19

If you imagine a Dirac delta impulse as the limit of a very narrow very high rectangular impulse with unit area centered at $t=0$, then it's clear that its derivative must be a positive impulse at $0^-$ (because that's where the original impulse goes from zero to a very large value), and a negative impulse at $0^+$ (where the impulse goes from a very large ...


14

First of all the dirac delta is NOT a function, it's a distribution. See for example http://web.mit.edu/8.323/spring08/notes/ft1ln04-08-2up.pdf Treating it as a conventional function can lead to misunderstandings. Example: "informally" the dirac delta is often defined as "infinity at x=0 and zero everywhere else". Now let's look at a ...


9

Maybe a picture is worth a thousand words? Here's how a Gaussian pulse of variable width and its derivatives look like: As others have said, Dirac is a distribution, hence the Gaussian pulse, and its width gets narrower and narrower. The derivative of $$\mathrm{e}^{-x^2}=-2x\mathrm{e}^{-x^2}$$ Which says that the derivative is the same as the function, ...


6

The estimation of derivative is straightforward: $$x'(n)~=\frac{x(n+1)-x(n-1)}{2}$$ $$x''(n)~={x(n+1)-2*x({n})+x({n-1})}$$ or if you have a signal sampled at $t_i=i\Delta t$, it is $$x'(t_{i})~=\frac{x(t_{i+1})-x(t_{i-1})}{2*\Delta t}$$ $$x''(t_{i})~=\frac{x(t_{i+1})-2*x(t_{i})+x(t_{i-1})}{(\Delta t)^2}$$ What you are interested in may be how to ...


6

Ideal derivative filter Let $f(x)$ be a signal bandlimited to frequencies $(-\pi,\, \pi)$. Given $f(x)$ as input, the same $f(x)$ is given as output by a system that has as its impulse response the sinc function: $$\operatorname{sinc}(x) = \left\{\begin{array}{ll}1&\text{if }x = 0,\\ \frac{\sin(\pi x)}{\pi x}&\text{otherwise.}\end{array}\right.\tag{...


6

That's a trick which you will also find in a DSP context, that's why I choose to provide an answer here. It is related to the Wirtinger derivative, and you can find more details about it in this answer over at math.SE. In practice this trick is often used to compute the extremum (minimum or maximum) of a real-valued function depending on a complex variable (...


6

Dirac's $\delta$ is a distribution. Distributions can be interpreted as limits of smooth functions under an integral or as operators acting on functions in ways which are defined by integrals. Both approaches have in common that basic properties of integrals are expected to work, partial integration in particular. Other answers have showed you the ...


5

I assume you mean the derivative with respect to $t$. In that case, the derivative of $\sin(\omega_0t)$ is not $\cos(\omega_0t)$ but $\omega_0\cos(\omega_0t)$. And luckily, this is also obtained via the Fourier transform relation you mentioned in your question: $$\begin{align}\mathcal{F}\left\{\frac{d}{dt}\sin(\omega_0t)\right\}&=j\omega\cdot \mathcal{F}...


5

Well, Since basically the Derivative Operation is Linear Filter you can chose your own optimal trade off between Noise Sensitivity and Bandwidth. If you look at Finite Differences Coefficients page at Wikipedia you can see you can chose higher "Accuracy" filters for 1st Derivative. You can also chose Forward / Central / Backward method.


5

$\delta(t)$ is a distribution, which means it is represented by a limitng set of functions. To find $\delta'(t)$, start with a limiting set of functions for $\delta(t)$ that at least have a first derivative. The triangle function of unit area is the simplest function to chose: $$\delta(t) = \lim_{\epsilon \to 0} \dfrac{\Lambda\left(\frac{t}{\epsilon }\right)...


4

2 point discrete differentiation is bound to produce highly noisy results. try the 5-points stencil. you can also generate coefficients (i.e. more points) yourself using derivation of Lagrange polynomials.


4

So, the delta function satisfies: $$\int_{-\infty}^{\infty} f(x) \delta(x - a)\, \mathrm{d}x = f(a)$$ Now, suppose we substitute $f(x) = x$ $$\int_{-\infty}^{\infty} f(x) \delta(x - a)\, \mathrm{d}x = a = \int_{-\infty}^{\infty} a \delta(x - a)\, \mathrm{d}x$$ This means that multiplying by $\omega$ is the same as multiplying by a constant. If you substitute ...


4

You'll be interested in Bernstein's inequality, which I first learned about in Lapidoth, A Foundation in Digital Communication (page 92). With a well-behaved signal $f(t)$ as you defined it above (in particular, $f(t)$ is integrable and bandlimited to $B\,\text{Hz}$, and $\text{sup}\,|f(t)| = A$), then $$\left|\frac{\text{d}f(t)}{\text{d}t}\right| \leq 2AB\...


4

This is a really nice problem. Problem Formulation I will formulate it as following: Let $ x \in \mathbb{R}^{n} $ be a signal. Given $ y \in \mathbb{R}^{n} $ which is a noisy measurement of $ x $ such that $ y = x + v $ and $ z $ be a noisy measurement of the derivative of $ x $ such that $ z = F x + w $ where $ F $ is the finite differences operator. It is ...


4

Your first suggestion is correct, the derivatives of the local polynomials are being sampled. From the horse's mouth (Savitzky & Golay 1964): Again, if we restrict ourselves to evaluating the function only at the center point of a set of equally spaced observations, then there esist sets of convoluting integers for the first derivative as well. (...


4

It's a matter of perspective if the bilinear transform "fails so miserably" when trying to approximate a derivative. First of all, it's not true that the approximation is quadratic for small $\omega$, it's linear as it should be: $$D(e^{j\omega})=\frac{2}{T}\frac{1-e^{-j\omega}}{1+e^{-j\omega}}=\frac{2j}{T}\tan\left(\frac{\omega}{2}\right)\approx \frac{j\...


4

In general, the time derivative property of the Fourier Transform is given as $$\mathscr{F}[\frac{d}{dt}x(t)] = j\omega X(j\omega) $$ Notice that we can simply multiply by the frequency index in the Fourier Transform result. For the 2D FT result: $$\mathscr{F}[f(x,y)]= F(u,v)$$ Using the same property results in: $$\mathscr{F}[\frac{d}{dx}f(x,y)]= uF(u,...


4

Simply put, $\delta'$ picks the opposite of the derivative of $f$ at the origin. Let us imagine that I can forget for a moment about that $\delta$ is not a function, that it should be defined in a strict mathematical sense (over compactly supported smooth test functions), etc. It can be simpler to consider that $\delta$ acts as an operator on (nice enough ...


4

Pay attention that convolution mean flipping the kernel both on the x and y axis. Hence the first element is a multiplication of $ -1 \cdot -1 $ which yields $ 1 $ as in the answer in the original post. Pay attention that this is a discrete approximation of the gradient based on Finite Differences. This specific one is based on the Sobel 1st derivative ...


4

First of all, you're trying to evaluate the derivative of an exponential. If the base of the exponential is positive, the derivative exists. However, if the base is negative, the derivative does not exist. It would be pointless to use a Kalman filter when you have an exponential with a negative base. Secondly, your model of the echo envelope is $$ A(t) = ...


3

Simply put: take an independent identically distributed Gaussian noise (one observation in blue, left) and its gradient (in green, right). The amplitude is roughly multiplied by the norm of the gradient filter (here $h=[1\,-1]$), which is $\sqrt{2}$, which you can see from the picture. Their average energy $E$ thus differ by a factor of $|h\|^2=\sqrt{2}^...


3

I will answer this with the understanding that $p$ is the Laplace variable (which I will call $s$). As you stated, differentiation in the time domain corresponds to multiplication by $s$ in the Laplace domain. So it seems that we could get what we want by just choosing $F(s)=s$; then the output is the derivative of the input, i.e., $$y(t) = \frac{du}{dt}$...


3

There is a good article in Wikipedia: Numerical Differentiation Since you'd expect differentiation to be (Anti?)symmetric, using IIR filters might not be the wisest of ideas.


3

This is not a standard 5-point derivative formula, the corresponding transfer function of which is $$H(z)=\frac{1}{12}\left(-z^2+8z-8z^{-1}+z^{-2}\right)\tag{1}$$ The figure below shows the magnitude responses of an ideal differentiator (red), of the standard 5 point approximation given by (1) (green), and of the approximation in your question (blue): As ...


3

So, first of all: A differentiator is really just a high-pass filter. Digitally (and this is dsp.SE, so I presume this is the case), that typically means a differentiator is really just FIR with taps $[1;-1]$. A butterworth filter being a low-pass IIR, the combination of both does sound like you might want to build a bandpass, or are doing something that ...


3

Differentiation in one axis will amplify high-frequency components. Assuming noise at Fs/2 (i.e maximum frequency), the gain at this frequency will be 6 dB. Intuitively, if you have a sequence alternating between +1 and -1 i.e (1, -1, 1, - 1), Then the difference will be +2, - 2, +2, -2, a gain of 6 dB Another reason is that by differentiating, you remove ...


3

Let's have a look at: $$\Delta I (x_1,x_2) = I_{actual}(x_2, y) - I_{actual}(x_1, y) = I_{ideal}(x_1, y) + I_{noise}(x_1, y) - I_{ideal}(x_2, y) - I_{noise}(x_2, y)$$ If you assume that for close $x_1$ and $x_2$ the $I_{ideal}$ doesn't change much so that $$ I_{actual}(x_2, y) \approx I_{ideal}(x_2, y)$$ then $$\Delta I (x_1,x_2) = I_{noise}(x_1, y) - ...


3

No, you are not increasing the sampling time. The sample rate stays the same. The Central difference derivative filter is simply longer than the First difference derivative filter. A longer filter just means more delay before you get your result. No, you should not low pass filter without a clear reason for why it is needed to meet your signal processing ...


3

I felt I needed to write an additional answer to try to clear my mind about the question. Here is the try, step by step. Caveat: for simplicity, I used the same notation $C$ of a function of reals $u$ and $v$, for its rewriting in $x=u+iv$ and $\bar x$, and on complex $x$ alone. I hope it is not confusing for the reader. Let $C(x)$ be a function (we don't ...


3

Practically calculating the derivative of a digital signal is straightforward, just convolve the signal samples with the taps of a(n approximate) derivative filter. MatLab has the functions filter() and conv() which can help you do this. The First Difference derivative filter has taps $$h[n] = [1.0, -1.0]$$ The Central Difference derivative filter has ...


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