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The short answer is: if you transmit one block of $N$ symbols (no data before it for at least $L$ symbol times) over a frequency-selective channel of length $L$, then the channel matrix will be of dimension $NN_TN_R\times N$. The long answer: start with $N_T=N_R=1$. Then the $n^{\text{th}}$ received sample can be written as $$y_n=\sum_{l=0}^Lh_lx_{n-l}+... 2 I doubt there is general closed-form solution. For the first equation, you essentially want to solve$$\mu=\sum_k \frac{1}{a_k+b_k \mu}$$for \mu which equals to search for the zeros of a k+1th order polynomial, which is impossible for k>3. So, I fear you need to do a numeric solution. Once you have \mu you can go for a simple calculation of \... 2 Yes OFDM-IM is somehow new and it's very hot topic of searching. That article you have shares is the basic of OFDM-IM. So if you are at the beginning to understand it, that's the right paper to to read and understand. OK let me explain that paper for you in easier way, consider we have block with length 32 bits you want to transmit it, in the traditional ... 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(... 1 You should double check the formula. The classic single input multiple output (SIMO) equation with N_T receive antennas is: \mathbf{y}=\mathbf{h}x+\mathbf{w}. Where x is the transmitted symbol (usually complex valued), \mathbf{w} \sim CN(0,N_0\mathbf{I}) is the complex noise, and \mathbf{y} is the received vector, which is a vector of size N_{R}\... 1 One of the most readable (imho) authors on the communication systems is Simon Haykin and the following book from him would probably address most of the issues you would encounter in a wireless communication system analysis. Modern Wireless Communication Systems Similar known authors do have related books, but I guess most compact academic begining would be ... 1 First you have to know that the vector in U and the vector in V are orthonomal,that is ,\vec u \vec u^H=I,and \vec v \vec v^H=I. So now H=U\Sigma V^H,and y=Hs+n=(U\Sigma V^H)s+n and U^H y=\Sigma V^Hs+n,see the \Sigma V^Hs,isn't it like " channel gain \times beamforming \times signal s "? Now you can also find that according to ... 1 The union bound can indeed result in a BER higher than 1. As you say, it is just an upper bound, and it becomes tighter as the SNR increases. For low SNR, the union bound can be quite loose. 1 Let us consider a very simple example of a frequency-flat quasi-static MIMO communication system. In such a setup, we can write the received signal from, say, M_{\rm R} receive antennas as$$\mathbf y(t) = \mathbf H \cdot \mathbf x(t) + \mathbf w(t),$$where \mathbf x(t) \in \mathbb{C}^{M_{\rm T}} is the vector of transmitted symbols from the M_{\rm T} ... 1 Frequency-flat fading essentially means that the channel's transfer function is (approximately) constant in your frequency band of interest, which means you can treat it as if it would not depend on frequency at all. This means two things: Your MIMO transmission model is simply \mathbf y(t) = \mathbf H \mathbf x(t) + \mathbf w(t) with a transmit signal \... 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 What you mention could be right in low modulation order, such as QAM but in 16-QAM and 32-QAM, the amplitude is changed also, I mean you will have different amplitude levels, so you can't implement that detector. 1 For 0 < K < \infty, the channel is a combination of both a deterministic component (i.e., LOS) and a fading component. As the K-factor is the ratio of the energy in the deterministic Line-of-Sight (LOS) component to the energy in the aggregation of the random scattered paths (i.e., the fading component), higher 𝐾 means that the channel is more ... 1 You must attention to the written text. It doesn't say the ML isn't optimal, what it says is that the problem isn't regular LS problem but Least squares problem with Constraints. The constraints make analytic solution infeasible and hence in order to solve it one must go through any signal in the space of valid signals and mark the one which minimizes the ... 1 If you have three or more antennas, simply don't put them in a linear array. Any other array setup doesn't suffer from the ambiguity. Linear array is the worst choice here. Of course, that breaks the periodicity constraint on the linear array that makes the autocorrelation space method MUSIC So, you'd need to change that algorithm. Luckily, you can simply ... 1 I think it should be:$$ \boldsymbol{y}_{i} = {H}_{i} \boldsymbol{x} + \boldsymbol{n}_{i}  For the $i$ -th antenna and ${H}_{i}$ being the convolution matrix of this specific channel. You could write this in a Matrix Form where $H$ becomes a tensor. In case $\forall i, j, \; {H}_{i} = {H}_{j}$ then it can be made simpler. You could also work on ...