Skip to main content

Optimal Precoding for Millimeter wave Massive MIMO Systems


 

Optimal Precoding for Millimeter wave Massive MIMO Systems

In case of MIMO system we deploy multiple transmitter antennas at receiver side and multiple receiver antennas at receiver side. MIMO technology was introduced to support multiple simultaneous data streams between transmitter and receiver to multiply the capacity of a system. But there is also interference between multiple data streams. Precoding technique minimizes the interference between multiple data streams. 



What Exactly Precoding Technique is

We all are familiar with the channel matrix of an MIMO system, that looks like, =


\      R1     R2     R3     R4

T1  h11    h12     h13   h14

T2  h21    h22     h23   h24

T3  h31    h32     h33   h34

T4  h41    h42     h43   h44


Here, in the above figure channel matrix, is shown. In channel matrix it shown different gains between different antennas. Now, we see in the above matrix for example, h11 represents the channel gain between transmitter antenna, T1 and receiver antenna, R1 and h11 also means connection between the antennas as well. R1 also receives the signals from T2, T3, and T4 antennas too. So, there is some kind of interface between multiple data streams when we process the signal at receiver side. Here, precoding help us to reduce interference between multiple data streams. 



Optimal Precoding in MIMO

Typically, received signal at receiver side is represented as,

y = Hx + n       .....(i)

Where, is channel matrix gain

y = Received signal vector 

= Transmitted signal vector 

= Additive white Gaussian noise

Here, in the above equation you can image channel matrix, as same as above channel matrix where we've shown channel gains between TX side antennas T1, T2, T3, T4, and receiver side antennas, R1, R2, R3, R4, respectively. We've also talked about interference with T1's signal at R1 antenna due to transmission from T2, T2, and T3. 

Now, let imagine your channel matrix looks like that, =


\       R1     R2     R3     R4

T1   h11     0        0         0

T2     0     h22      0        0

T3     0       0      h33      0

T4     0       0       0       h44


Now in equation (i), if you the put the above channel matrix value then you see there is no interference with T1' signal with T2, T3, and T4's transmission at receiver R1. 

Similar approach is performed for optimal precoding technique we channel matrix is decomposed in to two unitary matrix U, V, and one diagonal eigen value matrix, Î£. We've already talked about "Singular Value Decomposition in MIMO Channel" in a separate article. 

There is matrix, Î£we operate row and column matrix in a such way that Î£ becomes diagonal matrix where elements are in descending order. We do that by operating multiple operations in matrix as shown in the above mentioned article.

Generally, matrix is decomposed into, H = UΣVH

As and are unitary matrix so, multiplication of those matrix with its hermitian matrix itself are identity matrix. Alternatively, UUH = VVH = I



Signal Processing at Receiver Side for Optimal Precoding

During transmission we multiply original message signal vector with unitary matrix, V. So, now transmitted signal vector becomes, Vx. On the side at receiver side, received signal vector is multiplied with vector UH. So, as per above equation (i), received signal vector at receiver side as follows

y = UH (UΣVH) Vx + n

y= IΣIx + n

y = Î£x +n 

Now, you see Î£ is a diagonal matrix and signal vector, is multiplied with that diagonal matrix. So, you can observe there the simultaneous data streams between MIMO transmitter and receiver antennas without interference among them. Now we further do optimal power allocation to each antennas to maximize sum-rate or overall throughput as shown in a separate article. There is the URL link above.


# mimo beamforming

Why OFDM precoding modulation used in uplink?

Contact Us

Name

Email *

Message *

Popular Posts

UGC NET Electronic Science Previous Year Question Papers with Solutions

Home / Engineering & Other Exams / UGC NET 2026 PYQ ⬇️ Download Papers and Solutions 📋 Exam Pattern 💡 Preparation Tips ❓ FAQs 📊 Exam Highlights: Electronic Science (88) Feature Details Junior Research Fellowship (JRF) ₹37,000 + HRA per month Eligibility M.Sc/M.Tech in Electronics (55%) Validity of Certificate JRF (3 Years) | Lectureship (Lifetime) 📥 Download UGC NET Electronics PDFs Complete collection of previous year question papers, answer keys and explanations for Subject Code 88. Start Downloading 📂 View All Question Papers June 2025 - Question Paper Download PDF June 2025 - Solved Paper + Explanation ...

UGC NET Electronic Science June 2025 Question Paper with Answer Key & Detailed Solutions

Home / UGC NET PYQ / June 2025 Solved UGC NET Electronic Science June 2025 Question Paper with Answer Key and Full Explanations 📥 Download Question Paper (PDF) 2025 2024 2023 2022 2021 2020 Explanations 1.  Answer: Option (3) For forming a p-type semiconductor, the dopant must be a trivalent impurity (three valence electrons) so that it creates acceptor levels and holes become the majority carriers. Among the given elements, boron (B) is a group-III element (trivalent). Arsenic (As) and phosphorus (P) are group-V (pentavalent) donors that produce n-type material, and germanium (Ge) is a group-IV element usually used as the semiconductor, not as an acceptor dopant. Hence, doping an intrinsic semiconductor with B produces a p-type semiconductor. 2.  Answer: Option (4) The ohmic resistance of a JFET at zero gate bias is given by the standard relation: R DS(on) = V P / I DSS ...

BER vs SNR for M-ary QAM, M-ary PSK, QPSK, BPSK, ...(MATLAB Code + Simulator)

Bit Error Rate (BER) & SNR Guide Analyze communication system performance with our interactive simulators and MATLAB tools. 📘 Theory 🧮 Simulators 💻 MATLAB Code 📚 Resources BER Definition SNR Formula BER Calculator MATLAB Comparison 📂 Explore M-ary QAM, PSK, and QPSK Topics ▼ 🧮 Constellation Simulator: M-ary QAM 🧮 Constellation Simulator: M-ary PSK 🧮 BER calculation for ASK, FSK, and PSK 🧮 Approaches to BER vs SNR What is Bit Error Rate (BER)? The BER indicates how many corrupted bits are received compared to the total number of bits sent. It is the primary figure of merit f...

Q-function in BER vs SNR Calculation

Q-function in BER vs. SNR Calculation | Interactive Guide Q-function in BER vs. SNR Calculation In digital communications and signal processing, the Q-function plays a significant role in predicting system reliability. It allows engineers to quantify the probability that Gaussian noise will exceed a specific threshold, causing a bit error. What is the Q-function? The Q-function is a mathematical function representing the tail probability of the standard normal (Gaussian) distribution. It is the complementary cumulative distribution function (CCDF) of a standard Gaussian distribution. Q(x) = (1 / √(2Ï€)) ∫â‚“∞ e^(-t² / 2) dt Q-Function Interactive Simulator Move the slider to see how the "Tail Probability" (the area in red) changes. This area represents the Probability of Error (BER) . Threshold Distance ( x ) — (Simulates Increasing SNR) ...

MATLAB Code for ASK, FSK, and PSK (with Online Simulator)

MATLAB Code for ASK, FSK, and PSK Comprehensive implementation of digital modulation and demodulation techniques with simulation results. 📘 Theory 📡 ASK Code 📶 FSK Code 🎚️ PSK Code 🕹️ Simulator 📚 Further Reading Amplitude Shift Frequency Shift Phase Shift Live Simulator ASK, FSK & PSK HomePage MATLAB Code MATLAB Code for ASK Modulation and Demodulation COPY % The code is written by SalimWireless.Com clc; clear all; close all; % Parameters Tb = 1; fc = 10; N_bits = 10; Fs = 100 * fc; Ts = 1/Fs; samples_per_bit = Fs * Tb; rng(10); binar...

UGC NET Electronic Science December 2024 Question Paper with Answer Key & Detailed Solutions

Home / UGC NET PYQ / June 2025 Solved UGC NET Electronic Science December 2024 Question Paper with Answer Key and Full Explanations 📥 Download Question Paper (PDF) 2025 2024 2023 2022 2021 2020 Q.1 Answer: Option (3) Q.2 Answer: Option (3) Solution 1. JMP SHORT LABEL Intrasegment (within the same code segment). Direct jump. ❌ Not intersegment indirect. 2. JMP 5000H:2000H Intersegment (far jump because both CS and IP are specified). Direct jump (address is explicitly given). ❌ Not indirect. 3. JMP [2000H] The destination address is taken from memory location 2000H. This is indirect. In 8086, a far indirect jump can use a memory operand containing both IP and CS (depending on operand size), making it an intersegment indirect jump. ✅ Correct answer. 4. JMP [BX] Indirect jump through memory addressed by BX. Usually intrasegment (near indirect jump). ❌ Not in...

Which of the following statements are correct? A. If the intermediate frequency is too high, poor selectivity results even if sharp cutoff filters are used in the IF stage.

  61) Which of the following statements are correct?  A. If the intermediate frequency is too high, poor selectivity results even if sharp cutoff filters are used in the IF stage.  B. A high value of intermediate frequency increases tracking difficulties.  C. As the intermediate frequency is lowered, image frequency rejection becomes better.  D. A very low intermediate frequency can make the selectivity too sharp.  Choose the correct answer from the options given below:  1. A and B only [Option ID = 3073]  2. B and C only [Option ID = 3074]  3. C and D only [Option ID = 3075]  4. B and D only [Option ID = 3076 Answer: 4  Previous yr Question papers with Full Explanations → Electronics and Communiaction Study Materials → Try Interactive Online Simulator Run the Simulation The Superheterodyne Principle The...

Constellation Diagrams of ASK, PSK, and FSK (with MATLAB Code + Simulator)

Constellation Diagrams: ASK, FSK, and PSK Comprehensive guide to signal space representation, including interactive simulators and MATLAB implementations. 📘 Overview 🧮 Simulator ⚖️ Theory Q-function 📚 Resources BASK (Binary ASK) Modulation Transmits one of two signals: 0 or $\sqrt{E_b}$, representing binary 0 and 1. BFSK (Binary FSK) Modulation Transmits one of two signals: $\sqrt{E_b}$ on the Y-axis or $\sqrt{E_b}$ on the X-axis. These are orthogonal signals. BPSK (Binary PSK) Modulation Transmits $+\sqrt{E_b}$ or $-\sqrt{E_b}$ (antipodal signaling). Signal Space Simulator Visualize Constellation Diagrams with Noise Control. SNR (dB): 15 ...