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22

Say you're interested in $$M^{j2\pi f_0 t}. \tag{1}$$ Note that $$M = e^{\log M},$$ so $(1)$ can be written as \begin{align} M^{j2\pi f_0 t} &= \left( e^{\log M} \right) ^ {j2\pi f_0 t} \\ &= e^{j2\pi (f_0\log M) t} \\ &= \cos(2\pi (f_0\log M) t) + j \sin(2\pi (f_0\log M) t), \end{align} which is a complex sinusoid with frequency $f_0 \log M$. ...


13

Intuition: The intuition is this: Your noise is some event or events that are rare, and that when compared to other events, look like outliers that shouldn't really be there. For example, if you are measuring the speeds of every car on the highway as they pass by you and plot them, you will see that they are usually in the range of say, $50$ mph to $70$ ...


9

the general polynomial form is: $$\begin{align} f(u) &= \sum\limits_{n=0}^{N} \ a_n \ u^n \\ \\ &= a_{\small{0}} + \Bigg(a_{\small{1}} + \bigg(a_{\small{2}} + \Big(a_{\small{3}} + \,... \big(a_{\small{N-2}} + (a_{\small{N-1}} + a_{\small{N}} \,u \,)u \, \big)u \ ...\Big)u \, \bigg)u \, \Bigg)u\\ \end{align}$$ the latter form is using Horner's ...


9

\begin{align} (1-x)\sum_{k=0}^{n}x^k &= \sum_{k=0}^{n}x^k - x\sum_{k=0}^{n}x^k \\ \\ &= \sum_{k=0}^{n}x^k - \sum_{k=1}^{n+1}x^k \\ \\ &=1 + x+x^2+ \dots+x^n \ \\ & \qquad -x-x^2-\dots-x^n-x^{n+1}\\ \\ &= 1 - x^{n+1} \end{align} Thus if $x \ne 1$, dividing both sides by $(1-x)$ results in $$\sum_{k=0}^{n}x^k = \frac{...


8

If I draw a number uniformly between zero and one, what is the probability that they are equal? Mathematically, it should be zero but I don't recall why? Can somebody please help in explaining why the probability should be zero? Avoiding formal definitions of (Lebesgue) probability measure, an informal way is thinking the probability at a point of a ...


6

The most straightforward way to see this is to note that for $k=mN$ $$W_N^{kn}=e^{j2\pi mnN/N}=e^{j2\pi mn}=1$$ So the sum for the case $k=mN$ is simply $$\sum_{n=0}^{N-1}1=1+1+\ldots+1=N$$ Note that the solution using L'Hopital's rule is a bit dubious because for $k=mN$ the formula for the geometric series is not valid because the terms are all equal to ...


6

As you have already pointed out in your question, it is not possible (without using optimization methods) to compute an exact L2 solution for the frequency domain design problem of IIR filters due to the non-linear relationship between the filter coefficients and the error function. There is, however, a method which can come close and which transforms the ...


6

Assuming independent random variables with normal distributions, the probability that a value will fall beyond, say, 2 standard deviations will be about 0.01. If you have a median filter of width 3, that triplet must contain two outliers on the same side of the mean in order for an outlier to come through. This event has a probability of $2 \cdot 0.005^2 =$ ...


6

It just means "the transformation that turns $x$ into $y$." You might also see $\mathbf{T}^{-1}$ which means the inverse: turning $y$ into $x$.


5

Here's just a polynomial version: $$ \arcsin(x) = x + \frac{1}{2} \frac{x^3}{3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{x^5}{5} + \frac{1\cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{x^7}{7} $$ function y = arcsin_test3(x) y = x.*(1+x.*x.*(1/6+ x.*x.*(3/(2*4*5) + x.*x.*((1*3*5)/(2*4*6*7))))) endfunction which seems to have five multiplies (assuming you can ...


5

The filter you referenced is known as a Binomial filter. It is an approximation to a Gaussian, but for smaller filters it's a very crude one. The design is oriented more toward efficiency than accuracy. Rather than sampling a Gaussian directly, the idea behind the approximation is based on the Central Limit Theorem. In this case, it means that a small moving ...


5

Although the question could belong to SE.math, mastering inequalities for $\ell_p$ norms (for $p\ge 1$) or quasinorms (for $0<p< 1$), and their norm ratios and powers, is quite important in modern signal/image processing: starting from Cauchy-Bunyakovsky-Schwarz, to power means, Rogers-Hölder, Young, Minkowski, Dresher-Gini etc.) There are scalar, ...


4

Your algorithm, as described, is not correct. You are simply counting the number of negative numbers, not the number of zero crossings. For the particular data set that you use, B1, the two numbers are the same, but in general they won't be. The way to do it is to multiply each number by the number after it, and if that number is negative, then you have a ...


4

The key is in the last step of your work: $$ \frac{1 -e^{j2\pi k}}{1-e{\frac{j2\pi k}{N}}} $$ If $k$ is some integer multiple of $N$, then the exponents in the numerator and denominator are both some integer multiple of $j2\pi$. In this case, both exponential functions are equal to 1, meaning that the expression above is equal to $\frac{0}{0}$ for $k$ an ...


4

There are two questions here: Would studying functional analysis be useful and why don't more engineers study it. First I will say there's definitely no harm in studying it and if you find it an interesting subject and are of a more mathematical bent its probably quite useful. Now why don't most engineers study functional analysis. Well in short while its ...


4

Q1. As shown by Oppenheim's experiment, the phase spectrum contains most of the structural information about the image. In 2D this are things like lines and edges. In 3D it is things like lines and edges but also movement. Instead of 2D frames and time, imagine the video as a 3D solid where the z-axis is the frame number. If you took a slice along the z-axis ...


4

To complement my part to this question: Here is a somewhat shorted answer based upon a manual expansion of the odd function $f(x)$ \begin{align*} f(x)&=\ln\left(\arctan\left(\alpha e^x\right)\right)-\ln\left(\arctan\left(\alpha e^{-x}\right)\right)\\ &=f_1x+f_3x^3+O\left(x^5\right)\tag{1} \end{align*} into a series up to the third order. Some ...


4

i have no idea what the "reconstruction fidelity term" is or what it's about. Hermitian symmetry is a term usually applied to some form the Fourier Transform of a signal that is purely real. for continuous-time, continuous-frequency Fourier Transform: $$ X(f) = \int\limits_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \ dt $$ if $x(t)$ is purely real (that ...


4

Mathematically, an electrical circuit is as an operator, i.e., a function that takes a function and returns another function. Let this operator be denoted by $\mathcal T$, let $x : \mathbb R \to \mathbb R$ be the input signal and let $y := \mathcal T (x)$ be the output signal. Let $\mathcal D_{t_0}$ be the delay operator that delays its input by $t_0 > 0$....


4

Let's first consider periodic signals. Under rather mild conditions, a periodic signal with period $T$ can be represented by its Fourier series: $$f(t)=\sum_{k=-\infty}^{\infty}c_ke^{j2\pi kt/T}\tag{1}$$ Eq. $(1)$ shows that the periodic function $f(t)$ is represented by a sum of weighted complex exponentials with frequencies $\omega_k=2\pi k/T$, which are ...


3

Gaussian Kernel is made by using the Normal Distribution for weighing the surrounding pixel in the process of Convolution. Since we're dealing with discrete signals and we are limited to finite length of the Gaussian Kernel usually it is created by discretization of the Normal Distribution and truncation. I created a project in GitHub - Fast Gaussian Blur. ...


3

That's the old question: You say it's better, and wonder why, and everyone can only ask you to mathematically define better. As soon as you define a measure for processing quality, you will be able, typically with not too much linear algebra, derive based on the properties of your signal, your measure and your basis, why one is better than the other.


3

i have a pretty good implementation of $\arctan()$ here. i think you can use the identity: $$ \arcsin(x) = \arctan\left( \frac{x}{\sqrt{1-x^2}} \right) $$ to get what you want.


3

It depends on whether you intend to enter academia or industry. Academics write papers and can use all the maths they can get. In industry not so much, unless you intend to enter a research lab, which is like academia. I think the most important branches of mathematics for telecommunications are probability, statistics, linear algebra, and information ...


3

You don't need to use the Viterbi algorithm to do what you want to do. You have only two possible outcomes: T or H. So train a model on your training sequence (not sure if this is the numbers you give - it looks like emission probabilities for 3 states, but you don't mention a transition matrix and how you derived these probabilities...), and then compute ...


3

HINT If we have the diagonal matrix: $$ D = \left[\begin{array}{cccc} d_1&0&0&0\\ 0&d_2&0&0\\ 0&0&\ddots&0\\ 0&0&0&d_n \end{array}\right]$$ Multiplying another matrix $$M_r = \left[\begin{array}{c} r_1\\ r_2\\ \vdots\\ r_n \end{array}\right]$$to the left with it multiplies each row like this: $$DM_r = \left[\...


3

To answer your first question formulated in the first sentence: if $X(\omega)$ is the Fourier transform of the input signal, and $H(\omega)$ is the frequency response of the IIR filter, the Fourier transform ("frequency domain representation" as you call it) of the output signal is simply $Y(\omega)=X(\omega)H(\omega)$. This is valid in continuous time as ...


3

(Converting comment to answer.) Using Wolfram Alpha, $f'''(x)$ at $x=0$ evaluates to: $$\begin{align} \\ f'''(0) = & -\frac{6 \alpha^2}{(\alpha^2 + 1)^2 (\arctan(\alpha))^2} \ + \ \frac{2 \alpha}{(\alpha^2 + 1) \arctan(\alpha)} \\ & \quad \quad + \frac{16 \alpha^5}{(\alpha^2 + 1)^3 \arctan(\alpha)} \ + \ \frac{12 \alpha^4}{(\alpha^2 + 1)^3 (\arctan(\...


3

okay, i promised to put up bounty and i will keep my promise. but i have to confess that i might renege a little bit on being satisfied with just the third derivative of $f(x)$. what i really want are the two coefficients for $g(y)$. so i didn't realize that there was this Wolfram language as an alternative to mathematica or Derive and i didn't realize it ...


3

The problem as posed in the question appears to have no closed-form solution. As mentioned in the question and shown in other answers, the result can be developed into a series, which can be accomplished by any symbolic math tool such as Mathematica. However, the terms become quite complicated and ugly, and it is unclear how good the approximation is when we ...


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