. (24) If we consider the geometry of Figure 13 the signal returned from the n th scatterer located at x, y is

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1 .5 SAR SIGNA CHARACTERIZATION I order to formulate a SAR processor we first eed to characterize the sigal that the SAR processor will operate upo. Although our previous discussios treated SAR cross-rage imagig as a atea problem, for the rest of the developmet we will cast the problem i the Doppler domai. Past experiece i this course idicates that the Doppler formulatio is easier to uderstad tha the atea formulatio..5. Derivatio of the SAR Sigal As we have doe so far, we will iitially cosider oly the cross-rage problem. We will later exted the discussios to the dow-rage ad crossrage problem. Sice we are cosiderig the cross-rage problem we start by cosiderig a CW trasmit sigal of the form T v t e j fct. (4) If we cosider the geometry of Figure 3 the sigal retured from the th scatterer located at x, y is P P v t v t r t c e r t r t S S j tr t c (5) RF T r where r t x y d t 3 (6) ad dt is the y positio of the aircraft at some time t. If we assume that the aircraft is flyig at a costat velocity of V ad t 0 occurs at y 0 of Figure 3 we get d t Vt. (7) We assume that the total time for the aircraft to travel a distace of is T ad that the aircraft starts at whe t T. With this we get T V. (8) We ote that the area to be imaged has a cross-rage width of w ad a dow rage legth of l. The regio is cetered i cross-rage at y 0 ad i dowrage at x r. o I Equatio (5), P S is the ormalized sigal power associated with the th scatterer. It is determied from the radar rage equatio, without the term (sice it is icorporated ito Equatio (5)). 4 R 3 For ow we are assumig that the aircraft (platform) is at a altitude of zero. The extesio to o-zero altitude is straightforward. 04 M. C. Budge - merv@thebudges.com 4

2 r If we assume that w, l ad are small relative to r o we ca replace t i the deomiator of Equatio (5) by P vrf t e e r r o ad write S j fct j4r t (9) o where c fc is the wavelegth of the trasmit sigal. Figure 3 Geometry Used to Develop Sigal Represetatio Cross-rage Imagig Sice the iformatio eeded to form the image is i the secod expoetial term we will elimiate the first expoetial term by heterodyig (which is doe i the actual radar). We also elimiate the r term through ormalizatio to yield the basebad sigal j fcr j4r t. (30) v t r e v t P e o RF S If we have would be N N s scatterers i the image regio the resultig basebad sigal N s s j4r t S (3) v t v t P e o 04 M. C. Budge - merv@thebudges.com 5

3 .5. Examiatio of the Phase of the SAR Sigal Sice the iformatio we seek is i the phase of examie it. To proceed we eed to examie r v t. We ca write t we wat to r t x y y Vt V t r y Vt V t r y Vt r Vt r. (3) We ote that r r o, o Vt ro Vt,. This meas that the secod ad third terms of the last square root are small relative to. This, i tur, allows us to write y r ad r t r y Vt r Vt r r y Vt r V t r. (33) If we substitute this ito Equatio (30 we get j4r 4 j V r t j y V r t v t P e e e. (34) S.5.. iear Phase, or Costat Frequecy, Term The first expoetial is a phase caused by rage delay to the scatterer that we must live with. The secod term is a liear phase term or a term that we associate with frequecy. I fact, the frequecy represeted by this term is f y V r. (35) y This tells us that v t has a costat frequecy term that depeds upo the scatterer cross rage positio, y. f y also depeds upo the aircraft velocity, V, ad the radar wavelegth,. However both of these are kow (ad fixed). Fially, f y also depeds upo r. If we assume that all of the scatterers are at the same x ro (which we ca do sice we are oly cocered with the crossrage problem) ad ote that y r o we get the previous assertio that r x y r. (36) o This discussio tells us that, if we ca determie f y, we ca determie y from y fyro. (37) V.5.. Quadratic Phase, or FM, Term The third expoetial of Equatio (34) is a quadratic phase, or liear frequecy modulatio, term ad causes problems. We ca write the quadratic phase as 04 M. C. Budge - merv@thebudges.com 6

4 V Q t t. (38) r With the previous assumptio that r r o t Q is approximately the same for all scatterers. This meas that we ca remove it by a mixig or heterodyig process. 4 If we do this we will be left with oly the magitude, costat phase term ad the y -depedet frequecy term. This is what we wat.,.5.3 Extractig the Cross-Rage Iformatio Oce we remove the quadrtic phase we have j V r o t j4 r j f y v t e v t P e e t (39) I S for a sigle scatterer. For the more geeral case of N s scatterers we have Ns j V ro t j4 r j f yt v t e v t P e e. (40) I S The forms of Equatios (39) ad (40) tell us that we ca extract the iformatio we wat by takig the Fourier trasform of v t. v t or, more geerally, From our experiece with Fourier trasforms, this will give us a respose that peaks at the frequecies f. The heights of the peaks will be proportioal to y P S. If we use Equatio (37) to plot this as itesity ( P S ) vs. y we have a oe dimesioal image. 5 To verify the above, ad begi to uderstad SAR resolutio properties, we will fid the Fourier trasform of v t. Ideed j ft VI f vi te dt I. (4) To perform the itegratio we eed to realize that v t ad thus vi measured oly over t T, T,. Thus, we assume that v outside of these limits ad write I I I t is t is zero T T j ft j4r j f yt VI f vi te dt PS e e dt T T where P e sic f f T j4r S y (4) 4 Note that this is similar to stretch processig wherei we remove the quadratic phase i the mixer. 5 Agai, ote the similarity to stretch processig 04 M. C. Budge - merv@thebudges.com 7

5 sic x si xx. (43) Figure 4 cotais a plot I V f vs. y f f T. It will be oted that the respose has a peak at f f y 0, or at f fy ad that the peak has a height V f P of I S. The width of the peak is image will have a resolutio of T T which meas that the SAR f. (44) Figure 4 Plot of VI f P S vs. y f f T If we chage the horizotal axis to y usig the relatio (see Equatio (37)) fr y o (45) V VI y vs. y y with a height of V y P we get the plot of y y I Figure 4 is f T the resolutio of Figure 5 is of Figure 5. This plot has a peak at S. From Equatio (44), if the resolutio of ro ro ro y f (46) V T V sice T V. This is the same as the resolutio we obtaied from the liear array approach (see Equatio (8)). 04 M. C. Budge - merv@thebudges.com 8

6 Figure 5 Plot of V y P vs. I S y y y.6 PRACTICA IMPEMENTATION I the previous sectio we foud that the processig methodology we must use to form a image is to form a Fourier trasform of v t ). However, this approach makes the tacit assumptio that vi vi t (or I t is a cotiuous fuctio of time. Thikig ahead to whe we will cosider both cross-rage ad dow-rage imagig we realize that the SAR will trasmit a pulsed sigal rather tha a CW sigal. Because of this, we recogize that v t will ot be a cotiuous-time sigal but a discrete-time sigal with samples spaced by the radar PRI, T. If we also recogize that we will perform the processig with digital sigal processors (DSPs) we reiforce the realizatio that v t is, i fact, a discrete-time sigal, vi kt or I v k. I I.6. A Discrete-Time Model For a sigle scatterer (i.e. I v t ) we get (from Equatio (34)) I j4r 4 j V r j yv r kt kt. (47) v k P e e e S After we (digitally) remove the quadratic phase term, the sigal we process to form the image is I j4 r j4 yv rtk. (48) v k P e e S 04 M. C. Budge - merv@thebudges.com 9

7 Sice v k (ad v I I k for multiple scatterers) is a discrete-time sigal we use the discrete-time Fourier trasform (DFT) to fid its Fourier trasform. Specifically we fid I I fkt. (49) k j V f v k e As with the cotiuous-time Fourier trasform, we limit the sum by cosiderig that we gather data oly from to or for t T. If we use t kt the limits o k become k T T K (50) where it is uderstood that we roud, or trucate, T T to the earest iteger. get If we use this i Equatio (49), with Equatio (48) ad Equatio (35), we K j4r j f y f Tk VI f PS e e kk P e S j4r si si y y K f f T f f T We ote that this is similar to Equatio (4). Figure 6 cotais a plot of I. (5) V f vs. f f y T. As ca be see, it has a peak at f f y 0 as did Figure 4. However, it also has peaks at f f T. I fact, if we recall the theory associated with discrete-time y sigals ad the DFT we recogize that I V f will have peaks at f f y T where is a iteger. All peaks except the oe correspodig to 0 are ambiguities ad are udesirable. I terms of SAR, they result i what are termed ghost images. We wat to be sure we choose T, ad the characteristics of the SAR atea, to avoid these ghosts sice they ca result i misleadig SAR images. The SAR atea was metioed because it acts as a spatial atialiasig filter. 04 M. C. Budge - merv@thebudges.com 0

8 Figure 6 Plot of VI f P S vs. y f f T Usig a Discrete-time sigal with T 0.T.6. Other Cosideratios As we did before, we wat to chage the horizotal axis of Figure 6 to cross-rage distace rather tha frequecy. To do so we use Equatio (45). This results i the plot of V y show i Figure 7. The ambiguities are I show i this figure ad are located at y ambig ro y. (5) VT The above tells us that we wat to choose the PRI such that all scatterers lie withi ±/ ambiguity. That is we wat to choose the PRI such that all y satisfy y ro. (53) VT Sice all scatterers of iterest lie withi the imaged area this says that we wat to choose the PRI such that r o w. (54) VT I fact we usually choose the PRI such that w r o VT (55) 04 M. C. Budge - merv@thebudges.com

9 so as to be sure that the SAR atea beam adequately atteuates targets outside of the imaged regio. After we form the larger image, because we will be usig a larger effective w, we will trucate it to iclude oly the regio of iterest. Figure 7 Plot of V y P vs. I S y y y Usig a Discrete-time sigal with T 0.T We ca tur Equatio (54) aroud ad use it to fid a upper boud o PRI. Specifically we solve Equatio (54) for T to yield r o T (56) Vw or from Equatio (55) T r o. (57) Vw If we cosider a earlier example where ro 0 Km, 0.03 m ad w 50 m, ad cosider a aircraft velocity of V 50 m/s we get T 0 ms, (58) which is a easy costrait to satisfy. Whe we cosider dow-rage imagig we will impose a lower limit o T so as to satisfy uambiguous rage operatio. However, that lower limit is geerally well below the upper limit of Equatio (58). We ow wat to summarize the above as a algorithm that we ca implemet to form a cross-rage image. 04 M. C. Budge - merv@thebudges.com

10 .7 AN AGORITHM FOR CREATING A CROSS-RANGE IMAGE We assume we have a base-bad, CW sigal (see Equatio (30) ad Equatio (3)) We sample this sigal at itervals of T ad geerate K samples where o T r o Vw (Equatio (57)) o K T T (Equatio (50)) o T V (Equatio (8)) o r y (Equatio (46)) o I these equatios, r o, V, w, ad y are kow. The samples are take for kt betwee T ad T or for k betwee K ad K. Remove the quadratic phase by multiplyig the sampled sigal by h V j kt r o (59) v k e This gives I v k. Compute the DFT of v I k. This is most easily doe usig a FFT. The miimum FFT legth is K although we usually choose the FFT legth to be a power of greater tha K. I real applicatios we ofte adjust various SAR parameters so that K is close to a power of. For purposes of this class we choose a FFT legth much greater (4 to 6 times) tha K so that the resultig frequecy plot is smooth. If FFT is the legth of the FFT, the frequecy spacig betwee output FFT taps is f (60) T FFT After the frot ad rear halves of the FFT outputs are swapped (thik fftshift) the frequecies of the taps are f mf, f (6) FFT FFT Trasform the frequecy scale to cross-rage usig r y o f (Equatio (45)) V 04 M. C. Budge - merv@thebudges.com 3

11 ad plot the magitude of the FFT output vs. y. This does t produce a image but a liear plot as show i Figure 6. To geerate a pseudo image, create a array of zeros where the umber of colums is equal to the umber of samples eeded to cover the width, w, of the image area ad a comparable umber of rows. If there are N rows, replace row N row with the appropriate FFT outputs. Use the resultig array to create the pseudo image. row.8 EXAMPE To illustrate the above, we cosider a specific example. The parameters of this example are give i Table. Table Parameters Used i SAR Example Parameter Width of image area, w Depth of image area, l Rage to image area ceter, r o SAR wavelegth, Aircraft velocity, V Sythetic array legth, Number of scatterers, Value 50 m 50 m 0 Km 0.03 m 50 m/s 600 m N 3 s Scatterer locatios, x, y (m) ( r o,0), ( r o,0), ( r o,-5) Scatterer powers, P S (w), 0.5, 0.09 Give these we ca compute some of the SAR parameters idicated i the algorithm descriptio. Specifically: T T V s ad (6) 0.030, 000 r o 0 ms. (63) Vw We will choose a PRI of 50 ms, that is we choose T 50 ms. (64) This gives 04 M. C. Budge - merv@thebudges.com 4

12 K T 0. (65) T 0.05 With this the SAR starts samplig at t 6 s ad samples util t 6 s. The samples are take every T 50 ms ad a total of K 4 samples are used. This meas that we eed, as a miimum, a 56 poit FFT. I order to produce a smooth plot we will use a 048 poit FFT. We ote that, sice we chose icluded i the image is w actual 0.030, 000 T 50 ms the actual width of the area ro 0 m. (66) VT Oce we form the image we will eed to discard the FFT outputs outside of the rage of ±5 m (after the coversio from frequecy to y positio) The resolutio of the SAR image is 0, r o y 0.5 m. (67) 600 This meas that we ought to be able to distiguish scatterers separated by about m or greater, ad maybe dow to 0.5 m separatio if their relative powers ad phases allow this. Before processig the SAR sigal usig the previously discussed algorithm we eed to geerate the SAR sigal. To do so we use Equatio (3) with Ns 3. We geerate 4 samples of vt startig at t 6 s ad edig at t 6 s i steps of T 0.05 s. I my code I geerated (7). I the used these r t, alog with the S compute the three v t. Fially, I summed the three v t to form Figure 8 is a liear plot of the r t,,,3, usig Equatio (6) ad Equatio I P values i Equatio (30) to vt. V y for 5 m y 5 m ad Figure 9 is a pseudo image. To form the pseudo image I first created a array of zeros that had 0 rows (which is l y ) ad a umber of colums equal to the umber of y values i the liear plot. I ext loaded the liear plot i the 5 st row of the array. Next I used the commads imagesc(x, y, max(max( V y ))- V to geerate the image. The use of max(max( I I VI y );colormap gray V y ))- V I I y values from the y creates a image with a white backgroud, which is really a egative image. I did this to save o priter ik. 04 M. C. Budge - merv@thebudges.com 5

13 Figure 8 iear plot of VI y - Three Scatterers at -5, 0, 0 m I examiig Figure 8 we ote that VI y has three peaks at the y positios of the scatterers. Further, the heights of the peaks are P S. The image (Figure 9) shows three dots at the give scatterer positios. It will be oted that the dots are differet shades of gray, idicatig differet amplitudes. To check the aforemetioed resolutio statemet I moved the y positios of the scatterers to -, 0 ad m ad reformed the liear plot ad image. These are show i Figure 0 ad Figure. The liear plot clearly shows three peaks but the relative amplitudes are somewhat differet tha those of Figure 8. This is due to the sidelobes of the Fourier trasform respose fuctio. The presece of the three scatterers ca also be see i the image of Figure. 04 M. C. Budge - merv@thebudges.com 6

14 Figure 9 Image of VI y - Three Scatterers at -5, 0, 0 m Figure 0 iear plot of VI y - Three Scatterers at -, 0, m 04 M. C. Budge - merv@thebudges.com 7

15 Figure Image of VI y - Three Scatterers at -, 0, m As aother iterestig experimet I elimiated the quadratic phase removal step of the SAR processig algorithm. The results are show i Figure ad Figure 3 (I reverted to the origial locatios of the scatterers). As ca be see, the peaks are spread ad the image is blurred i the y directio. I SAR ligo, we say that the image is ot focused. I fact, the process of removig the quadratic phase is sometimes termed focusig of the SAR image. 04 M. C. Budge - merv@thebudges.com 8

16 Figure iear plot of VI y - Three Scatterers at -5, 0, 0 m Without Quadratic Phase Removal Figure 3 Image of VI y - Three Scatterers at -5, 0, 0 m Without Quadratic Phase Removal 04 M. C. Budge - merv@thebudges.com 9

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