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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}...


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

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 ...


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

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{...


4

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 (...


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

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\...


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

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

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

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

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

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

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 ...


3

As shown in the answers to this question, for continuous-time signals, the bound predicted by Bernstein's inequality is achieved with equality by a sinusoidal signal with a frequency equal to the upper frequency limit. In analogy to this, in this answer I'll show a bound on $\big|x[n]-x[n-1]\big|$ for a discrete-time sinusoidal signal $x[n]$ at angular ...


3

Your interpretation is correct: directional derivation operators highlight variation in a given direction. Here, you use the $2$-point discrete derivative in the $x$-direction (along image rows). It may emphasize vertical features. First, such operators indeed extend the initial image range. However, one often uses the absolute value of the derivative to ...


2

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.


2

You can represent the convolution as a matrix multiplication with a $N\times N$ Toeplitz matrix, where $N$ is the length of $\mathbf{w}$. Check out this explanation. So you can write $$ E(\mathbf{w})=\frac{1}{2}\|\mathbf{AHw}+\mathbf{b}\|^2_2 $$ where $\mathbf{H}$ is your Teoplitz matrix filter. Now if you define $$ \tilde{\mathbf{A}} = \mathbf{AH} $$ then $$...


2

I looked this up on Wikipedia: A complex Gaussian random variable $V = \mathfrak{Re}(V)+j\mathfrak{Im}(V)$ is said to be zero mean circularly symmetric $\mathcal{CN}(0,\Gamma)$ if the random vector $[\mathfrak{Re}(V),\mathfrak{Im}(V)]$ is a Gaussian random vector with mean $[0,0]$ and covariance matrix $ \frac{1}{2}\begin{bmatrix} \mathfrak{Re}(\Gamma) & ...


2

Hi: It's worded more clearly now in that you're estimating A and there's only one RV which makes more sense. Consider the first link I sent in the other message and go to where it says the "likelihood is". Your likelihood is quite similar except, since your variance is multiplied by 2, this causes 2 things to happen in terms of how would you change the ...


2

If computational efficiency is important to you, and your signal's bandwidth is not to large relative to the $f_s$ sample rate, three simple causal differentiators that tend to attenuate high-frequency noise are described at DSPRelated. A plot of the results of that article is below. Applying these three filters to your example signal (or an attempt to ...


2

Non-zero pixels after derivation provide you with two informations: a potential location (where the pixel is), and a potential strength (in magnitude) of an edge-pixel. Linear edge detectors are linear filters. A $3\times3$ kernel adds a smoothing effect in the orthogonal direction. It will favor edges with a certain spatial extend, over short-lengthed ...


2

In general you would get something like this, but it might not be tight: $$\begin{align}|f'(t)|&=\left|\frac{1}{2\pi}\int_{-\infty}^{\infty}j\omega F(j\omega)e^{-j\omega t}d\omega\right|\\&\le \frac{1}{2\pi}\int_{-\infty}^{\infty}|\omega||F(j\omega)|d\omega\\&=\frac{1}{2\pi}\int_{-\omega_c}^{\omega_c}|\omega||F(j\omega)|d\omega\\&\le \frac{|\...


2

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. ...


2

When I try this, the results looks as expected. So you really have to explain in more detail what exactly it is that you're doing, because it doesn't seem to be a property of the discrete-time derivative. EDIT: Now that I see your code, I'm convinced that the problem will disappear if you use dydt = zeros(1,sr); to initialize the derivative vector.


2

Even with @MattL.'s fix you are discarding typically non-zero parts of the discrete-time derivative by not including its first and last sample, which destroys its autocorrelation properties near the end points, typically resulting in the low-frequency plateau in the frequency spectrum as you have observed. We can add a bit of a zero-valued safety buffer at ...


1

The answer has already been given in Fat32's comment: if the solution manual says what you claim then it is wrong and you are right. What I want to add here is that the product rule for distributions is by no means some "engineering heuristic" or some dubious magic, but it can be proved in a rigorous way (see e.g. this document, p. 5) by treating the ...


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