斜距转为地距_resolution

# -*- coding: utf-8 -*- """ Created on Wed Sep 23 20:14:11 2020 @author: yang """ import numpy as np #matrix/array calculations etc def resolution (Naz,R0,H,daitah,Br,Fr,va,PRF,Kr,data): c = 2.99792458e8 # 光速 Nrg=data.shape[0] naz=data.shape[1] # 斜距分辨率 dr=c/2/Br #斜距坐标轴 rg=np.arange(1, naz+1, 1)*dr #计算距离向波数下视角角度 tr_RCMC = np.zeros((1,Naz), dtype=np.float ) #时间轴 for i in range(Naz): tr_RCMC[0,0]=2*R0/c-Naz/2/Fr tr_RCMC[0,i]=tr_RCMC[0,i-1]+Naz/2/Fr/Naz*2 R0_RCMC = (c/2)*tr_RCMC #延距离向变化的斜距，记为R0_RCMC r_open = int(input("请输入数据距离向起始位置:")) r_end = int(input("请输入数据距离向结束位置:")) r_range_A = R0_RCMC[0,r_open:r_end] # 截取所计算区域 cos_theta = (H-daitah)/r_range_A # 对应于每一个天线A的最近斜距时，波束下视角的余弦 #计算出波束下视角（弧度）； theta_range_A=np.arccos(cos_theta) #计算距离向分辨率 ran_re=dr/np.sinc(theta_range_A) #计算距离向坐标轴 ran_zhou = np.zeros((ran_re.shape[0]), dtype=np.float ) for i in range (1, ran_re.shape[0], 1): ran_zhou[0]=ran_re[0] ran_zhou[i]=ran_zhou[i-1] +ran_re[i] ###方位向坐标轴时间间隔 PRI=1/PRF ta=np.arange(1,Nrg+1,1)*PRI #方位向坐标轴 azi_zhou=va*ta return rg,ran_zhou,azi_zhou