CN103344958B - Based on the satellite-borne SAR high-order Doppler parameter evaluation method of almanac data - Google Patents

Abstract

The invention discloses a kind of satellite-borne SAR high-order Doppler parameter evaluation method based on almanac data, directly rely on the ephemeris parameter of echoed signal to estimate the instantaneous orbit radical of echoed signal, its error source only derives from the precision of ephemeris parameter, the Doppler's parameter estimate precision obtained determined by ephemeris parameter completely, and Processing Algorithm can not introduce additive error again; The present invention only utilizes the ephemeris parameter of echoed signal central instant satellite to estimate high-order Doppler parameter, effectively promotes the efficiency of parameter estimation; The present invention, directly for the demand of high-resolution imaging process to Doppler parameter, proposes to utilize ephemeris parameter to estimate the method for high-order Doppler parameter, meets the demand of high precision imaging processing, promote the practicality of imaging processing.

Description

Based on the satellite-borne SAR high-order Doppler parameter evaluation method of almanac data

Technical field

The invention belongs to signal transacting field, relate to a kind of parameter evaluation method, particularly a kind of satellite-borne SAR high-order Doppler parameter evaluation method based on almanac data.

Background technology

Synthetic-aperture radar (SyntheticApertureRadar, SAR) be a kind of active remote sensing device being operated in microwave frequency band, it has feature that is round-the-clock, round-the-clock work, can large area imaging, and the penetration capacity had some atural object, in some cases, the effect that other remote sensing does not have can be played.

High precision image-processing algorithms is the basis obtaining high-quality radar image, is also the prerequisite of SAR system application.In traditional satellite-borne SAR echoed signal imaging process, utilize the Doppler parameter of satellite position and the velocity information estimation echoed signal obtained in ephemeris parameter, and utilize the Doppler parameter obtained to carry out focal imaging process, and then obtain high-precision radar image.Owing to mostly just containing position and the velocity information of satellite in the ephemeris parameter of radar system, if directly utilize the position of satellite and velocity information front two rank Doppler parameters that estimation of Doppler parameter only can estimate echoed signal, i.e. doppler centroid f _dwith doppler frequency rate f _r.When radar system spatial resolution is not high, directly utilize second order oblique distance model just can describe the change of relative tertiary location, utilize second order Doppler parameter can realize the vernier focusing process of echoed signal accordingly.

Along with the continuous increase of radar system spatial resolution, the synthetic aperture time of radar system increases gradually, utilize second order oblique distance model to convert the error introduced also increase gradually to describe relative position, now, second order Doppler parameter no longer can meet the demand of high precision imaging processing.In order to complete the high precision imaging processing to echoed signal, requiring to utilize ephemeris parameter to estimate the high-order Doppler parameter of echoed signal, requiring the order motion parameter that satellite platform is provided in ephemeris parameter accordingly.But, by the restriction of measuring technique, current survey rail/orbit determination technology only can ensure positional accuracy measurement and the velocity survey precision of satellite, the acceleration of satellite or the precision measure of more order motion parameter cannot be obtained, be difficult to directly utilize the ephemeris parameter of satellite directly to calculate the high-order Doppler parameter of echoed signal, and then impact obtains the quality of radar image.Therefore, existing appearance rail measuring technique how is utilized to become the precondition obtaining High Accuracy Radar image to the accurate estimation completing high-order Doppler parameter, effectively can promote the efficiency of radar image quality and Image Acquisition, for the detectivity promoting radar system, there is important reusing.

Transformational relation between 6 coordinate systems of prior art culminant star ground space geometric relationship and coordinate system is:

1. the geocentric coordinate system E do not rotated ₀

True origin: the earth centre of sphere

Z axis: the axis of rotation along the earth points to the positive arctic

X-axis: under the line in plane, points to the first point of Aries

Y-axis: under the line in plane, makes this coordinate system form right hand rectangular coordinate system

The geocentric coordinate system do not rotated is the middle rectangular coordinate system set up in the equator, the earth's core reference frame (inertial reference system) of not rotating.

2. the geocentric coordinate system E rotated _g

True origin: the earth centre of sphere

Z axis: the axis of rotation along the earth points to the positive arctic

X-axis: under the line in plane, by first branch of Greenwich meridian

Y-axis: under the line in plane, makes this coordinate system form right hand rectangular coordinate system

3. satellite orbit plane coordinate system E _v

True origin a: focus (i.e. the earth centre of sphere) of elliptical orbit

Z axis: perpendicular to satellite orbit plane, forward points to the angular-momentum vector direction of satellite

Y-axis: in satellite plane, forward points to pericenter

X-axis: in satellite orbit plane, makes this coordinate system form right hand rectangular coordinate system

4. satellite platform coordinate system E _r

True origin: centroid of satellite

Z axis: perpendicular to satellite orbit plane, forward points to the angular-momentum vector direction of satellite

X-axis: in satellite orbit plane, the y direction (dividing property of the design direction of satellite) of gyropanel

Y-axis: in satellite orbit plane, makes this coordinate system form right hand rectangular coordinate system

5. satellite health coordinate system E _e

True origin: centroid of satellite;

X-axis: along satellite health y direction (the Live Flying direction of satellite)

Y-axis, Z axis: along two other principal axis of inertia direction of satellite health

6. antenna coordinate system E _a

True origin: antenna phase center point;

X-axis: forward points to the Live Flying direction of satellite;

Y-axis: along antenna boresight, refers to be earthwards forward;

Z axis: right hand rule provides, makes this coordinate system form right hand rectangular coordinate system.

Meet as follows between six coordinate systems:

1. the geocentric coordinate system E rotated _g/ geocentric coordinate system the E that do not rotate ₀

E _g＝A _go·E _o（1）

The geocentric coordinate system E do not rotated ₀the Greenwich hour angle (GHA) H in a first point of Aries is turned over counterclockwise around axle _gjust obtain the geocentric coordinate system E rotated _g.

$A_{\mathrm{go}}=\left[\begin{array}{ccc}\cos H_g& \sin H_g& 0\\ -\sin H_g& \cos H_g& 0\\ 0& 0& 1\end{array}\right]---\left(2\right)$

2. the geocentric coordinate system E do not rotated ₀/ orbit plane coordinate system E _v

E _o＝A _ov·E _v（3）

The geocentric coordinate system E do not rotated ₀orbit plane coordinate system E is obtained through three rotations _v.For the first time, the geocentric coordinate system do not rotated is rotated counterclockwise an angle Ω around Z axis; For the second time, the coordinate system obtained is rotated counterclockwise an angle i around X-axis again; For the third time, then gained coordinate system is rotated counterclockwise an angle ω around Z axis, finally obtains orbit plane coordinate system E _v.

$A_{\mathrm{ov}}=\left[\begin{array}{ccc}\cos \mathrm{\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]---\left(4\right)$

3. orbit plane coordinate system E _v/ satellite platform coordinate system E _r

E _v＝A _vr·E _r（5）

Satellite platform coordinate system E _rbe rotated counterclockwise an angle 90 ° of+θ-γ around Z axis and obtain satellite orbit plane coordinate system E _v.Wherein, θ is the very near heart angle of satellite, and γ is the flight-path angle of satellite.

To be looked for the truth nearly heart angle θ by the corresponding relation of very near heart angle and eccentric angle E:

$E-e\sin E=\sqrt{\frac{\mathrm{\μ}}{a^3}}(t-\mathrm{\τ})---\left(6\right)$

$\tan \frac{\mathrm{\θ}}{2}=\sqrt{\frac{1+e}{1-e}}\tan \frac{E}{2}---\left(7\right)$

Flight-path angle γ:

$\tan \mathrm{\γ}=\frac{e\sin \mathrm{\θ}}{1+e\cos \mathrm{\θ}}$ |γ|≤90°（8）

$A_{\mathrm{vr}}=\left[\begin{array}{ccc}-\sin (\mathrm{\θ}-\mathrm{\γ})& -\cos (\mathrm{\θ}-\mathrm{\γ})& 0\\ \cos (\mathrm{\θ}-\mathrm{\γ})& -\sin (\mathrm{\θ}-\mathrm{\γ})& 0\\ 0& 0& 1\end{array}\right]---\left(9\right)$

4. satellite platform coordinate system E _r/ satellite health coordinate system E _e

E _r＝A _re·E _e（10）

Satellite health coordinate system E _esatellite platform coordinate system E is obtained through three rotations _r.For the first time, by satellite health coordinate system E _eto turn clockwise a roll angle θ around X-axis _r; For the second time, the coordinate system obtained to be turned clockwise a luffing angle θ around Z axis _p; For the third time, then by obtained coordinate system an angle θ is rotated counterclockwise around Y-axis _y, finally obtain satellite platform coordinate system E _r.

$A_{\mathrm{re}}=\left[\begin{array}{ccc}\cos {\mathrm{\θ}}_y& 0& -\sin {\mathrm{\θ}}_y\\ 0& 1& 0\\ \sin {\mathrm{\θ}}_y& 0& \cos {\mathrm{\θ}}_y\end{array}\right]\left[\begin{array}{ccc}\cos {\mathrm{\θ}}_p& -\sin {\mathrm{\θ}}_p& 0\\ \sin {\mathrm{\θ}}_p& \cos {\mathrm{\θ}}_p& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos {\mathrm{\θ}}_r& -\sin {\mathrm{\θ}}_r\\ 0& \sin {\mathrm{\θ}}_r& \cos {\mathrm{\θ}}_r\end{array}\right]---\left(11\right)$

5. satellite health coordinate system E _e/ antenna coordinate system E _a

E _e＝A _ea·E _a（12）

Antenna coordinate system E _aan angle θ is rotated counterclockwise around X-axis _lobtain satellite health coordinate system E _e.

$A_{\mathrm{ea}}=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos {\mathrm{\θ}}_L& \sin {\mathrm{\θ}}_L\\ 0& -\sin {\mathrm{\θ}}_L& \cos {\mathrm{\θ}}_L\end{array}\right]---\left(13\right)$

Summary of the invention

The present invention proposes a kind of satellite-borne SAR high-order Doppler parameter evaluation method based on almanac data, in parameter estimation procedure, satellite orbit model being incorporated into Doppler parameter estimation process as priori, completing the estimation process of high-order Doppler parameter when only utilizing satellite position parameter and speed parameter.The method utilizes the position coordinates of almanac data inverting beam position point and the instantaneous orbit radical of this moment satellite of satellite, obtain the order motion parameter of beam position point and radar satellite on this basis, the order motion sense parameter of synthesized beam sensing point and radar satellite obtains the high-order Doppler parameter of corresponding echoed signal.

Based on the satellite-borne SAR high-order Doppler parameter evaluation method of almanac data, comprise the steps:

Step 1: according to echo data Store form, reads the auxiliary data corresponding to echo data;

Step 2: the geocentric coordinate that calculating is not rotated is tied to the transition matrix A between satellite platform coordinate system _or, satellite platform coordinate is tied to transition matrix A between satellite health coordinate system _re, satellite health coordinate is tied to transition matrix A between antenna coordinate system _ea;

Step 3: obtain terrestrial beam and point to point at the coordinate position not rotationally in spherical coordinate system;

Step 4: the orbital tracking determining satellite, forgives semi-major axis of orbit a, orbit inclination i, right ascension of ascending node Ω, eccentric ratio e, argument of perigee ω, very near heart angle θ;

Step 5: calculate orbit plane coordinate system and do not rotate the transition matrix A between geocentric coordinate system _ov;

Step 6: calculate the instantaneous acceleration vector A not rotating geocentric coordinate system Satellite _osand acceleration change vector

Step 7: ground is calculated and do not rotated in geocentric coordinate system, terrestrial beam points to the velocity V of point _op, acceleration A _opand acceleration change vector

Step 8: calculate Relative position vector R _{o_sp}, relative velocity vector V _{o_sp}, relative acceleration vector A _{o_sp}with relative acceleration rate of change vector

Step 9: in conjunction with Relative position vector, relative velocity vector, relative acceleration vector, relative acceleration rate of change vector, estimation echoed signal Doppler parameter;

The invention has the advantages that:

(1) precision is high.The present invention directly relies on the ephemeris parameter of echoed signal to estimate the instantaneous orbit radical of echoed signal, its error source only derives from the precision of ephemeris parameter, the Doppler's parameter estimate precision obtained determined by ephemeris parameter completely, and Processing Algorithm can not introduce additive error again;

(2) efficiency is high.The present invention only utilizes the ephemeris parameter of echoed signal central instant satellite to estimate high-order Doppler parameter, effectively promotes the efficiency of parameter estimation;

(3) practical.The present invention, directly for the demand of high-resolution imaging process to Doppler parameter, proposes to utilize ephemeris parameter to estimate the method for high-order Doppler parameter, meets the demand of high precision imaging processing, promote the practicality of imaging processing.

Accompanying drawing explanation

Fig. 1 the present invention is based on the process flow diagram of the satellite-borne SAR high-order Doppler's parameter estimate method of almanac data;

Fig. 2 determination radar antenna of the present invention beam position point is not rotating the process flow diagram of coordinate position of geocentric coordinate system;

Fig. 3 the present invention is based on the process flow diagram of almanac data determination instantaneous orbit radical;

The processing flow chart of Fig. 4 determination echo data of the present invention quadravalence Doppler parameter.

Embodiment

Below in conjunction with accompanying drawing, the present invention is described in further detail;

Satellite-borne SAR high-order Doppler parameter evaluation method based on almanac data of the present invention, flow process as shown in Figure 1, comprises the steps:

Step 1: according to echo data Store form, read the auxiliary data corresponding to echo data, auxiliary data comprises: wavefront oblique distance R _min, signal sampling rate f _s, pulse repetition rate f _pRF, transmit signal pulse width τ, satellite position vector R _os=(R _{o, sx}, R _{o, sy}, R _{o, sz}) ', the velocity V of satellite _os=(V _sx, V _sy, V _sz) ', the crab angle θ of satellite _y, satellite pitching angle theta _p, satellite roll angle θ _r, radar antenna work view angle theta _l.

Step 2: in conjunction with the position vector R of satellite _os=(R _{o, sx}, R _{o, sy}, R _{o, sz}) ', the velocity V of satellite _os=(V _sx, V _sy, V _sz) ', the crab angle θ of satellite _y, satellite pitching angle theta _p, satellite roll angle θ _r, calculate the transition matrix A between each coordinate system _or, A _re, A _ea.

$A_{\mathrm{or}}=\left[\begin{array}{ccc}1& 0& 0\\ 0& y_1/r_1& x_1/r_1\\ 0& -x_1/r_1& y_1/r_1\end{array}\right]\left[\begin{array}{ccc}{v}_{\mathrm{sx}}/v_1& -v_{\mathrm{sy}}/v_1& 0\\ v_{\mathrm{sy}}/v_1& v_{\mathrm{sx}}/v_1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}{v}_1/v& 0& v_{\mathrm{sz}}/v\\ 0& 1& 0\\ {-v}_{\mathrm{sz}}/v& 0& v_1/v\end{array}\right]---\left(14\right)$

Wherein: $r_1=\left[\begin{array}{ccc}{v}_1/v& 0& -v_{\mathrm{sz}}/v\\ 0& 1& 0\\ v_{\mathrm{sz}}/v& 0& v_1/v\end{array}\right]\left[\begin{array}{ccc}{v}_{\mathrm{sx}}/v_1& v_{\mathrm{sy}}/v_1& 0\\ -v_{\mathrm{sy}}/v_1& v_{\mathrm{sx}}/v_1& 0\\ 0& 0& 1\end{array}\right]\·(R_{\mathrm{os}}\×V_{\mathrm{os}})=(0,x_1,y_1);$

r ₁＝|r ₁|；

v＝|V _os|；

$v_1=\sqrt{v_{\mathrm{sx}}^2+v_{\mathrm{sy}}^2}.$

$A_{\mathrm{re}}=\left[\begin{array}{ccc}\cos {\mathrm{\θ}}_y& 0& -\sin {\mathrm{\θ}}_y\\ 0& 1& 0\\ \sin {\mathrm{\θ}}_y& 0& \cos {\mathrm{\θ}}_y\end{array}\right]\left[\begin{array}{ccc}\cos {\mathrm{\θ}}_p& -\sin {\mathrm{\θ}}_p& 0\\ \sin {\mathrm{\θ}}_p& \cos {\mathrm{\θ}}_p& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos {\mathrm{\θ}}_r& -\sin {\mathrm{\θ}}_r\\ 0& \sin {\mathrm{\θ}}_r& \cos {\mathrm{\θ}}_r\end{array}\right]$

$A_{\mathrm{ea}}=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos {\mathrm{\θ}}_L& \sin {\mathrm{\θ}}_L\\ 0& -\sin {\mathrm{\θ}}_L& \cos {\mathrm{\θ}}_L\end{array}\right]$

Step 3: as shown in Figure 2, after calculating the mutual transition matrix between each coordinate system according to step 2, further calculating do not rotate geocentric coordinate system Satellite and terrestrial beam point to a little between distance vector, distance vector between pointing to a little in conjunction with the position vector of satellite and satellite and terrestrial beam, obtains terrestrial beam by vector addition process and points to point and do not rotating the position vector in geocentric coordinate system.Consider that controlling antenna wave beam to point point is in the Y-axis of antenna coordinate system, meet following vector [0, R _p, 0], obtain terrestrial beam by vector superposed process and point to point at the coordinate position not rotationally in spherical coordinate system:

$R_{\mathrm{op}}=R_{\mathrm{os}}+R_{o\_\mathrm{sp}}$

$=\left[\begin{array}{c}{R}_{o,\mathrm{sx}}\\ R_{o,\mathrm{sy}}\\ R_{o,\mathrm{sz}}\end{array}\right]+A_{\mathrm{or}}{A}_{\mathrm{re}}{A}_{\mathrm{ea}}\left[\begin{array}{c}0\\ R_p\\ 0\end{array}\right]$

$=\left[\begin{array}{c}{R}_{o,\mathrm{sx}}\\ R_{o,\mathrm{sy}}\\ R_{o,\mathrm{sz}}\end{array}\right]+\left[\begin{array}{ccc}{A}_{\mathrm{oe}\_11}& A_{\mathrm{oe}\_12}& A_{\mathrm{oe}\_13}\\ A_{\mathrm{oe}\_12}& A_{\mathrm{oe}\_22}& A_{\mathrm{oe}\_23}\\ A_{\mathrm{oe}\_13}& A_{\mathrm{oe}\_23}& A_{\mathrm{oe}\_33}\end{array}\right]\left[\begin{array}{c}0\\ R_p\cos {\mathrm{\θ}}_L\\ -R_p\sin {\mathrm{\θ}}_L\end{array}\right]---\left(15\right)$

$=\left[\begin{array}{c}{R}_{o,\mathrm{sx}}\\ R_{o,\mathrm{sy}}\\ R_{o,\mathrm{sz}}\end{array}\right]+\left[\begin{array}{c}{A}_{\mathrm{oe}\_12}\·R_p\cos {\mathrm{\θ}}_L-A_{\mathrm{oe}\_13}\·R_p\sin {\mathrm{\θ}}_L\\ A_{\mathrm{oe}\_22}\·R_p\cos {\mathrm{\θ}}_L-A_{\mathrm{oe}\_23}\·R_p\sin {\mathrm{\θ}}_L\\ A_{\mathrm{oe}\_32}\·R_p\cos {\mathrm{\θ}}_L-A_{\mathrm{oe}\_33}\·R_p\sin {\mathrm{\θ}}_L\end{array}\right]=\left[\begin{array}{c}{R}_{o,\mathrm{nx}}\\ R_{o,\mathrm{ny}}\\ R_{o,\mathrm{nz}}\end{array}\right]$

Wherein: R _{o_sp}represent the distance vector between not rotating satellite and terrestrial beam in geocentric coordinate system points to a little, n _rrepresent that distance is to sampling number; C represents the light velocity.

Step 4: the orbital tracking determining satellite, forgives semi-major axis of orbit a, orbit inclination i, right ascension of ascending node Ω, eccentric ratio e, argument of perigee ω, very near heart angle θ.As shown in Figure 3, in conjunction with the mutual relationship between coordinate conversion matrix, have:

$A_{\mathrm{or}}=\left[\begin{array}{ccc}1& 0& 0\\ 0& y_1/r_1& x_1/r_1\\ 0& {-x}_1/r_1& y_1/r_1\end{array}\right]\left[\begin{array}{ccc}{v}_{\mathrm{sx}}/v_1& {-v}_{\mathrm{sy}}/v_1& 0\\ v_{\mathrm{sy}}/v_1& v_{\mathrm{sx}}/v_1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}{v}_1/v& 0& v_{\mathrm{sz}}/v\\ 0& 1& 0\\ -v_{\mathrm{sz}}/v& 0& v_1/v\end{array}\right]=\left[\begin{array}{ccc}{a}_{\mathrm{or}}11& a_{\mathrm{or}}21& a_{\mathrm{or}}31\\ a_{\mathrm{or}}12& a_{\mathrm{or}}22& a_{\mathrm{or}}32\\ a_{\mathrm{or}}13& a_{\mathrm{or}}23& a_{\mathrm{or}}33\end{array}\right]---\left(16\right)$

$=\left[\begin{array}{ccc}\cos \mathrm{\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}-\sin (\mathrm{\θ}-\mathrm{\γ})& -\cos (\mathrm{\θ}-\mathrm{\γ})& 0\\ \cos (\mathrm{\θ}-\mathrm{\γ})& -\sin (\mathrm{\θ}-\mathrm{\γ})& 0\\ 0& 0& 1\end{array}\right]$

Wherein: A _orin each coefficient a _orij is by satellite position vectors and velocity, and i shows that this coefficient is the i-th row coefficient in matrix, and j shows that this coefficient is the jth row coefficient in matrix, 1≤i≤3,1≤j≤3.γ represents flight-path angle, is calculated obtain by following formula:

$\tan \mathrm{\γ}=\frac{e\sin \mathrm{\θ}}{1+e\cos \mathrm{\θ}}---\left(17\right)$

Following equation can be obtained in conjunction with the relation between above-mentioned transition matrix:

$\begin{array}{c}\cos i=a_{\mathrm{or}}33\\ \sin i\·\sin \mathrm{\Ω}=a_{\mathrm{or}}31\end{array}---\left(18\right)$

By solving equation (18), after the orbit inclination i that can determine satellite orbit and right ascension of ascending node Ω, equation (16) can be modified to:

$\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}-\sin (\mathrm{\θ}-\mathrm{\γ})& -\cos (\mathrm{\θ}-\mathrm{\γ})& 0\\ \cos (\mathrm{\θ}-\mathrm{\γ})& -\sin (\mathrm{\θ}-\mathrm{\γ})& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}-\sin (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})& -\cos (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})& 0\\ \cos (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})& -\sin (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})& 0\\ 0& 0& 1\end{array}\right]$

$={\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]}^{-1}{\left[\begin{array}{ccc}\cos \mathrm{\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]}^{-1}\left[\begin{array}{ccc}{a}_{\mathrm{or}}11& a_{\mathrm{or}}21& a_{\mathrm{or}}31\\ a_{\mathrm{or}}12& a_{\mathrm{or}}22& a_{\mathrm{or}}32\\ a_{\mathrm{or}}13& a_{\mathrm{or}}23& a_{\mathrm{or}}33\end{array}\right]---\left(19\right)$

$=\left[\begin{array}{ccc}{b}_{\mathrm{or}}11& b_{\mathrm{or}}21& b_{\mathrm{or}}31\\ b_{\mathrm{or}}12& b_{\mathrm{or}}22& b_{\mathrm{or}}32\\ b_{\mathrm{or}}13& b_{\mathrm{or}}23& b_{\mathrm{or}}33\end{array}\right]$

Wherein, consider that the right side matrix in above formula is known matrix, can calculate accordingly:

$\begin{array}{c}\cos (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})=b_{\mathrm{or}}12\\ \sin (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})=-b_{\mathrm{or}}11\end{array}---\left(20\right)$

Angle (ω+θ-γ) can be calculated by solving above formula.

In conjunction with satellite at the position coordinates R not rotationally in spherical coordinate system _os=(R _{o, sx}, R _{o, sy}, R _{o, sz}) ', can calculate and vow that radius r is:

$r=\sqrt{R_{o,\mathrm{sx}}^2+R_{o,\mathrm{sy}}^2+R_{o,\mathrm{sz}}^2}---\left(21\right)$

Position coordinates meets following equation simultaneously:

$r\left[\begin{array}{ccc}\cos \mathrm{\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}\cos \left(\mathrm{\θ}\right)\\ \sin \left(\mathrm{\θ}\right)\\ 0\end{array}\right]=\left[\begin{array}{c}{R}_{o,\mathrm{sx}}\\ R_{o,\mathrm{sy}}\\ R_{o,\mathrm{sz}}\end{array}\right]---\left(22\right)$

Have after bringing arrow radius, orbit inclination and right ascension of ascending node into above-mentioned equation:

$\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}\cos \left(\mathrm{\θ}\right)\\ \sin \left(\mathrm{\θ}\right)\\ 0\end{array}\right]={\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]}^{-1}{\left[\begin{array}{ccc}\cos \mathrm{\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{x}_s\\ y_s\\ z_s\end{array}\right]$ (23)

$=\left[\begin{array}{c}{c}_{\mathrm{or}}1\\ c_{\mathrm{or}}2\\ c_{\mathrm{or}}3\end{array}\right]=\left[\begin{array}{c}\cos (\mathrm{\ω}+\mathrm{\θ})\\ \sin (\mathrm{\ω}+\mathrm{\θ})\\ 0\end{array}\right]$

Following equation can be obtained according to above formula:

$\begin{array}{c}\cos (\mathrm{\ω}+\mathrm{\θ})=c_{\mathrm{or}}1\\ \sin (\mathrm{\ω}+\mathrm{\θ})=c_{\mathrm{or}}2\end{array}---\left(24\right)$

Angle (ω+θ) can be calculated by solving above formula.

Composite type (20) and formula (24), obtain the flight-path angle γ of radar system.

Consider that satellite is at the velocity V not rotationally in spherical coordinate system _os=(V _sx, V _sy, V _sz) ', meet following equation relation:

$V_{\mathrm{sx}}^2+V_{\mathrm{sy}}^2+V_{\mathrm{sz}}^2=\frac{\mathrm{\μ}}{a(1-e^2)}[1+e^2+2e\cos \mathrm{\θ}]$

Following equation can be obtained accordingly:

$V_{\mathrm{sx}}^2+V_{\mathrm{sy}}^2+V_{\mathrm{sz}}^2=\frac{\mathrm{\μ}}{a(1-e^2)}[1+e^2+2e\cos \mathrm{\θ}]$

$=\frac{\mathrm{\μ}}{a(1-e^2)}[2(1+e\cos \mathrm{\θ})+e^2-1]$ (26)

$=\frac{2\mathrm{\μ}(1+e\cos \mathrm{\θ})}{a(1-e^2)}+\frac{\mathrm{\μ}}{a(1-e^2)}[e^2-1]$

$=\frac{2\mathrm{\μ}}{r}-\frac{\mathrm{\μ}}{a}$

Utilize equation (26), the semi-major axis a of track can be obtained.

In conjunction with Kepler's radar equation and flight-path angle accounting equation, following system of equations (27) can be obtained:

$\begin{array}{c}a(1-e^2)=r(1+e\cos \mathrm{\θ})\\ \tan \mathrm{\γ}=\frac{e\sin \mathrm{\θ}}{1+e\cos \mathrm{\θ}}\end{array}---\left(27\right)$

Have after abbreviation is carried out to formula (27):

${\tan }^2\mathrm{\γ}\·[1+2\frac{a(1-e^2)-r}{r}+{\left(\frac{a(1-e^2)-r}{r}\right)}^2]=e^2-{\left(\frac{a(1-e^2)-r}{r}\right)}^2---\left(28\right)$

Can obtain the eccentric ratio e of track by solving above-mentioned biquadratic equation group, the computing formula simultaneously orbital eccentricity being brought into flight-path angle can obtain the instantaneous very near heart angle θ of radar satellite, and convolution (24) can obtain the argument of perigee ω of satellite orbit.

Step 5: the transition matrix Α calculating orbit plane coordinate system and do not rotate between geocentric coordinate system _ov.Transform matrix calculations formula orbit inclination, right ascension of ascending node and argument of perigee being brought into orbit plane coordinate system and not rotating between geocentric coordinate system, calculates orbit plane coordinate system and does not rotate the transition matrix A between geocentric coordinate system _ov.

Step 6: calculate the instantaneous acceleration vector A not rotating geocentric coordinate system Satellite _osand acceleration change vector

$A_{\mathrm{os}}=\frac{\mathrm{\μ}{(1+e\cos \left(\mathrm{\θ}\right))}^2}{a^2{(1-e^2)}^2}{A}_{\mathrm{ov}}\left[\begin{array}{c}-\cos \left(\mathrm{\θ}\right)\\ -\sin \left(\mathrm{\θ}\right)\\ 0\end{array}\right]---\left(29\right)$

${\stackrel{\·}{A}}_{\mathrm{os}}=\frac{\mathrm{\μ}{(1+e\cos \left(\mathrm{\θ}\right))}^3}{a^3{(1-e^2)}^3}\sqrt{\frac{\mathrm{\μ}}{a(1-e^2)}}{A}_{\mathrm{ov}}\left[\begin{array}{c}\sin \left(\mathrm{\θ}\right)+3e\cos \left(\mathrm{\θ}\right)\sin \left(\mathrm{\θ}\right)\\ e(2{\sin }^2\left(\mathrm{\θ}\right)-{\cos }^2\left(\mathrm{\θ}\right))-\cos \left(\mathrm{\θ}\right)\\ 0\end{array}\right]---\left(30\right)$

Step 7: ground is calculated and do not rotated in geocentric coordinate system, terrestrial beam points to the velocity V of point _op, acceleration A _opand acceleration change vector

$V_{\mathrm{op}}=\left(\begin{array}{c}{V}_{o\_\mathrm{px}}\\ V_{o\_\mathrm{py}}\\ V_{o\_\mathrm{pz}}\end{array}\right)=\left(\begin{array}{c}-{\mathrm{\ω}}_e{R}_{o,\mathrm{ny}}\\ {\mathrm{\ω}}_e{R}_{o,\mathrm{nx}}\\ 0\end{array}\right)---\left(31\right)$

$A_{\mathrm{op}}=\left(\begin{array}{c}{A}_{o\_\mathrm{px}}\\ A_{o\_\mathrm{py}}\\ A_{o\_\mathrm{pz}}\end{array}\right)=\left(\begin{array}{c}-{\mathrm{\ω}}_e^2{R}_{o,\mathrm{nx}}\\ -{\mathrm{\ω}}_e^2{R}_{o,\mathrm{ny}}\\ 0\end{array}\right)---\left(32\right)$

${\stackrel{\·}{A}}_{\mathrm{op}}=\left(\begin{array}{c}{\stackrel{\·}{A}}_{o\_\mathrm{px}}\\ {\stackrel{\·}{A}}_{o\_\mathrm{py}}\\ {\stackrel{\·}{A}}_{o\_\mathrm{pz}}\end{array}\right)=\left(\begin{array}{c}{\mathrm{\ω}}_e^3{R}_{o,\mathrm{ny}}\\ -{\mathrm{\ω}}_e^3{R}_{o,\mathrm{nx}}\\ 0\end{array}\right)---\left(33\right)$

Step 8: calculate Relative position vector R _{o_sp}, relative velocity vector V _{o_sp}, relative acceleration vector A _{o_sp}with relative acceleration rate of change vector

R _{o_sp}＝R _os+R _op（34）

V _{o_sp}＝V _os+V _op（35）

A _{o_sp}＝A _os+A _op（36）

${\stackrel{\·}{A}}_{o\_\mathrm{sp}}={\stackrel{\·}{A}}_{\mathrm{os}}+{\stackrel{\·}{A}}_{\mathrm{op}}---\left(37\right)$

Step 9: in conjunction with Relative position vector, relative velocity vector, relative acceleration vector, relative acceleration rate of change vector, estimation echoed signal Doppler parameter.

$f_1=\frac{2}{\mathrm{\λ}}\frac{V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}}}{\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}---\left(38\right)$

$f_2=\frac{2}{\mathrm{\λ}}(\frac{V_{o\_\mathrm{spx}}{V}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{V}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{V}_{o\_\mathrm{spz}}}{\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}+\frac{A_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+A_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+A_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}}}{\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}$ （39）

$-\frac{{(V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})}^2}{{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{3/2}})$

$f_3=\frac{2}{\mathrm{\λ}}(\frac{{\stackrel{\·}{A}}_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+{\stackrel{\·}{A}}_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+{\stackrel{\·}{A}}_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}}}{6\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}+\frac{A_{o\_\mathrm{spx}}{V}_{o\_\mathrm{spx}}+A_{o\_\mathrm{spy}}{V}_{o\_\mathrm{spy}}+A_{o\_\mathrm{spz}}{V}_{o\_\mathrm{spz}}}{2\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}$

$-\frac{(A_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+A_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+A_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})(V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})}{{2(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{3/2}}$ （40）

$-\frac{(V_{o\_\mathrm{spx}}{V}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{V}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{V}_{o\_\mathrm{spz}})(V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})}{{2(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{3/2}}$

$-\frac{{(V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})}^3}{2{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{5/2}})$

$f_4=\frac{2}{\mathrm{\λ}}(\frac{{\stackrel{\·}{A}}_{o\_\mathrm{spx}}{V}_{o\_\mathrm{spx}}+{\stackrel{\·}{A}}_{o\_\mathrm{spy}}{V}_{o\_\mathrm{spy}}+{\stackrel{\·}{A}}_{o\_\mathrm{spz}}{V}_{o\_\mathrm{spz}}}{6\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}+\frac{A_{o\_\mathrm{spx}}{A}_{o\_\mathrm{spx}}+A_{o\_\mathrm{spy}}{A}_{o\_\mathrm{spy}}+A_{o\_\mathrm{spz}}{A}_{o\_\mathrm{spz}}}{8\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}$ （41）

$-\frac{{\left(\mathrm{\λ}{f}_2\right)}^2}{32{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{1/2}}-\frac{{\left(\mathrm{\λ}{f}_1{f}_3\right)}^2}{24{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{1/2}})$

Composite type (38) ~ (41) obtain the quadravalence Doppler parameter of echoed signal, and then meet the demand of high-resolution imaging process.

For whole invention, creatively achievement embodies in the following areas:

1, first time proposes only to utilize satellite position vectors and velocity to complete the computing of quadravalence Doppler, and gives the treatment scheme of the calculating of high-order Doppler parameter;

2, in step 4, give the method for estimation utilizing satellite position vectors and velocity to complete orbit elements of satellite;

3, quadravalence Doppler parameter is applied to and sets up on the oblique distance model of SAR, effectively can improve the imaging processing precision of high resolution SAR.

Embodiment:

When echo data is of a size of 65536 × 32768, the satellite-borne SAR high-order Doppler parameter evaluation method based on almanac data of the present invention, concrete treatment step is as follows:

Step 1: the auxiliary data reading echo data, comprising: wavefront oblique distance R _min=729616.950604m, echo signal sample rate f _s=1.2GHz, transmit signal pulse width τ=0.00002s,

The position vector R of satellite _os=(-3506792.3866m, 6058045.0564m ,-56558.4003m) ',

The velocity V of satellite _osthe crab angle θ of=(882.4660m/s, 571.8726m/s, 7472.3786m/s) ', satellite _y=3.876 °, the pitching angle theta of satellite _p=0.0 °, the roll angle θ of satellite _r=0.0 °, radar antenna work view angle theta _l=30.0 °.

Step 2: in conjunction with the positional information R of satellite _os, satellite velocity information V _os, satellite crab angle θ _y, satellite pitching angle theta _p, satellite roll angle θ _r, calculate the transition matrix A between each coordinate system _or, A _re, A _ea.

$A_{\mathrm{or}}=\left[\begin{array}{ccc}1& 0& 0\\ 0& y_1/r_1& x_1/r_1\\ 0& -x_1/r_1& y_1/r_1\end{array}\right]\left[\begin{array}{ccc}{v}_{\mathrm{sx}}/v_1& -v_{\mathrm{sy}}/v_1& 0\\ v_{\mathrm{sy}}/v_1& v_{\mathrm{sx}}/v_1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}{v}_1/v& 0& v_{\mathrm{sz}}/v\\ 0& 1& 0\\ {-v}_{\mathrm{sz}}/v& 0& v_1/v\end{array}\right]---\left(14\right)$

$=\left[\begin{array}{ccc}0.116945& 0.500850& 0.857597\\ 0.075785& -0.865505& 0.495134\\ 0.990243& 0.007089& -0.139173\end{array}\right]$

Wherein: $r_1=\left[\begin{array}{ccc}{v}_1/v& 0& -v_{\mathrm{sz}}/v\\ 0& 1& 0\\ v_{\mathrm{sz}}/v& 0& v_1/v\end{array}\right]\left[\begin{array}{ccc}{v}_{\mathrm{sx}}/v_1& v_{\mathrm{sy}}/v_1& 0\\ -v_{\mathrm{sy}}/v_1& v_{\mathrm{sx}}/v_1& 0\\ 0& 0& 1\end{array}\right]\·(R_{\mathrm{os}}\×V_{\mathrm{os}})=(0,x_1,y_1);$

r ₁＝|r ₁|；

v＝|V _os|；

$v_1=\sqrt{v_{\mathrm{sx}}^2+v_{\mathrm{sy}}^2}.$

$A_{\mathrm{re}}=\left[\begin{array}{ccc}{\cos \mathrm{\θ}}_y& 0& {-\sin \mathrm{\θ}}_y\\ 0& 1& 0\\ {\sin \mathrm{\θ}}_y& 0& {\cos \mathrm{\θ}}_y\end{array}\right]\left[\begin{array}{ccc}{\cos \mathrm{\θ}}_p& {-\sin \mathrm{\θ}}_p& 0\\ {\sin \mathrm{\θ}}_p& {\cos \mathrm{\θ}}_p& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& {\cos \mathrm{\θ}}_r& {-\sin \mathrm{\θ}}_r\\ 0& {\sin \mathrm{\θ}}_r& {\cos \mathrm{\θ}}_r\end{array}\right]$

$=\left[\begin{array}{ccc}0.998& 0& -0.066\\ 0& 1& 0\\ 0.066& 0& 0.998\end{array}\right]$

$A_{\mathrm{ea}}=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos {\mathrm{\θ}}_L& \sin {\mathrm{\θ}}_L\\ 0& -\sin {\mathrm{\θ}}_L& \cos {\mathrm{\θ}}_L\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0.866& 0.500\\ 0& -0.500& 0.866\end{array}\right]$

Step 3: calculate terrestrial beam sensing point and do not rotating the coordinate position in geocentric coordinate system.Consider that controlling antenna wave beam to point point is in the Y-axis of antenna coordinate system, meet following vector [0, R _p, 0], obtain terrestrial beam by vector superposed process and point to point at the coordinate position not rotationally in spherical coordinate system:

$R_{\mathrm{op}}=R_{\mathrm{os}}+R_{o\_\mathrm{sp}}$

$=\left[\begin{array}{c}{R}_{o,\mathrm{sx}}\\ R_{o,\mathrm{sy}}\\ R_{o,\mathrm{sz}}\end{array}\right]+A_{\mathrm{or}}{A}_{\mathrm{re}}{A}_{\mathrm{ea}}\left[\begin{array}{c}0\\ R_p\\ 0\end{array}\right]$

$=\left[\begin{array}{c}{R}_{o,\mathrm{sx}}\\ R_{o,\mathrm{sy}}\\ R_{o,\mathrm{sz}}\end{array}\right]+\left[\begin{array}{ccc}{A}_{\mathrm{oe}\_11}& A_{\mathrm{oe}\_12}& A_{\mathrm{oe}\_13}\\ A_{\mathrm{oe}\_12}& A_{\mathrm{oe}\_22}& A_{\mathrm{oe}\_23}\\ A_{\mathrm{oe}\_13}& A_{\mathrm{oe}\_23}& A_{\mathrm{oe}\_33}\end{array}\right]\left[\begin{array}{c}0\\ R_p\cos {\mathrm{\θ}}_L\\ -R_p\sin {\mathrm{\θ}}_L\end{array}\right]---\left(15\right)$

$=\left[\begin{array}{c}{R}_{o,\mathrm{sx}}\\ R_{o,\mathrm{sy}}\\ R_{o,\mathrm{sz}}\end{array}\right]+\left[\begin{array}{c}{A}_{\mathrm{oe}\_12}\·R_p\cos {\mathrm{\θ}}_L-A_{\mathrm{oe}\_13}\·R_p\sin {\mathrm{\θ}}_L\\ A_{\mathrm{oe}\_22}\·R_p\cos {\mathrm{\θ}}_L-A_{\mathrm{oe}\_23}\·R_p\sin {\mathrm{\θ}}_L\\ A_{\mathrm{oe}\_32}\·R_p\cos {\mathrm{\θ}}_L-A_{\mathrm{oe}\_33}\·R_p\sin {\mathrm{\θ}}_L\end{array}\right]=\left[\begin{array}{c}{R}_{o,\mathrm{nx}}\\ R_{o,\mathrm{ny}}\\ R_{o,\mathrm{nz}}\end{array}\right]=\left[\begin{array}{c}-3499675.184327\\ 5332207.232214\\ 22496.349019\end{array}\right]$

Wherein: n _rrepresent that distance is to sampling number; C represents the light velocity.

Step 4: the orbital tracking determining satellite, forgives semi-major axis of orbit a, orbit inclination i, right ascension of ascending node Ω, eccentric ratio e, argument of perigee ω, very near heart angle θ.In conjunction with the mutual relationship between coordinate conversion matrix, have:

$A_{\mathrm{or}}=\left[\begin{array}{ccc}{a}_{\mathrm{or}}11& a_{\mathrm{or}}21& a_{\mathrm{or}}31\\ a_{\mathrm{or}}12& a_{\mathrm{or}}22& a_{\mathrm{or}}32\\ a_{\mathrm{or}}13& a_{\mathrm{or}}23& a_{\mathrm{or}}33\end{array}\right]$ （16）

$=\left[\begin{array}{ccc}\cos \mathrm{\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}-\sin (\mathrm{\θ}-\mathrm{\γ})& -\cos (\mathrm{\θ}-\mathrm{\γ})& 0\\ \cos (\mathrm{\θ}-\mathrm{\γ})& -\sin (\mathrm{\θ}-\mathrm{\γ})& 0\\ 0& 0& 1\end{array}\right]$

Wherein: A _orin each coefficient a _orij, by satellite position vectors and velocity, obtains according to formula (14); γ represents flight-path angle, is calculated obtain by following formula:

$\tan \mathrm{\γ}=\frac{e\sin \mathrm{\θ}}{1+e\cos \mathrm{\θ}}---\left(17\right)$

Following equation can be obtained in conjunction with the relation between above-mentioned transition matrix:

By solving equation (18), after the orbit inclination i that can determine satellite orbit and right ascension of ascending node Ω, equation (16) can be modified to:

$\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}-\sin (\mathrm{\θ}-\mathrm{\γ})& -\cos (\mathrm{\θ}-\mathrm{\γ})& 0\\ \cos (\mathrm{\θ}-\mathrm{\γ})& -\sin (\mathrm{\θ}-\mathrm{\γ})& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}-\sin (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})& -\cos (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})& 0\\ \cos (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})& -\sin (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})& 0\\ 0& 0& 1\end{array}\right]$

$={\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]}^{-1}{\left[\begin{array}{ccc}\cos \mathrm{\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]}^{-1}\left[\begin{array}{ccc}{a}_{\mathrm{or}}11& a_{\mathrm{or}}21& a_{\mathrm{or}}31\\ a_{\mathrm{or}}12& a_{\mathrm{or}}22& a_{\mathrm{or}}32\\ a_{\mathrm{or}}13& a_{\mathrm{or}}23& a_{\mathrm{or}}33\end{array}\right]---\left(19\right)$

$=\left[\begin{array}{ccc}{b}_{\mathrm{or}}11& b_{\mathrm{or}}21& b_{\mathrm{or}}31\\ b_{\mathrm{or}}12& b_{\mathrm{or}}22& b_{\mathrm{or}}32\\ b_{\mathrm{or}}13& b_{\mathrm{or}}23& b_{\mathrm{or}}33\end{array}\right]=\left[\begin{array}{ccc}0.007159& -0.999974& 0\\ 0.999974& 0.007159& 0\\ 0& 0& 1\end{array}\right]$

Consider that the right side matrix in above formula is known matrix, can calculate accordingly:

$\begin{array}{c}\cos (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})=b_{\mathrm{or}}12\\ \sin (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})=-b_{\mathrm{or}}11\end{array}---\left(20\right)$

Angle (ω+θ-γ)=-0.410184 ° can be calculated by solving above formula.

In conjunction with satellite at the position coordinates R not rotationally in spherical coordinate system _os=(R _{o, sx}, R _{o, sy}, R _{o, sz}) ', can calculate and vow that radius r is:

$r=\sqrt{R_{o,\mathrm{sx}}^2+R_{o,\mathrm{sy}}^2+R_{o,\mathrm{sz}}^2}=7000050.114232m---\left(21\right)$

Position coordinates meets following equation simultaneously:

$r\left[\begin{array}{ccc}\cos \mathrm{\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}\cos \left(\mathrm{\θ}\right)\\ \sin \left(\mathrm{\θ}\right)\\ 0\end{array}\right]=\left[\begin{array}{c}{R}_{o,\mathrm{sx}}\\ R_{o,\mathrm{sy}}\\ R_{o,\mathrm{sz}}\end{array}\right]---\left(22\right)$

Have after bringing arrow radius, orbit inclination and right ascension of ascending node into above-mentioned equation:

$\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}\cos \left(\mathrm{\θ}\right)\\ \sin \left(\mathrm{\θ}\right)\\ 0\end{array}\right]{\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]}^{-1}{\left[\begin{array}{ccc}\cos \mathrm{\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{x}_s\\ y_s\\ z_s\end{array}\right]$ (23)

$=\left[\begin{array}{c}{c}_{\mathrm{or}}1\\ c_{\mathrm{or}}2\\ c_{\mathrm{or}}3\end{array}\right]=\left[\begin{array}{c}\cos (\mathrm{\ω}+\mathrm{\θ})\\ \sin (\mathrm{\ω}+\mathrm{\θ})\\ 0\end{array}\right]$

Following equation can be obtained according to above formula:

$\begin{array}{c}\cos (\mathrm{\ω}+\mathrm{\θ})=c_{\mathrm{or}}1\\ \sin (\mathrm{\ω}+\mathrm{\θ})=c_{\mathrm{or}}2\end{array}---\left(24\right)$

Angle (ω+θ)=-0.467488 ° can be calculated by solving above formula.

Composite type (20) and formula (24), obtain flight-path angle γ=-0.057294 ° of radar system.

Consider that satellite is at the velocity V not rotationally in spherical coordinate system _os=(V _sx, V _sy, V _sz) ', meet following equation relation:

$\left(\begin{array}{c}{V}_{\mathrm{sx}}\\ V_{\mathrm{sy}}\\ V_{\mathrm{sz}}\end{array}\right)=\sqrt{\frac{\mathrm{\μ}}{a(1-e^2)}}{A}_{\mathrm{ov}}\left(\begin{array}{c}-\sin \mathrm{\θ}\\ e+\cos \mathrm{\θ}\\ 0\end{array}\right)---\left(25\right)$

Following equation can be obtained accordingly:

$V_{\mathrm{sx}}^2+V_{\mathrm{sy}}^2+V_{\mathrm{sz}}^2=\frac{\mathrm{\μ}}{a(1-e^2)}[1+e^2+2e\cos \mathrm{\θ}]$

$=\frac{\mathrm{\μ}}{a(1-e^2)}[2(1+e\cos \mathrm{\θ})+e^2-1]$ (26)

$=\frac{2\mathrm{\μ}(1+e\cos \mathrm{\θ})}{a(1-e^2)}+\frac{\mathrm{\μ}}{a(1-e^2)}[e^2-1]$

$=\frac{2\mathrm{\μ}}{r}-\frac{\mathrm{\μ}}{a}=7546.007390$

Utilize equation (26), the semi-major axis a=7000000.0m of track can be obtained.

In conjunction with Kepler's radar equation and flight-path angle accounting equation, following system of equations (27) can be obtained:

$\begin{array}{c}a(1-e^2)=r(1+e\cos \mathrm{\θ})\\ \tan \mathrm{\γ}=\frac{e\sin \mathrm{\θ}}{1+e\cos \mathrm{\θ}}\end{array}---\left(27\right)$

Have after abbreviation is carried out to formula (27):

${\tan }^2\mathrm{\γ}\·[1+2\frac{a(1-e^2)-r}{r}+{\left(\frac{a(1-e^2)-r}{r}\right)}^2]=e^2-{\left(\frac{a(1-e^2)-r}{r}\right)}^2---\left(28\right)$

Eccentric ratio e=0.01 of track can be obtained by solving above-mentioned biquadratic equation group, the computing formula simultaneously orbital eccentricity being brought into flight-path angle can obtain instantaneous very near θ=-90.467488 °, heart angle of radar satellite, and convolution (24) can obtain argument of perigee ω=90.000000 ° of satellite orbit.

Step 5: the transition matrix Α calculating orbit plane coordinate system and do not rotate between geocentric coordinate system _ov.Transform matrix calculations formula orbit inclination, right ascension of ascending node and argument of perigee being brought into orbit plane coordinate system and not rotating between geocentric coordinate system, calculates orbit plane coordinate system and does not rotate the transition matrix A between geocentric coordinate system _ov.

$A_{\mathrm{ov}}=\left[\begin{array}{ccc}0.120527& 0.500000& 0.857597\\ 0.069587& -0.866025& 0.495134\\ 0.990268& 0.000000& -0.139137\end{array}\right]$

Step 6: calculate the instantaneous acceleration vector A not rotating geocentric coordinate system Satellite _osand acceleration change vector

$A_{\mathrm{os}}=\frac{\mathrm{\μ}{(1+e\cos \left(\mathrm{\θ}\right))}^2}{a^2{(1-e^2)}^2}{A}_{\mathrm{ov}}\left[\begin{array}{c}-\cos \left(\mathrm{\θ}\right)\\ -\sin \left(\mathrm{\θ}\right)\\ 0\end{array}\right]=\left[\begin{array}{c}4.075166m/s^2\\ -7.039921m/s^2\\ 0.065725m/s^2\end{array}\right]---\left(29\right)$

${\stackrel{\·}{A}}_{\mathrm{os}}=\frac{\mathrm{\μ}{(1+e\cos \left(\mathrm{\θ}\right))}^3}{a^3{(1-e^2)}^3}\sqrt{\frac{\mathrm{\μ}}{a(1-e^2)}}{A}_{\mathrm{ov}}\left[\begin{array}{c}\sin \left(\mathrm{\θ}\right)+3e\cos \left(\mathrm{\θ}\right)\sin \left(\mathrm{\θ}\right)\\ e(2{\sin }^2\left(\mathrm{\θ}\right)-{\cos }^2\left(\mathrm{\θ}\right))-\cos \left(\mathrm{\θ}\right)\\ 0\end{array}\right]$ (30)

$=\left[\begin{array}{c}-0.001012m/s^3\\ -0.000687m/s^3\\ -0.008683m/s^3\end{array}\right]$

Step 7: calculate and do not rotate in geocentric coordinate system, terrestrial beam points to the velocity V of point _op, acceleration A _opand acceleration change vector

$V_{\mathrm{op}}=\left[\begin{array}{c}{V}_{o\_\mathrm{px}}\\ V_{o\_\mathrm{py}}\\ V_{o\_\mathrm{pz}}\end{array}\right]=\left[\begin{array}{c}-{\mathrm{\ω}}_e{R}_{o,\mathrm{ny}}\\ {\mathrm{\ω}}_e{R}_{o,\mathrm{nx}}\\ 0\end{array}\right]=\left[\begin{array}{c}-387.768774m/s\\ -254.503379m/s\\ 0.000000m/s\end{array}\right]---\left(31\right)$

$A_{\mathrm{op}}=\left[\begin{array}{c}{A}_{o\_\mathrm{px}}\\ A_{o\_\mathrm{py}}\\ A_{o\_\mathrm{pz}}\end{array}\right]=\left[\begin{array}{c}-{\mathrm{\ω}}_e^2{R}_{o,\mathrm{nx}}\\ -{\mathrm{\ω}}_e^2{R}_{o,\mathrm{ny}}\\ 0\end{array}\right]=\left[\begin{array}{c}0.018508m/s^2\\ -0.028199m/s^2\\ 0.000000m/s^2\end{array}\right]---\left(32\right)$

${\stackrel{\·}{A}}_{\mathrm{op}}=\left[\begin{array}{c}{\stackrel{\·}{A}}_{o\_\mathrm{px}}\\ {\stackrel{\·}{A}}_{o\_\mathrm{py}}\\ {\stackrel{\·}{A}}_{o\_\mathrm{pz}}\end{array}\right]=\left[\begin{array}{c}{\mathrm{\ω}}_e^3{R}_{o,\mathrm{ny}}\\ -{\mathrm{\ω}}_e^3{R}_{o,\mathrm{nx}}\\ 0\end{array}\right]=\left[\begin{array}{c}2.050711e-006m/s^3\\ 1.345938e-006m/s^3\\ 0.000000m/s^3\end{array}\right]---\left(33\right)$

Step 8: calculate Relative position vector R _{o_sp}, relative velocity vector V _{o_sp}, relative acceleration vector A _{o_sp}with relative acceleration rate of change vector

$R_{o\_\mathrm{sp}}=R_{\mathrm{os}}+R_{\mathrm{op}}=\left[\begin{array}{c}-7117.202766m\\ 725837.823947m\\ -79054.749337m\end{array}\right]---\left(34\right)$

$V_{o\_\mathrm{sp}}=V_{\mathrm{os}}+V_{\mathrm{op}}=\left[\begin{array}{c}1270.234842m/s\\ 826.375982m/s\\ 7472.378664m/s\end{array}\right]---\left(35\right)$

$A_{o\_\mathrm{sp}}=A_{\mathrm{os}}+A_{\mathrm{op}}=\left[\begin{array}{c}4.056658m/s^2\\ -7.011721m/s^2\\ 0.065725m/s^2\end{array}\right]---\left(36\right)$

${\stackrel{\·}{A}}_{o\_\mathrm{sp}}={\stackrel{\·}{A}}_{\mathrm{os}}+{\stackrel{\·}{A}}_{\mathrm{op}}=\left[\begin{array}{c}-0.001014m/s^3\\ -0.000689m/s^3\\ -0.008683m/s^3\end{array}\right]---\left(37\right)$

Step 9: in conjunction with Relative position vector, relative velocity vector, relative acceleration vector, relative acceleration rate of change vector, estimation echoed signal Doppler parameter.

$f_1=\frac{2}{\mathrm{\λ}}\frac{V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}}}{\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}=4.328164\mathrm{Hz}---\left(38\right)$

$f_2=\frac{2}{\mathrm{\λ}}(\frac{V_{o\_\mathrm{spx}}{V}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{V}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{V}_{o\_\mathrm{spz}}}{\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}+\frac{A_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+A_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+A_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}}}{\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}$ （39）

$-\frac{{(V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})}^2}{{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{3/2}})=4839.947118\mathrm{Hz}/s$

$f_3=\frac{2}{\mathrm{\λ}}(\frac{{\stackrel{\·}{A}}_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+{\stackrel{\·}{A}}_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+{\stackrel{\·}{A}}_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}}}{6\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}+\frac{A_{o\_\mathrm{spx}}{V}_{o\_\mathrm{spx}}+A_{o\_\mathrm{spy}}{V}_{o\_\mathrm{spy}}+A_{o\_\mathrm{spz}}{V}_{o\_\mathrm{spz}}}{2\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}$

$-\frac{(A_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+A_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+A_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})(V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})}{2{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{3/2}}$ （40）

$-\frac{(V_{o\_\mathrm{spx}}{V}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{V}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{V}_{o\_\mathrm{spz}})(V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})}{2{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{3/2}}$

$-\frac{{(V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})}^3}{2{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{5/2}})=0.003546\mathrm{Hz}/s^2$

$f_4=\frac{2}{\mathrm{\λ}}(\frac{{\stackrel{\·}{A}}_{o\_\mathrm{spx}}{V}_{o\_\mathrm{spx}}+{\stackrel{\·}{A}}_{o\_\mathrm{spy}}{V}_{o\_\mathrm{spy}}+{\stackrel{\·}{A}}_{o\_\mathrm{spz}}{V}_{o\_\mathrm{spz}}}{6\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}+\frac{A_{o\_\mathrm{spx}}{A}_{o\_\mathrm{spx}}+A_{o\_\mathrm{spy}}{A}_{o\_\mathrm{spy}}+A_{o\_\mathrm{spz}}{V}_{o\_\mathrm{spz}}}{8\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}$ （41）

$-\frac{{\left(\mathrm{\λ}{f}_2\right)}^2}{32{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{1/2}}-\frac{{\left(\mathrm{\λ}{f}_1{f}_3\right)}^2}{24{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{1/2}})=0.051908\mathrm{Hz}/s^3$

Table 1 comparative analysis directly calculates Doppler parameter and parameter is generally strangled by oblique distance matching institute

From simulation result, calculate the Doppler parameter that obtains and the Doppler parameter error directly obtained based on distance matching all after radix point six based on ephemeris parameter, there is very high estimation precision, the demand of high precision imaging processing can be met.

Claims (1)

1., based on the satellite-borne SAR high-order Doppler parameter evaluation method of almanac data, comprise the steps:

Step 1: according to echo data Store form, read the auxiliary data corresponding to echo data, auxiliary data comprises: wavefront oblique distance R _min, signal sampling rate f _s, pulse repetition rate f _pRF, transmit signal pulse width τ, satellite position vector R _os=(R _{o, sx}, R _{o, sy}, R _{o, sz}) ', the velocity V of satellite _os=(V _sx, V _sy, V _sz) ', the crab angle θ of satellite _y, satellite pitching angle theta _p, satellite roll angle θ _r, radar antenna work view angle theta _l;

Step 2: in conjunction with the position vector R of satellite _os=(R _{o, sx}, R _{o, sy}, R _{o, sz}) ', the velocity V of satellite _os=(V _sx, V _sy, V _sz) ', the crab angle θ of satellite _y, satellite pitching angle theta _p, satellite roll angle θ _r, the geocentric coordinate that calculating is not rotated is tied to the transition matrix A between satellite platform coordinate system _or, satellite platform coordinate is tied to transition matrix A between satellite health coordinate system _re, satellite health coordinate is tied to transition matrix A between antenna coordinate system _ea;

$A_{\mathrm{or}}=\left[\begin{array}{ccc}1& 0& 0\\ 0& y_1/r_1& x_1/r_1\\ 0& -x_1/r_1& y_1/r_1\end{array}\right]\left[\begin{array}{ccc}{v}_{\mathrm{sx}}/v_1& -v_{\mathrm{sy}}/v_1& 0\\ v_{\mathrm{sy}}/v_1& v_{\mathrm{sx}}/v_1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}{v}_1/v& 0& v_{\mathrm{sz}}/v\\ 0& 1& 0\\ -v_{\mathrm{sz}}/v& 0& v_1/v\end{array}\right]---\left(14\right)$

Wherein: $r_1=\left[\begin{array}{ccc}{v}_1/v& 0& -v_{\mathrm{sz}}/v\\ 0& 1& 0\\ v_{\mathrm{sz}}/v& 0& v_1/v\end{array}\right]\left[\begin{array}{ccc}{v}_{\mathrm{sx}}/v_1& v_{\mathrm{sy}}/v_1& 0\\ -v_{\mathrm{sy}}/v_1& v_{\mathrm{sx}}/v_1& 0\\ 0& 0& 1\end{array}\right]\·(R_{\mathrm{os}}\×V_{\mathrm{os}})=(0,x_1,y_1);$

r ₁＝|r ₁|；

v＝|V _os|；

$v_1=\sqrt{v_{\mathrm{sx}}^2+v_{\mathrm{sy}}^2};$

$A_{\mathrm{re}}=\left[\begin{array}{ccc}\cos {\mathrm{\θ}}_y& 0& -\sin {\mathrm{\θ}}_y\\ 0& 1& 0\\ \sin {\mathrm{\θ}}_y& 0& \cos {\mathrm{\θ}}_y\end{array}\right]\left[\begin{array}{ccc}\cos {\mathrm{\θ}}_p& -\sin {\mathrm{\θ}}_p& 0\\ \sin {\mathrm{\θ}}_p& \cos {\mathrm{\θ}}_p& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos {\mathrm{\θ}}_r& -\sin {\mathrm{\θ}}_r\\ 0& \sin {\mathrm{\θ}}_r& \cos {\mathrm{\θ}}_r\end{array}\right]$

$A_{\mathrm{ea}}=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos {\mathrm{\θ}}_L& \sin {\mathrm{\θ}}_L\\ 0& -\sin {\mathrm{\θ}}_L& \cos {\mathrm{\θ}}_L\end{array}\right]$

Step 3: after calculating the mutual transition matrix between each coordinate system according to step 2, further calculating do not rotate geocentric coordinate system Satellite and terrestrial beam point to a little between distance vector, distance vector between pointing to a little in conjunction with the position vector of satellite and satellite and terrestrial beam, obtains terrestrial beam by vector addition process and points to point and do not rotating the position vector in geocentric coordinate system; Consider that controlling antenna wave beam to point point is in the Y-axis of antenna coordinate system, meet following vector [0, R _p, 0], obtain terrestrial beam by vector superposed process and point to point at the coordinate position not rotationally in spherical coordinate system:

$\begin{array}{c}{R}_{\mathrm{op}}=R_{\mathrm{os}}+R_{o\_\mathrm{sp}}\\ =\left[\begin{array}{c}{R}_{o,\mathrm{sx}}\\ R_{o,\mathrm{sy}}\\ R_{o,\mathrm{sz}}\end{array}\right]+A_{\mathrm{or}}{A}_{\mathrm{re}}{A}_{\mathrm{ea}}\left[\begin{array}{c}0\\ R_p\\ 0\end{array}\right]\\ =\left[\begin{array}{c}{R}_{o,\mathrm{sx}}\\ R_{o,\mathrm{sy}}\\ R_{o,\mathrm{sz}}\end{array}\right]+\left[\begin{array}{ccc}{A}_{\mathrm{oe}\_11}& A_{\mathrm{oe}\_12}& A_{\mathrm{oe}\_13}\\ A_{\mathrm{oe}\_12}& A_{\mathrm{oe}\_22}& A_{\mathrm{oe}\_23}\\ A_{\mathrm{oe}\_13}& A_{\mathrm{oe}\_23}& A_{\mathrm{oe}\_33}\end{array}\right]\left[\begin{array}{c}0\\ R_p\cos {\mathrm{\θ}}_L\\ -R_p\sin {\mathrm{\θ}}_L\end{array}\right]\\ =\begin{array}{c}\left[\begin{array}{c}{R}_{o,\mathrm{sx}}\\ R_{o,\mathrm{sy}}\\ R_{o,\mathrm{sz}}\end{array}\right]\left[\begin{array}{c}{A}_{\mathrm{oe}\_12}\·R_p\cos {\mathrm{\θ}}_L-A_{\mathrm{oe}\_13}\·R_p\sin {\mathrm{\θ}}_L\\ A_{\mathrm{oe}\_22}\·R_p\cos {\mathrm{\θ}}_L-A_{\mathrm{oe}\_23}\·R_p\sin {\mathrm{\θ}}_L\\ A_{\mathrm{oe}\_32}\·R_p\cos {\mathrm{\θ}}_L{-A}_{\mathrm{oe}\_33}\·R_p\sin {\mathrm{\θ}}_L\end{array}\right]\end{array}\left[\begin{array}{c}{R}_{o,\mathrm{nx}}\\ R_{o,\mathrm{ny}}\\ R_{o,\mathrm{nz}}\end{array}\right]\end{array}---\left(15\right)$

Wherein: R _{o_sp}represent the distance vector between not rotating satellite and terrestrial beam in geocentric coordinate system points to a little, n _rrepresent that distance is to sampling number; C represents the light velocity;

Step 4: the orbital tracking determining satellite, forgives semi-major axis of orbit a, orbit inclination i, right ascension of ascending node Ω, eccentric ratio e, argument of perigee ω, and very near heart angle θ, in conjunction with the mutual relationship between coordinate conversion matrix, has:

$\begin{array}{c}{A}_{\mathrm{or}}=\left[\begin{array}{ccc}1& 0& 0\\ 0& y_1/r_1& x_1/r_1\\ 0& -x_1/r_1& y_1/r_1\end{array}\right]\left[\begin{array}{ccc}{v}_{\mathrm{sx}}/v_1& -v_{\mathrm{sy}}/v_1& 0\\ v_{\mathrm{sy}}/v_1& v_{\mathrm{sx}}/v_1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}{v}_1/v& 0& v_{\mathrm{sz}}/v\\ 0& 1& 0\\ -v_{\mathrm{sz}}/v& 0& v_1/v\end{array}\right]=\left[\begin{array}{ccc}{a}_{\mathrm{or}}11& a_{\mathrm{or}}21& a_{\mathrm{or}}31\\ a_{\mathrm{or}}12& a_{\mathrm{or}}22& a_{\mathrm{or}}32\\ a_{\mathrm{or}}13& a_{\mathrm{or}}23& a_{\mathrm{or}}33\end{array}\right]\\ =\left[\begin{array}{ccc}\mathrm{soc\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}-\sin (\mathrm{\θ}-\mathrm{\γ})& -\cos (\mathrm{\θ}-\mathrm{\γ})& 0\\ \cos (\mathrm{\θ}-\mathrm{\γ})& -\sin (\mathrm{\θ}-\mathrm{\γ})& 0\\ 0& 0& 1\end{array}\right]\end{array}---\left(16\right)$

Wherein: A _orin each coefficient a _orij is obtained by satellite position vectors and velocity, and i shows that this coefficient is the i-th row coefficient in matrix, and j shows that this coefficient is the jth row coefficient in matrix, 1≤i≤3,1≤j≤3; γ represents flight-path angle, is calculated obtain by following formula:

$\tan \mathrm{\γ}=\frac{e\sin \mathrm{\θ}}{1+e\cos \mathrm{\θ}}---\left(17\right)$

Following equation is obtained in conjunction with the relation between above-mentioned transition matrix:

$\begin{array}{c}\cos i=a_{\mathrm{or}}33\\ \sin i\·\sin \mathrm{\Ω}=a_{\mathrm{or}}31\end{array}---\left(18\right)$

By solving equation (18), after the orbit inclination i determining satellite orbit and right ascension of ascending node Ω, equation (16) is modified to:

$\begin{array}{c}\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}-\sin (\mathrm{\θ}-\mathrm{\γ})& -\cos (\mathrm{\θ}-\mathrm{\γ})& 0\\ \cos (\mathrm{\θ}-\mathrm{\γ})& -\sin (\mathrm{\θ}-\mathrm{\γ})& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}-\sin (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})& -\cos (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})& 0\\ \cos (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})& -\sin (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})& 0\\ 0& 0& 1\end{array}\right]\\ ={\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]}^{-1}{\left[\begin{array}{ccc}\cos \mathrm{\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]}^{-1}\left[\begin{array}{ccc}{a}_{\mathrm{or}}11& a_{\mathrm{or}}21& a_{\mathrm{or}}31\\ a_{\mathrm{or}}12& a_{\mathrm{oa}}22& a_{\mathrm{or}}32\\ a_{\mathrm{or}}13& a_{\mathrm{or}}23& a_{\mathrm{or}}33\end{array}\right]\\ =\left[\begin{array}{ccc}{b}_{\mathrm{or}}11& b_{\mathrm{or}}21& b_{\mathrm{or}}31\\ b_{\mathrm{or}}12& b_{\mathrm{or}}22& b_{\mathrm{or}}32\\ b_{\mathrm{or}}13& b_{\mathrm{or}}23& b_{\mathrm{or}}33\end{array}\right]\end{array}---\left(19\right)$

Wherein, the right side matrix in above formula is known matrix, calculates accordingly:

$\begin{array}{c}\cos (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})=b_{\mathrm{or}}12\\ \sin (\mathrm{\ω}+\mathrm{\θ}-\mathrm{\γ})=-b_{\mathrm{or}}11\end{array}---\left(20\right)$

Angle (ω+θ-γ) can be calculated by solving above formula;

In conjunction with satellite at the position coordinates R not rotationally in spherical coordinate system _os=(R _{o, sx}, R _{o, sy}, R _{o, sz}) ', calculate and vow that radius r is:

$r=\sqrt{R_{o,\mathrm{sx}}^2+R_{o,\mathrm{sy}}^2+R_{o,\mathrm{sz}}^2}---\left(21\right)$

Position coordinates meets following equation simultaneously:

$r\left[\begin{array}{ccc}\cos \mathrm{\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}\cos \left(\mathrm{\θ}\right)\\ \sin \left(\mathrm{\θ}\right)\\ 0\end{array}\right]=\left[\begin{array}{c}{R}_{o,\mathrm{sx}}\\ R_{o,\mathrm{sy}}\\ R_{o,\mathrm{sz}}\end{array}\right]---\left(22\right)$

Have after bringing arrow radius, orbit inclination and right ascension of ascending node into above-mentioned equation:

$\begin{array}{c}\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}\cos \left(\mathrm{\θ}\right)\\ \sin \left(\mathrm{\θ}\right)\\ 0\end{array}\right]={\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]}^{-1}{\left[\begin{array}{ccc}\cos \mathrm{\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{x}_s\\ y_s\\ z_s\end{array}\right]\\ =\begin{array}{c}\left[\begin{array}{c}{c}_{\mathrm{or}}1\\ c_{\mathrm{or}}2\\ c_{\mathrm{or}}3\end{array}\right]=\left[\begin{array}{c}\cos (\mathrm{\ω}+\mathrm{\θ})\\ \sin (\mathrm{\ω}+\mathrm{\θ})\\ 0\end{array}\right]\end{array}\end{array}---\left(20\right)$

Following equation is obtained according to above formula:

$\begin{array}{c}\cos (\mathrm{\ω}+\mathrm{\θ})=c_{\mathrm{or}}1\\ \sin (\mathrm{\ω}+\mathrm{\θ})=c_{\mathrm{or}}2\end{array}---\left(24\right)$

Angle (ω+θ) can be calculated by solving above formula;

Composite type (20) and formula (24), obtain the flight-path angle γ of radar system;

Consider that satellite is at the velocity V not rotationally in spherical coordinate system _os=(V _sx, V _sy, V _sz) ', meet following equation relation:

$\left(\begin{array}{c}{V}_{\mathrm{sx}}\\ V_{\mathrm{sy}}\\ V_{\mathrm{sz}}\end{array}\right)=\sqrt{\frac{\mathrm{\μ}}{a(1-e^2)}}{A}_{\mathrm{ov}}\left(\begin{array}{c}-\sin \mathrm{\θ}\\ e+\cos \mathrm{\θ}\\ 0\end{array}\right)---\left(25\right)$

Obtain following equation accordingly:

$\begin{array}{c}{V}_{\mathrm{sx}}^2+V_{\mathrm{sy}}^2+V_{\mathrm{sz}}^2=\frac{\mathrm{\μ}}{a(1-e^2)}[1+e^2+2e\cos \mathrm{\θ}]\\ =\frac{\mathrm{\μ}}{a(1-e^2)}[2(1+\cos \mathrm{\θ})+e^2-1]\\ =\frac{2\mathrm{\μ}(1+e\cos \mathrm{\θ})}{a(1-e^2)}+\frac{\mathrm{\μ}}{a(1-e^2)}[e^2-1]\\ =\frac{2\mathrm{\μ}}{r}-\frac{\mathrm{\μ}}{a}\end{array}---\left(26\right)$

Utilize equation (26), the semi-major axis a of track can be obtained;

In conjunction with Kepler's radar equation and flight-path angle accounting equation, obtain following system of equations (27):

$\begin{array}{c}a(1-e^2)=r(1+e\cos \mathrm{\θ})\\ \tan \mathrm{\γ}=\frac{e\sin \mathrm{\θ}}{1+e\cos \mathrm{\θ}}\end{array}---\left(27\right)$

Have after abbreviation is carried out to formula (27):

${\tan }^2\mathrm{\γ}\·[1+2\frac{a(1-e^2)-r}{r}+{\left(\frac{a(1-e^2)-r}{r}\right)}^2]=e^2-{\left(\frac{a(1-e^2)-r}{r}\right)}^2---\left(28\right)$

Obtain the eccentric ratio e of track by solving above-mentioned biquadratic equation group, the computing formula simultaneously orbital eccentricity being brought into flight-path angle obtains the instantaneous very near heart angle θ of radar satellite, and convolution (24) obtains the argument of perigee ω of satellite orbit;

Step 5: the transition matrix Α calculating orbit plane coordinate system and do not rotate between geocentric coordinate system _ov, transform matrix calculations formula orbit inclination, right ascension of ascending node and argument of perigee being brought into orbit plane coordinate system and not rotating between geocentric coordinate system, calculates orbit plane coordinate system and does not rotate the transition matrix A between geocentric coordinate system _ov;

$A_{\mathrm{ov}}=\left[\begin{array}{ccc}\cos \mathrm{\Ω}& -\sin \mathrm{\Ω}& 0\\ \sin \mathrm{\Ω}& \cos \mathrm{\Ω}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos i& -\sin i\\ 0& \sin i& \cos i\end{array}\right]\left[\begin{array}{ccc}\cos \mathrm{\ω}& -\sin \mathrm{\ω}& 0\\ \sin \mathrm{\ω}& \cos \mathrm{\ω}& 0\\ 0& 0& 1\end{array}\right]---\left(4\right)$

Step 6: calculate the instantaneous acceleration vector A not rotating geocentric coordinate system Satellite _osand acceleration change vector

$A_{\mathrm{os}}=\frac{\mathrm{\μ}{(1+e\cos \left(\mathrm{\θ}\right))}^2}{a^2{(1-e^2)}^2}{A}_{\mathrm{ov}}\left[\begin{array}{c}-\cos \left(\mathrm{\θ}\right)\\ -\sin \left(\mathrm{\θ}\right)\\ 0\end{array}\right]---\left(29\right)$

${\stackrel{\·}{A}}_{\mathrm{os}}=\frac{\mathrm{\μ}{(1+e\cos \left(\mathrm{\θ}\right))}^3}{a^3{(1-e^2)}^3}\sqrt{\frac{\mathrm{\μ}}{a(1-e^2)}}{A}_{\mathrm{ov}}\left[\begin{array}{c}\sin \left(\mathrm{\θ}\right)+3e\cos \left(\mathrm{\θ}\right)\sin \left(\mathrm{\θ}\right)\\ e(2{\sin }^2\left(\mathrm{\θ}\right)-{\cos }^2\left(\mathrm{\θ}\right))-\cos \left(\mathrm{\θ}\right)\\ 0\end{array}\right]---\left(30\right)$

Step 7: calculate and do not rotate in geocentric coordinate system, terrestrial beam points to the velocity V of point _op, acceleration A _opand acceleration change vector

$V_{\mathrm{op}}=\left(\begin{array}{c}{V}_{o\_\mathrm{px}}\\ V_{o\_\mathrm{py}}\\ V_{o\_\mathrm{pz}}\end{array}\right)=\left(\begin{array}{c}-{\mathrm{\ω}}_e{R}_{o,\mathrm{ny}}\\ {\mathrm{\ω}}_e{R}_{o,\mathrm{nx}}\\ 0\end{array}\right)---\left(31\right)$

$A_{\mathrm{op}}=\left(\begin{array}{c}{A}_{o\_\mathrm{px}}\\ A_{o\_\mathrm{py}}\\ A_{o\_\mathrm{pz}}\end{array}\right)=\left(\begin{array}{c}-{\mathrm{\ω}}_e^2{R}_{o,\mathrm{nx}}\\ -{\mathrm{\ω}}_e^2{R}_{o,\mathrm{ny}}\\ 0\end{array}\right)---\left(32\right)$

${\stackrel{\·}{A}}_{\mathrm{op}}=\left(\begin{array}{c}{\stackrel{\·}{A}}_{o\_\mathrm{px}}\\ {\stackrel{\·}{A}}_{o\_\mathrm{py}}\\ {\stackrel{\·}{A}}_{o\_\mathrm{pz}}\end{array}\right)=\left(\begin{array}{c}-{\mathrm{\ω}}_e^2{R}_{o,\mathrm{ny}}\\ -{\mathrm{\ω}}_e^2{R}_{o,\mathrm{nx}}\\ 0\end{array}\right)---\left(33\right)$

Step 8: calculate distance vector R _{o_sp}, relative velocity vector V _{o_sp}, relative acceleration vector A _{o_sp}with relative acceleration rate of change vector

R _{o_sp}＝R _os+R _op(34)

V _{o_sp}＝V _os+V _op(35)

A _{o_sp}＝A _os+A _op(36)

${\stackrel{\·}{A}}_{o\_\mathrm{sp}}={\stackrel{\·}{A}}_{\mathrm{os}}+{\stackrel{\·}{A}}_{\mathrm{op}}---\left(37\right)$

Step 9: in conjunction with Relative position vector, relative velocity vector, relative acceleration vector, relative acceleration rate of change vector, estimation echoed signal Doppler parameter;

$f_1=\frac{2}{\mathrm{\λ}}\frac{V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}}}{\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}---\left(\text{38}\right)$

$\begin{array}{c}{f}_2=\frac{2}{\mathrm{\λ}}(\frac{V_{o\_\mathrm{spx}}{V}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{V}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{V}_{o\_\mathrm{spz}}}{\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}+\frac{A_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+A_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+A_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}}}{\sqrt{R_{o\_\mathrm{spx}}^2+R_{\mathrm{opy}}^2+R_{o\_\mathrm{spz}}^2}}\\ -\frac{{(V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})}^2}{{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{3/2}})\end{array}---\left(39\right)$

$\begin{array}{c}{f}_3=\frac{2}{\mathrm{\λ}}(\frac{{\stackrel{\·}{A}}_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+{\stackrel{\·}{A}}_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+{\stackrel{\·}{A}}_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}}}{6\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}+\frac{A_{o\_\mathrm{spx}}{V}_{o\_\mathrm{spx}}+A_{o\_\mathrm{spy}}{V}_{o\_\mathrm{spy}}+A_{o\_\mathrm{spz}}{V}_{o\_\mathrm{spz}}}{2\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}\\ -\frac{(A_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+A_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+A_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})(V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})}{2{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{3/2}}\\ -\frac{(V_{o\_\mathrm{spx}}{V}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{V}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{V}_{o\_\mathrm{spz}})(V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})}{2{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{3/2}}\\ -\frac{{(V_{o\_\mathrm{spx}}{R}_{o\_\mathrm{spx}}+V_{o\_\mathrm{spy}}{R}_{o\_\mathrm{spy}}+V_{o\_\mathrm{spz}}{R}_{o\_\mathrm{spz}})}^3}{2{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{5/2}})\end{array}---\left(40\right)$

$\begin{array}{c}{f}_4=\frac{2}{\mathrm{\λ}}(\frac{{\stackrel{\·}{A}}_{o\_\mathrm{spx}}{V}_{o\_\mathrm{spx}}+{\stackrel{\·}{A}}_{o\_\mathrm{spy}}{V}_{o\_\mathrm{spy}}+{\stackrel{\·}{A}}_{o\_\mathrm{spz}}{V}_{o\_\mathrm{spz}}}{6\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}+\frac{A_{o\_\mathrm{spx}}{A}_{o\_\mathrm{spx}}+A_{o\_\mathrm{spy}}{A}_{o\_\mathrm{spy}}+A_{o\_\mathrm{spz}}{A}_{o\_\mathrm{spz}}}{8\sqrt{R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2}}\\ -\frac{{\left(\mathrm{\λ}{f}_2\right)}^2}{32{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{1/2}}-\frac{{\left({\mathrm{\λf}}_1{f}_3\right)}^3}{24{(R_{o\_\mathrm{spx}}^2+R_{o\_\mathrm{spy}}^2+R_{o\_\mathrm{spz}}^2)}^{1/2}})\end{array}---\left(41\right)$

Composite type (38) ~ (41) obtain the quadravalence Doppler parameter of echoed signal, and then obtain satellite-borne SAR high-order Doppler parameter, are applied to Space-borne SAR Imaging process.