ECE Spring Prof. David R. Jackson ECE Dept. Notes 32

Transcription

1 ECE 6345 Spring 215 Prof. David R. Jackson ECE Dept. Notes 32 1

2 Overview In this set of notes we extend the spectral-domain method to analyze infinite periodic structures. Two typical examples of infinite periodic problems: Scattering from a frequency selective surface (FSS) Input impedance of a microstrip phased array 2

3 FSS Geometry Reflected plane wave z Incident plane wave ( 1, ) (, 1) (,) (,1) y ( 1, ) ( mn, ) L W x b a Metal patch Dielectric layer Transmitted plane wave 3

4 FSS Geometry (cont.) ψ inc = Ae ( θ, φ ) ( ) jk x+ k y + jk z e x y z = arrival angles k = k sinθ cosφ x k = k sinθ sinφ y k = k cosθ z Reflected plane wave Incident plane wave z L W (, 1 ) x (, ) ( 1, ) b ( 1, ) (,1 ) a ( mn, ) y Metal patch Note: ψ denotes any field component of interest. Dielectric layer Transmitted plane wave Note: We are following plane-wave convention for k x and k y, and transmission-line convention for k z. 4

5 Microstrip Phased Array Geometry Probe current mn : j( kxma kynb) I = I e + mn z Probe W L (,) a ( mn, ) y b Dielectric layer Metal patch x Ground plane The wavenumbers k x and k y are impressed by the feed network. 5

6 Microstrip Phased Array Geometry (cont.) j( kxma kynb) I = I e + mn ( θ, φ ) = radiation angles k = k sinθ cosφ x k = k sinθ sinφ y Note: If the structure is infinite, a plane wave get launched. Antenna beam ( θ, φ ) z Probe W L (,) a ( mn, ) y b Dielectric layer Metal patch x Ground plane 6

7 Floquet s Theorem Fundamental observation: If the structure is infinite and periodic, and the excitation is periodic except for a phase shift, then all of the currents and radiated fields will also be periodic except for a phase shift. This is sometimes referred to as Floquet s theorem. z y x 7

8 Floquet s Theorem (cont.) From Floquet s theorem: ( ) ( ) jk J r J r r e t rmn = kt = xkx + yk mn s s mn ˆ ˆ z Layered media (,) y L W a r mn ( mn, ) b x 8

9 Floquet s Theorem (cont.) If we know the current of field at any point within the (,) unit cell, we know the current and field everywhere. A = ab = area of unit cell S z Layered media (,) y S L W a b x 9

10 Floquet Waves Let ψ denote any component of the surface current or the field (at a fixed value of z). ψ ψ ( +, ) = ψ (, ) (, + ) = ψ (, ) x ay xye xy b xye jk jk x y a b ψ where jk ( xx+ kyy) xy, = e Pxy, ( ) ( ) ( +, ) = (, ) (, + ) = (, ) P x a y P x y Pxy b Pxy z Layered media (,) y L W a r mn ( mn, ) b x 1

11 Floquet Waves (cont.) ψ jk ( xx+ kyy) xy, = e Pxy, ( ) ( ) From Fourier-series theory, we know that the 2D periodic function P can be represented as (, ) 2π p 2πq j x+ y a b pq p= q= Pxy A e = Hence we have ψ (, ) 2π p 2πq j kx+ x+ ky+ y a b pq p= q= xy A e = 11

12 Floquet Waves (cont.) Hence, the surface current or field can be expanded in a set of Floquet waves: ψ (, ) = xy A e p= q= pq ( xp yq ) jk x+ k y k k xp yq k + x k + y 2π p a 2π q b ( xp, yq ) ( pq, ) k k = wavenumbers of Floquet wave k = xk ˆ + yk ˆ tpq xp yq 2 2 ( ˆ ˆ π p q k ) ˆ π ˆ tpq = xkx + yk y + x + y a b Incident part Periodic part 12

13 Floquet Waves (cont.) Note: Each Floquet wave repeats from one unit cell to the next, except for a phase shift that corresponds to that of the incident wave. ψ Hence pq (, ) x+ ay = e ψ pq = = = = e ( xp ( ) yq ) j k x+ a + k y ( xp ) ( xp yq ) e j k a j k x+ k y j a j( kx a) a j( kxp x+ kyq y) e e e e e ( ) ( x a) j( kxp x+ kyq y) e j k ( a) j k x ψ pq 2π p ( a) ( xy, ) ( ) x x+ ay, = e jk ψ xy, pq Similarly, ψ pq jk ( y b) xy, + b= e ψ xy, ( ) pq ( ) 13

14 Periodic SDI The surface current on the periodic structure is represented in terms of Floquet waves: J xy a e s ( ) jk, tpq = p= q= pq r r= xx ˆ + yy ˆ To solve for the unknown coefficients, multiply both sides by and integrate over the (,) unit cell S : e jk tp q r S s ( ) jk r jk r jk r = pq S p= q= tp q tpq tp q, J x y e ds a e e ds = 2π 2π j ( p p ) x+ ( q q ) y a b pq p= q= S a e ds Use orthogonality: ( ) S jk tp q J x, y e ds a A s r = pq 14

15 Periodic SDI (cont.) Hence, we have: Therefore we have: 1 jktpq r apq = J s ( x, y) e ds A S 1 jktpq r apq = J s ( x, y) e ds A S = = = 1 A 1 A 1 A jk ( xpx+ kyq y) J x y e ds (, ) (, ) s xp yq s s J k k J ( ktpq ) so The current J s is the current on the (,) patch. 1 a = J k k A (, ) pq s xp yq 15

16 Periodic SDI (cont.) Hence the current on the 2D periodic structure can be represented as 1 (, ) (, ) jktpq Js xy = J s kxp kyq e A p= q= r We now calculate the Fourier transform of the 2D periodic current J s (x, y) (this is what we need in the SDI method): jktpq r jkxpx jkyq y F e = F e e = = jkxpx jkyq y + jkx ( x + ky y ) e e e dxdy jk x jk y + jk y xp + jkxx yq y e e dx e e dy ( kx kxp ) πδ ( ky kyq ) = 2πδ 2 16

17 Periodic SDI (cont.) Hence 1 J k k J k k k k k k A (, ) = (, ) 2πδ ( ) 2πδ ( ) s x y s xp yq x xp y yq p= q= Next, we calculate the field produced by the periodic patch currents: 1 jkx ( x + ky y ) E x, y, z = G k, k ; z, z J k, k e dk dk ( ) E k k z G k k z z J k k (,, ) = (, ;, ) (, ) ( 2π ) x y x y s x y 2 ( ) ( ) x y s x y x y 17

18 Periodic SDI (cont.) Hence, we have 1 E xyz,, = Gk, k; zz, ( ) ( 2π ) 2 1 A ( ) x y p= q= e jkx+ ky J k k k k k k (, ) 2πδ ( ) 2πδ ( ) s xp yq x xp y yq ( x y ) dk dk x y 18

19 Periodic SDI (cont.) Therefore, integrating over the delta functions, we have: 1 ( ) ( ) E x, y, z = G k, ;, (, ) xp kyq z z J s kxp kyq e A p= q= ( xp yq ) jk x+ k y The field is thus in the form of a double summation of Floquet waves. 19

20 Periodic SDI (cont.) Compare: Single element (non-periodic): 1 jkx ( x + ky y ) E x, y, z = G k, k ; z, z J k, k e dk dk ( ) ( 2π ) 2 ( ) ( ) x y s x y x y Periodic array of phased elements: 1 ( ) ( ) E x, y, z = G k, ;, (, ) xp kyq z z J s kxp kyq e A p= q= ( xp yq ) jk x+ k y 2

21 Periodic SDI (cont.) Conclusion: ( ) ( 2π ) 2 ( ) F k, k dk dk F k, k A x y x y xp yq p= q= where ( ˆ ˆ ) k = xk + yk tpq xp yq ( ) 2π p 2πq = xk ˆ ˆ ˆ ˆ x + yk y + x + y a b The double integral is replaced by a double sum, and a factor is introduced. 21

22 Periodic SDI (cont.) Sample points in the (k x, k y ) plane k y ( kxp, kyq ) k x 2 π /b ( kx, ky) 2 π /a 22

23 Microstrip Patch Phased Array Example z Find E x (x,y,) y L W x a b k k = k sinθ cosφ x = k sinθ sinφ y Microstrip Patch Phased Array 23

24 Phased Array (cont.) Single patch: 1 1 E xy,, = J k, k ( ) ( 2π ) + + ( ) x 2 2 sx x y kt 2 2 k k x y + D k D k ( ) ( ) m t e t e ( x y ) j k x+ k y dk dk x y J sx π x, = cos L ( xy) L cos kx π W 2 J sx ( kx, ky ) = LW sinc k y π L kx

25 Phased Array (cont.) 2D phased array of patches: k xp k yq Ex ( xy,,) = J 2 sx ( kxp, kyq ) + e A p= q= ktpq Dm ( ktpq ) De ( ktpq ) ( xp yq ) jk x+ k y where J sx π x, = cos L ( xy) L cos k π W J k k LW k 2 2 π L kxp 2 2 xp 2 sx ( xp, yq ) = sinc yq 2 2 k = k + k 2 2 tpq xp yq 25

26 Phased Array (cont.) The field is of the form where x (,,) E xy = A e = = p= q= p= q= p= q= A A pq pq pq e ψ ( xp yq ) jk x+ k y jk pq tpq r ( xy, ) 2π p 2πq k ˆ ˆ ( ˆ ˆ ) ˆ ˆ tpq = xkxp + yk yq = xkx + yk y + x + y a b The field is thus represented as a sum of Floquet waves. 26

27 Scan Blindness in Phased Array This occurs when one of the sample points (p,q) lies on the surface-wave circle (shown for (-2, )). Scan blindness from (-2,) k y ( kxp, kyq ) β TM k x ( kx, ky) 2 π /a (E-plane scan) 2 π /b 27

28 Scan Blindness (cont.) The scan blindness condition is: k = k = β ( for some ( pq, )) tpq tpq TM k xp k yq Ex ( xy,,) = J 2 sx ( kxp, knq ) + e A p= q= ktpq Dm ( ktpq ) De ( ktpq ) ( xp yq ) j k x+ k y ( ) ( ) D k D β m tpq = m TM = The field produced by an impressed set of infinite periodic phased surface-current sources will be infinite. 28

29 Scan Blindness (cont.) Physical interpretation: All of the surface-wave fields excited from the patches add up in phase in the direction of the transverse phasing vector: Proof: AF = A e sw = mn, mn, mn, A A mn mn mn k = xk ˆ + yk ˆ e tpq xp yq Start with the surface-wave array factor: e ( βtm m cosφ βtm n sinφ) + j x + y kxp kyq + j βtm x m + βtm y n k tpq k tpq kxp kyq + j βtm x m + βtm y n β TM β TM k xp cosφ = k tpq y φ N elements x y m n x r = ma = nb 29

30 Scan Blindness (cont.) Hence we have, in this direction, that y AF = A e sw = = mn, mn, mn, A A mn mn mn e e ( xp m yq n ) + jk x + k y 2π p 2πq + j k + x + k + y a b x m y n 2π p 2πq + j kx+ ( x+ ma) + ky+ ( y+ nb) a b + jk x + jk y 2π p 2πq + j ma + j nb a b mn mn, xp yq + j( kxma+ kynb) = e e A e e e = + jk x + jk y mn, xp yq e e A e ( x y ) + jkxpx + jkyq y j( kxma+ kynb) + j( kxma+ kynb) e e A e e + jk x + jk y mn, xp yq e e A N mn + j k ma+ k nb φ N elements k xp cosφ = k tpq x r 3

31 Scan Blindness (cont.) y TM surface wave φ In the direction φ, the surface fields from each patch add up in phase. x k xp cosφ = β TM N elements sw + jkxpx + jkyq y ( ) AF = N A e e Note: There is also an element pattern as well, with the field decaying as 1/ρ 1/2, but this is ignored here. 31

32 Scan Blindness (cont.) Example p = 2, q= k k xp yq = β = TM β β TM cosφ = = 1 TM y k y β TM x β ( ) TM kx, ky k x 2 π /b 2 π /a E-plane scan 32

33 Grating Lobes Grating lobes occur when one or more of the higher-order Floquet waves propagates in space. For a finite-size array, this corresponds to a secondary beam that gets radiated. ( θg, φg) ( θ, φ ) z Probe W L (,) a ( mn, ) y b Dielectric layer Metal patch x Ground plane 33

34 Grating Lobes (cont.) ( for some ( )) ktpq < k ( pq, ), Grating wave for (-1,) k y ( ) k xp, k yq k x k 2 π /b ( kx, ky) 2 π /a 34

35 Pozar Circle Diagram u k = k sinθ cosφ x v k = k Define H-plane scan sinθ sinφ y v Visible space circle φ φ uv u u + v = k sin θ = 2 2 u v k θ sin k tan φ = v/ u = tanφ uv φ = φ uv E-plane scan 35

36 Pozar Circle Diagram (cont.) tpq ( ) k < k uv visible space circle for, Grating Lobes So, we require that k + k < k xp yq for u + v < k The first inequality gives us or 2 2 2π p 2πq 2 x + + y + < k k k a b 2 2 2π p 2πq 2 sinθcosφ + + sinθsinφ + < k k k a b or 2 2 2π p 2πq u+ + v+ < k a b 2 36

37 Pozar Circle Diagram (cont.) or 2 2 2π p 2πq u + v < k a b 2 ( p) ( q) or u u v v k pq + < (we are inside a shifted (, ) visible space circle) u p 2π p 2πq, vq a b Summary of grating lobe condition: ( p) ( q) 2 2 u u + v v < k u + v < k Part of the interior of the (p,q) circle is also inside the visible space circle. 37

38 Pozar Circle Diagram (cont.) Grating lobe region ( p) ( q) 2 2 u u + v v < k u + v < k v 2 u v p q 2π p = a 2π q = b (, 1) ( up, vq) + k ( 1, ) k ( 1, ) u (,1) 38

39 Pozar Circle Diagram (cont.) This diagram shows when grating lobes occur in the principal scan planes. u = k sinθ cosφ v = k sinθ sinφ v (, 1) H-plane scan ( 1, ) ( 1, ) u (,1) E-plane scan 39

40 Pozar Circle Diagram (cont.) To avoid grating lobes for all scan angles, we require: u v > 2k 1 > 2k 1 (The circles do not overlap.) or 2π > a 2π > b 2k 2k or ka kb < π < π Hence a < λ /2 b < λ /2 4

41 Pozar Circle Diagram (cont.) k tpq TM Scan Blindness ( uv, ) = β for visible space circle So, we require that k + k = β xp yq TM The equation gives us for u + v < k 2 2 2π p 2πq 2 x + + ky + = βtm k a b or 2 2 2π p 2πq 2 sinθcosφ + + ksinθsinφ + = βtm k a b or 2 2 2π p 2πq u+ + v+ = β a b 2 TM 41

42 Pozar Circle Diagram (cont.) or 2 2 2π p 2πq u + v = a b β 2 TM or ( u up ) + ( v vq ) = βtm Summary of scan blindness condition: ( u u ) ( v v ) p q TM + = β u + v < k where u v p q 2π p = a 2π q = b Part of the boundary of the (p,q) circle is inside the visible space circle. 42

43 Pozar Circle Diagram (cont.) Scan blindness curve ( u u ) ( v v ) p q TM + = β u + v < k v u v p q 2π p = a 2π q = b (, 1) ( up, vq) + β TM ( 1, ) k ( 1, ) u (,1) 43

44 Pozar Circle Diagram (cont.) To avoid scan blindness for all scan angles, we require or u v > β + 1 TM > β + 1 TM k k (The circles do not overlap.) 2π > βtm a + k 2π > βtm b + k or 2π ka 2π kb > β / k + 1 TM > β / k + 1 TM a / λ < b / λ < Hence β β TM TM 1 1 / k + 1 / k