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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}$$for \mu which equals to search for the zeros of a k+1th order polynomial, which is impossible for k>3. So, I fear you need to do a numeric solution. Once you have \mu you can go for a simple calculation of \... 2 I think it should be:$$ \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 ... 2 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 ... 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(... 2 For a given OFDM symbol, no subcarrier can be assigned to two users simultaneously. Each subcarrier is uniquely assigned to a unique user. Else you have to implement an Interference cancellation at receiver (which would require having channel knowledge of other users as well, meaning global CSI) or at the transmitter and extremely complex interference ... 2 You are right, coherence bandwidth is the frequency domain counterpart of delay spread. However, to achieve diversity across antennas the spacing between the antennas is important "relative to the environment".Let me explain that a bit more. In a user mobile phone you would typically find the order of wavelengths seperation between antennas sufficient to ... 2 MAI is when there is Interference due to multiple users accessing a common wireless media. So if you consider an OFDM system then as long as users are trasmiting in their allocated sub carriers there is no multi access interference irrespective of the MIMO technique used, delay spread etc. If each user is using the same frequencies to transmit ... 1 The point of increasing the distance between receive antennas is not to reduce the interference; it is to uncorrelate the channels. In the system you're studying, the signal x_1 travels by two paths. One has gain h_{11} and the other has gain h_{12}. Both of these gains are random. A channel is considered bad when the magnitude of its gain, |h|, is ... 1 Stop thinking about the maths and just think about the reality. You have 1 receiver. That means there is a signal coming from the receiver into the processor. There is only one signal. You can't measure it twice and get two different signals. You may be thinking of the channel matrix, which has dimension 1x4 (not 4x1, sorry). That's because the channel ... 1 Multi-input single-output (MISO) is describing the inputs and outputs of the system. In communication, this is the channel, so if you have a single-output it means that the received vector is 1 \times 1. The general model for a MISO system is: \begin{split} y &= \mathbf{h^Tx}+\text{noise} \\ &= (h_1x_1+h_2x_2+...+h_Nx_N)+\text{noise} \end{split} ... 1 In contrast to your previous question(Why multi-path channel has linear phase within the coherence bandwidth?), the coherence BW in mmWave may not be relevant considering delay spread alone. If you see eq (7) of the reference, the H has contribution from all the N_p paths, even though channel is mentioned as narrow band (flat fading). This is true till ... 1 Looking at the equation you have posted, it seems l is the variable for N_p multipaths and for each path, there will be a constant delay \tau_l, an attenuation \alpha_l, a doppler shift in the carrier \nu_l , angle of arrival at the receiver \mathbb a_{\mathbf R} and angle of departure at transmission \mathbb a_{\mathbf T}. And both of these ... 1 It is possible if the system supports MU-MIMO. 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 The answer is simply orthogonalization of radio resources. For any given OFDM symbol, exactly one user will be allocated certain sub carriers. This allocation is dynamic and informed to the user in what is known as the DCI, in the downlink control channel. Each user will decode data only on the subcarriers on which it has a grant allocated by the base ... 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 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 The Channel Impulse Response (CIR) for a beamforming vector \mathbf{w} \in \mathbb{C}^{N_T \times 1} (which corresponds to your spatial streams) is the following$$\mathbf{c}(n) = \sum\limits_{k=1}^q \alpha_k \mathbf{a}_{R,k}\mathbf{a}_{T,k}^\top \mathbf{w} \delta(n - \tau_k) \tag{1}$$where q is the number of paths (is 2 due to the two-ray ... 1 Let us consider a very simple example of a frequency-flat quasi-static MIMO communication system. In such a setup, we can write the received signal from, say, M_{\rm R} receive antennas as$$\mathbf y(t) = \mathbf H \cdot \mathbf x(t) + \mathbf w(t),$$where \mathbf x(t) \in \mathbb{C}^{M_{\rm T}} is the vector of transmitted symbols from the M_{\rm T} ... 1 In frequency-selecting model, the received samples for SISO channel are given by$$y_n = \sum_{l=0}^Lh_lx_{n-l}+z_n$$while in flat-fading channel the received sample are given by$$y_n = h_n x_n +z_n$$where \{h_l\} are the channel coefficients, \{x_n\} are the data symbols, and \{z_n\} are AWGN samples. In the first case you need equalization ... 1 Frequency-flat fading essentially means that the channel's transfer function is (approximately) constant in your frequency band of interest, which means you can treat it as if it would not depend on frequency at all. This means two things: Your MIMO transmission model is simply \mathbf y(t) = \mathbf H \mathbf x(t) + \mathbf w(t) with a transmit signal \... 1 There is no mention of light-of-sight (LOS) on the time you mentioned. There is a mention of point-to-point, however. Point-to-point channels could be LOS or multipath. MU-MIMO is better than point-to-point MIMO because users are usually well-separated in space. MU-MIMO is particularly better when the channels are correlated at the base station (BS). The ... 1 You should double check the formula. The classic single input multiple output (SIMO) equation with N_T receive antennas is: \mathbf{y}=\mathbf{h}x+\mathbf{w}. Where x is the transmitted symbol (usually complex valued), \mathbf{w} \sim CN(0,N_0\mathbf{I}) is the complex noise, and \mathbf{y} is the received vector, which is a vector of size N_{R}\... 1 I'm not to familiar with the HT-LTF, but OFDM channel est. is done as follows: rx'd long training symbol (LTF_{rx}) is the tx'd LTF (LTF_{tx}) convolved with the channel:$$LTF_{rx}(n) = \sum_{m=-\infty}^{\infty}h(m)LTF_{tx}(n-m)$$translate this to frequency domain:$$LTF_{rx}(w) = H(w) LTF_{tx}(w)$$and the channel inverse is then:$$H(w)^{-1} = ...