ECE 538/635 Antenna Engineering Spring 006 Dr. Stuart Lng Chapter 6 Part 7 Schelkunff s Plynmial 7
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
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
Schelkunff s Can write: AF AN ( )( )( 3) ( N-) Where n are the rts f the plynmial 75
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
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
Schelkunff s Take the prduct f tw f these arrays AF ( + ) ( + ) + + 3-element array with magnitudes f : : Sharper beam, but still n sidelbes A, A, A 78
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
BINOMIAL COEFFICIENTS Ν Ν Ν 3 Ν 4 Ν 5 Ν 6 Ν 7 Ν 8 3 3 4 6 4 5 0 0 5 6 5 0 5 6 7 35 35 7 (Add the tw numbers abve each entry) 80
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
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
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 ) + cs0.45 0.33 AF A + A + A Schelkunff s Example Amplitude cefficients A ; A.33; A 83
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
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
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
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
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
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
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
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
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
Schelkunff s GRAPHICAL calculatin f AF frm rts d d d 3 - - - 3 d d e jψ ψ ) d 4-4 3 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
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
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 ± 5 3 7 7 ( 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
Schelkunff s AF Example 3 ψ ) 4π 5 π 5 π 0 π 5 4π 5 π θ 90 θ 80 θ ) 96
Schelkunff s Example 3 Abslute maxima at ψ 0 44 44 d d 3 7 d d 4 7 0 ψ ) d d d 7 sin 4 44 sin d3 AF d d.75.90 d d3 4 4.999 97
44 d Example 3 08 d Sidelbe maxima halfway between 7 and 44 7 0 Schelkunff s ψ ) d d d 36 sin d 08 sin 80 sin 3 4 SLL d d d 3.68.0 d 0.68 4.36 44 d 3 d 4 SLR.36 5.0 0.47 7 SLR.4 [db] 98
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
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
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 44 44 08 d d 3 d 7 d 4 0 ψ ) 7 0
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 88 49 6 ψ ) 6 Re 47 59 ψ ) 59 Re 3 49 6 3 47 59 4 87 4 88 0
Schelkunff s Cnvergence Example 3 Im 89 Trial & errr : try ± 89, ± 45.5 s rts are 56.5 and 69 apart 45.5 3 45.5 56.5 56.5 ψ ) 56.5 Re SLL SLL -0 db 89 4 03
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 ) + ) 4 3 +.6 +.95 +.6 + 04
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 ) + ) 4 3 + + 05
Schelkunff s Example 3 Nte that Relative current cefficients Center element strngest.6.95.6 Symmetric taper N lnger unifrm array Magnitudes tapered t give lwer SLL at expense f brader beamwidth 06
Schelkunff s Example 3 Summary UNIFORM REDUCED SIDELOBES BINOMIAL Cefficients.6.95.6 4 6 4 Sidelbe Level -3.5 db -0 db N sidelbes 07
ECE 538/635 Antenna Engineering Spring 006 Dr. Stuart Lng Chapter 6 Part 8 Chebyshev s Arrays 08
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
CHEBYSHEV POLYNOMIALS T ( ) T ( ) T T T T ( ) 3 ( ) 4 3 3 4 ( ) 8 8 4 + ( ) 5 3 6 0 5 5 + 0
CHEBYSHEV Recurrence relatin Or functinally : T m eg.. T4 ( ) T ( ) T ( ) m m ( ) ( 3 ) ( 4 3 ) 8 4 8 + e. g. m n ( ) T [ T ( ) ] T [ T ( ) ] T T 4 m n ( ) T [ T ( ) ] n ( ) 4 ( 4 + ) 4 8 4 8 T + m ( )
CHEBYSHEV T m ( ) Fig. 6.
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
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
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
CHEBYSHEV is largest er f T m is the value f that gives T m R T m ( ) R.0 6
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
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
CHEBYSHEV Frm text (pp. 36-38) Fr even number f elements M ( AF) nm M n an cs kdcsθ n 9
CHEBYSHEV Frm text (pp. 36-38) Fr dd number f elements M+ M + ( AF) ( ) nm+ n ( ) a cs n kdcsθ n With amplitude f center elements a 0
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
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 +
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
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
Example 6.9 CHEBYSHEV d d d d λ 4 0.767 λ 0 3λ 4 0.767 λ.085 Fig. 6. Tschebyscheff plynmial f rder 9 (a) amplitude (b) magnitude 5
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
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
CHEBYSHEV Example Substitute csu / ( AF ) n 5 a + a / a + 8a 3 4 / 4 + 8a 3 / + a 3 4 4 ( AF ) ( a a + a ) + ( a 8a )( / ) + ( 8a )( / ) n 5 3 3 3 4 using T ( ) 8 8 4 + and equating cefficients [6.69] 8
9 CHEBYSHEV CHEBYSHEV 4 4 3 3 8 8 a a 4-4 4 4 4 8 8 3 3 a a a a ( ) 4 4 4 4 4 3 4 4 3 3 + + + + + a a a a a a Example Example
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
Example CHEBYSHEV fr ur case m 4 csh csh 4 0.93 r withut using csh functin (where p #elem -) ( ) ( ) p R + R + R R ( ) ( ) 4 0 + 99 + 0 99 4. 93 p 3
Example CHEBYSHEV Calculate excitatin cefficients nrmalie such that a 3 a3.798 4 a 3 a 3.0 a 4a 3 4 4.50 a + a a3 a 5.407.704 fr dd # elements, center element a a 3 a a 3.609.93 3
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
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
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
CHEBYSHEV Examples d d d λ 0.707 4 λ 0 3 λ 0.707 4 T m πd extends frm cs λ which crrespnds t 0 θ 90 kd vs. S visible range n plt ψ 0 36
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
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 +. 443 L + d L λ + d r use Fig. 6- t apprximate [6.a] 38
CHEBYSHEV.) Calculate the beam bradening factr f f + 0.636 R csh [ ] (csh R ) π [6.4a] r read frm Fig. 6-4a 3.) BWCHEBY f BWUNIFORM 39
CHEBYSHEV Fig. 6-4a Beam bradening factr fr Chebyshev arrays. 40
CHEBYSHEV Fig. 6.4b Directivity f Chebyshev arrays. 4
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
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.433.5 00. 79.8 0.4 43
CHEBYSHEV fr 0 [ db] SLL f.0 Θ 0. 4 h D + R λ ( R ) f + ( 99) L + d 0 ().5 4.93 r D 0.5 Θ h 4.97 Cmpare with 5-element BINOMIAL HPBW.06.53 rad N 30.4 D.77 N 3.96 44
CHEBYSHEV Example [Prblem 6.47 Balinis Textbk] Chebyshev 3-element array dλ SLL - 0 [db] R 0 ( ) ( ) 0 + 99 + 0 99. 345 45
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
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
CHEBYSHEV Example Example Or nrmalied : Excitatin Cefficients a a 0.88.0.0 /.636 /.0 Center element excitatin a 48
CHEBYSHEV Example Example thus ARRAY FACTOR ( AF ) + ( ) θ + 3 πd πd 0.88.0 cs cs 0.8 cs csθ λ λ ( AF ) + ( ) ( ) + a a u a a a u cs 3 cs 49
CHEBYSHEV Example Example T find nulls fr d λ ( AF ) ( π θ ) 3 0.88 + 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. ± 3.754 [ rad] als at ± 3.754 ± π n [ rad] 50
5 CHEBYSHEV CHEBYSHEV 66.3 3.754 cs 3.7 3.754 cs 6.7, 53.3 3.754 cs 53.3.59 cs 6.7.59 cs 3.7, 66.3.59 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 6 3 66 53
CHEBYSHEV Example Maxima ( AF ).88 0.88 + 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
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
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
CHEBYSHEV Example [use knwn facts abut Cheby plynmials] (alternate methd) 3-element array 0 0 uses Cheby T ers at ± 0. 707-3 - - 0 3 -.345 -.707.707.345 55
CHEBYSHEV Example csu π cs d csθ λ fr cs d λ; ( π csθ ) n ±.345; 0.707.345 nulls at ± 0.305 ± 0. 707 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 ± 07.5 53.3 r 3.7 r 6.7 ±.6 [ rad] ±.877 [ rad] 56
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
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
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
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
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
Example 3 CHEBYSHEV θ fr n n m 0 3λ d 4 + cs ± 3 θ θ n n ( n ) cs cs ± 3 n 0,,, 48. 3.9 ( ± ) DOES NOT EXIST, -π π θ kd kd 3π 6
DOLPH-TSCHEBYSCHEFF (N. f Elements:0, Spacing:0.5 Wavelength) 0.9 0.8 NORMALIZED EXCITATION COEFFICIENTS 0.7 0.6 0.5 0.4 0.3 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 0. 0. 0 - -0.8-0.6-0.4-0. 0 0. 0.4 0.6 0.8 ARRAY LENGTH (λ) 63
8 DOLPH-TSCHEBYSCHEFF (N. f Elements:0, Spacing:0.5 Wavelength) DIRECTIVITY HPBW 50 7 40 DIRECTIVITY (db) 6 30 HPBW (degrees) 5 0 4 0 0 0 30 SIDE LOBE LEVEL (db) 64