Assuming there are M transmitting antennas, we can correspond to a vector for each transmitting angle:
[ ej 0 ψ ej 2 ψ . . . ej ( M − 1 ) ψ ] − − − Formula ( 3 ) [e^{j0 \psi}\;\;e^{j2\psi}\;\;...\;\;e^{j(M-1)\psi}]\;\;---Formula (3)[ej 0 pej2ψ...ej(M−1)ψ]−−−Official ( 3 )
We know that the above is an M-dimensional vector, which can be considered as a vector in the M-dimensional space. Since there are infinitely many angles to choose, infinitely many vectors can be generated. Since it is an M-dimensional space, we should be able to find M orthogonal vectors constitute a basis, and all other vectors can be generated by linear combination of M vectors in this basis.
So, the question is, are we guaranteed to find M orthogonal vectors? Of course, if there is no restriction, there must be M orthogonal vectors forming a basis in the M-dimensional space. However, if we impose constraints on the orthogonal vectors and cannot choose them arbitrarily, it may not be possible to find them.
The condition we add is that the vector of the form (3), ψ \psiWhen ψ takes different values, different vectors are obtained. Under such conditions, the two vectors are orthogonal, requiring twoψ \psiThe value of ψ differs from 2 π / M 2\pi/Mwhich is an integer multiple of non-zero2π / M . _ It can be shown (see Appendix) that such two vectors are orthogonal.
If the first vector, we take ψ = 0 \psi=0p=0
The second vector takes ψ = 2 π / M \psi=2\pi/Mp=2π/M
And so on, the last vector takes
ψ = 2 π M ∗ ( M − 1 ) \psi=\frac{2\pi}{M} * (M-1)p=M2 p.m∗(M−1)
Therefore, ψ \psiThe value range of ψ must be greater than or equal to2 π 2\pi2 π , otherwise, M orthogonal vectors cannot be obtained.
From formula (2), we can see thatψ \psiThe range width of ψ
is 2 π d λ ∗ 2 \frac{2\pi d}{\lambda} * 2l2πd∗2
Let:
2 π d λ ∗ 2 ≥ 2 π \frac{2\pi d}{\lambda}*2\geq 2\pil2πd∗2≥2 p.m
则:
d λ ≥ 1 2 \frac{d}{\lambda}\geq \frac{1}{2}ld≥21
Based on the above derivation, then:
d λ = 1 2 \frac{d}{\lambda}= \frac{1}{2}ld=21
And the M orthogonal vectors are:
This results in a basis consisting of M orthogonal vectors.
如果
d λ < 1 2 \frac{d}{\lambda}< \frac{1}{2} ld<21
Then the M-dimensional space cannot be formed, and it will become a subspace of the M-dimensional space (this is a bit abstract, and I don’t know how to make it clearer). From the perspective of "pointing where to hit", it cannot Do a very precise "point where to hit".
import numpy as np
from matplotlib import pyplot as plt
N = 32 #天线数量
theta = np.arange(-np.pi,np.pi-0.0000001,0.01)
d_vs_lambda = 0.01 # 请修改这个值，即使用不同的 d/lambda
psi = 2 * np.pi * d_vs_lambda * np.cos(theta)
r = np.abs(np.sin(N * psi/2)/np.sin(psi/2))/N
plt.figure()
plt.polar(theta,r)
plt.show()
plt.figure()
plt.plot(theta/np.pi,r)
plt.ylim([0,1.1])
plt.show()
Modify the value of the variable d_vs_lambda in the program and gradually decrease from 0.5 to see the effect:
It is equivalent to only one antenna, which indeed radiates signals to the surroundings.
In the M-dimensional space, if M orthogonal vectors satisfying the above conditions cannot be found to form a basis, then it is the loss of precision of "pointing where to hit" in the space, which is fully explained in the video .
Free:Specify the infinitesimal∑
n = 0 M − 1 e − j ψ nej ( ψ + 2 π M k ) n = ∑ n = 0 M − 1 ej 2 π M kn \sum_{n=0}^{M -1}e^{-j\psi n}e^{j(\psi+\frac{2\pi}{M}k)n}=\sum_{n=0}^{M-1}e^{ j\frac{2\pi}{M}kn}n=0∑M−1e−jψnej ( ψ +M2 p.mk)n=n=0∑M−1ejM2 p.mkn
It can be considered as summing in one cycle, and the result is 0.
Note: complex numbers are orthogonal, <A takes the conjugate, B>=0, then A and B are orthogonal.