New answers tagged continuous-signals
2
Agreed to others. You need to narrow down your interest to a frequency band as PLC can be achieved as narrow-band or wide-band. Your channel and noise model would depend on the frequency band you are interested in. For instance, if you are interested in broadband communication (e.g. OFDM), then you need to take transmission/reflection coefficients of the ...
1
The secret to proving the limit is to not convert the $sin$ to $sinc$. This leaves the $T$ in the argument and not outside. No need to worry about the complex conjugate as the values are real. The $sin$ values are bounded by -1 and 1.
At that point it becomes:
$$
\begin{align}
0 \le S_{xx}(f) &= \lim_{T\rightarrow\infty}\frac{1}{T}\left|X_{T}(f) \...
1
You don't need to make it so complicated as to use two variables $t$ and $t^\prime$.
\begin{align} X_T(f) &= \int_{-T/2}^{T/2} x(t)\exp(-j2\pi ft)\,\mathrm dt\\
&= \int_{-T/2}^{T/2}\frac{\exp(j2\pi f_0t)+\exp(-j2\pi f_0t)}{2}\exp(-j2\pi ft)\,\mathrm dt\\
&= \frac 12\int_{-T/2}^{T/2} \exp(j2\pi (f_0-f)t)+\exp(-j2\pi (f_0+f)t)\,\mathrm dt\\
&= ...
1
The OP is correct in their dimensional analysis
$|X(f)|^2$ is NOT the power spectral density, despite what other authors might claim. Other authors probably call this the power spectral density because it is close to right and it captures most of the important features without having to delve into technicalities.
Power has dimensions of $[\text{signal}^2]$. ...
1
Starting with from the linked question:
$$S_{xx}(\omega)=\lim\limits_{T\to \infty}\mathbf{E} \left[ | \hat{x}_T(\omega) |^2 \right] $$
$$ = \lim\limits_{T\to \infty}\mathbf{E} \left[ \frac{1}{T} \int\limits_0^T x^*(t) e^{i\omega t}\, dt \int\limits_0^T x(t') e^{-i\omega t'}\, dt' \right] = \lim\limits_{T\to \infty}\frac{1}{T} \int\limits_0^T \int\limits_0^T \...
1
This is not really an answer, but a different angle:
Physically speaking, the power of a single signal is not well defined. Physical power (or intensity) is always the product of two root-power quantities (which used to be called filed quantities). Voltage times current, force times velocity, etc.
Let's say you have an impedance with a circuit with a ...
2
I am curious about the same issue the OP raised: sometimes the units of power spectral density seem not quite right. I assume it is just me, so I always go back to my touchstone reference 1. Blinchikoff and Zverev use the definitions of Fourier transform and inverse transforms I have always used and preferred [1, p. 294]:
and they give the units of the ...
2
THIS IS NOT YET A COMPLETE ANSWER BUT THE CONTINUATION OF THE OP's QUESTION IN MY ATTEMPT TO ANSWER and like the OP, I would welcome a short concise bottom line answer that addresses this.
Update: After the OP pushed me through the mud on this I ended up siding with his same level of questioning and concluding that in a strict sense $|X(f)|^2$ is en energy ...
1
It has more to do with the simple nature of analysis of the data present. For example: given a sequence, it is easier to calculate the DTFT to get the Fourier transform then to calculate the autocorrelation and calculate the Fourier transform of the autocorrelation to get the PSD.
One could easily interpolate the DTFT, assuming a sampling time to get the ...
0
That seems fine. If you integrate your PSD over all frequencies you get a $1$ at $-f_0$ and $+f_0$ and zero everywhere else. $1+1 = 2$ so the total integral will come out to be $A^2/2$ which matches your time domain number.
Yes, the PSD is also the magnitude squared of the Fourier Transform, i.e.
$$PSD(f) = X(f) \cdot X^*(f)$$
where $X(f)$ is the Fourier ...
0
Cleanliness of the eye pattern sample point is also used for symbol synchronization for NDA timing recovery.
My communications book essentially ignored this, I don't know why. But carrier and symbol synchronization seems to be completely ignored in many newer digital communications texts.
4
It is the objective of the receiver to make the best estimate for each symbol as to what was transmitted. This is often done by ultimately determining a decision time in each sample (through timing recovery) on the waveform after it has been processed by the receiver (equalization and matched filtering) in which to sample the waveform and make a decision as ...
1
Welcome! Regarding your questions:
a) Its value is not finite. $A$ (or $2\pi A$, which is the correct answer) is not its value, it's the delta function's "area". $\delta(k)$ is not a proper function, it's a distribution or a generalized function. When we sketch such functions, we denote their area as a scaling factor. Indeed the actual value of the delta ...
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