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????????%????????CROSS-RANGE?IMAGING?????????%
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colormap（gray（256））
cj=sqrt（-1）;
pi2=2*pi;
%
c=3e8;??????????????%?propagation?speed
fc=200e6;???????????%?frequency
lambda=c/fc;????????%?Wavelength
k=pi2/lambda;???????%?Wavenumber
Xc=1.e3;????????????%?Range?distance?to?center?of?target?area
%
%?Case?1:
L=400;??????????????%?synthetic?aperture?is?2*L
Y0=100;?????????????%?target?area?in?cross-range?is?within?[Yc-Y0Yc+Y0]
Yc=0;???????????????%?Cross-range?distance?to?center?of?target?area
%
%?Case?2:?這時候需要進行zero-padding
%?L=150;
%?Y0=200;
%?Yc=.5e3;
%
theta_c=atan（Yc/Xc）;??%?squint?angle?to?center?of?target?area
Rc=sqrt（Xc^2+Yc^2）;???%?squint?range?to?center?of?target?area
kus=2*k*sin（theta_c）;?%?Doppler?frequency?shift?in?ku?domain?due?to?squint
??????????????????????%?正側(cè)視時候的正對著的目標斜視角為零?多普勒偏移也就為零
??????????????????????%?但正側(cè)視情況下每一個目標都有被正視的情況?這個事情暫時是沒有考慮的
??????????????????????%?畢竟這里只研究了方位向成像?且只是研究了一個合成孔徑內(nèi)部發(fā)生的事情
%
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%???Program?performs?slow-time?compression?to?save?PRF???%
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%
Xcc=Xc/（cos（theta_c）^2）;???????%?redefine?Xc?by?Xcc?for?squint?processing
%
du=（Xcc*lambda）/（4*（Y0+L））;????%?sample?spacing?in?aperture?domain?回波信號的理論的采樣間隔
duc=（Xcc*lambda）/（4*Y0）;???????%?sample?spacing?in?aperture?domain?回波信號的壓縮形式的采樣間隔
???????????????????????????????%?for?compressed?signal?這個和書上的公式不太一樣？
DY=（Xcc*lambda）/（4*L）;?????????%?Cross-range?resolution
%
L_min=max（Y0L）;???????????????%?Zero-padded?aperture
%這樣為什么相當于補零呢？如果Y0較大，那么后面計算壓縮信號的采樣個數(shù)的時候利用較大的Y0
%　而不是Ｌ，但是信號其實在Ｌ長度之外是沒有值的，也就是為零

%?u?domain?parameters?and?arrays?for?compressed?signal
%
mc=2*ceil（L_min/duc）;?????????????%?number?of?samples?on?aperture?
??????????????????????????????????%?用duc去采樣回波信號?會aliased
??????????????????????????????????%?所以才要進行壓縮處理
uc=duc*（-mc/2:mc/2-1）;????????????%?synthetic?aperture?array?壓縮信號的下標索引
dkuc=pi2/（mc*duc）;????????????????%?sample?spacing?in?ku?domain
kuc=dkuc*（-mc/2:mc/2-1）;??????????%?kuc?array?壓縮信號的頻域序列索引
%
dku=dkuc;?????????????????????????%?sample?spacing?in?ku?domain
%不確定這么理解對不對：頻域的補零對應時域的內(nèi)插，所以時域的信號長度是不變的，那么頻域的間隔也就不變
%
%?u?domain?parameters?and?arrays?for?Synthetic?aperture?signal??下面的參數(shù)對應經(jīng)壓縮-升采樣-解壓縮后的回波信號
%
m=2*ceil（pi/（du*dku））;????????????%?number?of?samples?on?aperture?pi/du?對應頻域的范圍?范圍除以間隔得到采樣個數(shù)?
%這里的意思就是說回波信號（未壓縮）的理論采樣間隔決定了其頻域的范圍?這個是確定的
%然后，回波信號按照duc采樣后混疊了，解決辦法是壓縮?上采樣?再解壓縮

du=pi2/（m*dku）;%頻域的而采樣間隔乘以采樣的個數(shù)是頻域的范圍?他的倒數(shù)對應時域的采樣間隔

u=du*（-m/2:m/2-1）;????????????????%?synthetic?aperture?array?信號的下標索引
??????????????????????????????????%?這個是回波信號（經(jīng)壓縮-升采樣-解壓縮后的）的索引?注意不是直接采樣得到的
ku=dku*（-m/2:m/2-1）;??????????????%?ku?array?
%
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%%%%%??????????SIMULATION??????目標的基本信息?%%%%%%%??????????
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%
ntarget=4;?????????????%?Number?of?targets
%
%?Targets‘?coordinates?and?reflectivity

yn（1）=0;?????????????fn（1）=1;

?屬性????????????大小?????日期????時間???名稱
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?????文件???????9130??2016-05-07?12:47??crange.m

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?????????????????9130????????????????????1