# Tag Info

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Parseval's theorem will hold, but take into account that your signal in the time domain will no longer be $x[n]$. Namely, if you have that $$\sum_{n=0}^{N-1} \Big| x[n] \Big|^2 = \frac{1}{N} \sum_{k=0}^{N-1} \Big| X[k] \Big|^2$$ then, if you window the signal $x[n]$ with a window $w[n]$, your signal will now be $\hat{x}[n]=x[n]w[n]$, and the theorem will ...

7

[Added a reference on Schwartz's impossibility theorem for products of distribution] The continuous Dirac delta $\delta$ is not considered a true function or signal, but a distribution. From its wikipedia page: The delta function can also be defined in the sense of distributions exactly as above in the one-dimensional case. However, despite ...

7

If you multiply a continuous-time finite energy signal $f(t)$ with an impulse train you get $$\tilde{f}(t)=\sum_{n=-\infty}^{\infty}f(nT)\delta(t-nT)\tag{1}$$ where $T$ is the sampling interval and $\delta(t)$ is the Dirac delta impulse. Note that the "signal" $\tilde{f}(t)$ is a mathematical fiction, it cannot exist in practice, and it cannot even be ...

5

Note that the condition $$\int_{-\infty}^{\infty}|f(t)|^2dt<\infty\tag{1}$$ (i.e., that the signal $f(t)$ has finite energy) is very restrictive when we try to model signals, even though obviously any actually occurring signal must have finite energy. Modeling signals as random processes means that we ignore condition $(1)$. Models are always ...

5

You simply have to apply the definition of energy. Assuming all signals involved are real, the energy of $g_3(t)$ is given by \begin{align} E &= \int_{-\infty}^\infty g_3^2(t) \, dt \\ &= \int_{-\infty}^\infty \big(g_1(t) + g_2(t)\big)^2 \, dt \\ &= \int_{-\infty}^\infty g_1^2(t) \, dt + 2\int_{-\infty}^\infty g_1(t)g_2(t) \, dt + \int_{-\infty}^...

5

Yes indeed! In theory as long as the wavelet is orthogonal, the sum of the squares of all the coefficients should be equal to the energy of the signal. In practice, one should be careful that: the decomposition is not "expansive", i.e. the number of samples and of coefficients is the same. wavelet filter coefficients are not re-scaled, as happens in some ...

4

The typical inverse tangent function maps the input range of $t \in (-\infty,\infty)$ into an output range of $(-\pi/2, \pi/2)$ as in the figure below: Based on this, its values are bounded for all $t$. Yet, since the intergal of its square is unbounded, then it cannot be an energy signal;i.e., $$\int_{-\infty}^{\infty} |\tan^{-1}(t)|^2 dt ~~~~~\to \... 3 Your result is correct but note that for complex signals, the even and odd parts are defined by$$x_e(t)=\frac12\left[x(t)+x^*(-t)\right]\tag{1}$$and$$x_o(t)=\frac12\left[x(t)-x^*(-t)\right]\tag{2}$$where ^* denotes complex conjugation. From (1) and (2) it follows that the real part of x_e(t) is even and its imaginary part is odd, whereas x_o(... 3 [EDIT: 20180307, expanded some details] Globaly no, windowing does NOT affect Parseval's theorem (because theorems are only affected, more precisely not applicable, when their hypotheses are not met), in the sense that the equality in energy relates a signal (windowed x_w or not, x, resp.) and its Fourier transform (from a windowed signal, X_w or ... 3 While the Fourier transform, discrete or continuous, can be regarded as unitary transform i.e a naturally norm preserving change between orthonormal bases in a normed complex vector space, the windowed FT does not in general possess this quality. And non-unitary operators cannot be turned into unitary ones by re-scaling. Here is why: Unitary operators can ... 3 windowing your data, x[n], with window w[n] is equivalent to windowing the square of your data (or square-magnitude), \Big|x[n]\Big|^2, with the square of the window w^2[n]. so think of this square of the window w^2[n] as just another window. now stochastically your typical squared sample x^2[n] will have some expected mean-square. the ... 3 \delta(x) doesn't really exist at all for any particular x. Like Laurent Duval said, Dirac is not an \mathbb{R}\to\mathbb{R} function, rather the whole mapping$$\backslash f \mapsto f(a) \equiv \int_\mathbb{R}\!\!\mathrm{d}t\: f(t) \cdot \delta(t-a)"$$is a functional, mapping functions to values of the function evaluated at some particular point. ... 3 You're right that the square of a Dirac delta impulse is undefined, so energy and power cannot be defined in the usual way for signals containing Dirac impulses. However, in analogy with discrete-time signals, it is common to define energy and power of a signal consisting of Dirac impulses in the following way. If a signal x(t) is given by$$x(t)=\sum_{n=...

3

one thing about a non-minimum phase system (with a rational transfer function), is that it can be thought of as the series concatenation (or cascade) of a minimum-phase system, having identical magnitude response as the given non-min-phase filter, with an all-pass filter. the APF will have a poles that cancels specific zeros of the min-phase system that are ...

3

I think simple. We want to model a random physical phenomenon for analysis purpose. One way is to model it by a stochastic process $X(t)$, i.e. a time series of random variables $\left\lbrace X(t_k) = X(t=t_k), t_k \in \mathbb{R} \right\rbrace$. The random variable $X(t_k)$ is associated with a probability distribution function (PDF) with some finite ...

3

The signal, whose total energy you want to calculate, is periodic therefore it will have infinite energy... To see that note, the following Fourier transform pair: $$x(t) = \cos(2\pi f_0 t) \longleftrightarrow X(f) = 0.5 \delta(f + f_0) + 0.5 \delta(f - f_0)$$ And based on Parseval's relation, $$E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt = \int_{-\... 3 Without information on Φ, you can obtain almost anything, since \lambda Φ could be a valid CS matrix as well. Generally, one imposes structure contraints, such as unit energy for their rows or columns. This being said, compressive sensing does not compress data in a strict sense, and energy is a poor measure of compressibility. Entropy and norm ratios ... 3 The procedure is always the same. You need to compute the expectation E\{|A_k|^2\}, where A_k are the complex symbols of the constellation:$$E\{|A_k|^2\}=\sum_kP_k|A_k|^2\tag{1}$$P_k is the probability that the k^{th} symbol is chosen. Usually you can assume that all symbols are equally likely, i.e., P_k=1/M, where M is the number of symbols. 2 In addition to Marcus Müller comment, If a signal has finite energy then the signal value must reach zero after long enough time, but for random signals your signals generally don't have such restriction. 2 If x(t) and X(f) are Fourier transform pairs then the value$$\int_{-\infty}^{\infty}|X(f)|df$$can serve as an upper bound on the magnitude of x(t):$$|x(t)|=\left|\int_{-\infty}^{\infty}X(f)e^{j2\pi f t}df\right|\le \int_{-\infty}^{\infty}|X(f)|df$$2 The energy of x(t) is given by$$E_x=\int_{-\infty}^{\infty}x^2(t)dt=\int_{-\infty}^{\infty}\frac{\sin^2(\pi t)}{(\pi t)^2}dt\tag{1}$$If we may assume that we know that x(t) is the impulse response of an ideal low pass filter, the integral (1) can be computed without using the Fourier transform and Parseval's theorem by noticing that it can be ... 2 No it is not. Total Variation is like the amount of changes in the signal. Though changes require energy it doesn't mean they are proportional. For instance, imagine that during a Window we see a constant signal of high value. Clearly this high energy signal (Unless energy for you is the Variance, usually it is the 2nd moment) yet its Total Variation is ... 2 Neither the integral$$I_1=\int_{0}^{\infty}\left|\frac{\sin(x)}{x}\right|dx\tag{1}$$nor the sum$$S_1=\sum_{n=1}^{\infty}\left|\frac{\sin(n)}{n}\right|\tag{2} converge. So the sinc function (sequence) is not in $L_1$ ($l_1$). A proof of the fact that $I_1$ diverges can be found here. For the sum $S_1$, note that since $|\sin(n)|\ge \sin^2(n)$ (for $n\... 2 In the additive model$y=s+n$, when the signal is deterministic, it adds coherently over the "realizations". Hence, its variance$V(\sum s_n) = V(N s) = N^2 V( s)$. And when the noise$w$is independent identically distributed (IID), then$V(\sum w_n) = NV( n) \$. This is a classical result on the Variance of Uncorrelated Variables. You can find this ...

1

Pick up an apple and raise it one meter. You have just spent (approximately) one joule of energy. You had to spend energy because the action (raising the apple) goes against a force (Earth's gravity). The same is true for electrons. An electrical signal is nothing but electrons moving in an electric field. Causing these electrons to move takes energy, ...

1

Matt L.'s answer is great because it uses an insight from signal processing. Here's a purely "turn the crank" method that uses no signal processing: \begin{align} \int_{-\infty}^\infty \frac{\sin^2(\pi x)}{(\pi x)^2} dx &= \frac{1}{\pi}\int_{-\infty}^\infty \frac{\sin^2 x}{x^2} dx \tag{1} \\ &= \frac{1}{\pi}\left[ \sin^2x \int \frac{1}{x^2} dx\...

1

It seems that you are looking for books to learn about multiple topics. Skimming through your question I think these are the topics: Aspects of signals (signal mean, its power,amplitude,frequency,etc ) Signals propagation in space or any other medium I would recommend parts of the following books : Bernard Sklar - Digital Communications Fundamentals and ...

1

Tendero's answer clearly explains why Parseval's theorem is still valid so I'll just address the question of getting the correct magnitudes (normalization). First off, it should be emphasized that windowing always results in energy loss (except in the case of a rectangular window). In order to compute the correct magnitudes we must compensate for this energy ...

1

Hi: In statistics we call that the probability of type I error ( rejecting when true ) and type II error ( accepting when false ). The way it's done there is that, once you make an assumption about the distribution of the data ( i.e: normal, t, whatever ), you decide on the null and the alternative, along with what you want the P(type I error ) to be ( say 0....

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