Tag Info

Hot answers tagged filtering

3

If there is no good reason to choose otherwise, you would initialize with zeros, so in your case $y[-1]=0$. Other initializations may be useful in certain situations, e.g., when processing blocks of data to avoid transients between blocks. Note that if the filter is stable (which it has to be in almost all useful applications), the influence of the initial ...

3

Another way to simply get your result for this kind of a problem (where $h[n]$ is very short) is to use the following method : Let your output of the discrete convolution sum be $y[n]$ : $$y[n] = x[n] \star h[n]$$ Then by expanding $h[n]$ into impulses, the convolution will be distributed over addition ( using the highpass filter {-1,2,-1}; k = -1,0,1; ...

2

I'm not sure what's you model is. Let's say it is something like: $$y = H x + n$$ Now, using the Least Squares model is optimal (In the MSE sense) when $n$ is AWGN (It is the linear optimal estimator if the noise is white). So unless the noise in your model is colored, no gain by filtering the data before applying the Least Squares method. Now, what ...

2

The lower the cut-off frequency of the anti-aliasing filter, the higher the delay thus degrading your phase margin. You need to be really careful when implementing an anti-aliasing filter in control loops applications. First question, what is the amount of noise above 500 Hz? Based on the picture, you have noise at about 3 kHz, the main purpose of your anti-...

2

A personal rule: in general, it can be useful to perform non-linear operations before linear ones. One reason behind that is that a lot of practical concerns are related to outliers or suspect behavior, which can easily be smoothed by linear filters. Let me reformulate. If $f_i$ denote filters, and $s_i$ signals, should one do $f_0 \ast(s_1 .s_2)$ or $(... 2 The discrete convolution of a signal$x[k]$with an impulse response$h[k]$, where$k$is the discrete time index, is computed as $$y[k] = x[k] \ast h[k] = \sum_{n=-\infty}^{\infty} x[n] \cdot h[k-n].$$ for a given sample index$k_0$, e.g.$k_0 = 0$, you can compute the first sample of your exercise as $$y_\text{hp}[k_0] = \sum_{n=-1}^1 x[n] \cdot g[k_0-... 2 Classic filteration is indeed done using convolution. Though I have seen broader definition of filtering as shaping the signal in its frequency domain which can be done in many other methods as well. Of course you can create meaningful operations using element wise operations and even specifically multiplication. Think of the case you have a noise with the ... 2 I figured it ouy. In order to display DFT image properly they need to be log scaled. I used the log transformation s = c*log(1+r) where r is your normalized input image, and c is a constant. Plot s and you will be able to see the DFT correctly. %Log transform of DFT Image normalized_dftImage_F_uv = F_uv_dftImage/255; c1 = 1; s_LogTransform_F_uv_dftImage = ... 1 There are a lot of misconceptions in the way that the problem has been posed (in particular, X(\omega) as defined by the OP in the first version of his question -- he has since then deleted the definition -- ) has nothing to do with the matter), but when \{x(t)\} is a wide-sense-stationary (WSS) process, then the processes \{y_1(t)\} and \{y_2(t)\} ... 1 The sharper the filter is in the frequency domain, the longer the impulse response will be. This typically leads to "time blur" or "ringing" in the time domain. In addition, a zero phase filter is non-causal, so you get "pre-ringing" and any sharp onsets or transients in the time domain get degraded. The long impulse response also leads to a long "... 1 DCTs (they exist under different flavors) map data along some cosine-related frequency axis, from 0 to... the maximum frequency of your data, which we don't know, since we don't have its sampling frequency. So the realtive frequency index is implicit, between 0 and 255/256 in relative frequency. 2.669825 , mostly probably, is the DC component, here ... 1 The sharp peaks are actually due to segmentation and concatenation, you need to have overlapping segmentations. The peaks occur on the edges of segments mostly. In following figures I tried to show what I mean, blue curve are Hanning function coefficients. I did not understood why your application is so, but if it is essential to process your signal in ... 1 In general you will need to multiply first and then low pass filter. You also have to make sure that your sample rate is high enough so the multiply doesn't create aliasing. Let's look at a simple example: feed a 1kHz signal into a loudspeaker and measure current and voltage to determine the average (thermal) power with maybe a 100ms time constant. The ... 1 Plus is the simplest operation : z = x+y. Fourier is inherently linear, and good at addressing it. However, most processes and data combination are nonlinear, and they should be dealt with. The second simplest operation is multiplication. Homomorphic filtering deals with z = x\times y. It is more complicated, especially because of zeroes. Logarithms ... 1 In image processing, an image f can be seen as a function from positions (or coordinates) to values (the pixel values). If the image is grayscale, the values are just real numbers in the range [0, 1] (or, equivalently, in the range [0, 255]). However, each pixel of an image can also have a color different than grey (e.g. red). In that case, we need "... 1 First of all, I want to thank Fat32 and Hilmar for answering the question. I'm sorry I couldn't answer before, because I had some external problems to solve. Finally I was able to solve the problem by using the reverse time property of the Zeta-Transform (despite having some difficulties with the initial conditions). Consider a non-minimum phase system ... 1 Here is one way to think about it: Let's say you have an unstable transfer function, which is unstable because of a single pole outside the unit circle. We can write this as$$H_0(z)=H_{stable}(z) \cdot \frac{1}{1-z^{-1} \cdot p}$$That's a cascade of the stable and unstable part. Now we can multiply this with a unity filter that has a pole and zero at the ... 1 An unstable system can be stablized by all-pass decomposition for exampe, but simply constructing the anti-causal impulse respone extension will not provide the same output; so I think it's not possible, but I'm not rigorous at this point. Following just shows why. Assume that your unstable and causal IIR filter has the impulse response$h_+[n]$and by ... 1 Zero-phase filtering is a non-causal procedure, so it cannot be done in real time, only offline (or pseudo real-time, i.e., with a sufficient delay). A zero-phase filter needs to have a purely real-valued frequency response, and, consequently, it must have an impulse response that is even with respect to the time index$n=0\$, i.e., it is non-causal. Zero ...

1

If allowed, I would view the signal in the frequency domain (2048 pt FFT for a single symbol, Fs=30.72Mhz). Anything outside the central 1200 carriers in the symbol is "noise" and shouldn't be there. You can filter the noise by setting the outer FFT bin amplitudes to 0, then convert back to the time-domain or perform other signal processing as desired. ...

1

Median filter is considered good because unlike averaging filter which ruins the edges of an image by blurring it to remove the noise, median filter removes only the noise without disturbing the edges. Well, median filter is the best and only filter to remove salt and pepper noise. Hope this helps:) Thank you!!!

Only top voted, non community-wiki answers of a minimum length are eligible