8

Your mathematical derivation is correct, your $H[k]$ is the single-tap equalizer (i.e. one tap for each subcarrier, and the subcarriers do not mix with each other. That's the orthogonal in OFDM). Let me try to explain this a bit more general, without going into coherence bandwidth and flat fading. To my understanding, explaining it with $B_c$ and flat ...


6

By using cyclic prefix and its associated demodulation techniques, LTE OFDM (nearly) eliminates the inter-symbol inteference (ISI). The intra-block inteference, if I understand well your question, refers to intra-carrier inteference (ICI) when the orthogonality between sub-carriers is not maintained due to Doppler, delay spread too large, synchronisation ...


6

I am a little late but I post my answer anyway so that someone having the same question will find it interesting and discuss. The discrete baseband multipath channel can be modeled as a FIR, i.e. $$y[n] = \sum_{l=0}^{L-1} x[n-l] h_l + w[n]$$ where $L$ is the number of channel taps. $L$ depends on the relation between the bandwidth of basis waveform and the ...


6

Lets say we want to transmit a sequence of discrete data $\left\lbrace x[n] \right\rbrace$. But because we are living in analog world, the sequence must be modulated. Call $T_s$ is symbol duration and use a set of orthonormal waveforms $\left\lbrace p_n(t) = p(t-nT_s), n \in \mathbb{Z} \right\rbrace$, (baseband) signal $x(t)$ can be written as \begin{...


4

The strategy depends heavily your "realistic scenario", e.g. which estimator, which equalizer, which estimation error, etc. In general scenario, you cannot trust the estimate of $h$, it means that you cannot use coheent detection. Thus try non-coherent detection (detection by energy) : to send bit $0$, use 2 channel uses (2 symbols) $x_0 = [x[0], x[1]] = [1,...


4

A discrete-time baseband model of a multi-propagation channel can be written as $$y[m] = \sum_l h_l[m] x[m-l] + w[m]$$ where $l$ is the index of channel taps if the channel is modeled as a FIR $\{h_l, 0\leq l < L-1\}$. $w[m]$ is AWGN sample. The channel tap $h_l[m] = \sum_i \alpha_i(m/W) \times \mathrm{sinc}(l-W\times\tau_i(m/W))$ where $i$ is the ...


4

There are too many things named after Rayleigh floating around in this problem. Consider first additive white Gaussian noise with two-sided power spectral density $\frac{N_0}{2}$. This is generally taken as the model for channel noise, though in fact the source of the noise actually is thermal noise in the front-end of the receiver. NASA once even ...


4

Your approach will not model the actual channel but can be used to create the received waveform as having passed through a Rayleigh fading channel. I describe considerations to actually create the received waveform using your approach in either flat and frequency selective conditions below, which should give insight into why it would not be a channel ...


3

The fundamental idea to keep in mind is that in a wireless channels with reflections, if you transmit $s(t),$ you'll receive $$r(t)=\sum_{i=1}^Na_is(t-\tau_i).$$ Another important idea is that whether the channel is flat or not depends only on $s(t)$. For instance, let's say that the symbol time is $T_s$. Let's call the longest delay in the channel $\tau_{...


3

Question 1: Yes, it is a multipath Rayleigh Channel with $ 1000 $ taps. You need to know that when you generate complex coefficients, and the real part and the imaginary part of this coefficients are randomly chosen according to a Gaussian distribution, the magnitude will have a Rayleigh distribution and the phase will have a uniform distribution between $ ...


3

In brief, we consider the channel as frequency-selective channel if the frequency of the signal is larger than then frequency of channel No offense, but that ought to win the prize for the least-accurate definition of frequency-selectivity I heard ;-) What you mean is channel is frequency-selective if the bandwidth of the signal is larger than the ...


2

There's a lot of very valid aspects that you touch, but from what I've learned (and experienced having fun with OFDM SDR transceivers), the main reasoning to do OFDM is having a narrow-channel multicarrier system with low complexity. Let me elaborate: Multi-carrier is, as you mentioned, very handy because you get a flat channel. That means that you can do ...


2

The path loss exponent does not depend on the reference distance. There are two mistakes in your path loss formula. First, path loss does not include the gains of transmit or receive antennas. Second, the path loss exponent must also be part of the expression for the path loss at the reference distance. According to the Friis transmission equation the ...


2

IEEE 802.15.4a defines multiple bands from 2.4 to 10 GHz for operation. You should focus on one. Not sure you're actually looking for diffraction and not more for backscatter from surrounding objects and the ground. I'd recommend not trying to come up with a physical channel model first. Do measurements, find a statistical model that works reasonably well. ...


2

In a wireless system, most of the noise is added by the receiver itself. The received signal is very weak (-100 dB or less is typical), and its power is comparable to the thermal noise of the receiver's amplifiers. That is the reason why noise is modeled as being added to the signal present in the antenna. The noise still affects the estimated signal, of ...


2

Reed-Solomon codes are good for correcting bursts of errors (such as those that often occur in computer memory systems such as RAM, or on less volatile media such as CDs and DVDs) or at the output of block decoders -- either a received word is completely correct, or it has multiple errors all within the codeword length -- or at the output of convolutional ...


2

The channel which you have created is having 4-Taps and all taps are one after the other, meaning roughly there are 4 multi-paths and they are very close to each other. How close depends on what is the Sub-Carrier Spacing you would have assumed. Anyway, the point is only 3 samples of Cyclic Prefix would be enough and even with 0 or 2 samples of CP will not ...


2

Quasi-static channels can be said as "block-wise" time (in)variant. For example, your channel has no variations over time (not in delay domain) for some time-period, e.g., 1 ms, but the channel may change after 1 ms.


2

Quasi-static is almost-static. In other words, for a block (or window) period of time, you could assume that your channel is static. Below, i attach a figure that depicts this scenario. As you can see the channel could be assumed static for around 100 ms.


2

Here is a graphic I have that demonstrates ISI due to multipath propogation: Notice the eye diagram in the upper right hand corner showing what we should receive if there was no ISI, or other noise for that matter. (This eye diagram is showing the waveform vs time over 2 symbols again and again so that we can see the effects of ISI clearly. If you have any ...


2

First it is useful to understand what a Rayleigh and Ricean distribution is, and then we can more easily see how that statistic can be used to describe certain fading channels. A zero-mean complex signal that has an identical Gaussian distribution on both the in-phase and quadrature components will have a magnitude that is Rayleigh distributed. A useful ...


2

Most likely people use different words to describe the same concept. This happens quite often. The TEQ is the more general term and also used to equalize other methods that OFDM, while CS is related to OFDM and tries to make the channel shorter so that it fits within the CP duration. You can check online, you will find that almost all articles related to CS ...


2

Yes, you can have time variant channel in MATLAB either by using comm.RayleighChannel() , you can read about it in help of matlab. or the other way you use the same command you have mentioned but with each symbol the channel should be changed. so it will be like you will generate a channel which is Quasi-static. You above command will become: h = 1/sqrt(...


2

To answer the literal questions you asked: yes see above To Frame Challenge your question and give you what you were probably looking for: The path loss equation you cited: $$ 128+37\log_{10}(d) , $$ is equivalent to: $$ \frac{P_r}{P_t} = \frac{10^{-12.8}}{d^{3.7}}, $$ where $P_r$ is the received power in Watts, and $P_t$ is the transmit power in Watts. ...


2

OK, let me explain that in easier way, As you know, in a communication system, to transmit information from transmitter to receiver, we use the binary information system. which means, for example, this message will be transmitted 1000101010. But in receiver, this message will not be received correctly because of noise of the channel, so some bits will be ...


2

For BPSK it's $$P_e = \frac{1}{2}\left[1 - \sqrt{\frac{\gamma}{\gamma+1}}\right]$$ where $\gamma$ is the average SNR. For 16-QAM, I am not sure if there is a closed-form solution. Check this book. The authors use the moment generation function to find the ABER for all modulation schemes.


1

The channel coding used has little to do with your source coding. So, this depends on how you model your channel (and noise, and interferers), and what kind of errors you're willing to tolerate. Saying "SPIHT-compressed image" answers neither!


1

Let's say we have a function that we transmit which is: $$x(t)$$ Transmitting the function is kinda easy part. We assume there is just the function we are trying to send. However things are a little bit ugly in the air. Let's look at the receiver part. What we generally consider first is the AWGN. Which is: $$r(t) = y(t) + n(t)$$ $r(t)$ is our received ...


1

Your question is quite unclear, but it seems like you're trying to simulate the continuous-time radio channel. In this model, you do not multiply by complex coefficients; rather, you add delayed and attenuated copies of the transmitted signal. Say you transmit $s(t)$. Then, the received signal is $$r(t)=\sum_k g_k s(t-\tau_k) + n(t),$$ where $g_k$ is the ...


1

Because fading and noise affect the received signal in different ways. The received matched-filter output is $$r = hs + n,$$ where $s$ is the transmitted symbol, $h$ is the fading coefficient, and $n$ is Gaussian noise. All numbers are complex and both $h$ and $n$ are complex Gaussian random variables. To estimate $s$, the receiver calculates \begin{align} ...


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