In the previous article, we explained the basic mathematical principles of beamforming in plain language. In that article, we let all the transmitting antennas send the same signal , and then measured the strength of the signal received by the receiving antenna at different positions. The result is that the signal strength is the largest in the direction perpendicular to the transmitting antenna array.
So, how to make the receiving antenna in the specified direction receive the signal with the maximum strength? A very stupid method is to adjust the physical position of the antenna array so that the vertical direction and the direction of the transmitting antenna array just point to the receiver. Although this method is very stupid, it can facilitate our understanding of the underlying principles. In fact, from the perspective of the vertical and transmitting antenna array, because the receiver is in the vertical direction, the electromagnetic waves emitted by each transmitting antenna arrive at the same phase, which can just produce the effect of signal superposition ( no signal cancel each other), then, when the receiver is not in the vertical direction, can we create the same effect: that is, let the electromagnetic waves emitted by each transmitting antenna arrive at the receiving antenna with the same phase?
as the picture shows:
Receivers that are not in the vertical direction receive electromagnetic waves from different transmitting antennas, because the distances traveled by the electromagnetic waves are different, resulting in phase differences. Then, we can let the signals sent by the antennas take the phase differences into account in advance and change the transmission signal to offset the phase difference.
We use the graph in the previous article to analyze:
Taking the first antenna as the reference, the signal transmitted on the first antenna is S (complex number, including amplitude and phase), then the signal sent by the second transmitting antenna, after arriving at the receiver, is the same as the signal transmitted by the first antenna The phase difference between arriving at the receiving end is ej ψ e^{j\psi}ej ψ . So we can cancel the phase of the signal sent by the second transmitting antenna in advance, and multiply the signal of the first antenna bye − j ψ e^{-j\psi}eAfter − j ψ , it is transmitted from the second transmitting antenna. Similarly, for the third transmitting antenna, the original signal S is multiplied bye − j 2 ψ e^{-j2\psi}e− j 2 ψ and then transmit it from the third transmitting antenna, and so on until the Nth transmitting antenna, multiply the original signal S bye − j ( N − 1 ) ψ e^{-j(N- 1)\psi}e− j ( N − 1 ) ψ and then transmit from the Nth transmitting antenna.
From the above analysis, it can be seen that the receiving antenna should be able to receive the signal with the maximum strength at the specified azimuth. Next, use mathematical formulas to analyze.
The signal sent by N transmitting
antennas is: S , S ∗ e − j ψ 0 , S ∗ e − j 2 ψ 0 , . . . , S ∗ e − j ( N − 1 ) ψ 0 S, \;S* e^{-j\psi_{0}}, \;S*e^{-j2\psi_{0}}, ..., \;S*e^{-j(N-1)\psi_{0 }}S,S∗e−jψ0,S∗e−j2ψ0,...,S∗e− j ( N − 1 ) p0
For a given current, the orientation is ψ 0 = 2 π dcos ( θ 0 ) λ \psi_{0}=\frac{2\pi dcos(\theta_{0})}{\lambda};p0=l2 π d cos ( i0),define d = λ / 2 d=\lambda/2d=λ /2 ,则ψ 0 = π cos ( θ 0 ) \psi_{0}=\pi cos(\theta_{0})p0=π cos ( i0) , whereθ 0 \theta_{0}i0is the azimuth where the receiving antenna is located.
Then, at any specified azimuth θ \thetaOn θ , the signals receivedfrom N transmitting antennas, due to the difference in the path length of the radio waves, the signals transmitted in the air have different phase delays, and the delayed phases are:
0 , ψ , 2 ψ , . . . , ( N − 1 ) ψ 0,\;\psi,\;2\psi,\;...\;,\;(N-1)\psi0,ps ,2 ψ ,...,(N−1 ) ψ
whereψ= π cos ( θ ) \psi=\pi cos(\theta)p=π cos ( θ ) , due to the analysis done in the frequency domain, is equivalent to the signal transmitted from each transmitting antenna, which is multiplied by:
ej 0 , ej ψ , ej 2 ψ , . . . , ej ( N − 1 ) ψ e^{j0},\;e^{j\psi},\;e^{j2\psi},\;...\;,\;e^{j(N-1)\ psi}ej 0 ,ejψ,ej2ψ,...,ej ( N − 1 ) p
Then the signals received by the receiving end
from each transmitting antenna are: S ej 0 , S e − j ψ 0 ej ψ , S e − j 2 ψ 0 ej 2 ψ , . . . , S e − j ( N − 1 ) ψ 0 ej ( N − 1 ) ψ Se^{j0},\;Se^{-j\psi_{0}}e^{j\psi},\;Se^{-j2\psi_{0} }e^{j2\psi},\;...\;,\;Se^{-j(N-1)\psi_{0}}e^{j(N-1)\psi}With ej 0 ,With e−jψ0ejψ,With e−j2ψ0ej2ψ,...,With e− j ( N − 1 ) p0ej ( N − 1 ) p
整多下：
S , S ej ( ψ − ψ 0 ) , S ej 2 ( ψ − ψ 0 ) , . . . , S ej ( N − 1 ) ( ψ − ψ 0 ) S,\;Se^{j(\psi-\psi_{0})},\;Se^{j2(\psi-\psi_{0}) },\;...\;,\;Se^{j(N-1)(\psi-\psi_{0})}S,With ej ( ψ − ψ0),With ej 2 ( ψ − ψ0),...,With ej ( N − 1 ) ( ψ − ψ0)
Then the signal finally received by the receiving antenna is the sum of the above signals (divided by N, which is energy normalization and does not affect the analysis):
1 N ∑ k = 0 N − 1 S ejk ( ψ − ψ 0 ) = S 1 N ∑ k = 0 N − 1 ejk ( ψ − ψ 0 ) \frac{1}{N}\sum_{k=0}^{N-1}Se^{jk(\psi-\psi_{0}) }=S\frac{1}{N}\sum_{k=0}^{N-1}e^{jk(\psi-\psi_{0})}N1k=0∑N−1With ejk ( ψ − ψ0)=SN1k=0∑N−1ejk ( ψ − ψ0)
对增益设计做等比数列求和上整多：
1 N ∑ k = 0 N − 1 eq ( ψ − ψ 0 ) = { ∣ sin ( N ( ψ − ψ 0 ) / 2 ) N sin ( ( ψ − ψ 0 ) / 2 ) ∣ , if ψ − ψ 0 ≠ 0 1 , if ψ − ψ 0 = 0 \frac{1}{N}\sum_{k=0}^{N-1}e^{jk( \psi-\psi_{0})} = \begin{cases} |\frac{sin(N(\psi-\psi_{0})/2)}{Nsin((\psi-\psi_{0}) /2)}|, & \text{if $\psi-\psi_{0} \neq 0$ } \\ 1, & \text{if $\psi-\psi_{0}=0$} \\ \ end{cases}N1k=0∑N−1ejk ( ψ − ψ0)={
∣N s in (( ψ − ψ0)/2)s in ( N ( ψ − ψ0)/2)∣,1,if p−p0=0 if p−p0=0
It can be seen that when the transmitted signal is multiplied by the phase is ψ = ψ 0 \psi=\psi_{0}p=p0When , the strength of the signal is the largest, that is, θ = θ 0 \theta=\theta_{0}i=i0When , the signal strength in the direction of the specified receiving end is the largest, so that the receiving antenna azimuth angle θ 0 \theta_{0}i0In the direction, the signal strength is the highest, which realizes "pointing where to hit".
Level θ 0 = π 6 \theta_{0} = \frac{\pi}{6}i0=6pFor example, let θ ∈ [ 0 , 2 π ] \theta \in [0,2\pi]i∈[0,2 π ] , plot the signal strength on polar coordinates:
The Python code is as follows:
import numpy as np
from matplotlib import pyplot as plt
N = 16 #天线数量
theta0 = np.pi/6
psi0 = np.pi * np.cos(theta0)
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-psi0)/2)/np.sin((psi-psi0)/2))/N
plt.figure()
plt.polar(theta,r)
plt.show()