Research Article Joint Two-Dimensional Ambiguity Resolving ...

Research ArticleJoint Two-Dimensional Ambiguity Resolving Based onSpace-Time Filtering for MIMO-SAR

Ping-ping Huang1 Hua-sheng Li1 Sheng Zhang2 and Guang-Cai Sun2

1 College of Information Engineering Inner Mongolia University of Technology Hohhot 010051 China2The National Key Laboratory of Radar Signal Processing Xidian University Xirsquoan 710071 China

Correspondence should be addressed to Ping-ping Huang cimhwangpp163com

Received 16 March 2014 Revised 17 June 2014 Accepted 1 July 2014 Published 12 August 2014

Academic Editor Shengqi Zhu

Copyright copy 2014 Ping-ping Huang et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

In order to resolve the range and azimuth ambiguity in MIMO-SAR imaging this paper proposed a joint two-dimensionalambiguity resolving method based on space-time filtering which according to the ambiguity in different subscene is relativeto different nearest echo range and different azimuth time is relative to different instantaneous incidence angle to make two-dimensional space-time steering vector for resolving ambiguity of the echo which also can finish the imaging processing at thesame time Simulation results validate the proposed processing approach for the MIMO-SAR system

1 Introduction

Multiple-input and multiple-output spaceborne syntheticaperture radar (MIMO-SAR) as a high resolution and wideswath microwave remote sense system [1ndash5] is proved to bean extremely useful surveillance tool for moving target indi-cation wide swath imaging and three-dimensional imaging

ScanSAR also can obtain wide unambiguous swath cov-erage at the cost of a degraded azimuth resolution [6]Multichannel line arrays can be arranged by range or azimuth[7] for the range multichannel can obtain wide swathcoverage utilizing digital beamforming to resolve rangeambiguity but there is still blind zone [8 9] With theemergence of techniques such as displaced phase center [10]and multiple aperture reconstruction [11 12] a number ofdisplaced azimuth subapertures are used to receive echoesin one pulse repetition interval (PRI) thereby increasingthe effective azimuth sampling rate without changing thepulse repetition frequency (PRF) In this way one can lowerthe requirement on PRF to achieve wider swath imagingwithout impairing azimuth resolution in other words onecan broaden the Doppler spectrum to obtain finer azimuthresolution without reducing the swath coverage

To obtain wide swath coverage and high resolution onecan utilize two-dimensional array to transmit and receive

SAR signals and then resolve the range and azimuth ambi-guity [13] However this method makes range point targetdownlooking angle as range steer vector in approximate treat-ment and does not consider the changing of the downlookingangle with the azimuth time

In this paper a joint two-dimensional ambiguity resolv-ingmethod forMIMO-SAR is proposed which uses deferenttargets relative to deferent range as steer vector which isindependent of the downlooking angle Firstly the azimuthambiguity is resolved Then the range ambiguity resolvingand azimuth match filtering are performed instantaneouslyto get target image which makes the imaging algorithmssimpler

This paper is organized as follows Section 2 reviews thesignal model of MIMO-SAR The two-dimensional ambigu-ity resolving algorithm is analyzed in Section 3 Section 4gives the simulations and validates the proposed processingapproach for the MIMO-SAR system Finally Section 5concludes this paper

2 MIMO-SAR System Mode and Signal Model

Figure 1 presents the geometry of a multichannel SAR Itsupposed that the whole antenna aperture was divided into

Hindawi Publishing CorporationInternational Journal of Antennas and PropagationVolume 2014 Article ID 452141 8 pageshttpdxdoiorg1011552014452141

2 International Journal of Antennas and Propagation

x

y

z

H

d2

d1

Figure 1 MIMO-SAR model

119875 times 119876 subapertures where 119875 is the number of azimuthsubapertures and its length is 119889

1119876 is the number of azimuth

subapertures and its height is 1198892 We denoted the pth column

and the qth row as 119883119901= 1198830+ (119901 minus 1)119889

1 119901 = 1 119875

with 1198830the azimuth beginning coordinate (119883

119901 119884119902 119885) and

119884119902= 1198840+ (119902minus1)119889

2 119902 = 1 119876 with 119884

0the range beginning

coordinateUse the subcaptures on the same column range dimension

transmit signals at the same time through adjusting phaseweighting to control the wave beam the center of transmitwave beam is 1198841015840

119902= (119884119902+ 1198841199020)2 119884

1199020= 1198840+ 1198892(119876 minus 1)2 and

the whole antenna receives the echo signal According to theprinciple of equivalent phase center this mode can be knownas each equivalent phase center transmitting and receivingitself The equivalent phase centers located at (1198831015840

119901 1198841015840

119901 119885)

where1198831015840119901= (1198830+119883119901)2 and 1198841015840

119901= (119884119902+1198841199020)2 The deferent

range subapertures light the deferent subswaths from far tonearby adjusting phaseweightingsThe ellipses in Figure 1 arethe subswathes

Assume that during the Lth pulse period the first rangedimension subapertures transmit signal in the same timeand make the wave beam light the Lth subswath throughcontrolling phase weighting The transmitting signal can beexpressed as follows

119904 (119905119896) =

sin (1205871198761198892(sin 120579119897minus sin 120579) 2120582)

sin (1205871198892(sin 120579119897minus sin 120579) 2120582)

times rect((119905119896minus Δ119879)

119879119901119897

)

times exp (1198952120587119891119888(119905 minus Δ119879

119897) + 119895120587120574(119905

119896minus Δ119879119897)2

)

(1)

where 120579119897is the downlooking angle of the center line of

lth subswath 119891119888is the carrier frequency 120582 = 119888119891

119888is the

wavelength Δ119879119897= 1198791199011+ sdot sdot sdot + 119879

119901119897minus1is the delay time of the

lth subpulse 119905 = 119905119896+119898119879119903is the whole time 119905

119896is the fast time

119898 is a integer119879119903is pulse repetition interval and 120574 is the chirp

rate

The baseband signal received by the pth row and the qthcolumn subaperture was

119904119901119902(119905119896 119905119898) =

119871

sum

119897=1

119860 (120579119897) rect(

(119905119896minus Δ119879119897minus 2119877119901119902119897

(119905119898) 119888)

119879119901119897

)

times rect((1198831015840

119901+ 119909 minus 119909

119897)

119871119886

)

times exp(1198952120587119891119888(minusΔ119879

119897minus

2119877119901119902119897

(119905119898)

119888))

times exp(119895120587120574(119905119896minus Δ119879119897minus

2119877119901119902119897

(119905119898)

119888)

2

)

(2)

where 119860(120579119897) = sin(120587119876119889

2(sin 120579119897minus sin 120579)2120582) sin(120587119889

2(sin 120579119897minus

sin 120579)2120582) 119905119898

= 119896119879119903is slow time 119896 is a integer and

the value scope of 119897 is from 1 to 119871 corresponding to theswath from far to near 119909 = V119905

119898 V is the flat velocity

and 119871119886is the length of synthetic aperture 119877

119901119902119897(119905119898) =

radic(1198831015840119901+ V119905119898minus 119909119897)2+ (1198841015840119901minus 119910119897)2+ 1198672 is the instantaneous

slant range (119909119897 119910119897 119911119897) is the coordinates of one scatter point

in the Lth scene and119867 = 1198850minus119911119897 Assume that scene is plane

119911119897= 0Performing the range match filtering to (2) we can gain

119904119901119902(119905119896 119905119898) asymp

119871

sum

119897=1

1198601015840(120579119897) sin 119888 (119905

119896minus Δ119879119897minus2119877119897(119905119898)

119888)

times rect(1198831015840

119901+ 119909 minus 119909

119897

119871119886

) exp (minus1198954120587120582119877119901119902119897

(119905119898))

(3)

where 1198601015840(120579119897) = 119860(120579

119897) exp(minus1198952120587119891

119888Δ119879119897) 119877119897(119905119898) =


distance from the scatter point on the lth subswath to(1198830 1198840 1198850)

As we know theMIMO-SAR can be considered as a spacesignal sampling which should satisfy the sampling equationas follows

119889 le120582

2 (4)

where 119889 denotes the position change of antenna element and120582 denotes the wavelength of the signal

So the time delay should be

119879119889=119889

119888le120582

2119888=

1

2119891119888

(5)

where 119891119888denotes the carrier frequency

International Journal of Antennas and Propagation 3

fa

PRF2

minusPRF2

cos120593

(a) The Doppler is not ambiguity

PRF2

minusPRF2Band width

120579Kminus2 1205793 1205792 1205791120579K 120579Kminus1

famiddot middot middot

middot middot middot

cos120593

(b) The Doppler is ambiguity

Figure 2 The relationship between cosine of cone angle and Doppler

And the time resolution can be calculated by

119879119903=

1

2119861 (6)

where 119861 denotes the bandwidthIn a narrow band system 119891

119888is much larger than 119861 so we

can get 119879119889much smaller than 119879

119903 which means that the time

delay made by the position change of antenna element is verysmall compared with the time resolution

The time delay made by the position change of antennaelement can be ignored which is reasonable because thedistance between the elements is very small

This system uses low PRF to get wide swath but thelow PRF can cause azimuth Doppler ambiguity Figure 2illuminates this

After azimuth fast Fourier transforms (FFT) we can gainthe Doppler ambiguity signal 119904

119901119902(119905119896 119891119886)

119904119901119902(119905119896 119891119886) =

119871

sum

119897=1

119870

sum

119896=1

1198601015840(120579119897) sin 119888 (119905

119896minus Δ119879119897minus2119877119897(119891119886)

119888)

times rect(1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)

times exp(minus1198952120587119877119861119902

Vradic1198912119886119898minus 1198912119886)

times exp(11989541205871198831015840

119901cos120593119896(119891119886)

120582)

(7)

where119891119886isin (minusPRF2PRF2) 119877

119861119902= radic(119884

119902minus 1199101)2+ 1198672 is the

vertical distance form scatter point (1199091 1199101 1199111) to flight track

1198771199011199021

(119891119886) is the instantaneous distance of scatter point 120593

119896is

the angle of instantaneous slant distance and flight track andcos120593119896(119891119886) = 119891119886119891119886119898 119891119886119898= 2V120582

When1198771= 11988811987911990112+1198772= sdot sdot sdot = 119888(119879

1199011+sdot sdot sdot+119879

119901119871minus1)2+119877

119871is

satisfied 119871 different scatter points on 119871 subswathes can makealiasing and then cause range ambiguity But the differentscatter points are corresponding to different downlookingangles as shown in Figure 3

According to the forgoing analysis if the scatter instanta-neous slant distance satisfies the preceding equation (7) canbe shown as

119904119901119902(119905119896 119891119886) =

119871

sum

119897=1

119870

sum

119896=1

1198601015840(120579119897) sin 119888 (119905

119896minus21198771

119888)

times rect(1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)

times exp(minus1198952120587119877119861119902

Vradic1198912119886119898minus 1198912119886)

times exp(minus11989541205871198831015840

119901cos120593119896(119891119886)

120582)

(8)

where exp(minus119895(2120587119877119861119902V)radic1198912

119886119898minus 1198912119886) and exp(119895(41205871198831015840

119901cos120593119896

(119891119886)120582)) are caused by the position of azimuth and range

subapertures and they are the steer vector for resolvingambiguity One can know that there is the azimuth frequencyin the range steer vector so one needs to resolve the azimuthambiguity firstly and then resolve the range ambiguity


Range dimension

Rt

1cos120579

band width Δ120579

(a) The range is not ambiguity

Range dimension

Rt

1cos1205791205791

1205792

120579L

band width Δ120579

(b) The range is ambiguity

Figure 3 The relationship between downlooking angle and instantaneous slant distance

3 Two-Dimensional Ambiguity Resolving

31 Resolving Azimuth Ambiguity Because (8) is range andDoppler ambiguity signal we express thematrix vector z

119886(120593119896)

asz119886(120593119896)

= [exp(11989541205871198831015840

1cos120593119896

120582) exp(

11989541205871198831015840

119875cos120593119896

120582)]

119879

(9)

In this paper we use joint 2-dimensional method toresolve ambiguity based on static weighting vector It isassumed that the vector matrix isW

119875times119870and the 119899th column

weighting vector is shown as

W119899= [1199081119899 119908

119901119899 119908

119875119899]119879

(10)

It means that the corresponding downlooking angle andazimuth angle position is 1 and other ambiguities are zeroConsider

W119867119899119911119886= H119899 (11)

where ()119867 expresses matrix conjugate transpose operator andH119899= [ℎ1119899 ℎ

119870119899]119879 119899 = 1 119870 when 119896 = 119899 ℎ

119896119899= 1

Consider

W119899= (H119899119911+

119886)119867

(12)

where ()+ denotes the pseudoinverse operation 119911+

119886=

pinv(119911119886) Consider

S = [11990411 119904

1198751 11990412 119904

1198752 119904

1119876 119904

119875119876]119879

= sinc(119905119896minus21198771

119888) rect(

1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)

times exp(minus119895

2120587119877119861119902radic1198912119886119898minus 1198912119886

V)119911119886

(13)

Using the weighting vector to resolve ambiguity thefollowing is gained

S1015840 =W119867119899S = sinc(119905

119896minus21198771

119888)

times rect(1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)W119867119899119911119886

= sinc(119905119896minus21198771

119888) rect(

1198911198861198771198611

radic1198912119886119898minus 1198912119886

)

times exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)H119899

(14)

After this the resolving azimuth ambiguity operation isfinished

According to array signal theory the Doppler ambiguitytime 119870 must satisfy the following formula 119870 le 119875 Tothe member of azimuth subaperture 119875 it can make onerestriction direction at most and the other 119875 minus 1 is zero soonly 119875 timersquos ambiguity can be resolved on azimuth Dopplerdimension With the same principle only 119876 time ambiguitycan be resolved on azimuth Doppler dimension

After resolving the azimuth ambiguity the signal iscombined according to the Doppler spectrum The rangeresolving ambiguity is followed


32 Resolving Range Ambiguity According to (8) the rangesteering vector and the azimuth matching function areconcerned We expressed matrix vector z

119887(119877119897) as follows

z119886(119877119897) =

[[

[

exp(minus119895

21205871198771198611radic1198912119886119898minus 1198912119886

V)

exp(minus119895

2120587119877119861119876radic1198912119886119898minus 1198912119886

V)]]

]

119879

(15)

It is assumed that the vector matrix is W119876times119871

and the 119898thcolumn weighting vector is shown as

W119898= [1199081119898 119908

119902119898 119908

119876119898]119879

(16)


W119867119898119911119887= H119898 (17)


119871119898]119879 119898 = 1 119871 when 119897 = 119898 ℎ

119897119898= 1

Consider

W119898= (H119898119911+

119887)119867

(18)


119887=


S1015840 = [1199041 119904

119876]119879

= sinc(119905119896minus21198771

119888) rect(

1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)119911119887

(19)

x

y

z

H

1205792

1205791

1205932

1205932

h

120601

A

A998400

R

Figure 4 The geometry of terrain height variance

Utilizing theweighting vector to resolve the ambiguity wecan gain

S10158401015840 =W119867119898S1015840 = sinc (119905

119896minus21198771

119888)

times rect(119891119886119877119861119902

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)W119867119898119911119887

= sinc(119905119896minus21198771

119888) rect(

119891119886119877119861119902

radic1198912119886119898minus 1198912119886

)

times exp(minus1198952120587119891119886119909119897

V)

(20)

At this time the range ambiguity was resolved completelyThe azimuth was finished when range was resolved Thenafter the azimuth inverse FFT (IFFT) we can finish theazimuth pulse compression

For different point target slant distance 119877119897 the subpulse

time delay should be added For the 119897th subswath 11990410158401015840(119877119897)

added the time delay Δ119879119897= 1198791199011+ sdot sdot sdot + 119879

119901119897minus1to the range

envelope which make the real position of scatter point(119909119897 119910119897 119911119897) was get Then the whole wide swath signal was

obtained

33 Impact of Terrain Height Variance Forenamed staticweighting vector is based on the assumption that the terrain isflat Here we can analyze the impact of terrain height varianceon the weighting vector


minus4000 minus3000 minus2000 minus1000 0 1000 2000 3000 4000

2000

2050

2100

2150

2200

2250

2300

2350

2400

2450

2500

(a) The image with 2D ambiguity

minus40 minus20 0 20 40 60

500

1000

1500

2000

2500

3000

3500

4000

(b) The image after two-dimensional ambiguity resolving

minus5000 00

5000

01

02

03

04

05

06

07

08

09

1

(c) Azimuth profile of far scatter point

01

02

03

04

05

06

07

08

09

1

00

500 1000 1500 2000 2500 3000 3500 4000 4500

(d) Range profile of middle azimuth cell

Figure 5 The image of joint two-dimensional processing

As shown in Figure 4 there is a point 119860 with thedownlooking angle 120579

1and azimuth angle 120593

1 we assume there

is another point 1198601015840 with a height difference ℎ between 119860 sothe phase difference of the weighting vector pointing to119860 and1198601015840 is

ΔΦ =21205871198891(cos120593

1minus cos120593

2)

120582minus21205871198892(sin 1205791minus sin 120579

2)

120582

(21)

where 1205792 1205932are the downlooking angle and azimuth angle

in the position of 1198601015840 From the geometry configuration ofFigure 4 we can get that cos120593

1= sin 120579

1sin120601 and cos120593

2=

sin 1205792sin120601 so the phase difference can be written as

ΔΦ =2120587 (sin 120579

1minus sin 120579

2) (1198891sin120601 minus 119889

2)

120582

=

2120587 (1198891sin120601 minus 119889

2) (radic1198772 minus (119867 minus ℎ)

2minus radic1198772 minus 1198672)

120582119877

(22)

Table 1 Phase difference in the terrain with different height

Height ℎ (m) Phase differenceΔΦ (rad)

Remainedclutter (dB) Performance

352 minus12058732 minus202 Effective940 minus12058712 minus117 Worse1408 minus1205878 minus82 Much worse3735 minus1205873 0 Ineffective

The clutter which remained caused by the phase differ-ence is

20 log 10 [1 minus exp (119895ΔΦ)] (23)

With the simulating parameter of 119867 = 700 km 119877 =

880 km 120582 = 003m 1198891= 1m and 119889

2= 08m we get that the

simulation result of the clutter which remained varied withthe height difference shown in Table 1

FromTable 1 we can see that if the change of terrain heightis below 1 km our proposed method can still be effective


Table 2 Simulation parameters

Parameters ValueHeight of satellite119867 (km) 700Satellite velocity V (ms) 7200Carrier frequency 119891

119888(GHz) 2

System PRF (Hz) 1200Stepped-frequency signal band width 119861 (MHz) 100Transmitted pulse width 119879

Δ(120583s) 30

Azimuth subaperture length 1198891(m) 4

Elevation subaperture height 1198892(m) 08

Coordinates of the three scatter points (m)(0 427831 0)(0 367035 0)(0 398310 0)

4 Simulation Experiment

To validate the proposed approach a simulation experimenton point targets is carried out The main system parametersare listed in Table 2 In this simulation the whole antennaaperture was divided into 3 times 3 subapertures

It is supposed that these scatter points have the samebackscattering coefficient Through calculation one can get1198771

= 119888119879Δ2 + 119877

2= 119888119879

Δ+ 1198773 so the three scatter

points are range-ambiguous with each other The Dopplerband width is 119861

119886= 2V119863 = 3600Hz and the azimuth

sample frequency is 1200Hz which makes the azimuthDoppler spectrum ambiguous with three times Figure 5(a)shows the image of an original single channel The azimuthand the range both have ambiguity Range ambiguity makesthree scatter points with different range positions mixedtogether so only one point can be seen in Figure 5(a) becausethe system Doppler ambiguity three times three focusedpoints can be seen along azimuth Both sides were ambiguitypoints and their energy was smaller than the center pointobviously Through the joint two-dimensional ambiguityresolving method proposed in this paper the unambiguityimage can be obtained as shown in Figure 5(b) After two-dimensional ambiguity resolving the scatter points werelocated in their real positions Figure 5(c) is azimuth profile ofthe far scatter point In the figure the standard sinc functionwas obtained after the azimuth pulse compressed whichshows the unambiguous Doppler spectrum was recovered bythe joint two-dimensional processing Figure 5(d) is rangeprofile of the three range points From the figure one can notethat the range ambiguity is also resolved

5 Conclusion

In this paper a novel joint two-dimensional ambiguity resolv-ing method based on space-time filtering was proposedwhich can resolve the range and azimuth ambiguity ofMIMO-SAR for high-resolution and wide swath imagingThis method does not use any approximate operation whichcan finish resolving ambiguity and two dimensions focusedinstantaneously Imaging results of the simulated point targetvalidate the proposed approach Because this method does

not depend on SAR system work mode it has certain generaluse

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China under Grant 61201433 and supportedby Program for Young Talents of Science and Technology inUniversities of Inner Mongolia Autonomous Region (NJYT-14-B09)

References

[1] G Krieger N Gebert M Younis and A Moreira ldquoAdvancedsynthetic aperture radar based on digital beamforming andwaveform diversityrdquo in Proceedings of the IEEE Radar Confer-ence (RADAR rsquo08) pp 767ndash772 Rome Italy May 2008

[2] AOssowska J H Kim andWWiesbeck ldquoModeling of nonide-alities in receiver front-end for a simulation of multistatic SARsystemrdquo in Proceedings of the 4th European Radar Conference(EURAD 07) pp 13ndash16 Munich Germany October 2007

[3] G Krieger N Gebert and A Moreira ldquoMultidimensionalwaveform encoding a new digital beamforming technique forsynthetic aperture radar remote sensingrdquo IEEE Transactions onGeoscience and Remote Sensing vol 46 no 1 pp 31ndash46 2008

[4] W Q Wang ldquoSpace-time coding MIMO-OFDM SAR forhigh-resolution imagingrdquo IEEE Transactions on Geoscience andRemote Sensing vol 49 no 8 pp 3094ndash3104 2011

[5] W Wang ldquoMitigating range ambiguities in high PRF SARwith OFDM waveform diversityrdquo IEEE Geoscience and RemoteSensing Letters vol 10 no 1 pp 101ndash105 2013

[6] PHuang andWXu ldquoAnew spaceborne burst synthetic apertureradar imaging mode for wide swath coveragerdquo Remote Sensingvol 6 no 1 pp 801ndash814 2014

[7] P Huang S Li and W Xu ldquoInvestigation on full-apertureon multichannel azimuth data processing in TOPSrdquo IEEEGeoscience and Remote Sensing Letters vol 4 no 11 pp 728ndash732 2014

[8] G D Callaghan and I D Longstaff ldquoWide-swath space-borneSAR and range ambiguityrdquo in Proceedings of the IEEE RadarConference pp 248ndash252 Edinburgh Scotland October 1997

[9] G D Callaghan and I D Longstaff ldquoWide-swath space-borneSAR using a quad-element arrayrdquo IEE Proceedings Radar Sonarand Navigation vol 146 no 3 pp 159ndash165 1999

[10] G Krieger N Gebert and A Moreira ldquoUnambiguous SARsignal reconstruction from nonuniform displaced phase centersamplingrdquo IEEE Geoscience and Remote Sensing Letters vol 1no 4 pp 260ndash264 2004

[11] G Krieger N Gebert and A Moreira ldquoSAR signal reconstruc-tion from non-uniform displaced phase centre samplingrdquo inProceedings of the IEEE International Geoscience and RemoteSensing Symposium Proceedings Science for Society ExploringandManaging a Changing Planet (IGARSS rsquo04) vol 3 pp 1763ndash1766 Anchorage AK USA September 2004


[12] N Gebert G Krieger and A Moreira ldquoSAR signal reconstruc-tion from non-uniform displaced phase centre sampling in thepresence of perturbationsrdquo in Proceedings of the IEEE Interna-tional Geoscience and Remote Sensing Symposium (IGARSS rsquo05)pp 1034ndash1037 Seoul Korea July 2005

[13] P Huang W Xu and W Qi ldquoTwo dimension digital beam-forming pre processing in multibeam Scansarrdquo Progress inElectromagnetics Research vol 136 pp 495ndash508 2013

International Journal of

AerospaceEngineeringHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

RoboticsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014


Active and Passive Electronic Components

Control Scienceand Engineering

Journal of



RotatingMachinery


Hindawi Publishing Corporation httpwwwhindawicom

Journal ofEngineeringVolume 2014

Submit your manuscripts athttpwwwhindawicom

VLSI Design



Shock and Vibration


Civil EngineeringAdvances in

Acoustics and VibrationAdvances in



Electrical and Computer Engineering

Journal of

Advances inOptoElectronics


Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

SensorsJournal of


Modelling amp Simulation in EngineeringHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Chemical EngineeringInternational Journal of Antennas and

Propagation




Navigation and Observation



DistributedSensor Networks



x

y

z

H

d2

d1

Figure 1 MIMO-SAR model

119875 times 119876 subapertures where 119875 is the number of azimuthsubapertures and its length is 119889

1119876 is the number of azimuth

subapertures and its height is 1198892 We denoted the pth column

and the qth row as 119883119901= 1198830+ (119901 minus 1)119889

1 119901 = 1 119875

with 1198830the azimuth beginning coordinate (119883

119901 119884119902 119885) and

119884119902= 1198840+ (119902minus1)119889

2 119902 = 1 119876 with 119884

0the range beginning

coordinateUse the subcaptures on the same column range dimension

transmit signals at the same time through adjusting phaseweighting to control the wave beam the center of transmitwave beam is 1198841015840

119902= (119884119902+ 1198841199020)2 119884

1199020= 1198840+ 1198892(119876 minus 1)2 and

the whole antenna receives the echo signal According to theprinciple of equivalent phase center this mode can be knownas each equivalent phase center transmitting and receivingitself The equivalent phase centers located at (1198831015840

119901 1198841015840

119901 119885)

where1198831015840119901= (1198830+119883119901)2 and 1198841015840

119901= (119884119902+1198841199020)2 The deferent

range subapertures light the deferent subswaths from far tonearby adjusting phaseweightingsThe ellipses in Figure 1 arethe subswathes

Assume that during the Lth pulse period the first rangedimension subapertures transmit signal in the same timeand make the wave beam light the Lth subswath throughcontrolling phase weighting The transmitting signal can beexpressed as follows

119904 (119905119896) =

sin (1205871198761198892(sin 120579119897minus sin 120579) 2120582)

sin (1205871198892(sin 120579119897minus sin 120579) 2120582)

times rect((119905119896minus Δ119879)

119879119901119897

)

times exp (1198952120587119891119888(119905 minus Δ119879

119897) + 119895120587120574(119905

119896minus Δ119879119897)2

)

(1)

where 120579119897is the downlooking angle of the center line of

lth subswath 119891119888is the carrier frequency 120582 = 119888119891

119888is the

wavelength Δ119879119897= 1198791199011+ sdot sdot sdot + 119879

119901119897minus1is the delay time of the

lth subpulse 119905 = 119905119896+119898119879119903is the whole time 119905

119896is the fast time

119898 is a integer119879119903is pulse repetition interval and 120574 is the chirp

rate

The baseband signal received by the pth row and the qthcolumn subaperture was

119904119901119902(119905119896 119905119898) =

119871

sum

119897=1

119860 (120579119897) rect(

(119905119896minus Δ119879119897minus 2119877119901119902119897

(119905119898) 119888)

119879119901119897

)

times rect((1198831015840

119901+ 119909 minus 119909

119897)

119871119886

)

times exp(1198952120587119891119888(minusΔ119879

119897minus

2119877119901119902119897

(119905119898)

119888))

times exp(119895120587120574(119905119896minus Δ119879119897minus

2119877119901119902119897

(119905119898)

119888)

2

)

(2)

where 119860(120579119897) = sin(120587119876119889

2(sin 120579119897minus sin 120579)2120582) sin(120587119889

2(sin 120579119897minus

sin 120579)2120582) 119905119898

= 119896119879119903is slow time 119896 is a integer and

the value scope of 119897 is from 1 to 119871 corresponding to theswath from far to near 119909 = V119905

119898 V is the flat velocity

and 119871119886is the length of synthetic aperture 119877

119901119902119897(119905119898) =


slant range (119909119897 119910119897 119911119897) is the coordinates of one scatter point

in the Lth scene and119867 = 1198850minus119911119897 Assume that scene is plane

119911119897= 0Performing the range match filtering to (2) we can gain

119904119901119902(119905119896 119905119898) asymp

119871

sum

119897=1

1198601015840(120579119897) sin 119888 (119905

119896minus Δ119879119897minus2119877119897(119905119898)

119888)

times rect(1198831015840

119901+ 119909 minus 119909

119897

119871119886

) exp (minus1198954120587120582119877119901119902119897

(119905119898))

(3)

where 1198601015840(120579119897) = 119860(120579

119897) exp(minus1198952120587119891

119888Δ119879119897) 119877119897(119905119898) =


distance from the scatter point on the lth subswath to(1198830 1198840 1198850)

As we know theMIMO-SAR can be considered as a spacesignal sampling which should satisfy the sampling equationas follows

119889 le120582

2 (4)

where 119889 denotes the position change of antenna element and120582 denotes the wavelength of the signal

So the time delay should be

119879119889=119889

119888le120582

2119888=

1

2119891119888

(5)

where 119891119888denotes the carrier frequency


fa

PRF2

minusPRF2

cos120593


PRF2

minusPRF2Band width

120579Kminus2 1205793 1205792 1205791120579K 120579Kminus1



cos120593




119879119903=

1

2119861 (6)









119901119902(119905119896 119891119886)

119904119901119902(119905119896 119891119886) =

119871

sum

119897=1

119870

sum

119896=1

1198601015840(120579119897) sin 119888 (119905

119896minus Δ119879119897minus2119877119897(119891119886)

119888)

times rect(1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)

times exp(minus1198952120587119877119861119902

Vradic1198912119886119898minus 1198912119886)

times exp(11989541205871198831015840

119901cos120593119896(119891119886)

120582)

(7)


119861119902= radic(119884

119902minus 1199101)2+ 1198672 is the


1198771199011199021


119896is




119901119871minus1)2+119877

119871is



119904119901119902(119905119896 119891119886) =

119871

sum

119897=1

119870

sum

119896=1

1198601015840(120579119897) sin 119888 (119905

119896minus21198771

119888)

times rect(1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)

times exp(minus1198952120587119877119861119902

Vradic1198912119886119898minus 1198912119886)

times exp(minus11989541205871198831015840

119901cos120593119896(119891119886)

120582)

(8)


119886119898minus 1198912119886) and exp(119895(41205871198831015840

119901cos120593119896




Range dimension

Rt

1cos120579

band width Δ120579


Range dimension

Rt

1cos1205791205791

1205792

120579L

band width Δ120579





119886(120593119896)

asz119886(120593119896)

= [exp(11989541205871198831015840

1cos120593119896

120582) exp(

11989541205871198831015840

119875cos120593119896

120582)]

119879

(9)




W119899= [1199081119899 119908

119901119899 119908

119875119899]119879

(10)


W119867119899119911119886= H119899 (11)


119870119899]119879 119899 = 1 119870 when 119896 = 119899 ℎ

119896119899= 1

Consider

W119899= (H119899119911+

119886)119867

(12)


119886=


S = [11990411 119904

1198751 11990412 119904

1198752 119904

1119876 119904

119875119876]119879

= sinc(119905119896minus21198771

119888) rect(

1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)119911119886

(13)


S1015840 =W119867119899S = sinc(119905

119896minus21198771

119888)

times rect(1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)W119867119899119911119886

= sinc(119905119896minus21198771

119888) rect(

1198911198861198771198611

radic1198912119886119898minus 1198912119886

)

times exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)H119899

(14)






119887(119877119897) as follows

z119886(119877119897) =

[[

[

exp(minus119895

21205871198771198611radic1198912119886119898minus 1198912119886

V)

exp(minus119895

2120587119877119861119876radic1198912119886119898minus 1198912119886

V)]]

]

119879

(15)



W119898= [1199081119898 119908

119902119898 119908

119876119898]119879

(16)


W119867119898119911119887= H119898 (17)


119871119898]119879 119898 = 1 119871 when 119897 = 119898 ℎ

119897119898= 1

Consider

W119898= (H119898119911+

119887)119867

(18)


119887=


S1015840 = [1199041 119904

119876]119879

= sinc(119905119896minus21198771

119888) rect(

1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)119911119887

(19)

x

y

z

H

1205792

1205791

1205932

1205932

h

120601

A

A998400

R



S10158401015840 =W119867119898S1015840 = sinc (119905

119896minus21198771

119888)

times rect(119891119886119877119861119902

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)W119867119898119911119887

= sinc(119905119896minus21198771

119888) rect(

119891119886119877119861119902

radic1198912119886119898minus 1198912119886

)

times exp(minus1198952120587119891119886119909119897

V)

(20)







obtained




2000

2050

2100

2150

2200

2250

2300

2350

2400

2450

2500


minus40 minus20 0 20 40 60

500

1000

1500

2000

2500

3000

3500

4000


minus5000 00

5000

01

02

03

04

05

06

07

08

09

1


01

02

03

04

05

06

07

08

09

1

00

500 1000 1500 2000 2500 3000 3500 4000 4500





1 we assume there


ΔΦ =21205871198891(cos120593

1minus cos120593

2)


2)

120582

(21)



1= sin 120579

1sin120601 and cos120593

2=


ΔΦ =2120587 (sin 120579

1minus sin 120579

2) (1198891sin120601 minus 119889

2)

120582

=

2120587 (1198891sin120601 minus 119889



120582119877

(22)






20 log 10 [1 minus exp (119895ΔΦ)] (23)


880 km 120582 = 003m 1198891= 1m and 119889







119888(GHz) 2


Δ(120583s) 30







= 119888119879Δ2 + 119877

2= 119888119879





5 Conclusion





Acknowledgments


References

















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










fa

PRF2

minusPRF2

cos120593


PRF2

minusPRF2Band width

120579Kminus2 1205793 1205792 1205791120579K 120579Kminus1



cos120593




119879119903=

1

2119861 (6)









119901119902(119905119896 119891119886)

119904119901119902(119905119896 119891119886) =

119871

sum

119897=1

119870

sum

119896=1

1198601015840(120579119897) sin 119888 (119905

119896minus Δ119879119897minus2119877119897(119891119886)

119888)

times rect(1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)

times exp(minus1198952120587119877119861119902

Vradic1198912119886119898minus 1198912119886)

times exp(11989541205871198831015840

119901cos120593119896(119891119886)

120582)

(7)


119861119902= radic(119884

119902minus 1199101)2+ 1198672 is the


1198771199011199021


119896is




119901119871minus1)2+119877

119871is



119904119901119902(119905119896 119891119886) =

119871

sum

119897=1

119870

sum

119896=1

1198601015840(120579119897) sin 119888 (119905

119896minus21198771

119888)

times rect(1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)

times exp(minus1198952120587119877119861119902

Vradic1198912119886119898minus 1198912119886)

times exp(minus11989541205871198831015840

119901cos120593119896(119891119886)

120582)

(8)


119886119898minus 1198912119886) and exp(119895(41205871198831015840

119901cos120593119896




Range dimension

Rt

1cos120579

band width Δ120579


Range dimension

Rt

1cos1205791205791

1205792

120579L

band width Δ120579





119886(120593119896)

asz119886(120593119896)

= [exp(11989541205871198831015840

1cos120593119896

120582) exp(

11989541205871198831015840

119875cos120593119896

120582)]

119879

(9)




W119899= [1199081119899 119908

119901119899 119908

119875119899]119879

(10)


W119867119899119911119886= H119899 (11)


119870119899]119879 119899 = 1 119870 when 119896 = 119899 ℎ

119896119899= 1

Consider

W119899= (H119899119911+

119886)119867

(12)


119886=


S = [11990411 119904

1198751 11990412 119904

1198752 119904

1119876 119904

119875119876]119879

= sinc(119905119896minus21198771

119888) rect(

1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)119911119886

(13)


S1015840 =W119867119899S = sinc(119905

119896minus21198771

119888)

times rect(1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)W119867119899119911119886

= sinc(119905119896minus21198771

119888) rect(

1198911198861198771198611

radic1198912119886119898minus 1198912119886

)

times exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)H119899

(14)






119887(119877119897) as follows

z119886(119877119897) =

[[

[

exp(minus119895

21205871198771198611radic1198912119886119898minus 1198912119886

V)

exp(minus119895

2120587119877119861119876radic1198912119886119898minus 1198912119886

V)]]

]

119879

(15)



W119898= [1199081119898 119908

119902119898 119908

119876119898]119879

(16)


W119867119898119911119887= H119898 (17)


119871119898]119879 119898 = 1 119871 when 119897 = 119898 ℎ

119897119898= 1

Consider

W119898= (H119898119911+

119887)119867

(18)


119887=


S1015840 = [1199041 119904

119876]119879

= sinc(119905119896minus21198771

119888) rect(

1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)119911119887

(19)

x

y

z

H

1205792

1205791

1205932

1205932

h

120601

A

A998400

R



S10158401015840 =W119867119898S1015840 = sinc (119905

119896minus21198771

119888)

times rect(119891119886119877119861119902

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)W119867119898119911119887

= sinc(119905119896minus21198771

119888) rect(

119891119886119877119861119902

radic1198912119886119898minus 1198912119886

)

times exp(minus1198952120587119891119886119909119897

V)

(20)







obtained




2000

2050

2100

2150

2200

2250

2300

2350

2400

2450

2500


minus40 minus20 0 20 40 60

500

1000

1500

2000

2500

3000

3500

4000


minus5000 00

5000

01

02

03

04

05

06

07

08

09

1


01

02

03

04

05

06

07

08

09

1

00

500 1000 1500 2000 2500 3000 3500 4000 4500





1 we assume there


ΔΦ =21205871198891(cos120593

1minus cos120593

2)


2)

120582

(21)



1= sin 120579

1sin120601 and cos120593

2=


ΔΦ =2120587 (sin 120579

1minus sin 120579

2) (1198891sin120601 minus 119889

2)

120582

=

2120587 (1198891sin120601 minus 119889



120582119877

(22)






20 log 10 [1 minus exp (119895ΔΦ)] (23)


880 km 120582 = 003m 1198891= 1m and 119889







119888(GHz) 2


Δ(120583s) 30







= 119888119879Δ2 + 119877

2= 119888119879





5 Conclusion





Acknowledgments


References

















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










Range dimension

Rt

1cos120579

band width Δ120579


Range dimension

Rt

1cos1205791205791

1205792

120579L

band width Δ120579





119886(120593119896)

asz119886(120593119896)

= [exp(11989541205871198831015840

1cos120593119896

120582) exp(

11989541205871198831015840

119875cos120593119896

120582)]

119879

(9)




W119899= [1199081119899 119908

119901119899 119908

119875119899]119879

(10)


W119867119899119911119886= H119899 (11)


119870119899]119879 119899 = 1 119870 when 119896 = 119899 ℎ

119896119899= 1

Consider

W119899= (H119899119911+

119886)119867

(12)


119886=


S = [11990411 119904

1198751 11990412 119904

1198752 119904

1119876 119904

119875119876]119879

= sinc(119905119896minus21198771

119888) rect(

1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)119911119886

(13)


S1015840 =W119867119899S = sinc(119905

119896minus21198771

119888)

times rect(1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)W119867119899119911119886

= sinc(119905119896minus21198771

119888) rect(

1198911198861198771198611

radic1198912119886119898minus 1198912119886

)

times exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)H119899

(14)






119887(119877119897) as follows

z119886(119877119897) =

[[

[

exp(minus119895

21205871198771198611radic1198912119886119898minus 1198912119886

V)

exp(minus119895

2120587119877119861119876radic1198912119886119898minus 1198912119886

V)]]

]

119879

(15)



W119898= [1199081119898 119908

119902119898 119908

119876119898]119879

(16)


W119867119898119911119887= H119898 (17)


119871119898]119879 119898 = 1 119871 when 119897 = 119898 ℎ

119897119898= 1

Consider

W119898= (H119898119911+

119887)119867

(18)


119887=


S1015840 = [1199041 119904

119876]119879

= sinc(119905119896minus21198771

119888) rect(

1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)119911119887

(19)

x

y

z

H

1205792

1205791

1205932

1205932

h

120601

A

A998400

R



S10158401015840 =W119867119898S1015840 = sinc (119905

119896minus21198771

119888)

times rect(119891119886119877119861119902

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)W119867119898119911119887

= sinc(119905119896minus21198771

119888) rect(

119891119886119877119861119902

radic1198912119886119898minus 1198912119886

)

times exp(minus1198952120587119891119886119909119897

V)

(20)







obtained




2000

2050

2100

2150

2200

2250

2300

2350

2400

2450

2500


minus40 minus20 0 20 40 60

500

1000

1500

2000

2500

3000

3500

4000


minus5000 00

5000

01

02

03

04

05

06

07

08

09

1


01

02

03

04

05

06

07

08

09

1

00

500 1000 1500 2000 2500 3000 3500 4000 4500





1 we assume there


ΔΦ =21205871198891(cos120593

1minus cos120593

2)


2)

120582

(21)



1= sin 120579

1sin120601 and cos120593

2=


ΔΦ =2120587 (sin 120579

1minus sin 120579

2) (1198891sin120601 minus 119889

2)

120582

=

2120587 (1198891sin120601 minus 119889



120582119877

(22)






20 log 10 [1 minus exp (119895ΔΦ)] (23)


880 km 120582 = 003m 1198891= 1m and 119889







119888(GHz) 2


Δ(120583s) 30







= 119888119879Δ2 + 119877

2= 119888119879





5 Conclusion





Acknowledgments


References

















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation











119887(119877119897) as follows

z119886(119877119897) =

[[

[

exp(minus119895

21205871198771198611radic1198912119886119898minus 1198912119886

V)

exp(minus119895

2120587119877119861119876radic1198912119886119898minus 1198912119886

V)]]

]

119879

(15)



W119898= [1199081119898 119908

119902119898 119908

119876119898]119879

(16)


W119867119898119911119887= H119898 (17)


119871119898]119879 119898 = 1 119871 when 119897 = 119898 ℎ

119897119898= 1

Consider

W119898= (H119898119911+

119887)119867

(18)


119887=


S1015840 = [1199041 119904

119876]119879

= sinc(119905119896minus21198771

119888) rect(

1198911198861198771198611

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)119911119887

(19)

x

y

z

H

1205792

1205791

1205932

1205932

h

120601

A

A998400

R



S10158401015840 =W119867119898S1015840 = sinc (119905

119896minus21198771

119888)

times rect(119891119886119877119861119902

radic1198912119886119898minus 1198912119886

) exp(minus1198952120587119891119886119909119897

V)


2120587119877119861119902radic1198912119886119898minus 1198912119886

V)W119867119898119911119887

= sinc(119905119896minus21198771

119888) rect(

119891119886119877119861119902

radic1198912119886119898minus 1198912119886

)

times exp(minus1198952120587119891119886119909119897

V)

(20)







obtained




2000

2050

2100

2150

2200

2250

2300

2350

2400

2450

2500


minus40 minus20 0 20 40 60

500

1000

1500

2000

2500

3000

3500

4000


minus5000 00

5000

01

02

03

04

05

06

07

08

09

1


01

02

03

04

05

06

07

08

09

1

00

500 1000 1500 2000 2500 3000 3500 4000 4500





1 we assume there


ΔΦ =21205871198891(cos120593

1minus cos120593

2)


2)

120582

(21)



1= sin 120579

1sin120601 and cos120593

2=


ΔΦ =2120587 (sin 120579

1minus sin 120579

2) (1198891sin120601 minus 119889

2)

120582

=

2120587 (1198891sin120601 minus 119889



120582119877

(22)






20 log 10 [1 minus exp (119895ΔΦ)] (23)


880 km 120582 = 003m 1198891= 1m and 119889







119888(GHz) 2


Δ(120583s) 30







= 119888119879Δ2 + 119877

2= 119888119879





5 Conclusion





Acknowledgments


References

















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation











2000

2050

2100

2150

2200

2250

2300

2350

2400

2450

2500


minus40 minus20 0 20 40 60

500

1000

1500

2000

2500

3000

3500

4000


minus5000 00

5000

01

02

03

04

05

06

07

08

09

1


01

02

03

04

05

06

07

08

09

1

00

500 1000 1500 2000 2500 3000 3500 4000 4500





1 we assume there


ΔΦ =21205871198891(cos120593

1minus cos120593

2)


2)

120582

(21)



1= sin 120579

1sin120601 and cos120593

2=


ΔΦ =2120587 (sin 120579

1minus sin 120579

2) (1198891sin120601 minus 119889

2)

120582

=

2120587 (1198891sin120601 minus 119889



120582119877

(22)






20 log 10 [1 minus exp (119895ΔΦ)] (23)


880 km 120582 = 003m 1198891= 1m and 119889







119888(GHz) 2


Δ(120583s) 30







= 119888119879Δ2 + 119877

2= 119888119879





5 Conclusion





Acknowledgments


References

















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation












119888(GHz) 2


Δ(120583s) 30







= 119888119879Δ2 + 119877

2= 119888119879





5 Conclusion





Acknowledgments


References

















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation














RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation









Research Article Joint Two-Dimensional Ambiguity Resolving ...

Documents

Transcript of Research Article Joint Two-Dimensional Ambiguity Resolving ...