ECE 5318/6352 Antenna Engineering. Spring 2006 Dr. Stuart Long. Chapter 6. Part 7 Schelkunoff s Polynomial
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1 ECE 538/635 Antenna Engineering Spring 006 Dr. Stuart Lng Chapter 6 Part 7 Schelkunff s Plynmial 7
2 Schelkunff s Plynmial Representatin (fr discrete arrays) AF( ψ ) N n 0 A n e jnψ N number f elements in array Nte: each element can nw have a different excitatin cefficient A n ψ kd csθ + β 73
3 Schelkunff s let jnψ e AF( ) A + A + A + + A N N S, array factr fr a N-element N array is a plynmial f degree N- it has N- rts (ers) 74
4 Schelkunff s Can write: AF AN ( )( )( 3) ( N-) Where n are the rts f the plynmial 75
5 Schelkunff s Y Advantage -plane Shws explicitly the ers in the radiatin pattern (nulls) Can als think f it as the prduct f (N-) tw element arrays ψ X 76
6 Schelkunff s Hw t make this representatin useful??? Lk at -element array AF jψ + e + AF Fr kd π d λ There are n sidelbes! -π π ψ A, A 77
7 Schelkunff s Take the prduct f tw f these arrays AF ( + ) ( + ) element array with magnitudes f : : Sharper beam, but still n sidelbes A, A, A 78
8 Schelkunff s Cntinue AF ( + n ) Cefficients f this plynmial are the BINOMIAL COEFFICIENTS AF Ak k Gives an N-element array w/ n sidelbes A k nck n! k!( n k)! Similar t filters r transfrmers f type Butterwrth, Maximally flat, r Binmial 79
9 BINOMIAL COEFFICIENTS Ν Ν Ν 3 Ν 4 Ν 5 Ν 6 Ν 7 Ν (Add the tw numbers abve each entry) 80
10 Schelkunff s Placement f nulls Use factred frm f AF Example Place null at θ 6º In phase, β 0 3-element array λ/4 spacing (but nt unifrm) Just use amplitude shaping fr null placement ψ kd csθ (since β 0) 8
11 Schelkunff s Example ψ kd csθ S ne rt f plynmial at ψ ψ. 45π e j.45π Let ψ π λ cs 6 λ 4 π (.9).45π One mre er arbitrary make it symmetric ψ ψ. 45 π j.45 π e 8
12 AF ( - )( - ) j. 45π j. 45π ( e )( e ) j.45π j.45π ( e + e ) + ( π j π π j π) cs0.45 sin0.45 cs0.45 sin0.45 ( π ) + cs AF A + A + A Schelkunff s Example Amplitude cefficients A ; A.33; A 83
13 N 3 d λ/4 kd π/ β 0 NOTES Schelkunff s Example There is really a cheat here.. Allwing the center element t have a negative value is like having a β 80 Als, can chse nd. er in ther manners: e.g. if yu nly want ne null Chse ψ t be utside visible range, s then it will nt shw up as a null in the pattern θ 54 θ 6 A A -.33 A 84
14 Graphical aid Schelkunff s Example j.45π e Since e jψ All s have unit magnitude and are n unit circle in the cmplex plane (nly differ in phase) - - e j.45π jψ e VR ψ ) AF VISIBLE RANGE kd ψ kd π π ψ 85
15 Schelkunff s Example As mves arund unit circle frm ψ - π/ t π/ AF is prduct f lengths f the tw chrds - and - Gives maximum at ψ 0 and nulls at ψ ±.45 π VR e j.45 π - - e VR ψ ) jψ VISIBLE RANGE kd ψ kd π π ψ e j.45 π 86
16 Schelkunff s Example j.55π e - - j.45π e jψ e VR ψ ) Previus example, but want nly ne null at 6 ψ Chse.45π VR and ψ.55π (utside VR) θ 6 87
17 Schelkunff s Example Relatin between θ and As θ ψ kd csθ + β e jψ varies frm t θ 0 ψ s kd + β e j ( kd +β ) θ π ψ kd + β f e j ( kd +β ) 88
18 Schelkunff s Example Relatin between θ and Angular range f ψ kd is θ 0 ψ s Im kd + β ψ kd + β is VISIBLE RANGE VR jψ e ψ ) Re part f circle that crrespnds t physical angles 0 θ π θ π ψ f 89
19 Bradside VISIBLE RANGE kd ψ kd β 0 Schelkunff s kd kd VR ψ ) Arbitrary Phase Shift β kd + β VR ψ ) Same extent f VISIBLE RANGE but rtated by β kd + β kd + β ψ kd + β 90
20 Schelkunff s Visible Range and Spacing Extent f VR depends n spacing kd VR kd VR ψ ) ψ ) kd kd Smaller spacing bradside Larger spacing bradside (can verlap) 9
21 Schelkunff s Unifrm Array An (all A n ' s equal) AF N e n j ( n ) ) ψ e e jnψ jψ N ( but nt ) - because f factr Has N rts f N in denminatr m m ± j π N e m,,3, 9
22 Schelkunff s GRAPHICAL calculatin f AF frm rts d d d d d e jψ ψ ) d d 3 d 4 AF d d d3 d4 4 As pint mves arund the unit circle, yu get nulls in the AF when it hits the rts; abslute max at ; side lbe max. abut half way between rts. ψ 0 93
23 Schelkunff s GRAPHICAL calculatin f AF frm rts Can apprximate by measuring r calculating actual chrd lengths α α / d sin d α α ( ) d sin( ) s length f chrd can be calculated if yu knw the angular extent 94
24 Example 3 5-element unifrm array (4 th. rder plynmial with 4 rts) Schelkunff s GRAPHICAL calculatin f AF frm rts 7 Im 7 ψ ) 7 Re m π ± 5 e, e 4π ± ( ψ ± 7, ± 44 ) 4 d λ π λ kd π λ ψ s kd π ψ kd π f π π 4π 5 4π 5 π 5 π 5 VR ψ ) 95
25 Schelkunff s AF Example 3 ψ ) 4π 5 π 5 π 0 π 5 4π 5 π θ 90 θ 80 θ ) 96
26 Schelkunff s Example 3 Abslute maxima at ψ d d 3 7 d d ψ ) d d d 7 sin 4 44 sin d3 AF d d d d
27 44 d Example 3 08 d Sidelbe maxima halfway between 7 and Schelkunff s ψ ) d d d 36 sin d 08 sin 80 sin 3 4 SLL d d d d d 3 d 4 SLR SLR.4 [db] 98
28 Schelkunff s Example 3 As we saw in ne f the special cases n slide 6.47, t large a spacing gives t large a VR t prduce a secnd main beam (GRATING LOBE) Fr n prtin f Grating Lbe extend VR nly t last rt 4π 5 4π 5 π 5 π 5 VR ψ ) 99
29 Schelkunff s Example 3 VR 8π r 5 π 5 t 8π r 5 π 5 Fr mre general N-element array 8π π kd d 5 λ 4 dmax λ 0. 8λ 5 ( fr N 5) max d N λ N max 00
30 Schelkunff s Example 3 Reduced Sidelbes Since height f sidelbe is determined by distances t rts, if we mve rts clser tgether, the sidelbe level will g dwn. Fr bradside must cluster rts clser t ψ ±π. This enlarges main beam area and bradens the beamwidth. Purpse Try t reduce SLL (side lbe level) t -0 db 5-element unifrm array (rts at ± 7, ± 44 ) st. Sidelbe at -3.5 db nd. Sidelbe at -7.9 db d d 3 d 7 d 4 0 ψ ) 7 0
31 Trial & errr : try ± 87, ± 49 s rts are 6 apart instead f 7 SLL -8.5 db SLL.3 db Im 87 Schelkunff s Example 3 Trial & errr : This suggests that and 3 are t clse, but and are nt clse enugh try ± 88, ± 47 s rts are 59 and 66 apart Im ψ ) 6 Re ψ ) 59 Re
32 Schelkunff s Cnvergence Example 3 Im 89 Trial & errr : try ± 89, ± 45.5 s rts are 56.5 and 69 apart ψ ) 56.5 Re SLL SLL -0 db
33 AF Schelkunff s Example 3 Rts at: ±89 ±.55 rad ±45.5 ±.54 rad ( - )( - )( - )( - ) 3 4 ( j.55 )( j.54 )( j.55 e e e )( e j.54 ) ( cs( 89 ) + ) ( cs( 45.5 ) + )
34 Schelkunff s Example 3 Rts at: ±7 ±.56 rad ±44 ±.53 rad AF ( - )( - )( - )( - ) 3 4 j.56 j.53 j.56 j.53 ( e )( e )( e )( e ) ( cs( 7 ) + ) ( cs( 44 ) + )
35 Schelkunff s Example 3 Nte that Relative current cefficients Center element strngest Symmetric taper N lnger unifrm array Magnitudes tapered t give lwer SLL at expense f brader beamwidth 06
36 Schelkunff s Example 3 Summary UNIFORM REDUCED SIDELOBES BINOMIAL Cefficients Sidelbe Level -3.5 db -0 db N sidelbes 07
37 ECE 538/635 Antenna Engineering Spring 006 Dr. Stuart Lng Chapter 6 Part 8 Chebyshev s Arrays 08
38 CHEBYSHEV ARRAYS Ideally wuld like t design fr narrw beamwidth and lw sidelbe level, unfrtunately, nrmally ding ne results in the deteriratin f the ther. Frming the AF frm the Chebyshev plynmials ptimies the relatinship between beamwidth and sidelbe level. 09
39 CHEBYSHEV POLYNOMIALS T ( ) T ( ) T T T T ( ) 3 ( ) ( ) ( )
40 CHEBYSHEV Recurrence relatin Or functinally : T m eg.. T4 ( ) T ( ) T ( ) m m ( ) ( 3 ) ( 4 3 ) e. g. m n ( ) T [ T ( ) ] T [ T ( ) ] T T 4 m n ( ) T [ T ( ) ] n ( ) 4 ( 4 + ) T + m ( )
41 CHEBYSHEV T m ( ) Fig. 6.
42 Prperties f CHEBYSHEV Plynmials T m ( ). is f degree m in. Fr m even (dd) cntains in even (dd) pwers nly 3. All plynmials pass thrugh the pint, and either, r, (specifically the pint [, ( ) ] m ) ( ) ( ) ( ) [ ] ( ) 4. Over the interval + all plynmials m > have values between - and + ( extrema ±) ( ) 5. All rts (ers) m > 0 ccur within [ +] 3
43 Prperties f CHEBYSHEV Plynmials 6. Extrema at π p cs m p,,3, in [,] ers at ( p ) π + cs p 0,,, in m [,] 7. Fr > ; T ~ m m 8. T m ( ) cs( mcs ) + ( ) ( ) T csh mcsh < r >+ m [Nte: csh ln ± ( ) 4
44 CHEBYSHEV Chebyshev plynmial is ptimum in the sense that it minimies the distance between the last er and the value f that crrespnds t a certain value f the plynmial while keeping < ver + T m 5
45 CHEBYSHEV is largest er f T m is the value f that gives T m R T m ( ) R.0 6
46 CHEBYSHEV S fr a given value f R, the distance, is the smallest pssible value fr all plynmials f rder m Basic idea is t use the plynmial ver the range (-, +) as the side lbes and as far as necessary int the > regin fr the main lbe 7
47 CHEBYSHEV Side lbe levels are all equal Can either specify () Side lbe level belw main beam (in db) () Angular width f main beam Need t fit this plynmial with the general AF fr a symmetric, nn-unifrm array 8
48 CHEBYSHEV Frm text (pp ) Fr even number f elements M ( AF) nm M n an cs kdcsθ n 9
49 CHEBYSHEV Frm text (pp ) Fr dd number f elements M+ M + ( AF) ( ) nm+ n ( ) a cs n kdcsθ n With amplitude f center elements a 0
50 CHEBYSHEV let π d u csθ λ k d csθ (fr even) (fr dd) M ( AF) a cs ((n ) u) nm n M + ( AF) cs ( ) nm+ n n n ( ) a n u
51 CHEBYSHEV Rewriting each harmnic in terms f csu ( 0 u) cs ( u) csu cs cs cs cs ( u) cs u 3 ( 3 u) 4cs u 3csu 4 ( 4 u) 8cs u 8cs u +
52 CHEBYSHEV csu If let, then Chebyshev Plynmials cs cs cs cs cs ( 0) ( ) T ( u ) T ( ) ( u ) T ( ) 3 ( 3u ) 4 3 T ( ) 3 ( ) ( ) mu T m 3
53 CHEBYSHEV Want t match ( ) cs Chebyshev plynmials T t series Example 6.9 ( p 335 Balinis) 0-element Chebyshev w/ 6 [db] sidelbes, even # elements 4
54 Example 6.9 CHEBYSHEV d d d d λ λ 0 3λ λ.085 Fig. 6. Tschebyscheff plynmial f rder 9 (a) amplitude (b) magnitude 5
55 Example CHEBYSHEV 5-element Chebyshev w/ 0 [db] sidelbes (dd # f elements) M M + 5 ( ) ( ) a cs ((n ) u) n M+ n 5 n n ( AF ) n 5 cs u + a cs 4u ( ) ( ) ( 4 AF a + a cs u + a 8cs u 8cs u + ) n 5 a + a 3 AF AF [6.6] 3 3 6
56 Example CHEBYSHEV ( ) Find the pint such that T m R(sidelbe rati) where m # elements - want t use regin ( ) as the sidelbes null nearest + ( and the regin as the mainlbe ) 7
57 CHEBYSHEV Example Substitute csu / ( AF ) n 5 a + a / a + 8a 3 4 / 4 + 8a 3 / + a ( AF ) ( a a + a ) + ( a 8a )( / ) + ( 8a )( / ) n using T ( ) and equating cefficients [6.69] 8
58 9 CHEBYSHEV CHEBYSHEV a a a a a a ( ) a a a a a a Example Example
59 Example CHEBYSHEV -0 [db] sidelbes 0lg R lg R R 0 0 definitin R csh csh[ mcsh R mcsh csh R csh m csh csh R m ] (want inverse) 30
60 Example CHEBYSHEV fr ur case m 4 csh csh r withut using csh functin (where p #elem -) ( ) ( ) p R + R + R R ( ) ( ) p 3
61 Example CHEBYSHEV Calculate excitatin cefficients nrmalie such that a 3 a a 3 a 3.0 a 4a a + a a3 a fr dd # elements, center element a a 3 a a
62 Example CHEBYSHEV like reduced side lbe example (-0 db) with mving rts (trial and errr) 5-element array with rts at ± 89 ; ± 45.5 cefficients.0 /.6 /.95/.6 /.0 33
63 Example CHEBYSHEV T 4 ( ) R 5-element T 4 ( ) Depending n spacing d, different prtins f T 4 ( ) will be used fr the AF. largest used will be smallest will be θ 0 λ d + ψ ) 34
64 Example CHEBYSHEV AF Redraw in rectangular crdinates π d cs u cs csθ λ 4π 5 π 5 π 0 π 5 4π 5 π ψ ) θ 0 π d cs λ kd cs csθ θ 80 θ 90 θ ) 35
65 CHEBYSHEV Examples d d d λ λ 0 3 λ T m πd extends frm cs λ which crrespnds t 0 θ 90 kd vs. S visible range n plt ψ 0 36
66 CHEBYSHEV fr Chebyshev w/ largest pssible spacing w/ GRATING LOBES want t utilie Chebyshev plynmial up t, but nt beynd - with want d max λ π cs cs cs - u π d λ max cs π d λ cs θ 37
67 CHEBYSHEV Apprximatins fr Beamwidth and Directivity (fr Chebyshev nt near endfire and SLL -0 t -60 db).) Calculate beamwidth f unifrm array f same # elements and same spacing Θ h cs cs λ θ.443 cs cs θ L + d L λ + d r use Fig. 6- t apprximate [6.a] 38
68 CHEBYSHEV.) Calculate the beam bradening factr f f R csh [ ] (csh R ) π [6.4a] r read frm Fig. 6-4a 3.) BWCHEBY f BWUNIFORM 39
69 CHEBYSHEV Fig. 6-4a Beam bradening factr fr Chebyshev arrays. 40
70 CHEBYSHEV Fig. 6.4b Directivity f Chebyshev arrays. 4
71 CHEBYSHEV Directivity D R λ + ( R ) f L + d Or use Fig. 6-4b D 0.5 Θ h Even mre apprximately [like a gain/bandwidth prduct] Rule f thumb : HPBW 00 D 4
72 CHEBYSHEV Previus example Fr Unifrm 5-element, λ d L + d.5λ 5-element Cheby 0 [db] λ d R SLL 0 bradside Θ h cs cs λ.433 L + d.433 cs cs.5 λ.433 L + d
73 CHEBYSHEV fr 0 [ db] SLL f.0 Θ 0. 4 h D + R λ ( R ) f + ( 99) L + d 0 () r D 0.5 Θ h 4.97 Cmpare with 5-element BINOMIAL HPBW rad N 30.4 D.77 N
74 CHEBYSHEV Example [Prblem 6.47 Balinis Textbk] Chebyshev 3-element array dλ SLL - 0 [db] R 0 ( ) ( )
75 CHEBYSHEV Example ( AF ) a cs( ( n ) u) 3 n ( AF ) a + a cs( u) 3 n a+ a u ( AF ) ( ) 3 + ( ) cs a a acs u a a + acs csθ π d λ 46
76 CHEBYSHEV Example csu / and equate t T ( AF ) ( ) / 3 + ( ) n a a a equating ceff. a a a a a 4.5 and a / a a 5.5 (.345) 47
77 CHEBYSHEV Example Example Or nrmalied : Excitatin Cefficients a a /.636 /.0 Center element excitatin a 48
78 CHEBYSHEV Example Example thus ARRAY FACTOR ( AF ) + ( ) θ + 3 πd πd cs cs 0.8 cs csθ λ λ ( AF ) + ( ) ( ) + a a u a a a u cs 3 cs 49
79 CHEBYSHEV Example Example T find nulls fr d λ ( AF ) ( π θ ) cs cs 0 ( π θ ) cs cs 0.88 π θ ( ) cs n cs 0.88 n n ± 44.9 ±.59 [ rad] als at ±.59 ± π n [ rad] ± 5. ± [ rad] als at ± ± π n [ rad] 50
80 5 CHEBYSHEV CHEBYSHEV cs cs 6.7, cs cs cs 3.7, cs + ± ± ± + ± ± ± π π θ π π θ π θ π π θ π π θ π θ n n n n n n nulls at Example Example 6.7, 3.7, 66.3, 53.3 θ n
81 CHEBYSHEV Example Maxima ( AF ) cs( π csθ ) 3 max m ( π θ ) cs cs m π csθm cs () 0 ± 360 θ θ θ m m m 0 π π π π π cs 90 cs 0 cs 80 5
82 CHEBYSHEV Example relative maxima als exist between 53.3 and 66.3 and between 3.7 and 6.7 take derivative f (AF)3 and set 0 d ( AF) dθ 3 sin sinθ θ m ( π csθ )( π sinθ ) m 0 0,, m 80 sin m ( π csθ ) m 0 0 (abs. max. frm abve) 53
83 CHEBYSHEV Example Example sin m m ( π csθm ) 0 π csθm sin ( 0) 0 csθ 0 θ cs ( 0) 90 m m π csθ m π csθ m m ± π θ m ± π θ cs m cs ± ± mπ, 60 ( ± ) 0,80 m,0 s we find relative max. at 60 and 0 This prcedure des nt yield any nulls because the nulls ccur at axis crssings [Fr larger arrays this prcedure will be mre difficult] (need alternative ) 54
84 CHEBYSHEV Example [use knwn facts abut Cheby plynmials] (alternate methd) 3-element array 0 0 uses Cheby T ers at ±
85 CHEBYSHEV Example csu π cs d csθ λ fr cs d λ; ( π csθ ) n ±.345; nulls at ± ± π csθ cs n π csθ cs n n θ cs n θ cs ( ± 0.305).6 ± π ( ± 0.305).877 ± π 66.3 ± 7.45 ± r 3.7 r 6.7 ±.6 [ rad] ±.877 [ rad] 56
86 CHEBYSHEV Example frm graph max. at 0, ± ±. 345 fr ( π θ ) 0 cs cs 0 m + π θ ( ) π π m ± ± 3 m θm cs ± DOES NOT EXIST cs m cs 0 ± m 0 π csθm θm cs 60, 0 57
87 Example fr cs ± CHEBYSHEV ( π csθ ) ± π csθ cs ( ± ) m ±.345 m ± mπ m m m 0 π csθ π csθ m m π csθ m 0 θ m ± π θ m ± π θ m cs cs ( 0) 90 ( ± ) 0, 80 ( ± ) DOES NOT EXIST cs (same answers as befre) 58
88 Example 3 CHEBYSHEV [Prblem 6.33 Balinis Textbk] Binmial 4-element array d3λ/4 excit. ceff.,3,3, π 3λ kd λ 4 a 3 a N M 3π 4 M 59
89 Example 3 CHEBYSHEV ( AF) ( ) 4 M π d an cs (n ) u ; u csθ λ n ( AF) + 4 a csu a cs 3u 3csu+ cs3u 3 3csu+ 4cs u 3csu 3 3 π d 4cs u 4cs csθ λ 60
90 Example 3 CHEBYSHEV nulls when ( AF ) 4 π d θ λ 3 4cs cs 0 π d λ θ θ ( ) ( n + ) cs n cs 0 ± 0,,, cs n ± (n + ) λ d π n 6
91 Example 3 CHEBYSHEV θ fr n n m 0 3λ d 4 + cs ± 3 θ θ n n ( n ) cs cs ± 3 n 0,,, ( ± ) DOES NOT EXIST, -π π θ kd kd 3π 6
92 DOLPH-TSCHEBYSCHEFF (N. f Elements:0, Spacing:0.5 Wavelength) NORMALIZED EXCITATION COEFFICIENTS Side lbe level 80 db Side lbe level 60 db Side lbe level 40 db Side lbe level 30 db Side lbe level.05 db Side lbe level 0 db ARRAY LENGTH (λ) 63
93 8 DOLPH-TSCHEBYSCHEFF (N. f Elements:0, Spacing:0.5 Wavelength) DIRECTIVITY HPBW DIRECTIVITY (db) 6 30 HPBW (degrees) SIDE LOBE LEVEL (db) 64
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