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 ...


6

Would the definition be equally intuitively valid? No. Using the expectation operator $\mathbb{E}$ implies that both $P$ and $N$ are functions of time. The problem with the second definition is that $P(t)$ can be occasionally very small or zero in which case the quotient gets very large or infinite. These "low noise" instances will dominate the ...


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 ...


5

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 ...


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

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 ...


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

If $x(t)$ is a finite-energy signal with Fourier transform $X(f)$, then $x(t)\cos(2\pi f_c t)$ is also a finite-energy signal with Fourier transform $\left.\left.\frac 12 \right[X(f-f_c) + X(f+f_c)\right]$. This is just the modulation theorem of Fourier transform theory. The energy spectral density of $x(t)$ is $S_x(f) = |X(f)|^2$ while the energy spectral ...


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

Could somebody explain what does this mean? Why does a window leak power into adjacent frequency bins? You can think of frequency bins as imposing an artificial frequency grid which allows only certain frequencies to fit nicely within it (producing a sharp peak) and all other frequencies are smeared across adjacent bins thus producing spectral leakage. ...


3

If the $x[n]$ are i.i.d. random variables with mean $\mu$ and variance $\sigma^2$, then the power spectral density of this discrete-time random process (which is effectively white noise plus a possibly nonzero mean $\mu$) is what you have calculated. The shape of the common pdf of the random variables is irrelevant in all this except insofar as the shape ...


3

The spectrum is that of the base pulse used in the modulation, so in this case a rectangular pulse. A single rectangular pulse in time, as given by the Fourier Transform, is a Sinc in Frequency with the first nulls spaced at $1/T$ away from the center of the Sinc, where $T$ refers to the duration of the pulse in time. (Just as that shown by the OP). If the ...


3

Given that RMS means Root Mean Square the answer is rather obvious: it's a root power quantity. The RMS of a signal has the same units that the signal itself. Saying that "RMS is proportional to power" is just sloppy phrasing. Correct would be "the square of the RMS is proportional to power". If you by "power" you mean "...


3

In principle, you can always test your channel and adjust your power according to the channel state before sending any signal. Each channel is always unique regardless of being in a city or in more open area, so testing the channel is always ideal. Generally, two crucial factors are affecting your transmission; path-loss and fading. First, you want to know ...


3

This sounds like something from a datasheet. It means, that a dBFS-Value is stored (in some register). So, not a linear value, but a dB-Value. The resolution in dB/LSB can be thought of as a multiplicator. With 1dB/LSB (Least Significant Bit), the dBFS-Value can be directly obtainend by reading the register. So, an 8Bit-Register holding the value 0x20 with ...


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