# Tag Info

8

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 ...

8

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.[25] However, despite ...

6

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 zero....

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

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 ...

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 ...

5

I think you are correct. People are being fast and loose with the expression in your Eq. (2), but it captures the behaviour of the energy of the signal up to a constant $T_s$ factor (the sampling period), which is maybe why they do it. As you say, the signal energy is given by $$E_s = \int_{-\infty}^{+\infty} |x(t)|^2\;dt \tag{1}$$ If you want to ...

4

[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 ...

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., $$\int_{-\infty}^{\infty} |\tan^{-1}(t)|^2 dt ~~~~~\to \... 4 The procedure is always the same. You need to compute the expectation E\{|A_k|^2\}, where A_k are the complex symbols of the constellation:$$E\{|A_k|^2\}=\sum_kP_k|A_k|^2\tag{1}$$P_k is the probability that the k^{th} symbol is chosen. Usually you can assume that all symbols are equally likely, i.e., P_k=1/M, where M is the number of symbols. 4 By the Convolution Theorem multiplication in Time / Spatial domain is equivalent of Convolution in the Frequency Domain. The sampling rate (In its classic interpretation) is proportional to the support of a function in the frequency domain. So if a function has a certain support in frequency, what would be its support after convolution with itself? Indeed ... 3 Your result is correct but note that for complex signals, the even and odd parts are defined by$$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}$$where ^* denotes complex conjugation. From (1) and (2) it follows that the real part of x_e(t) is even and its imaginary part is odd, whereas x_o(... 3 windowing your data, x[n], with window w[n] is equivalent to windowing the square of your data (or square-magnitude), \Big|x[n]\Big|^2, with the square of the window w^2[n]. so think of this square of the window w^2[n] as just another window. now stochastically your typical squared sample x^2[n] will have some expected mean-square. the ... 3 \delta(x) doesn't really exist at all for any particular x. Like Laurent Duval said, Dirac is not an \mathbb{R}\to\mathbb{R} function, rather the whole mapping$$\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

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....

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, $$E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt = \int_{-\... 3 Without information on Φ, you can obtain almost anything, since \lambda Φ could be a valid CS matrix as well. Generally, one imposes structure contraints, such as unit energy for their rows or columns. This being said, compressive sensing does not compress data in a strict sense, and energy is a poor measure of compressibility. Entropy and norm ratios ... 3 The OP is correct in their dimensional analysis |X(f)|^2 is NOT the power spectral density, despite what other authors might claim. Other authors probably call this the power spectral density because it is close to right and it captures most of the important features without having to delve into technicalities. Power has dimensions of [\text{signal}^2]. ... 3 Constellation diagrams exist in what is called signal space which is an abstraction used to describe finite-energy signals. The coordinate axes, even if they are marked x and y as in Marcus Muller's answer, really represent unit-energy signals such as$$s_I(t) = \sqrt{\frac 2T}\cos(2\pi f_c t)\mathbf 1_{t \in [0,T)}(t),s_Q(t) = -\sqrt{\frac 2T}\sin(...

3

I guess this probably just a mistake in your analysis code. Quantization noise is white and for a noise signal it's uncorrelated to the original signal, so spectrum of the original noise doesn't matter. I did repeat your steps and saw exactly what I expected: The quantization noise is white and the spectrum of the quantized signal follows the original signal ...

3

Potentially, no. I will attempt to find that $L_2$ is not closed under convolution. In other words, a convolution of two $L_2$ functions (or energy functions) is not necessarily in $L_2$. Convolution is a complicated operation, so working in the Fourier domain with products of functions is tempting. However, if Fourier provides an isometry between the primal ...

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

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\...

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