I am working in digital image restoration field. According to it the image degradation model is defined like that:
g(x,y) = h(x,y)*f(x,y) + n(x,y)
many times i have studied that the blur-kernel is shift-invariant, can anyone please explain that what does it mean.? i have already searched it on Google, but i did not get satisfactory answer that i can tell during my presentation.
Shift invariant means that if some arbitrary value x is added to (or subtracted from) every element of a sample (in this case the pixels covered by the kernel), then the result of the kernel is also affected by the addition (or subtraction) of the value x.
It's most easily understood if you consider a blur kernel as a simple average (mean) rather than a Gaussian or whatever.
So if you have pixels with values v1, v2, v3 ... vn
with an average, A = (v1 + v2 + v3 + ... vn) / n,
then if you add some value x to each pixel (i.e. v1 + x, v2 + x, v3 + x ... vn + x),
the new average will simply be A + x.
So the output of the convolution is shifted by the same amount as each of the inputs. Hence shift invariant.
Related
I am a student working with time-series data which we feed into a neural network for classification (my task is to build and train this NN).
We're told to use a band-pass filter of 10 Hz to 150 Hz since anything outside that is not interesting.
After applying the band-pass, I've also down-sampled the data to 300 samples per second (originally it was 768 Hz). My understanding of the Shannon Nyquist sampling theorem is that, after applying the band-pass, any information in the data will be perfectly preserved at this sample-rate.
However, I got into a discussion with my supervisor who claimed that 300 Hz might not be sufficient even if the signal was band-limited. She says that it is only the minimum sample rate, not necessarily the best sample rate.
My understanding of the sampling theorem makes me think the supervisor is obviously wrong, but I don't want to argue with my supervisor, especially in case I'm actually the one who has misunderstood.
Can anyone help to confirm my understanding or provide some clarification? And how should I take this up with my supervisor (if at all).
The Nyquist-Shannon theorem states that the sampling frequency should at-least be twice of bandwidth, i.e.,
fs > 2B
So, this is the minimal criteria. If the sampling frequency is less than 2B then there will be aliasing. There is no upper limit on sampling frequency, but more the sampling frequency, the better will be the reconstruction.
So, I think your supervisor is right in saying that it is the minimal condition and not the best one.
Actually, you and your supervisor are both wrong. The minimum sampling rate required to faithfully represent a real-valued time series whose spectrum lies between 10 Hz and 150 Hz is 140 Hz, not 300 Hz. I'll explain this, and then I'll explain some of the context that shows why you might want to "oversample", as it is referred to (spoiler alert: Bailian-Low Theorem). The supervisor is mixing folklore into the discussion, and when folklore is not properly-contexted, it tends to telephone tag into fakelore. (That's a common failing even in the peer-reviewed literature, by the way). And there's a lot of fakelore, here, that needs to be defogged.
For the following, I will use the following conventions.
There's no math layout on Stack Overflow (except what we already have with UTF-8), so ...
a^b denotes a raised to the power b.
∫_I (⋯x⋯) dx denotes an integral of (⋯x⋯) taken over all x ∈ I, with the default I = ℝ.
The support supp φ (or supp_x φ(x) to make the "x" explicit) of a function φ(x) is the smallest closed set containing all the x-es for which φ(x) ≠ 0. For regularly-behaving (e.g. continuously differentiable) functions that means a union of closed intervals and/or half-rays or the whole real line, itself. This figures centrally in the Shannon-Nyquist sampling theorem, as its main condition is that a spectrum have bounded support; i.e. a "finite bandwidth".
For the Fourier transform I will use the version that has the 2π up in the exponent, and for added convenience, I will use the convention 1^x = e^{2πix} = cos(2πx) + i sin(2πx) (which I refer to as the Ramanujan Convention, as it is the convention I frequently used in my previous life oops I mean which Ramanujan secretly used in his life to make the math a whole lot simpler).
The set ℤ = {⋯, -2, -1, 0, +1, +2, ⋯ } is the integers, and 1^{x+z} = 1^x for all z∈ℤ - making 1^x the archetype of a periodic function whose period is 1.
Thus, the Fourier transform f̂(ν) of a function f(t) and its inverse are given by:
f̂(ν) = ∫ f(t) 1^{-νt} dt, f(t) = ∫ f̂(ν) 1^{+νt} dν.
The spectrum of the time series given by the function f(t) is the function f̂(ν) of the cyclic frequency ν, which is what is measured in Hertz (Hz.); t, itself, being measured in seconds. A common convention is to use the angular frequency ω = 2πν, instead, but that muddies the picture.
The most important example, with respect to the issue at hand, is the Fourier transform χ̂_Ω of the interval function given by χ_Ω(t) = 1 if t ∈ [-½Ω,+½Ω] and χ_Ω(t) = 0 else:
χ̂_Ω(t) = ∫_[-½Ω,+½Ω] 1^ν dν
= {1^{+½Ω} - 1^{-½Ω}}/{2πi}
= {2i sin πΩ}/{2πi}
= Ω sinc πΩ
which is where the function sinc x = (sin πx)/(πx) comes into play.
The cardinal form of the sampling theorem is that a function f(t) can be sampled over an equally-spaced sampled domain T ≡ { kΔt: k ∈ ℤ }, if its spectrum is bounded by supp f̂ ⊆ [-½Ω,+½Ω] ⊆ [-1/(2Δt),+1/(2Δt)], with the sampling given as
f(t) = ∑_{t'∈T} f(t') Ω sinc(Ω(t - t')) Δt.
So, this generally applies to [over-]sampling with redundancy factors 1/(ΩΔt) ≥ 1. In the special case where the sampling is tight with ΩΔt = 1, then it reduces to the form
f(t) = ∑_{t'∈T} f(t') sinc({t - t'}/Δt).
In our case, supp f̂ = [10 Hz., 150 Hz.] so the tightest fits are with 1/Δt = Ω = 300 Hz.
This generalizes to equally-spaced sampled domains of the form T ≡ { t₀ + kΔt: k ∈ ℤ } without any modification.
But it also generalizes to frequency intervals supp f̂ = [ν₋,ν₊] of width Ω = ν₊ - ν₋ and center ν₀ = ½ (ν₋ + ν₊) to the following form:
f(t) = ∑_{t'∈T} f(t') 1^{ν₀(t - t')} Ω sinc(Ω(t - t')) Δt.
In your case, you have ν₋ = 10 Hz., ν₊ = 150 Hz., Ω = 140 Hz., ν₀ = 80 Hz. with the condition Δt ≤ 1/140 second, a sampling rate of at least 140 Hz. with
f(t) = (140 Δt) ∑_{t'∈T} f(t') 1^{80(t - t')} sinc(140(t - t')).
where t and Δt are in seconds.
There is a larger context to all of this. One of the main places where this can be used is for transforms devised from an overlapping set of windowed filters in the frequency domain - a typical case in point being transforms for the time-scale plane, like the S-transform or the continuous wavelet transform.
Since you want the filters to be smoothly-windowed functions, without sharp corners, then in order for them to provide a complete set that adds up to a finite non-zero value over all of the frequency spectrum (so that they can all be normalized, in tandem, by dividing out by this sum), then their respective supports have to overlap.
(Edit: Generalized this example to cover both equally-spaced and logarithmic-spaced intervals.)
One example of such a set would be filters that have end-point frequencies taken from the set
Π = { p₀ (α + 1)ⁿ + β {(α + 1)ⁿ - 1} / α: n ∈ {0,1,2,⋯} }
So, for interval n (counting from n = 0), you would have ν₋ = p_n and ν₊ = p_{n+1}, where the members of Π are enumerated
p_n = p₀ (α + 1)ⁿ + β {(α + 1)ⁿ - 1} / α,
Δp_n = p_{n+1} - p_n = α p_n + β = (α p₀ + β)(α + 1)ⁿ,
n ∈ {0,1,2,⋯}
The center frequency of interval n would then be ν₀ = p_n + ½ Δp₀ (α + 1)ⁿ and the width would be Ω = Δp₀ (α + 1)ⁿ, but the actual support for the filter would overlap into a good part of the neighboring intervals, so that when you add up the filters that cover a given frequency ν the sum doesn't drop down to 0 as ν approaches any of the boundary points. (In the limiting case α → 0, this produces an equally-spaced frequency domain, suitable for an equalizer, while in the case β → 0, it produces a logarithmic scale with base α + 1, where octaves are equally-spaced.)
The other main place where you may apply this is to time-frequency analysis and spectrograms. Here, the role of a function f and its Fourier transform f̂ are reversed and the role of the frequency bandwidth Ω is now played by the (reciprocal) time bandwidth 1/Ω. You want to break up a time series, given by a function f(t) into overlapping segments f̃(q,λ) = g(λ)* f(q + λ), with smooth windowing given by the functions g(λ) with bounded support supp g ⊆ [-½ 1/Ω, +½ 1/Ω], and with interval spacing Δq much larger than the time sampling Δt (the ratio Δq/Δt is called the "hop" factor). The analogous role of Δt is played, here, by the frequency interval in the spectrogram Δp = Ω, which is now constant.
Edit: (Fixed the numbers for the Audacity example)
The minimum sampling rate for both supp_λ g and supp_λ f(q,λ) is Δq = 1/Ω = 1/Δp, and the corresponding redundancy factor is 1/(ΔpΔq). Audacity, for instance, uses a redundancy factor of 2 for its spectrograms. A typical value for Δp might be 44100/2048 Hz., while the time-sampling rate is Δt = 1/(2×3×5×7)² second (corresponding to 1/Δt = 44100 Hz.). With a redundancy factor of 2, Δq would be 1024/44100 second and the hop factor would be Δq/Δt = 1024.
If you try to fit the sampling windows, in either case, to the actual support of the band-limited (or time-limited) function, then the windows won't overlap and the only way to keep their sum from dropping to 0 on the boundary points would be for the windowing functions to have sharp corners on the boundaries, which would wreak havoc on their corresponding Fourier transforms.
The Balian-Low Theorem makes the actual statement on the matter.
https://encyclopediaofmath.org/wiki/Balian-Low_theorem
And a shout-out to someone I've been talking with, recently, about DSP-related matters and his monograph, which provides an excellent introductory reference to a lot of the issues discussed here.
A Friendly Guide To Wavelets
Gerald Kaiser
Birkhauser 1994
He said it's part of a trilogy, another installment of which is forthcoming.
Given input signal x (e.g. a voltage, sampled thousand times per second couple of minutes long), I'd like to calculate e.g.
/ this is not q
y[3] = -3*x[0] - x[1] + x[2] + 3*x[3]
y[4] = -3*x[1] - x[2] + x[3] + 3*x[4]
. . .
I'm aiming for variable window length and weight coefficients. How can I do it in q? I'm aware of mavg and signal processing in q and moving sum qidiom
In the DSP world it's called applying filter kernel by doing convolution. Weight coefficients define the kernel, which makes a high- or low-pass filter. The example above calculates the slope from last four points, placing the straight line via least squares method.
Something like this would work for parameterisable coefficients:
q)x:10+sums -1+1000?2f
q)f:{sum x*til[count x]xprev\:y}
q)f[3 1 -1 -3] x
0n 0n 0n -2.385585 1.423811 2.771659 2.065391 -0.951051 -1.323334 -0.8614857 ..
Specific cases can be made a bit faster (running 0 xprev is not the best thing)
q)g:{prev[deltas x]+3*x-3 xprev x}
q)g[x]~f[3 1 -1 -3]x
1b
q)\t:100000 f[3 1 1 -3] x
4612
q)\t:100000 g x
1791
There's a kx white paper of signal processing in q if this area interests you: https://code.kx.com/q/wp/signal-processing/
This may be a bit old but I thought I'd weigh in. There is a paper I wrote last year on signal processing that may be of some value. Working purely within KDB, dependent on the signal sizes you are using, you will see much better performance with a FFT based convolution between the kernel/window and the signal.
However, I've only written up a simple radix-2 FFT, although in my github repo I do have the untested work for a more flexible Bluestein algorithm which will allow for more variable signal length. https://github.com/callumjbiggs/q-signals/blob/master/signal.q
If you wish to go down the path of performing a full manual convolution by a moving sum, then the best method would be to break it up into blocks equal to the kernel/window size (which was based on some work Arthur W did many years ago)
q)vec:10000?100.0
q)weights:30?1.0
q)wsize:count weights
q)(weights$(((wsize-1)#0.0),vec)til[wsize]+) each til count v
32.5931 75.54583 100.4159 124.0514 105.3138 117.532 179.2236 200.5387 232.168.
If your input list not big then you could use the technique mentioned here:
https://code.kx.com/q/cookbook/programming-idioms/#how-do-i-apply-a-function-to-a-sequence-sliding-window
That uses 'scan' adverb. As that process creates multiple lists which might be inefficient for big lists.
Other solution using scan is:
q)f:{sum y*next\[z;x]} / x-input list, y-weights, z-window size-1
q)f[x;-3 -1 1 3;3]
This function also creates multiple lists so again might not be very efficient for big lists.
Other option is to use indices to fetch target items from the input list and perform the calculation. This will operate only on input list.
q) f:{[l;w;i]sum w*l i+til 4} / w- weight, l- input list, i-current index
q) f[x;-3 -1 1 3]#'til count x
This is a very basic function. You can add more variables to it as per your requirements.
The lens model in OpenCV is a sort of distortion model which distorts an ideal position to the corresponding real (distorted) position:
x_corrected = x_distorted ( 1 + k_1 * r^2 + k_2 * r^4 + ...),
y_corrected = y_distorted ( 1 + k_1 * r^2 + k_2 * r^4 + ...),
where r^2 = x_distorted^2 + y_distorted^2 in the normalized image coordinate (the tangential distortion is omitted for simplicity). This is also found in Z. Zhang: "A Flexible New Technique for Camera Calibration," TPAMI 2000, and also in "Camera Calibration Toolbox for Matlab" by Bouguet.
On the other hand, Bradski and Kaehler: "Learning OpenCV" introduces in p.376 the lens model as a correction model which corrects a distorted position to the ideal position:
x_distorted = x_corrected ( 1 + k'_1 * r'^2 + k'_2 * r'^4 + ...),
y_distorted = y_corrected ( 1 + k'_1 * r'^2 + k'_2 * r'^4 + ...),
where r'^2 = x_corrected^2 + y_corrected^2 in the normalized image coordinate.
Hartley and Zisserman: "Multiple View Geometry in Computer Vision" also describes this model.
I understand the both correction and distortion models have advantages and disadvantages in practice. For example, the former makes correction of detected feature point locations easy, while the latter makes the undistortion of the entire image straightforward.
My question is, why they share the same polynomial expression, while they are supposed to be the inverse of each other? I could find this document evaluating the inversibility, but its theoretical background is not clear to me.
Thank you for your help.
I think the short answer is: they are just different models, so they're not supposed to be each other's inverse. Like you already wrote, each has its own advantages and disadvantages.
As to inversibility, this depends on the order of the polynomial. A 2nd-order (quadratic) polynomial is easily inverted. A 4th-order requires some more work, but can still be analytically inverted. But as soon as you add a 6th-order term, you'll probably have to resort to numeric methods to find the inverse, because a 5th-order or higher polynomial is not analytically invertible in the general case.
According to taylor expansion every formula in world can be written as c0 + c1*x + c2*x^2 + c3*x^3 + c4*x^4...
The goal is just discover the constants.
In our particular case the expression must be symmetric in x and -x (even function) so the constants in x, x^3, x^5, x^7 are equal to zero.
I understand well perceptron so put accent only on kernel but I am not familiar with matemathic expressions so please give me an numerical example and a guide on kernel.
For example:
My hyperplane of perceptron is x1*w1+x2*w2+x3*w3+b=0; The RBF kernel formula: k(x,z) = exp((-|x-z|^2)/2*variance^2) where takes action the radial basis function kernel here. Is x an input and what is z variable here?
Or of what I have to calculate variance if it is variance in the formula?
Somewhere I have understood so that I have to plug this formula in perceptron decision function x1*w1+x2*w2+x3*w3+b=0; but how does it look look like If I plug in?
I would like to ask a numerical example to avoid confusion.
Linear Perceptron
As you know linear perceptrons can be trained for binary classification. More precisely, if there is n features, x1, x2, ..., xn in n-dimensional space, Rn, and you want to label them in 2 categories, y1 & y2 (usually -1 and +1), you can use linear perceptron which defines a hyperplane w1*x1 + ... + wn*xn + b = 0 to do so.
w1*x1 + ... + wn*xn + b > 0 or W.X + b > 0 ==> class = y1
w1*x1 + ... + wn*xn + b < 0 or W.X + b < 0 ==> class = y2
Linear perceptron will work well, only if the problem is linearly separable in Rn. For example, in 2D space, this means that one line can separate the 2 sets of points.
Algorithm
One common algorithm to train the perceptron, i.e., find weights and bias, w's & b, based on N data points, X1, ..., XN, and their labels, Y1, ..., YN is the following:
Initialize: W = zeros(n,1); b = 0
For i=1 to N:
Calculate F(Xi) = W.Xi + b
If F(Xi)*Yi <= 0:
W <--- W + Xi*Yi
b <--- b + Yi
This will give the final value for W & b. Besides, based on the training, W will be a linear combination of training points, Xi's, more precisely, the ones that were misclassified. So W = a1*X1 + ... + ...aN*XN where a's are in {0,y1,y2}.
Now, if there is a new point, let's say Z, to label, we check the sign of F(Z) = W.Z + b = a1*(X1.Z) + ... + aN*(XN.Z) + b. It is interesting that only the inner product of new point and training points take part in it.
Kernel Perceptron
Now, if the problem is not linearly separable, one may try to go to a higher dimensional space in which a hyperplane can do the classification. As an example, consider a circle in 2D space. The points inside and outside of the circle can't be separated by a line. However, if you find a transformation that can take the points to 3D space such that the first 2 coordinates remain the same for all points, and the 3rd coordinate become +1 and -1 for the points inside and outside of the circle respectively, then a plane defined as 3rd coordinate = 0 can separate the points.
Finding such transformations can be difficult and computationally heavy, so the kernel trick is introduced. Notice that we only used the inner product of new points with the training points. Kernel trick employs this fact and defines the inner product of the transformed points without actually finding the transformation.
If the unknown transformation is P(X) then Kernel function will be:
K(Xi,Xj) = <P(Xi),P(Xj)>. So instead of finding P, kernel functions are defined which represent the scalar result of the inner product in high-dimensional space. There are also theorems about what functions can be kernel functions, i.e., correspond to inner product in another space.
After choosing a kernel function, the algorithm will be modified as follows:
Initialize: F(X) = 0
For i=1 to N:
Calculate F(Xi)
If F(Xi)*Yi <= 0:
F(.) <--- F(.) + K(.,Xi)*Yi + Yi
At the end, F(.) = a1*K(.,X1) + ... + ...aN*K(.,XN) + b where a's are in {0,y1,y2}.
RBF Kernel
Radial basis function is one type of kernel function that is actually computing the inner product in an infinite-dimensional space. It can be written as
K(Xi,Xj) = exp(- norm2(Xi-Xj)^2 / (2*sigma^2))
Sigma is some parameter that you can work with to find an optimum value for. For example, you can train the model with different values of sigma and then find the best value based on the performance. You can start with sigma = 1
After training the model to find F(.), for a new data Z, the sign of F(Z) = a1*K(Z,X1) + ... + ...aN*K(Z,XN) + b will determine the class.
Remarks:
Regarding to your question about variance, you don't need to find any variance.
About x and z in your question, in each iteration, you should find the kernel output for the current data point and all the previously added points (the points that were misclassified and hence were added to F).
I couldn't come up with a simple instructive numerical example.
References:
I borrowed some notation from
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwjVu-fXo8DOAhVDxCYKHQkcDDAQFggoMAE&url=http%3A%2F%2Falex.smola.org%2Fteaching%2Fpune2007%2Fpune_3.pdf&usg=AFQjCNHlxy9TnY8xNe2-QDERipN_GycSqQ&bvm=bv.129422649,d.eWE
I'm trying to minimize my function "FunctionToMinimize", which is defined as follows:
FunctionToMinimize[a_, b_, c_, d_] := (2.35*Sqrt[
Variance[1/2*
(a*#1 + b*#2 + c*#3 + d*#4)
]
]
/Mean[1/2*(a*#1 + b*#2 + c*#3 + d*#4)])
&[DataList1[[1 ;; 1000]],DataList2[[1 ;; 1000]],
DataList3[[1 ;; 1000]], DataList4[[1 ;; 1000]]]
The four parameters a,b,c and d are restricted to be somewhere between 0.5 and 1.5. My Problem is now, that if I call
NMinimize[{Funktion[w, x, y, z],
0.75 < w < 1.25 && 0.75 < y < 1.25 && 0.75 < x < 1.25 && 0.75 < z < 1.25},
{w, x, y, z}]
the Mathematica kernel shuts down because it has not enough memory. If I use only the first 100 entries in my DataLists, it will find me results (in 4.1 sec), but if I use DataList[[1;;1000]] or even more entries, the kernel crashes.
Has anybody an idea, why the NMinimize function uses so much memory? I would need to have the minimization for 150'000 events in each list...
Thanks for your answer,
Cheers,
Andreas
I would guess (but haven't in any way checked) that the problem is that on each call to your function, Mathematica is trying to construct a symbolic expression derived from all your data and that occupies much more memory than you'd expect.
Regardless, the good news -- if you haven't long since moved on and forgotten about this problem -- is that you can turn the function into something much simpler.
So, first of all, the 2.35 and the 1/2s just change your function by a constant factor and don't affect where the minimum is, so let's ignore them. Next, your function is always non-negative, so minimizing it is the same as minimizing its square, so let's do that.
So now you're trying to minimize var(aw+bx+cy+dz)/mean(aw+bx+cy+dz)^2 where w,x,y,z are (perhaps quite long) vectors.
Now your numerator and denominator are both just quadratic forms in a,b,c,d whose coefficients depend (in fixed ways) on those vectors. Specifically, suppose your vectors have length N. Then your function is just
[sum(aw+bx+cy+dz)^2/N - sum(aw+bx+cy+dz)^2/N^2] / (sum(aw+bx+cy+dz)^2/N^2)
which you might prefer to write as N sum(aw+bx+cy+dz)^2 / sum(aw+bx+cy+dz)^2 - 1
and in that fraction, e.g., the coefficient of bc in the numerator is 2 sum(xy), and the coefficient in the denominator is 2 sum(x) sum(y).
So you can take your big vectors, compute the relevant coefficients once, and then just ask Mathematica to optimize a function of the form (quadratic / quadratic), which should be pretty painless.