New answers tagged digital-communications
1
I'm guessing it's used in the following context.
Assuming you have a physical source with some of an amplitude distribution in a plane: you can calculate the polar pattern of that source simply as the Fourier Transform of the amplitude distribution as a function of space (not time!).
Fourier Beamforming would be the inverse process: you start with the ...
1
My personal point of view: what matters much more than the distribution is the second-order statistics of the channel. In particular things like delay spread / coherence bandwidth and Doppler spread / coherence time. These parameters tell you a thing or two that you must take into account when estimating your channels:
The coherence bandwidth (roughly equal ...
2
Let $p_k(t) = p(t-kT)$ an orthonormal signal set for integer $k$ and $T>0$. In other words, we require that:
$$
\int_{-\infty}^\infty p(t-\alpha T) p(t - \beta T) dt =
\begin{cases}0, \text{ if $\alpha \neq \beta$}\\1, \text{ if $\alpha = \beta$}\end{cases} $$
The most common example of an orthonormal set is the square-root raised cosine pulse (SRRC). ...
1
Lets say you had two signals, assume that they use the same pulse shaping filter $p(t)$, and assume that their codes, $\mathbf{s}_1$ and $\mathbf{s}_2$, are orthogonal ($<\mathbf{s}_1,\mathbf{s}_2>=0$) and length $N$. The two pulse shaped signals are $x_1(t)=\sum_n s_1[n]p(t-nT)$ and $x_2(t)=\sum_ks_2[k]p(t-kT)$, where the notation $s_1[n]$ means the $...
0
Based on my experience in 802.11ad/11ay standard, I tried to check whether Golay codes used in this standard satisfy this criterion. https://en.wikipedia.org/wiki/Binary_Golay_code
The binary golay sequences comprising of +/-1 are used in the 802.11ad/ay standard for preamble transmission as well as spreading. The 32 and 64 length golay sequence are listed ...
1
In general for a pulse shaping filter and any orthogonal code family: No.
UPDATE: See @Engineer's answer which I confirmed by excluding the first row in the set of Walsh codes (the all 1 row) the remaining rows which have a balance of 1's and 0's are indeed all orthogonal.
*SECOND UPDATE: @MBaz has pointed out here that with Root-Raised Cosine filtering ...
2
So, I myself have taught based on material that uses that scheme, "P/S" and "S/P" after and before the transforms.
Personally, it's nonsense.
What the author tries to say is:
The IFFT is a mapping of sample vectors to vectors. So, you need a vector as input, not a stream of samples.
What they instead say is:
We use the terminology from very basic ...
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.
5
In digital communications, information is transmitted in quantized form. We are constrained to transmit items that belong to a finite, discrete set. The items are called "symbols", and the set is called a "constellation". The symbols may be transmitted by changing a signal's frequency, phase and/or amplitude.
As an example, pulse amplitude modulation (PAM) ...
0
Let us say you detected noise level power at -100dBm. You keep additional 10dB threshold to detect energy. Means if you detect >= -90dBm power, you try to detect preamble of Wifi packet. If you detect -95dBm power, you ignore it assuming it may not be a valid Wifi packet. This really works assuming you are very far from base station/access point(AP). Now ...
0
When considering a transceiver system, if we do a certain filtering
operation on the signal, will the BER of the system be affected?
Yes it does. The whole of equalization concept in digital communication is to mitigate the effect of transmit signal getting filtered by channel (which includes right from the Transmit chain components, the communication ...
1
What you mean might be circulant matrix instead of toeplitz matrix. See section 3.4.4 in https://web.stanford.edu/~dntse/Chapters_PDF/Fundamentals_Wireless_Communication_chapter3.pdf about how the circular convolution in OFDM is represented by matrix operations (eq 3.130 onwards).
First, in almost all standard OFDM systems, you can assume $D \le L$. The ...
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 ...
0
Help design my system would be far more detail than what we typically provide here, but I can give you some suggestions that may help or at least lead to a more concise question.
I do not get correct BER, which means my signal is getting aliased as
I receive through the DDC.
Not getting the correct BER does not necessarily conclude this.
In your ...
0
The answer to your direct question is no because the channel will not pass the frequencies where the signal is.
To put in some practical sense, it will happen in the high frequency (HF = 3-30 MHz) regime where many factors affect whether or not certain frequencies propagate well. For example, a signal may propagate at a given frequency during the certain ...
0
Slide 3, the $-70.5 dBm$ and not $dB$
0
Yes you can shift its spectrum to fit into the band of $[1k, 5k]$ Hz.
The following will do:
$$y(t) = \cos(2\pi ~5000~ t) \cdot x(t)$$
A new band of will appear at $[11k, 15k]$ Hz, that must either be filtered out by you before adding to the channel, or the channel itself will do it for you.
1
If you're familiar with LaPlace transforms, you can see the Z transform by analogy.
The unit circle is equivalent to the jw axis, with zero frequency at 1+j0 and the Nyquist rate at -1+j0.
1
BIBO stability of LTI systems implies that their impulse response is absolutely summable, that is,
\begin{equation}
\sum_{n=-\infty}^{+\infty}|h(n)| < +\infty
\end{equation}
That exact same relationship is a sufficient condition for the Fourier Transform of the impulse response - the so-called Frequency Response - to converge.
Convergence of the ...
0
We get the Fourier Transform of a signal at the unit circle. If the ROC does not include the unit circle, that means that the Fourier transform does not converge which means that the system is unstable. Also, please read bores signal processing basics website and Alan Oppenheim. Its explained really well there.
0
Your receiver does not know what BW you are using. Hence it will always be tuned to receive 20MHz BW (ch 36 or 40). So for the preamble part you need to transmit in channel bonding (2 x 20MHz) mode. Once the receiver has detected preamble it decodes the header to find out that data part is 40MHz. You need to turn on a bit in header for this (forgot the exact ...
0
As far as the recovery time and results for the timing recovery, this is a loop implementation and you would need to review the complete loop for stability and gain parameters to balance loop bandwidth & convergence time, stability and noise performance.
Even with direct board to board communication you will have a static timing offset as the receiver ...
0
If your frequency offset is small you can do timing synchronization and then carrier recovery which is only based on the rotation of the IQ samples.
I found PLL confusing until I understood them in a statistical sense with the filter being an averaging mechanism.
Timing synchronization can be done with variance minimization or power maximization, etc.
2
One approach not yet mentioned is frequency multiplication by a factor of the number of phase positions used followed by a PLL for noise reduction and then a frequency divider by then same factor. The reason for this is when you multiply the frequency by N, you also multiply the phase by N. If the phase positions are selections of $2\pi/N$ (or those with ...
1
If you know the pulse shape and bandwidth, a band-edge detector would be a classical first step; see fred harris' talk on that (+slides).
After doing that, you'd have a good carrier frequency and even phase estimate.
You'd do that before estimating whether it's PSK or QAM, because the latter becomes much easier after.
You could then use a PLL to track and ...
0
interference intensity is different at each sub-channel but varies with same distribution
You're already using the term intensity, so this sounds like you're most likely implicitly using a Poisson Process, which is a stochastic typical model for "arrival processes", e.g. packets appearing in a network. It's based on the idea that the probability of an event ...
0
The difference between linear and non linear for DFE equalizers for instance is due to the fact that there is a decision device in the equalizer, so the output is no more proportional to the input .
1
When we talk about Doppler, we typically refer to time-varying systems.
A popular model in system theory is to assume systems to be LTI (linear time-invariant). This means that beside being linear, their input-output transformation does not change in time, it does not matter when we apply a certain input signal $x(t)$, we'll always get the same $y(t)$. ...
1
I can't provide a lot of theory but I can provide this qualitative description of the problem:
Normally the output of your digital filter sampled every symbol period gives a point on your I/Q plane. Decision feedback usually selects the constellation point closest to this point and modifies your filter to make the output(given the same input) closer to the ...
-1
It seems that this PhD dissertation covers part of your question:
« Improved receivers for digital High Frequency communications: Iterative channel estimation, equalization, and decoding (adaptive turbo equalization)»
https://ntnuopen.ntnu.no/ntnu-xmlui/bitstream/handle/11250/249838/126139_FULLTEXT01.pdf?sequence=1&isAllowed=y
Edit:
4.2 tapped delay ...
1
Below 1D argumentation also explains the 2D case.
First consider the DTFT property for the pair $x[n] \longleftrightarrow X(e^{j\omega})$
$$ e^{j\omega_0 n} \cdot x[n] \longleftrightarrow X(e^{j(\omega - \omega_0)}) $$
Then recognise that $(-1)^n = e^{j \pi n} $ which yields:
$$ e^{j\pi n} \cdot x[n] \longleftrightarrow X(e^{j(\omega - \pi)}) $$
The ...
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