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...my question is, is there an elegant notation to express such operation? If you know that these operations are to act on orthogonal dimensions, you can define their 1D version and then use indexing to express the order by which the function is applied to the multidimensional signal. But if you are working with multidimensional signals over an orthogonal ...

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Of course you could. But I cannot find a useful manner. Because of duality, product or convolution in the image domain can be re-expressed as a convolution or product in the frequency domain. Also because if you apply it locally, components of a frequency transformation generally produce a value akin to intensity. In the extreme and trivial case, a one-...

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Dechirping is a bit different than matched filtering, but perhaps that not important here. The code you have looks like it should work if the variables are as you describe. Here is a hypothetical example that range compresses an LFM pulse. The reflected signal from an actual target in the scene is a just a delayed version of the waveform, so it will just be ...

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Okay. I think i got it after some clarification from Dan. Thanks Dan. ! I think what i was confused with was the step size in my frequency vector.. i had thought to have the final result in db/mhz.. the steps should be in 1 Mhz.. but i now realize that the freq vector could be in whatever steps i like but if its in Hz then i need to have my final answer ...

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You can't implement the transfer function H(z) directly, you need to convert it to a difference equation. However, the process is trivial, so once you understand it you'll see the connection between the diagram and transfer function better. First, we need to unroll the summation. For example, we get this with m=2, for a second-order equation: $H(z) = \... 0 You can try more smoothly windowing the impulse response of your filter so that it fits in half your current FFT length instead of getting sharply truncated. Or you can lengthen your FFT so the it is longer than the sum of the data window length plus however long it takes your filter's impulse response to fade low enough not to be heard. Or a combination ... 0 Have you seen the Spectral Filter supplied in the M4L Pluggo for Live packs in Ableton? It may be worth testing signals with both your implementation and that one to see whether these distortions you talk about are normal. You'll probably make more progress this way as only you really know what you're implementation is. You will also be able to open the Max ... 1 It would help to apply some specificity to the general equation:$H(z) = \frac{b_0 + b_1z^{-1}+b_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}$Rearrange into a difference equation, skipping the steps for brevity:$y[n] = b_0x[n] +b_1x[n-1]+b_2x[n-1]-a_1y[n-1]-a_2y[n-2]$Does the difference equation form show now that the Direct Form structure is derived purely by ... 1 Roger R Labbe Jr explains this in his fantastic book "Kalman and Bayesian Filters in Python" as You can not throw out any information even if how noisy it is. Two Gaussians are always better than one in Bayesian framework. If you multiply two Gaussians you will have smaller covariance matrix. 0 The Least Mean Square solution to find the "channel" or response of the filter is provided by the following MATLAB/Octave Code using the input to the filter as tx and the output of the filter as rx. For more details on how this works, see this post: Compensating Loudspeaker frequency response in an audio signal: function coeff = channel(tx,rx,ntaps) % ... 2 You can describe a frequency response in terms of its real-valued amplitude and its phase: $$H(f)=A(f)e^{j\phi(f)}\tag{1}$$ Note that$A(f)$is not the magnitude, but a bipolar amplitude function. Equivalently, you can express$H(f)$in terms of its magnitude and its phase: $$H(f)=M(f)e^{j\tilde{\phi}(f)}\tag{2}$$ Now we have$M(f)=|A(f)|\ge 0$, and, ... 2 That's a simple process, where$y(k)=u(k-3)$, as the$z^{-1}$blocks just perform a delay by one sample.$x_1(k)$,$x_2(k)$and$x_3(k)$are just arbitrary names. In other words: output will be exactly like input, only delayed by three samples. 1 First, all of the "elemental" filters will have the same corner or center frequency. The differences between them will be in their Q. Consider a second-order Butterworth lowpass filter with a corner frequency of 1 kHz, with a unity-gain passband. At 1 kHz, the response of the filter is -3 dB. The filter slope reaches -12 dB/octave. The response of a forth-... 1 First, the equations you solve for the coefficients depend on what type of filter you're after. They also depend on how the equations are derived. For instance, a common case would be a Butterworth lowpass, converted from an analog prototype in the s-domain via the bilinear z transform. Converting a higher order lowpass with the bilinear transformation would ... 1 The impulse response of two filters in series is the convolution of each filters impulse response. You can multiply two polynomials by convolving their coefficients. Likewise you can factor or combine filters by factoring or multiplying their transfer functions described as polynomials. For example, a fourth order filter can be factored into two 2nd order ... 0 A linear time-invariant (LTI) system is indeed completely described by its frequency response. Note that the frequency response is the Fourier transform of the impulse response, which also completely describes the system. So if the Fourier transform of the impulse response exists, then the resulting frequency response must represent a complete ... 0 https://se.mathworks.com/help/comm/examples/raised-cosine-filtering.html “ Ideal raised cosine filters have an infinite number of taps. Therefore, practical raised cosine filters are windowed. The window length is controlled using the FilterSpanInSymbols property. In this example, we specify the window length as six symbol durations, i.e., the filter spans ... 0 Was my prof wrong? No. You're right, you can find multiple systems with the same frequency response, or the same transfer function. But if you're using something as a (linear, which is often implied) filter, then all you care about is the frequency response (otherwise, you'd probably not be calling the thing "filter"). So, when I define a filter, it's ... 1 An ideal raised cosine impulse is band-limited with cut-off frequency$f_c=(1+\beta)/(2T)$, where$\beta$($0\le\beta\le 1$) is the roll-off factor. Since it is band-limited, it must have infinite support in the time domain. The important difference with a sinc impulse is that its envelope decays much faster (for$\beta>0$, of course). Note that even ... 0 So we have 2 neighborhoods (Which are Matrices)$ v \left( {\mathcal{N}}_{i} \right) $and$ v \left( {\mathcal{N}}_{j} \right) . How could we calculate the distance between them? One option would be the Euclidean Distance: $${\left\| v \left( {\mathcal{N}}_{i} \right) - v \left( {\mathcal{N}}_{j} \right) \right\|}_{2}^{2} = \sum_{k} {\left( {v \left( {\... 2 Your analog transfer function looks OK. For the sake of clarity - and to reduce the chance of making errors - I'd just rewrite it as$$H_a(s)=G\cdot\frac{s^2+as + b}{s^2+cs + d}\tag{1}$$with$$\begin{align}G&=\frac{2R_g}{R_d+2R_g}\\a&=\frac{R_d}{L}\\b&=\frac{1}{LC}\\c&=G\left(a+\frac{1}{2R_gC}\right)\\d&=G\cdot b\frac{}{}\end{align}$$... 3 The reason for this is that the maths centres on a false assumption: We know that the zeroes need to be complex conjugates Do the zeros need to be complex conjugates? Can they not be 2 real zeros? I sympathise here because I also read this book and the maths portion ended up utterly confusing me because of very odd methods. A more general method would ... 3 Does mixing brown noise and white noise create pink noise? No. Pink noise has a spectrum of that falls with 3dB/octave (or 10dB/decade). The spectrum of the sum of white and brown noise will be "brown" at low frequencies and "white" at high frequencies. The spectrum will have two slopes: below the transition frequency it will be -6dB/octave and above it, ... 0 If the signal is positive (or at least non-negative, with no long runs of zeros that could make the sum in denominator zero), then what you have is a basic filter with kernel w[-n], combined with a simple automatic gain control. Note that the term in the denominator could be implemented recursively:$$\bar{x}[i]=\sum_{k=-L}^{L}x[i+k]=\bar{x}[i-1]+x[i+L]-... 0 It is not really a convolution, more a weighting. Your expression is0$homogeneous in$x$, as you divide$x$s values by$x$s. Specifically: it is not defined if$\sum_{-K,K} x[i+k] = 0$if$x[k]$is locally a constant, the output does not depend on it (because it will be equal to$\sum h[k]$) if$h[k]$is an averaging filter ($h[k] = \frac{1}{2K+1}$), ... 1 Okay, when discussing white noise or pink noise (or red noise or brown noise or flicker) or some other random process, there is this property called the power spectrum, in which white noise has a constant value for all frequencies. But we integrate the power spectrum over all frequencies (from$-\infty$to$+\infty$) to get power. Integrating a constant, ... 0 What format is your input data in? It sounds like you have a phase history signal as it's often called. Assuming that the transmitted waveform was a linear FM chirp, has it already been deramped? Some receivers perform deramp-on-receive processing, which removes the chirped nature of the waveform. In that case, range compression is as simple as performing a ... 0 If the acceleration readings are performed in an inertial (non rotating in particular) frame then yes it's sufficient to use the accelerations, but if the acceleration measurements are performed on the moving craft then no it's not possible alone by accelerometer readings. Eventhough it's sufficient to use those accelerations$a_x$,$a_y$, and$a_z$on a ... 1 The following is what I believe to be an optimized approach for performance in the presence of additive white noise when no other information is known about quantity of pulses or their amplitude distributions, beyond that they are 100 or 200 us long rectangular pulses and repeat at the 1KHz and 1.4KHz rates. This can be even further improved if any other ... 0 I understand, that this involves a lot of math. Not so much, in principle. The basic idea behind linear (or nonlinear) filtering is to remplace a inaccurate or noisy sample$s[n]$by a combination of other samples, assuming that their values or location is somehow close to$s[n]$(cf. local vs non-local filters). At a low level, when the filter is both ... 2 Consider a moving average over N samples- this is a simple FIR filter where each new output is the average of the past N samples. It is easy to see how high frequency noise can be filtered out (so is a low pass filter), and the longer time duration we include in the averaging window the lower will be the frequency cut off (just compare a stock market 30 day ... 2 Since you are seeing your filter coefficients and much larger than your signal, this is indicative that you have a very large "impulse" at the start of your signal. The coefficients of your filter is the impulse response for the filter, so that is exactly what you are seeing: the response to an impulse. Review the start of your time domain data for a very ... 0 This is probably going to take multiple steps. As a start, here is what I'd do: You want a filter that retains pulse information, but filters noise. The most extreme you can do with this is a rectangular filter that's$100\mu\mathrm{s}\$ long, or possibly a raised-cosine filter that's about the same length. Assuming that the pulse timing is very consistent,...

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