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, H =


\      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, H 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, H is channel matrix gain

y = Received signal vector 

x = Transmitted signal vector 

n = Additive white Gaussian noise

Here, in the above equation you can image channel matrix, H 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, H =


\       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, Σ. We've already talked about "Singular Value Decomposition in MIMO Channel" in a separate article. 

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

Generally, H matrix is decomposed into, H = UΣVH

As U and V 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 = Σx +n 

Now, you see Σ is a diagonal matrix and signal vector, x 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

Download Papers and Solutions Exam Pattern Preparation Tips FAQs More Home / Engineering & Other Exams / UGC NET 2026 PYQ 📊 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 2026 - Question Paper Download PDF June 202...

The output voltage V0 (in Volt) for the network given in the Figure is ___

  Find the Output Voltage V o Solve the given circuit using Kirchhoff's Current Law (KCL), Ohm's law, and the supernode method. 1. Given Circuit Parameters Current source 1 mA Resistors 1 kΩ Each resistor Voltage source 6 V Required V o 2. Understand the Circuit The circuit contains a 1 mA current source, four 1 kΩ resistors, and a 6 V source connected between the two midpoint nodes. The output voltage is measured between the top and bottom rails. Reference node Take the bottom rail as ground: 0 V . Let the top rail voltage be V T , and...

GMSK BER vs SNR (with Simulation)

  GMSK Performance: Why the BER vs. SNR Curve Matters If you’ve ever wondered why GMSK (Gaussian Minimum Shift Keying) became the backbone of the global GSM revolution, the answer lies in its unique balance of spectral efficiency and power performance. But there is no free lunch in engineering. To get that sleek, narrow spectrum, you have to pay a price in Bit Error Rate (BER) . In this guide, we break down the theoretical BER vs. SNR performance of GMSK, the "magic" of the BT product, and how to optimize your link budget for real-world conditions. What is GMSK and Why Should You Care? GMSK is a derivative of MSK (Minimum Shift Keying) that uses a Gaussian pre-modulation filter. This filter "smooths" the phase transitions, significantly reducing sideband power. The result? A constant envelope signal that allows you to use high-efficiency Class-C amplifiers without distorting the signal. This is the secr...

Parametric Down-Conversion of Photons

  Nonlinear Optics One incoming photon is converted into two lower-energy photons, typically in a nonlinear optical medium. Energy is conserved: E in = E 1 + E 2 E in ​ = E 1 ​ + E 2 ​ . Parametric Down-Conversion A nonlinear optical process in which one pump photon is converted into two lower-energy photons, known as the signal and idler photons. 1 Energy Conservation The energy of the pump photon equals the sum of the energies of the signal and idler photons. \[ \boxed{\hbar\omega_p=\hbar\omega_s+\hbar\omega_i} \] \[ \boxed{\omega_p=\omega_s+\omega_i} \] ωₚ Pump angular frequency ...

SRAM, DRAM, EPROM, MRAM, and DDRAM Explained

Computer Fundamentals Types of Memory and Their Technologies Understand the fundamental relationship between different types of computer memory and the technologies used to implement them. Fundamental Matching Memory type, technology, and key characteristics Memory Type Technology Key Characteristic Cache Memory SRAM Very fast; stores frequently accessed data Main Memory DRAM Holds programs and data currently in use Au...

The cutoff frequency (in GHz) for the dominant TE10 mode of an air-filled rectangular waveguide of ...

  Q. The cutoff frequency (in GHz) for the dominant TE10 mode of an air-filled rectangular waveguide of inner dimension 0.28 inch×0.14 inch is __. (rounded off to two decimal places). Given Information Waveguide dimensions = 0.28 inch × 0.14 inch Wider dimension (a) = 0.28 inch Smaller dimension (b) = 0.14 inch Medium = Air Dominant mode = TE 10 Speed of light (c) = 3 × 10 8 m/s Step 1: Cutoff Frequency Formula The cutoff frequency of a rectangular waveguide is: f c = (c / 2) √[(m/a) 2 + (n/b) 2 ] For the dominant TE 10 mode: m = 1, n = 0 f c = c / 2a Step 2: Convert Inch into Meter We know that 1 inch = 0.0254 meter. a = 0.28 × 0.0254 a = 0.007112 m ...

A binary ripple counter is designed to count (0)₁₀ to (64)₁₀. Which of the following is/are the number of flip-flops required ...

  Q. A binary ripple counter is designed to count  (0)₁₀ to (64)₁₀ . Which of the following is/are the number of flip-flops required to design the counter?   (A) 6 (B) 7 (C) 4 (D) 5 Solution A binary ripple counter must count from (0)₁₀ to (64)₁₀ , including both endpoints. Step 1: Calculate the total number of states Since both 0 and 64 are included, add 1 to the difference: N = Last value − First value + 1 N = 64 − 0 + 1 = 65 states The counter visits every integer from 0 through 64: 0, 1, 2, 3, ..., 63, 64 There are 65 distinct states in total. Step 2: Calculate the number of flip-flops Each flip-flop stores one bit. With n flip-flops, the counter can represent 2ⁿ distinct states. 2⁶ = 64 states (not enough) 2⁷ = 128 states (eno...

The electric field of a monochromatic plane wave travelling in a lossless isotropic and homogenous medium is given by E(z, t) = ...

  Q. The electric field of a monochromatic plane wave travelling in a lossless isotropic and homogenous medium is given by E(z, t) = E₀ [x̂ cos(ωt − kz) + ŷ sin(ωt − kz)] in a right-handed orthogonal co-ordinate system. Which of the following is the correct polarization of the electromagnetic wave? (A) Right-handed circularly polarized Correct answer under the IEEE convention. (B) Left-handed circularly polarized (C) Linearly polarized (D) Linearly polarized at −45° to x̂ Solution 1. Identify the electric-field components Let ψ = ωt − kz. The components along the x and y directions are: Eₓ = E₀ cos ψ Eᵧ = E₀ sin ψ Both components have equal amplitudes, the same frequency, and a phase difference of 90°. These are the conditions for circular polarization. ...