If we specify a direction, the energy received by the receiving antenna in each direction satisfies:
∣ G ( ψ ) ∣ = { ∣ sin ( N ψ / 2 ) N sin ( ψ / 2 ) ∣ , if ψ ≠ 0 1 , if ψ = 0 − − − − − − Formula ( 1 ) |G(\psi)|= \begin{cases} |\frac{sin(N\psi/2)}{Nsin(\psi/ 2)}|, & \text {if $\psi \neq 0$} \\ 1, & \text{if $\psi=0$} \end{cases} ------Formula (1)∣ G ( ψ ) ∣={
∣N s in ( ψ /2 )s in ( N ψ /2 )∣,1,if p=0if p=0−−−−−−Formula ( 1 ) _
For example:
ψ = 2 π d λ cos ( θ ) − 2 π d λ cos ( θ 0 ) \psi=\frac{2\pi d}{\lambda}cos(\theta)-\frac{2\pi d }{\lambda}cos(\theta_{0})p=l2πdcos ( θ )−l2πdcos ( i0)
θ 0 \theta_{0} i0Indicates the direction we want to point, θ \thetaθ is a variable quantity, traversing the entire− π -\pi− π toπ \piSuch a circle of π . For simplicity, we might as well setθ 0 = π / 2 \theta_{0}=\pi/2i0=π /2 ,for:
ψ = 2 π d λ cos ( θ ) − 2 π d λ cos ( θ 0 ) = 2 π d λ cos ( θ ) − 公式 ( 2 ) \psi=\frac{2\pi d }{\lambda}cos(\theta)-\frac{2\pi d}{\lambda}cos(\theta_{0})=\frac{2\pi d}{\lambda}cos(\theta)-公式(2)p=l2πdcos ( θ )−l2πdcos ( i0)=l2πdcos ( θ )−formula ( 2 )
In order to have only one specified beam direction (referring to where to hit, here should be understood as a mathematical maximum point, and there should be only one beam maximum point), then ψ / 2 \psi /2ψ /2 should be between[ − π / 2 , π / 2 ] [-\pi/2,\pi/2][ − π /2 ,π /2 ] with, in2π d λ cos ( θ ) \frac{2\pi d}{\lambda}cos(\theta)l2πdcos ( θ )在[ − π , π ] [-\pi,\pi][ − π ,π ] , so
d λ ≤ 1 2 \frac{d}{\lambda}\leq \frac{1}{2}ld≤21
We can draw a picture to feel, if
d λ = 1 > 1 2 \frac{d}{\lambda}=1> \frac{1}{2}ld=1>21
ψ = 2 π d λ cos ( θ ) ∈ [ − 2 π , 2 π ] \psi=\frac{2\pi d}{\lambda}cos(\theta)\in[-2\pi,2\ pi]p=l2πdcos ( θ )∈[ − 2 p ,2 p ]
Then, for formula (1), we draw the range according to the above formula:
import numpy as np
from matplotlib import pyplot as plt
N = 32 #天线数量
psi = np.arange(-2*np.pi, 2*np.pi-0.0000001, 0.01)
r = np.abs(np.sin(N * psi/2)/np.sin(psi/2))/N
plt.figure()
plt.plot(psi/np.pi, r)
plt.grid()
plt.show()
at [ − π , π ] [-\pi,\pi][ − π ,π ] there is only one highest point, and beyond this, the maximum point begins to increase gradually, that is, in some other directions, there is also greater energy radiated to that direction.
Therefore, when d λ \frac{d}{\lambda}ldIn the process of gradually increasing from 0.5 to 1, you can see that the extra orientation gradually emerges:
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)
d_vs_lambda = 0.5
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.title("0.5")
plt.show()
d_vs_lambda = 0.6
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.title("0.6")
plt.show()
d_vs_lambda = 0.7
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.title("0.7")
plt.show()
d_vs_lambda = 0.8
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.title("0.8")
plt.show()
d_vs_lambda = 0.9
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.title("0.9")
plt.show()
d_vs_lambda = 1.0
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.title("1.0")
plt.show()