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Most certainly not. While there has been some claims to break Shannon here and there, it usually turned out that the Shannon theorem was just applied in the wrong way. I've yet to see any such claim to actually prove true. There are some methods known that allow for transmission of multiple data streams at the same time on the same frequency. The MIMO ...

16

The capacity of a channel should be viewed as analogous to the speed limit on a highway. It is possible to travel at a speed greater than the posted limit on a highway but it is not possible to achieve good gas mileage while doing so. Similarly, it is possible to transmit data at rates higher than the capacity of the channel (in fact, unlike highways, ...

9

Like other comments, I don't really understand your question. What I am trying to do is to list basic things so that other ones can suggest edit because I find it is much easier to write in answer part :). Please let me know if it makes you less confused. The general communication system is At transmitter, before modulator is digital, modulator is the ...

7

This is a behaviour that is commonly outlined in some textbooks and tutorials on FEC, but usually in the form of an observation. For instance, Turbo Codes are sometimes characterised by their Threshold $E_b/N_o$, which can be considered as the minimum SNR per bit from where the BER starts to decrease. I have not seen a formal explanation for this property, ...

6

A binary symmetric channel (BSC) can be characterized by its complemented probability $p$. Its well-known capacity is $$C = 1 - H(p) = 1 - (-p\log(p) - (1-p)\log(1-p))$$ where $H(p)$ is binary entropy function: A $L-$concatenated BSC, which is also a BSC characterized by $p_L$, can be visualized as in the figure below The complemented probability $p_L$ ...

5

There are, a few discrepancies that might be making a difference here. My suggestion would be to edit the question for clarity. There are quite a few assumptions that lead to non-straightforward thinking about the problem which I have tried to address to an extent and I would be happy to modify the response in light of more information. In machine ...

5

There are two different kinds of channel models that are being confused here. In the binary symmetric channel, the inputs and the outputs are constrained to be $0$ or $1$ and the key parameter is the transition probability: the probability that an input $0$ is changed to an output $1$ or vice versa. The channel capacity $C$ is a number between $0$ and $1$ ...

5

There are two main factors when figuring out how many bits are transmitted per symbol (or "channel use"): the modulation and the error correction encoding. For instance, BPSK modulation with no encoding transmits 1 bit/symbol, while QPSK with no encoding transmits 2 bits/symbol. Higher order modulation schemes (e.g. 8-PSK, 16-QAM, 32-QAM, etc.) can ...

4

It is not clear whether you want a derivation of the formula for the error probability of a wireless OOK system, or ideas about how the formula might apply to an optical communication system. For a wireless OOK system, there are two main types of receivers called coherent receivers and noncoherent receivers and these have different error probabilities. ...

4

The Berlekamp-Massey algorithm and the extended Euclidean algorithm (both further extended with the Chien search and the Forney calculation of error values) for decoding BCH and RS codes (hereinafter referred to as the decoder) have the following characteristic that is not well understood by many people: If there is a codeword $\hat{\mathbf C}$ that ...

4

I start with BPSK achieving 1 bits/sec/Hz over passband AWGN. Factoring in 1/3 rate coding this becomes 0.333 bits/sec/Hz. This is not correct. The Shannon noisy channel coding theorem states that the reliable discrete-time rate $r$ (whose unit is bits per symbol, or bits per channel-use, or bpcu) is upper-bounded $$r \lt \frac{1}{2}\log_2\left(1 + \frac{S}{... 3 What channel model fits data transmission over air best, I guess distortion generated by speakers should also be take under account? Disclaimer: I'm really not an Audio person. There's a lot of audio DSP engineers on here that probably have a far better idea of what's happening on an acoustic channel. But: Frequency selectiveness will be pretty important. ... 3 The energy per bit, E_b, is independent of the coding rate. Note that E_b measures the energy per transmitted information bit, not per transmitted symbol. Let's say you're willing to spend one joule per information bit, so E_b=1. You use uncoded BPSK, so that each transmitted symbol carries one bit of information and so it also has energy one. Let us ... 3 The relationship between SNR and E_b/N_0 is independent of the code rate. Note that E_b is the energy per data bit (not the coded bits), and R_b is the (uncoded) data bit rate. As long as you keep using these values, you can use the formula given in your question. Of course, when going from an uncoded system to a coded system, the values of E_b and ... 3 They don't get unified. Think of the transmitter pipeline (data source, source encoder, channel coder, modulator, etc) as a sequence of independent blocks. Blocks don't assign any particular meaning or order to their input: they regard the input as just a stream of bits. So, the output of the Huffman encoder can be regarded as a stream of 0s and 1s. The ... 3 I don't think you cannot state generally that y_1 and y_2 will be orthogonal. I'll try to sketch out my thinking: Since the Hamming code is a linear code, each parity bit can be represented as a linear combination of the information bits; that is, each bit can be represented as:$$ p_i = \mathbf{g_i^T}\mathbf{x} $$where \mathbf{x} is the ... 3 Because the entropy represents information quantity, or if being measured in bit, the smallest number of bits per symbol we need to represent a source. The source A contains 4 equiprobable symbols hence it is obvious that we need \log_2(4) = 2 bits per symbol to represent the source. The source C is simply A with its parity bits hence no new ... 3 First I found the sine wave extrema (black markers at the bottom) and stretched and superimposed a sine wave graph (red) on top of the signal graph with the extrema in the same locations. The signal matches (0) the sine wave over some of the non-zero segments, and over the other non-zero segments it matches the sine wave with a sign flip (1). This gives a ... 3 The IQ data looks like a sinusoidal modulated pulse train with binary phase shift keying (BPSK) superimposed. The BPSK symbol rate appears to be at the same rate as the pulse repetition frequency (PRF). I was bored, so I went ahead and modeled the data in MATLAB. clear; clc; % Measured quantities from IQ data pw = 2.1771e-4; % Pulse Width pri = ... 3 I think you're confusing two different (but related) terms. Nyquist says that in a channel of bandwidth B you can transmit up to 2B orthogonal pulses per second. So, R_p \leq 2B, where R_p is the pulse rate. To achieve R_p = 2B, the pulses need to be sinc-shaped. Other, more practical pulses achieve slightly less than that. For example, raised ... 3 As specified in documentation, using 'ParityCheckMatrix' you can configure the Parity Check Matrix (PCM) during the constructions of the encoder/decoder objects. The syntax is encoder = comm.LDPCEncoder('ParityCheckMatrix',pcm) or simply encoder = comm.LDPCEncoder(pcm); where pcm is the desired PCM which must be sparse type. An example for (probably poor ... 3 Yes, it is possible to determine the path that the Viterbi algorithm found through the trellis. Just apply the encoding algorithm to what you are calling Y_dec and you will get the corresponding codeword of length 2048. You can then compare it to Y, the transmitted codeword, to see where the channel made errors. Additional notes: If the data to be ... 2 I know of 3 ways to exceed Shannon - 1) MIMO exceeds Shannon. Technically each MIMO channel is limited by Shannon, but the sum of the channels exceeds the limit. The practical limit is the ability to distinguish each MIMO channel. 2) Dr. Solyman Ashrafi (CTO at MetroPCS) owns a patent for a technique using naturally orthogonal wavelets (or Hermite ... 2 If you want to transmit a symbol sequence A_k using baseband pulse amplitude modulation (PAM), the transmitted signal is$$s(t)=\sum_kA_kp(t-kT) where $p(t)$ is the transmit pulse, and $T$ is the symbol interval. If you want to avoid intersymbol interference (ISI), the pulse function $p(t)$ must satisfy the Nyquist criterion, which says that its value ...

2

The answer to the question depends on the fact if the code rate implies a change in the bit rate f the system. The bit rate would be the rate of bits that come out of the system per second, and the code rate the rate between an uncoded length of data and the coded length of data. If the coding of the data does not imply a change in the bit rate of the ...

2

This is solved slightly differently if you're simulating a continuous-time or discrete-time AWGN channel. I'll assume you're simulating a discrete-time channel; that is, you're simulating the output of the receiver's matched filter sampled at correct time instants. First consider the case with no noise. In this scenario, the matched filter's output is an ...

2

I don't know about optical, but I can give an electrical interpretation. In a noiseless OOK system, for each transmitted bit, the receiver will get either $P_r$ or 0. When you add AWGN, the numbers become $P_r+n$ and $n$, where $n$ is a sample of a Gaussian random variable with zero mean and variance $\sigma^2$. Consider transmission of a 0. There will be a ...

2

We have different interleaving techniques, and matrix interleaving is one of them. But at the end all of them do one thing: interleaving is a technique to protect against burst errors (no matter how we do it). To make it more clear, you should consider the reason a packet cannot be decoded (and is failed at the receiver). Each packet usually contains a ...

2

A few ideas to complement Marcus' answer: It is relatively easy to find an approximation to the channel's impulse response: send audio consisting of a single positive sample surrounded by zeros, and record the result. If the channel response is not flat, you can either transmit slowly over a narrow frequency band (so that the channel looks flat), or ...

2

"The best channel code" is really vague. We can compare channel codes under power, bandwidth, encoder or decoder computational complexity criteria. We can compare minimum hamming distance, correction capability, coding gain, spectral efficiency, error probability, ... Optimality can be analyzed in different contexts as well. RS codes achieve the Singleton ...

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