency domain as having: relatively high magnitude near the 0-frequency or DC, relatively low magnitude near the "Nyquist" or highest frequency, not too high magnitude "on average" in the middle. In other words, we expect that the lower frequency are better preserved (little attenuation) than ...

1

In essence you are trying to build an audio synthesizer that creates sounds with a given pitch. Over the last 50 years there have been dozens of technologies developed to do this: you can start here with an overviews: https://en.wikipedia.org/wiki/Synthesizer To get reasonably natural sounding instrument sound, your best bet is probably to build a sample ...

Top 50 recent answers are included