12

If the multiplier takes two voltages as input and returns a voltage as output, then there is necessarily a constant involved, with units of [1/V]. Take for example, AD633 (which was the first search result). The output is the product of the 2 inputs times a constant: $V_{out} = \frac{V_1 \times V_2}{10V}$ So the output units are Volts.


11

I mostly agree with @PeterK.'s answer; in practical terms, all signals are energy signals. However, a signal's energy does have important practical significance. Producing a signal of a given energy consumes at least that same amount of energy, for which you have to pay. In a battery-operated device, this becomes very important, because batteries store ...


9

An energy signal is defined as one with a finite energy: $$ E = \int_{-\infty}^{\infty} \left|s(t)\right|^2 dt \lt \infty $$ A power signal is defined as one with a finite power or energy per unit time: $$ P = \lim_{T \rightarrow \infty} \frac{1}{T}\int_{-T/2}^{T/2} \left|s(t)\right|^2 dt \lt \infty $$ This question over on Electronics.SE has a nice image ...


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 ...


5

I'd like to show you a more formal derivation. Note that the first formula for arbitrary (non-periodic) signals could be rewritten as $$P_x=\lim_{M\rightarrow\infty}\frac{1}{(2M+1)N}\sum_{n=-MN}^{(M+1)N-1}|x[n]|^2\tag{1}$$ for some integer $N\ge 1$. For $N=1$, Eq. $(1)$ is identical with the first formula in your question. If you choose $N>1$ you simply ...


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

The Decibel Miliwatts Scale $dBm$ is the power ratio in Decibels, considering a reference of $P_0=1mW$. $$P[dB]=10\text{log}_{10}(P/P_0)=10\text{log}_{10}(P[W]/0.001)$$ The general context is considering the signal amplitude $V$ in $volts$ and the power as an impedance load system $R$ in $\Omega$: $$P=\frac{1}{R}V^2$$ Moreover, sinusoidal excitations are ...


4

The basic trick is to bound the series above and below. Let us do it on one side, for positive indices. For any $N> 0$, you can write $N=kN_0+r_N$, with $0\le r_N< N_0$. Then if $a_n$ (here $a_n = |x_n|^2$) is positive, $\sum_{n=0}^{N-1} a_n$ is increasing. Now $kN_0 \le N< (k+1)N_0$, hence you have: $$ \sum_{n=0}^{kN_0-1} a_n \le \sum_{n=0}^{...


4

Variance is never defined as power. For a wide-sense stationary random process $X(t)$ with zero mean $$\mu_X=E\{X(t)\}=0\tag{1}$$ the variance of $X(t)$ equals its power. The autocorrelation of $X(t)$ is defined by $$R_X(\tau)=E\{X^*(t)X(t+\tau)\}\tag{2}$$ The power of $X(t)$ is $$P_X=E\{|X(t)|^2\}=R_X(0)\tag{3}$$ The variance of $X(t)$ is $$\sigma^...


4

Energy can be a genuine motivation for real systems, when actual power is used. As said before by @Peter K. and @MBaz, in signal processing, most practical signals are time-limited, and thus energy-signals. But we use energy because this is almost the only quantity for which we can solve stuff efficiently. In signal processing (and other domains as well), ...


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

According to this source, the sound power received by an aperture is proportional to the pressure squared, i.e.: $$ P = \frac{A p^2}{\rho c} \cos \theta, $$ where: $P$ is the received power $A$ is the surface area of the receiving aperture $p$ is the pressure $\rho$ is the density of medium $\theta$ is the angle between the propagation direction and the ...


4

In addition to Juancho's answer for the general mixer, I would like to give an example for a more simpler frequency mixer most commonly used in communication systems to shift the frequency spectrum of a message signal up or down for transmission or reception etc. The simplest understanding of a physical realisation of a mixer assumes an on-off switching ...


3

The first definition works for deterministic as well as for random signals. For random signals we define the autocorrelation by $$R_x(\tau)=E\{x^*(t)x(t+\tau)\}\tag{1}$$ where $E\{\cdot\}$ is the expectation operator. For deterministic power signals (i.e., signals with finite but non-zero power and infinite energy) we define $$R_x(\tau)=\lim_{T\to\infty}\...


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

Depends on your signal. Typically we try to sample at the Nyquist rate, which is equal to two times the maximum frequency of your signal. If you sample less often than this, you will lose information and your answer will not be correct, I think.


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

To get the total power across bins, sum the power in each bin. Also you need to compensate for your window loss if you want an accurate result. For a rectangular window, the power in each DFT bin is given as: $$P = \frac{Y[k]Y[k]^*}{N^2} = \bigg(\frac{|Y[k]|}{N}\bigg)^2$$ and $$P_{dB} = 20Log_{10}\bigg(\frac{|Y[k]|}{N}\bigg)$$ The total power over a ...


3

First of all, note that your formula for signal power is only valid for real-valued $x(t)$ that satisfy $x(t)=0$ for $t<0$. A more general formula is $$P_x=\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}|x(t)|^2dt\tag{1}$$ For signals with finite energy $$E_x=\int_{-\infty}^{\infty}|x(t)|^2dt\lt\infty\tag{2}$$ the power $P_x$ defined by $(1)$ is clearly ...


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 $[\text{signal}^2]$. ...


2

Another intuitive way: $P_{\text{x}}= \lim\limits_{N \to \infty}\frac{1}{2N + 1}\sum_{n=-N}^{N}|x(n)|^2$ Now the period is $N_0$ and it extends from -N to N, so total number of periods between $-N$ to $N$ are $2N/N_0$. $P_{\text{x}}= \lim\limits_{N \to \infty}\frac{2N}{N_0(2N + 1)}\sum_{n=0}^{N_0-1}|x(n)|^2$ $P_{\text{x}}= \lim\limits_{N \to \infty}\frac{...


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

No need to do the convolution to find the output. Just use the fact that $z^n$ is an eigenfunction of the system with $z=e^{j\omega_0}$. This will lead to the following rule $$x[n]=A\sin(\omega_0 n+\phi) \Rightarrow y[n]=|H(e^{j\omega_0})|A \sin(\omega_0 n+\phi+\angle H(e^{j\omega_0}))$$ That is, the output to a sinusoidal signal with angular frequency $\...


2

A signal either has finite energy, finite power or even infinite power. If it has finite energy, it will have zero average power, according to your definition $$P_x=\lim_{T_0\rightarrow\infty}\frac{1}{T_0}\int_{-\frac{T_0}{2}}^{\frac{T_0}{2}}|x(t)|^2dt.$$ Knowing that the sincs are orthogonal to each other, as well as the cos function, the average power is ...


2

The first definition you wrote corresponds to the average power of a discrete signal $x[n]$. If the signal is periodic, that means that $$\exists N\in\mathbb{Z}:x[n] = x[n+N] \ \forall n$$ So if you find the period of the discrete signal, $N$, then if you use your first formula you would be doing the average between a lot of quantities that are all exactly ...


2

The formula for signal power is the total energy of the signal, divided by the length of the signal: $$ P = \frac{E}{N},$$ where $E$ is the energy of the signal, and $N$ is the length in samples. The total power of the signal is given in both of your equations in the summation. The summation adds together the squared values of all the samples. That value is ...


2

Discrete signals are by definition periodic and you need to calculate the power in both domains over one period with proper normalization. In practice this means, you need to divide the spectrum by the FFT length. See https://en.wikipedia.org/wiki/Parseval%27s_theorem


2

Solve the integral: $$ P = \lim_{T \to \infty} \frac{1}{T}\int_{-T/2}^{T/2} |x(t)|^2 dt $$ This is usually unwieldy to solve directly; it can be shown that if a signal is periodic, you only need to integrate over the fundamental period $T_0$: $$ P = \frac{1}{T_0}\int_{-T_0/2}^{T_0/2} |x(t)|^2 dt $$ EDIT: Another "trick" is to note that sinusoidal signals ...


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