This article is written in Beamforming's own study notes
In simple language, the mathematical principle of beam assignment is introduced, and a simple python code simulation exercise is given at the same time.
What we are going to discuss is how to shape the beam to achieve "where to hit where", so that radio waves can hit in the specified direction.
The common antenna array is ULA (Uniform Linear Array, Uniform Linear Array), as shown in the following figure:
A group of antennas arranged uniformly horizontally, the distance between the antennas is recorded as d.
If the attenuation of the signal is not considered, the signal data we transmit is s (a complex number, with amplitude and phase), and the distance between the receiving end and the transmitting end is far enough, the connection between each antenna of the transmitting end and the single antenna of the receiving end, from the transmitting end The vicinity of the end antennas seems to be parallel (in fact, it is not parallel, but close to parallel, which is a simplification of the calculation. If it is not simplified, the calculation will be too complicated).
Assume that the angle between the antenna array connection and the receiving end connection is θ \thetaθThen
the wireless signals received by the receiving end from different transmitting antennas will have different time delays, and thus have different phase rotations
The transmitting antenna at the bottom of the above figure is used as a reference. We choose the antenna with the shortest straight-line distance as a reference. Then the transmitted radio waves of the first transmitting antenna reach the receiving antenna, and the radio waves of the second transmitting antenna next to it will reach the receiving antenna. Compared with the radio waves emitted by the first transmitting antenna, the distance traveled is dcos ( θ ) dcos(\theta)d cos ( θ )
Then the radio wave emitted by the kth antenna is( k − 1 ) dcos ( θ ) (k-1)dcos(\theta)(k−1 ) d cos ( θ )
Then the extra time is: the extra distance divided by the speed of light c
Δ t = ( k − 1 ) dcos ( θ ) c \Delta t=\frac{(k-1)dcos(\theta)}{c}Δt=c(k−1 ) d cos ( θ ) .
Discussing the single-frequency electromagnetic wave, assuming that the frequency is f, the phase deviation due to the extra travel time is:
2 π f Δ t = 2 π f ( k − 1 ) dcos θ c = 2 π ( k − 1 ) dcos ( θ ) λ 2\pi f\Delta t=\frac{2\pi f(k-1)dcos\theta}{c}=\frac{2\pi(k-1)dcos(\theta)}{ \lambda}2 p f D t=c2πf(k−1)dcosθ=l2π(k−1 ) d cos ( θ ) .
Form: λ = cf \lambda=\frac{c}{f}l=fcfor the wavelength.
Random voltage = 2 π dcos ( θ ) λ \psi=\frac{2\pi dcos(\theta)}{\lambda}p=l2 π d cos ( θ )
Then for N antennas, the corresponding phase deviations are:
0 , ψ , 2 ψ , 3 ψ , . . . , ( N − 1 ) ψ 0,\;\psi, \;2\psi,\;3\psi,\;...,\;(N-1)\psi0,ps ,2 ψ ,3 ps ,...,(N−1 ) p
In the frequency domain analysis, it is equivalent to the transmitted signal although the signal from each transmitting antenna is the same, but the signal received by the receiving end is multiplied by :
ej 0 , ej ψ , ej 2 ψ , ej 3 ψ , . . . , ej ( N − 1 ) ψ e^{j0},\;e^{j\psi},\;e^{j2\psi},\;e^{j3\psi},\;..., \;e^{j(N-1)\psi}ej 0 ,ejψ,ej2ψ,ej 3 ψ ,...,ej ( N − 1 ) p
Then the received signal is:
1 N ∑ k = 0 N − 1 sejk ψ = s 1 N ∑ k = 0 N − 1 ejk ψ \frac{1}{N}\sum_{k=0}^{N- 1}se^{jk\psi}=s\frac{1}{N}\sum_{k=0}^{N-1}e^{jk\psi}N1k=0∑N−1with ejkψ=sN1k=0∑N−1ejkψ
Note: Dividing the above formula by N is energy normalization and does not affect performance analysis.
Then the received signal is equivalent to the transmitted signal, and its change is:
1 N ∑ k = 0 N − 1 ejk ψ \frac{1}{N}\sum_{k=0}^{N-1}e^{jk\psi}N1k=0∑N−1ejkψ
If only energy gain is considered, then:
G ( ψ ) = 1 N ∣ ∑ k = 0 N − 1 ejk ψ ∣ G(\psi)=\frac{1}{N}|\sum_{k=0}^{ N-1}e^{jk\psi}|G ( ψ )=N1∣k=0∑N−1ejkψ∣
After some derivations (see appendix), the above formula can be summarized as:
G ( ψ ) = { ∣ sin ( N ψ / 2 ) N sin ( ψ / 2 ) ∣ , if ψ ≠ 0 1 , if ψ = 0 G(\ psi)= \begin{cases} |\frac{sin(N\psi/2)}{Nsin(\psi/2)}|, & \text {if $\psi \neq 0$} \\ 1, & \text{if $\psi=0$} \end{cases}G ( ψ )={
∣N s in ( ψ /2 )s in ( N ψ /2 )∣,1,if p=0if p=0
ψ
= 2 π dcos ( θ ) λ \psi=\frac{2\pi dcos(\theta)}{\lambda}p=l2 π d cos ( θ )
After substitution, we can calculate the different gains corresponding to different angles at the receiving end , when ψ \psiψ is0 0When 0 , take the maximum value, that is,θ = π 2 \theta=\frac{\pi}{2}i=2por 3 π 2 \frac{3\pi}{2}23 p.mWhen taking the maximum value.
Python program, the gain curve drawn on the polar coordinates is as follows:
Suppose d λ = 1 2 \frac{d}{\lambda}=\frac{1}{2}ld=21
The first picture: 8 antennas in a row
16 antennas in a row
It can be seen that the more antennas, the more concentrated and the narrower the beam.
import numpy as np
from matplotlib import pyplot as plt
N = 8 #天线数量
theta = np.arange(0.000001,2*np.pi-0.0000001,0.01)
psi = np.pi * np.cos(theta)
r = np.abs(np.sin(N * psi/2)/np.sin(psi/2))/N
plt.figure()
plt.polar(theta,r)
plt.show()
Form:
1 N ∑ i = 0 N − 1 ejk ψ = 1 N 1 − ej N ψ 1 − ej ψ = 1 N ej N ψ / 2 ej ψ / 2 ( e − j N ψ / 2 − ej N ψ / 2 ) ( e − j ψ / 2 − ej ψ / 2 ) = ej N ψ / 2 ej ψ / 2 sin ( N ψ / 2 ) N sin ( ψ / 2 ) \frac{1}{N}\sum_{ i=0}^{N-1}e^{jk\psi}=\frac{1}{N}\frac{1-e^{jN\psi}}{1-e^{j\psi}} =\frac{1}{N}\frac{e^{jN\psi/2}}{e^{j\psi/2}}\frac{(e^{-jN\psi/2-e^{ jN\psi/2}})}{(e^{-j\psi/2}-e^{j\psi/2})}=\frac{e^{jN\psi/2}}{e^ {j\psi/2}}\fraction{sin(N\psi/2)}{Nsin(\psi/2)}N1i=0∑N−1ejkψ=N11−ejψ1−ej N ψ=N1ejψ/2ej N ψ / 2(e−jψ/2−ejψ/2)(e− j N ψ /2 − ej N ψ /2 )=ejψ/2ej N ψ / 2N s in ( ψ /2 )s in ( N ψ /2 )
If only the magnitude is considered, of the two fractions in the above derivation, the modulus of the first fraction is 1, which can be ignored.
∣ 1 N ∑ k = 0 N − 1 ejk ψ ∣ = ∣ sin ( N ψ / 2 ) N sin ( ψ / 2 ) ∣ |\frac{1}{N}\sum_{k=0}^{N- 1}e^{jk\psi}|=|\frac{sin(N\psi/2)}{Nsin(\psi/2)}|∣N1k=0∑N−1ejkψ∣=∣N s in ( ψ /2 )s in ( N ψ /2 )∣