8

There are very few m-sequences of any given length with good cross-correlation properties. Their autocorrelation properties are excellent, but the cross-correlation properties are variable. For example, there are 18 m-sequences of period 127 but to have good cross-correlation properties, one must choose a set of no more than $\require{cancel}\cancel{\text{...


8

Direct-sequence spread spectrum (DSSS) is a technique that is used to generate a modulated signal that occupies more bandwidth than would be implied by its information content alone. A DSSS signal is generated via multiplication of a (typically digitally-modulated) baseband signal by another spreading code waveform. The spreading code waveform is constructed ...


5

disclaimer Because of the lack of document about LoRa CSS in Internet, the analysis below may be wrong with respect to LoRa system, not to the (linear) CSS principle. Any comment or update are appriciated. Thanks. LoRa CSS modulation LoRa CSS uses linear chirp spread spectrum. To transmit $\log_2 M$ bits, LoRa CSS modulation divides bandwidth $[-\frac{B}{...


4

Let's start by fixing a symbol rate $R_s$ symbols per second. To modulate $R_s$ symbols per second without ISI, Nyquist says that we need a bandwidth at least $BW_0=1/R_s$ Hz. With spread spectrum, we use more bandwidth, say $M$ times, than what Nyquist advised. The used bandwidth is $BW=M\times BW_0=M/R_s$. In LoRa specifications, $BW=125, 250, 500 \...


4

Yes, it is possible to hide data inside audio files using FSSS technique. Have a look at this publication and this one. Since HAS (Human Auditory System) is more sensitive than HVS (Human Visual System), you are usually limited with audio steganography comparing to image steganography. There are not many detalied publications with "Audio Steganography" tag....


3

You are mixing up two different notions that have little to do with each other. The use of spread-spectrum signaling is not in an effort to achieve (or even approach) the capacity of the (wideband) channel. Indeed, the spread signal uses only a small fraction of the capacity of the wideband channel, and the rest of the capacity is available for use by ...


3

If I'm only using bandwidth B1, doesn't that mean I can filter some of the noise out? And if so, would I be able to get S/N down? Wouldn't that be an alternative to occupying the full channel bandwidth? Capacity increases linearly with bandwidth, but only logarithmically with SNR. So, increasing SNR by decreasing bandwidth is a bad idea if your noise isn't ...


3

CDMA means that multiple signals are spread and "stacked" on top of the same wide-band channel (each signal takes the entire bandwidth). Each original signal can be obtained because the spreading sequences are orthogonal. Check out http://en.wikipedia.org/wiki/Code_division_multiple_access#Steps_in_CDMA_Modulation for a good example of how this signal "...


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

For a single polynomial generator, the minimum number of samples is 2N+1 where N is the order of the polynomial. If noise free this would result in 10 equations with 10 unknowns, sufficient to solve the linear equations involved. For the case of the Gold Code given above, the generator shown could also be done with a single polynomial which is determined by ...


2

There are many ways to do this. A simple method is to take the current value of the register (not just its output) as the carrier frequency selector. For instance, a (6,1) LFSR with 63 states can select between 63 different channels. Other methods are based on cubic and quadratic congruence codes. I don't know much about them myself, but look up the papers ...


2

If your wireless channel has bandwidth $B$ but your signal has bandwidth $B_1 < B$, there are a few things you can do: If your data rate is satisfactory, don't use a channel of bandwidth $B$, use one of bandwidth $B_1$! Assuming you're paying for your bandwidth, you'll save money. Note that Shannon's theorem also applies when you use $B_1$ instead of $B$....


2

This can be accomplished by changing the carrier frequency using a Numerically Controlled Oscillator (NCO) which maintains an accurate and continuous phase versus time trajectory via the phase accumulator. This is markedly different than changing the frequency with a classical PLL where we would typically break and reacquire lock to change frequency ...


2

Assuming one antenna transmits one symbol per time unit, then 16 symbols require 4 time units to be out. Then it is simply that r_1 = H_1 * x_1(1:4) r_2 = H_2 * x_1(5:8) r_3 = H_3 * x_1(9:12) r_4 = H_4 * x_1(13:16) If channel H is fixed during these 4 time units, [r_1 r_2 r_3 r_4] = H * [x_1(1:4) x_1(5:8) x_1(9:12) x_1(13:16)]; or r = reshape(H*reshape(...


2

The disadvantage is that the signal V21 = V1 + V2 = [2, 0, 2, 0] disappears every other chip interval which makes maintaining carrier and phase synchronization more difficult. Also, while the total energy in V21 is equals to the sum of the energies of V1 and V2, V21 uses 4 times the instantaneous power needed by either V1 or V2 and so the transmitter needs ...


1

OK, let me explain that for you. First, if you want to understand anything in engineering, it's recommended to write it in mathematical form, then try to solve it theoretically and build its code accordingly. Second, regarding your question, is it possible to use MRC for MIMO - CDMA, Yes, that's possible, why not? Now, The code you provided is not right. ...


1

This is may be the effect of the interference by other symbols


1

That's right, you can do spread using the kroon function, but be careful what you spread is not the data, it's the whole possibilities of data which is mod.The data is one of them, which is either, 0.7+0.7i or 0.7-0.7i ...etc. then you should know how you can do despread in the receiver side too. Good luck


1

Walsh-Hadamard matrix is usually used for DS CDMA. In this case, $\mathbf{x}$ contains symbols from different users. Since your matrix $\mathbf{H}$ is $4\times 4$, then the vector $\mathbf{x}$ must be of length of $4\times 1$. This means that you can support $4$ users at any given time. So, the answer to your question is no, you cannot spread a signal of ...


1

Yes you can, because you do spreading for each bit of data. So, regardless the length of your data, you are going to spread every bit on your code. In case if you have $x = 10$ you will have data after spreading with length 8, and if $x=101101$ you will have data to send with length 4 x 6 = 24. Just try to understand the process of spreading (I think it's ...


1

I totally agree with the above comments, but I'd like just to add a small note: CDMA is a special case of spread sequence, I mean using spread sequence doesn't mean that you are using CDMA. Spread sequence is using the sequence to spread your signal. for example if your signal is 1, then you can use the sequence 0101001 in order to spread the signal 1 ...


1

When you multiply the real and imaginary parts by $\mathbf{W}_1$ and $\mathbf{W}_2$, respectively, the resulting signal is no longer $S$. It will be a matrix. Regarding your questions: 1- Yes, the signal after spreading has the same duration. 2- One of the disadvantages of Walsh-Hadamard codes is that they don't have clear auto-correlation peak, which ...


1

$$A\cos(2\pi f+\theta_1)+B\cos(2\pi f+\theta_2)=C\cos(2\pi f+\theta_3)$$ where $$C=|u|\quad\textrm{and}\quad \theta_3=\arg\{u\}$$ with $$u=Ae^{j\theta_1}+Be^{j\theta_2}$$ The constant $C$ can be written as $$C=\sqrt{A^2+2AB\cos(\theta_1-\theta_2)+B^2}$$


1

Your query is mixing up ideas and formulas from very different systems to arrive at very questionable answers. The expression $\frac 12 \log_2(1+\text{SNR})$ is the capacity (measured in bits per use) of a discrete-time Gaussian channel. The model for this channel is that the $i$-th use of the channel consists of the transmission of a single real number ...


1

As the answer by @MBaz points out, one can use the contents of a $n$-bit maximal-length binary LFSR to select from $2^n-1$ different frequencies to hop to, thus creating a frequency-hopping sequence of period $2^n-1$. The difficulty is in creating more than one frequency-hopping sequence for use in a multiple-access scheme: the FH/CDMA scheme that the OP ...


1

Mathematically, you correlate your received signal with the (known) spreading sequence. That means that a single output of that correlator contains energy from all the sequence, not only from a few signal samples. That means that if there was signal energy of $e$ in each single sample, and the sequence is $N$ samples long, then the energy of each symbol is $...


1

I don't have too much familiarity with LoRa which is a particular form of chirp spread spectrum. However, in general, spread spectrum systems use code division for multiplexing. Code division is the CD in CDMA. For example, a frequency hopping spread spectrum signal like Bluetooth will have different frequency hopping patterns for each user/channel in the ...


1

The military GPS signals are crypto grade CDMA signals. One can use the civilian CA signal to recover the P(Y) military signal carrier because they use a common carrier and some receivers derive some clues from P(Y) but you need to know the sequence to recover the signal. The new M code is much harder because carrier recovery is much harder and the bit ...


1

(This does not answer your question which is a good one, so I may move this to a comment below your question) Warning! Typical Direct Sequence Spread Spectrum that uses linear spreading codes (based on linear feedback shift registers implemented from irreducible and primitive polynomials) are NOT a good cryptographic technique. See this post where the ...


1

To me the transmitters' information is divided in code, but codes can only be separated once they are bits not while they are "pass band symbols", right? Wrong! I'm not quite sure how to continue this answer, other than saying, "please re-read the CDMA introduction you've been reading". The whole idea is that you receive a sequence that is a ...


Only top voted, non community-wiki answers of a minimum length are eligible