Directional Coupler. 4-port Network

Transcription

1 Diretional Coupler 4-port Network 3 4 A diretional oupler is a 4-port network exhibiting: All ports mathed on the referene load (i.e. S =S =S 33 =S 44 =0) Two pair of ports unoupled (i.e. the orresponding S i,j parameters are zero). Typially the unoupled ports are (,3) and (,4) or (,4) and (,3)

2 Charateristi Parameters It is here assumed that the unoupled ports are (-4) and (-3). The S parameters the ideal oupler must exhibit are: S =S =S 33 =S 44 =0, S 4 =S 4 =S 3 =S 3 =0. In this ase the ports (-), (-3), (3-4), (-4) are oupled. Let define C (Coupling) as: C= S 3, C db =-0 log( S 3 ) If the network is assumed lossless and reiproal, the unitary ondition of the S matrix determines the following relationships: S + S + S + S = S + C = S = C 3 4 S + S + S + S = S + S = S = S = C S + S + S + S = C + S = S = S = C Then, exiting the network at port (), the output power is divided between ports 3 and (4 and ) aording the fators C and -C. No power omes out of port 4 (3)

3 Further impliations of lossless ondition (symmetri struture) ( ) j( + ) * * j 4 S S S S S S e S S e + = 0 = = We assume that the network is symmetri: S =S 34, S 3 =S 4 Then = 34 and 3 = 4. From the previous ondition: = 4 ±/ This means that the outputs are in quadrature eah other. It an be demonstrated that it is suffiient to impose the mathing ondition to the four ports of a lossless and reiproal network to get a diretional oupler

4 Parameters of a real diretional oupler In a real oupler the mathing at the ports is not zeros at all the frequenies. It is then speified the minimum Return Loss in the operating bandwidth. The oupling parameter C, it is in general referred to the port with the smallest oupling. Moreover, in a real devie a not zero power arrives to the unoupled port. To haraterize this unwanted effet the isolation parameter (I) is introdued: I=Power to the oupled port/power to the unoupled port For the oupler onsidered in the previous slides (assuming port 3 that with the lowest oupling): I = S S 3 4 Note that for the ideal oupler I is infinite

5 Use of the diretional oupler () Measure of the refletion oeffiient V + V + L Line, Z G L C, S = C V V + + L V i V - C V L 3 4 V r V V S = j C V, V V S = j C V + + i 3 L r L 4 L V j C V V = = = G V r L L + + i j C V V L L L

6 Use of the diretional oupler () Power divider (C=3 db) P in P out = P in / Ports losed on the referene load P out = P in / 3 4 C=3dB + V = V = V S = V P = P, V = V = V S = V P = P

7 Use of the diretional oupler (3) Power ombiner (C=3 db) V V + =jv g / jv g V g V + 3=V g / 3 4 P S S = P = = S = = S = V g R 0 j Vg V g V + + g V = V = V S + V3 S3 = j + j = j, V4 = 0 V Vg = = = + 3 R0 4 R0 P P P Only if P e P 3 are equal

8 Use of the diretional oupler (4) Sum and differene of voltages (C=3 db) 0 V out V A = 3 = 4 =0 34 = 0 0 Ports losed on the referene load V B C=3dB V out V = V = V S + V S = V + V ( ) out A B 3 A B V = V = V S + V S = V V ( ) out 4 A 4 B 34 A B,

9 Gain: Refletion: Use of the diretional oupler (5) V + a a 0 90 C=3 db Balaned Amplifier V + a a a V + a 3a V + 3a = V V 3a = G in G in jv A A V + a jv b a 3b a a A A b 3b C=3 db V = A V, V = j A V V V P V = j + = j A V = A b 3b + out 4b a Pin V = V, V = G V, V = jv, a a a in a 3a a + a V3 a = jg inv a, V a = ( jv3 a + V a ) = G inv a + GinV a V G in = = 0 V a + a b V 4b

10 Coupled TEM lines When two transmission lines are plaed lose together, the propagation in eah line is influened on the other. We talk in this ase of oupled-line propagation. There are two possible modes of propagation whih are alled even and odd (symmetrial struture). Eah mode is haraterized by its harateristi impedane whose meaning is illustrated in the next slide

11 Modes in TEM oupled lines Even mode (Z e ) Symmetry axis Magneti wall (open) Odd mode (Z o ) Z e, Z o Symmetry axis Example: Eletri wall (short) open short Coupled oax Even mode Line (Z e ) Odd mode Line (Z o )

12 Ciruit model of two oupled lines with finite length bl Z e, Z o 3 4 Goal: ompute the 4-port Z matrix (or Y, or S). Hypothesis: equal lines (two symmetry axis) Evaluation method: matrix eigenvalues

13 Eigenvalues and Eigenvetors of a matrix The eigenvalues S l of a square matrix S are the solutions of the equation: det S U = 0 The eigenvetors x l assoiated to S represent the solution of the homogenous system of equations: S l S x = S x l l l A matrix of order n has n eigenvalues and n eigenvetors (eah eigenvetors has n elements). The eigenvetors are defined up to a onstant. Properties If a N-port network is exited with a vetors of urrents representing an eigenvetor of Z, you see the same impedane at all ports, and its value is just the eigenvalue. The same holds for all the other matries (Y, S, ) If there are symmetry axis in the network, the eigenvalues an be derived by on suitably defined iruits (eigeniruits). The eigenvetors are obtained by indution (the exitations must determine either an open or a short along the symmetry axis). One the eigenvalues and eigenvetors are known, simple equations define the elements of the orresponding matries.

14 Example: -port network -port Network S S S S Let assume that the two eigenvetors are known (we have assigned arbitrarily the largest element of eah x i equal to Eigenvetor : x + = -port G Network G Eigenvetor : x + = -port G Network G b b b b = = G (Eigenvalue ) b b b b = = G (Eigenvalue )

15 -port network (ont.) Eigenvetor : b = s + s b = s + s b b = = G Eigenvetor : b = s + s b = s + s b b = = G s G G, s G G G G = =, s = s = Relationship between eigenvalues of S, Y, Z S l Zl Z0 Y0 Y = = Z + Z Y + Y l 0 0 l l Z l + Sl = Z0 = S Y l l

16 Evaluation of the Eigenvetors In general there is no way to derive the eigenvetors of a generi N-port network. Only is the network is symmetri we an dedut the eigenvetors by exploiting the effet produed by the eigenvetors exitation. Let s onsider for instane the ase of a symmetri -port (S=S). Being the network symmetri we an obtain the same refletion oeffiient at port and port only when the inident waves at the two ports have the same magnitude. As a onsequene the only possible values for and are and -: Symmetri -port - Symmetri -port G e G e G o G o Eigenvetor (Even): x e + = + Eigenvetor (Odd ): x e + = s s G + G G G = =, s = s =, G = s + s, G = s s e o e o e o

17 -port symmetri network (ont.) If the first eigenvetor is onsidered, two equal exitations (amplitude and phase) at the two ports determines, for the network symmetry, an open iruit along the symmetry axis. The first eigenvalue is the imput refletion oeffiient (or impedane or admittane) of the even eigenetwork (one port): G e Eigenetwork (even) Open iruit The seond eigenvetor onsists in two equal but opposite exitations: a short iruit is then determined along the symmetry axis. The seond eigenvalue an be omputed from the odd eigenetwork (one port): G o Eigenetwork (odd) Short iruit

18 Eigenvalues evaluation of -oupled lines Symmetry axis Symmetry axis 3 4 Exiting the network with an eigenvetor, an eletri or a magneti wall is obtained along the two symmetry axis. With referene to Z matrix, the exiting urrents for eah eigenvetor result: I I I I l l l3 l 4 = +, +, +, + = +,, +, = +, +,, = +,,, + Axis : Magneti, Axis : Magneti Axis : Magneti, Axis : Eletri Axis :Eletri, Axis : Magneti Axis : Eletri, Axis : Eletri

19 Evaluation of the eigenvalues using the eigenetwork Eigenvalue Z l : /, Z e OPEN Z = jz ot ( l e ) Z l OPEN Eigenvalue Z l : /, Z e SHORT Zl = jz e tan ( ) Z l OPEN Eigenvalue Z l3 : /, Z o OPEN Zl 3 = jz o ot ( ) Z l3 SHORT Eigenvalue Z l4 : /, Z o SHORT Zl 4 = jz o tan ( ) Z l4 SHORT

20 Evaluation of Z Matrix From the definition of Z, imposing eah eigenvetor as exitation, the four independent elements of Z are obtained: V Z = = Z + Z + Z + Z l 3 4 I0 V Z = = Z Z + Z Z l 3 4 I0 V Z = = Z + Z Z Z l3 3 4 I0 V Z = = Z Z Z + Z l I0 Z = 4 Z + Z + Z + Z Z = 4 Z Z + Z Z Z = 4 Z + Z Z Z Z = 4 Z Z Z + Z ( ) l l l3 l 4 ( ) l l l3 l 4 ( ) 3 l l l3 l 4 ( ) 4 l l l3 l 4 Using Z li, the eigenvalues of the other matries (Y, S) an be obtained. The above formulas an be then used for omputing the elements of also these matries

21 Expression of Z matrix elements j Z = ot tan ot tan 4 Ze + Ze Zo + Zo j Z = Ze ot Ze tan Zo ot Zo tan 4 j Z3 = Ze ot + Ze tan + Zo ot Zo t 4 an j Z4 = Ze ot Ze tan Zo ot Zo tan 4 + +

22 Compat expressions of Z and Y elements Z Z Z Z 3 4 ( Z + Z ) ( Z + Z ) ( Z Z ) ( Z Z ) ( ) e o = j ot = j e o sin ( ) ( ) e o = j ot = j e o sin ( ) Y Y Y Y 3 4 ( Y + Y ) ( Y + Y ) ( Y Y ) ( Y Y ) ( ) e o = j ot = j e o sin ( ) ( ) e o = j ot = j e o sin ( )

23 =bl=80 Speial ases ( ) ( ) ( ) ( ) Z = j Z ot, Z tan, l Z ot, Z tan e e o o The eigenvalues of Z are [0,, 0, ], so those of S result: S li = [-,, -, ]. The sattering matrix elements are then: S = 0, S =, S = 0, S = Note that line is ompletely unoupled from line! Perfet mathing at the four ports There is a value of the load Z 0 for whih the ports are all mathed (S =S =S 33 =S 44 =0) independently on bl: Z 0 = Z e Z Note that the mathing at the ports does not imply the absene of refleted waves along the two lines. It means that the refleted waves are aneled only at the ports o

24 Derivation of the mathing ondition Eigenvalues of S S li = jx jx li li Z + Z Parameter S : S = ( Sl + Sl + Sl3 + Sl 4 ) = 0 4 There are only two ases where the above ondition an be satisfied independently on =bl, i.e.: ( Sl Sl) ( S S ) + = 0 + = 0 l3 l4 ( ) ( ) ( ) ( ) Z = j Z ot, Z tan, Z ot, Z tan = jx l e e o o li 0 0 X X = Z l l 0 X X = Z l3 l4 0 Z Z = Z e 0 = Z o 0 NOT Admissible ( Sl Sl4) ( S S ) + = 0 + = 0 l l3 X X = Z l l4 0 X X = Z l3 l 0 Z Z = Z e e o Z Z = Z o 0 0 Admissible

25 Coupled TEM lines as diretional oupler bl Z e, Z o 3 4 Using the admissible ondition we obtain the math at all port imposing: ( Sl Sl4) ( S S ) + = 0 + = 0 l l3 Z 0 = Z e This ondition also implies that port 4 is unoupled : Z o S4 = ( Sl Sl Sl3 + Sl 4 ) = 4 0 The maximum value S 3 defines the oupling: C=( S 3 ) max It is obtained for bl=/ : S = S + S S S = S + S = ( ) ( ) 3 max l l l3 l 4 l l 4 max max Z Z e e Z + Z o o

26 C/Cmax Variation of frequeny Mathing and Isolation are frequeny independent and equal to zero and infinity respetively (ideal lossless TEM line). The oupling varies with the frequeny aording to the following expression: C ( ) C Ze Z + C Z + Z max =, C max = ( max ) ot ( ) e o o B 0.5 db Cmax= For C max <0. the variation of C is pratially independent on C max. Note that the band for C/C max < 0.5 db is about 0.44 f f/f0 f 0 is the frequeny for whih bl=/

27 Pratial Restritions They onern mainly the maximum value of C max. In fat, inreasing C max, the lines beome loser and loser until the pratial implementation is no more possible with suffiient auray. Typially the maximum value of C max must be lower than 0. (C db =0) Example: Design a stripline oupler with C=0. with Z0=50 W using the Z following figures reporting the values of e Zo Cmax =, Z0 = Ze Zo Ze + Zo as a funtion of S and W (lines separation and width). Frequeny: GHz W=9.5 W= W=.8 S=0.95 Cmax Re(Eqn(Cmax)) Output Equations W=.8 S=0.95 Z0 Re(Eqn(Z0)) Output Equations W= S (mm) S (mm)

28 Solution: We draw the line C=0. on the first graph and the line Z 0 =50 on the seond graph. A point on eah of these lines has to be then found, whih is haraterized by the same pair of values (S, W). e r =.8 mm.8 mm mm 0 mm L L = 75 mm 0 b L = = Note: If a oupler dimensioned for the requested C but with a different value of Z 0 is available, impedane transforming networks an be used in plae of redesigning a new oupler. Z 0 Z 0 Z 0 C=0. Z 0 Z 0 Z 0 Z 0 Z 0 Z 0

29 Couplers with quasi-tem Lines If quasi-tem oupled lines are onsidered (Mirostrip), the phase veloity of the even and odd modes are not exatly oinident. Stritly speaking, that would not allow to apply the model here assumed for the haraterization of the diretional oupler. In the pratie, until the differene between the two veloities is not too large, the same phase veloity an be assumed for both modes (equal to the average of the atual values), assuming the lines as TEM. There are however some differenes omparing the performanes with the ideal TEM oupler (a perfet mathing is no more possible) mm mm 0-5 S Mirostrip Coupler 000. MHz db e r = mm.575 mm S3 S MHz db MHz db S Frequeny (MHz) m= 0. S = mm Z = Z Z = 74. e 0 p d =.97, e =.7 eff, p eff, d 0L e eff, medio =.84, b L = e eff, medio = L = 55.4 mm

30 Y, Y, Coupler with lumped ouplings To realize ouplers with a large oupling (C<0 db) struture with lumped ouplings are used. In planar tehnology two kinds of suh ouplers are employed, the branh-line and the rat-rae. Branh-line Y, 3 Y, 4 l 4 0 = L=

31 Evaluation of the eigenvalues Y, Y, Y, B = Y+ Y Bd = Y Y s Y, l tan ( 45 )( ) ( ) ( ) ( ) Y = j Y + Y = j Y + Y S = Y jb Y + jb l 0 s 0 s l3 tan ( 45 ) ot ( 45 ) ( ) ( ) ( ) Y = jy jy = j Y Y S = Y jb Y + jb l3 0 d 0 d Y, Y, l Y, tan ( 45 ) tan ( 45 ) ( Y ) ( ) ( ) Y = jy + jy = j Y S = Y + jb Y jb l 0 d 0 d Y, l 4 tan ( 45 ) ot ( 45 ) ( Y ) ( ) ( ) Y = jy jy = j Y + S = Y + jb Y jb l 4 0 s 0 s

32 Coupling onditions All ports mathed, ports -3/-4 unoupled: S =0 and S 3 =S 4 =0 S = ( Sl + Sl + Sl3 + Sl 4 ) = 0 4 S3 = ( Sl + Sl Sl3 Sl 4 ) = 0 4 S + S = 0, S + S = 0 l l l3 l 4 This ondition is atually feasible, giving in fat: BB Y s d 0 = Y Y = Y 0

33 For S e S 4 we have: jb S = ( S S + S S ) = ( S S ) = + s l l l3 l 4 l l 4 4 bs b S = S S S + S = S + S =, b = s 4 ( l l l3 l 4 ) ( l l 4 ) 4 + bs s B Y s 0 Moreover, the unitary ondition of S implies the following relation between the phases of S and S 4 : f f4 = Then, being f =-/ f 4 =. The parameter b s must be then > for having S 4 negative. By imposing now the requested oupling ( S 4 =C), we an obtain the requested value of b s : b + C Y+ Y S = = C b = = s 4, s + bs C Y0 Y Y Y0 Taking into aount the first found ondition ( finally obtain the expressions of Y e Y : = ) we Y = Y, Y = Y C 0 0 C C

34 Branh-line oupler with C=0.5 (3 db) The ouplers with C=3 db are identified with the word Hybrid. To realize an hybrid of branh-line type the harateristi impedanes of the lines must be: 0.5 Z = Z0 0.5 = W, Z = Z0 = 50 W Z0 = 50 W 0.5 Pratial restritions ( ) One an easily verify that, with C tending to 0: Z Z 0 and Z. In the pratie, even with on C=0. (0 db) the orresponding value of Z is very diffiult to realize (Z =3Z 0 ). Usually, C must be between 3 and 6 db. Frequeny dependene For this devie, both mathing and isolation vary with frequeny (the nominal value is obtained at the frequeny where the length of the lines is l 0 /4). Also the oupling depends on f (the max is again at f 0 ). The bandwidth for a given value of maximum oupling inreases with C max ; Usually, the frequeny variation of mathing and isolation is more pronouned than that of the oupling.

35 Branh Line: a summary Conditions to be imposed: S = S = S = S = 0, S = S = S = S = C 4 3 S = S = C 34 Y Y Y -90 Design equations: Y = Y, Y = Y C 0 0 C C =0 Y S parameters obtained: S = S = C, S = S = j C For C=0.5 (3dB), it has (assuming Z 0 =/Y 0 =50W): 0.5 Z = Z0 0.5 = W, Z = Z0 = 50 W Z0 = 50 W 0.5 ( )

36 0-5 Aoppiamento Branh-line C=3dB Adattamento Isolamento DB( S(,) ) Shemati DB( S(,) ) Shemati DB( S(3,) ) Shemati DB( S(4,) ) Shemati Frequeny (MHz)

37 Rat-rae oupler 3 l 0 4 Y l 0 4 Y Y l 0 4 P P Y Y P3 Y P4 Y 4 Y 3l0 4 There is only one symmetry axis (vertial) It is anyhow possible to still use the eigenvalues method for finding the dimensioning equations

38 Conditions to be imposed: S = S = S = S = 0, S = S = S = S = C Y Y 4 3 Y -90 =0 S = S = C Y Design equations: Y = Y C, Y = Y C Y Y 4 3 =0 Y 90 S parameters obtained: Y S = j C, S = j C, S = S = j C Z For C=0.5 (3dB), we obtain (assuming Z 0 =/Y 0 =50W): = Z = Z = W, Z Z Z = = = W C 0.5 C 0.5

39 Parameters dependene on f 0-5 Rat-rae C=3dB Aoppiamento Adattamento Isolamento DB( S(,) ) RatRae DB( S(,4) ) RatRae DB( S(,3) ) RatRae DB( S(,) ) RatRae Frequeny (GHz) The bandwidth is larger than that of the branh-line With the dereasing of the oupling the bandwidth inreases The pratial feasibility limits the oupling between about 3 and 8 db

40 Br Br Br Br Couplers with lumped elements In some ases, the pratial implementation of planar ouplers with C in the range 0-5 db may be diffiult using distributed elements. A possible alternative is represented by a struture similar to the branh line but employing lumped omponents (for the most ritial elements). Le onsider the following (symmetrial) 4-port onfiguration, onstituted by lumped suseptanes (also the eigen-networks are shown in the figure): Br Br Bb S l Ba S l3 Bb Bb Ba 3 Ba 4 Bb Ba S l Br Br S l 4

41 Eigenvalues expressions (b i =B i /Y o ): ( ) ( ) ( ) ( ) ( ) ( ) jb j b + b j b + b j b + b + b S =, S =, S =, S =, r r a r b r a b l l l3 l 4 + jbr + j br + ba + j br + bb + j br + ba + bb Now impose S =0 and S 3 =0 (unoupled ports -3): S = ( Sl + Sl + Sl3 + Sl 4 ) = 0 4 S3 = ( Sl + Sl Sl3 Sl 4 ) = 0 4 Imposing the oupling C= S 4 : S = S S S + S = S S = ( ) ( ) S + S = 0, S + S = 0 l l l3 l 4 ( ) b = b + b r a b j4b b l l l3 l 4 l l3 4 ba + bb jba If we assume S real, b a -b b =, and: 3 Finally: b S = C = b ( ) b =, b C, b b b C = C = + b a ( ) a b r a b

42 Example of implementation We observe that Ba and Bb are always positive, so an be implemented with apaitanes. Br is instead negative and an be realized with a short iruited stub with harateristi impedane Z 0. Assume CdB=0 db (C=0.), f0=945 MHz and Y 0 =/50. From the previous formulas we get: b a =.054, b b =0.3333, b r = The apaitanes implementing Ba and Bb are given by: Ca=3.55 pf, Cb=. pf. The suseptane B r is realized with a short iruited stub with Z =50 W and bl=atan(-/b r )= The final iruit is shown below together with the expressions of S and S4 + ( Ca) ( Cb) ( ) ( ) ( ) ( ) S =, for = 0 S = j C + C jc 0C S a b a jc C =, for = S = C + C jc C b a b a b a a =0 C a C b C b 3 4 C a 80 90

43 Frequeny response of the lumped oupler S4 Lumped Coupler S S DB( S(,) ) Test DB( S(,) ) Test DB( S(,3) ) Test DB( S(,4) ) Test Frequeny (MHz) S

44 3-port lossless networks A 3-port reiproal lossless network annot be mathed at the 3 ports (i.e. S =S =S 33 =0 not possible). In fat, imposing the unitary of S: S + S = 3 S + S = S = S = S = 0.5 S S = 3 3 These onditions are inompatible so it is impossible to have S =S =S 33 =0 S S * 3 3 * * 3 = 0 S S = 0 S =0 or S = 0 or S = 0 S S = 0

45 Is it still possible to realize a power splitter with a 3-port? Possible solution: a 3 db hybrid with the unoupled port losed on a mathed load P in Z 0 C=3dB Porta 4 3 P out= P in / disaoppiata P out = P in / The 3-port network is lossy due to the presene of Z 0. This resistor however does not dissipate power beause the port 4 is unoupled. Pin is then split between ports and 3 without losses Ciruito a 3 porte

46 A 3-port divider: the Wilkinson network Z 0 l 0 4 Z Z l 4 0 R W 3 Z 0 Z 0 Due to the presene of R W the 3- port network is lossy, so the ondition S =S =S 33 =0 an be imposed -port network obtained by losing port with Z 0 : Z Z 3 l 4 0 R W Z 0 l 0 4 S =S 33 and S 3 an be omputed thorugh the eigenvalues of this network

47 R W R W Even Network Z Z 3 l 4 0 Z l 4 0 Z 0 Z 0 Odd Network Z e Z l 0 4 Z 0 Z o R W Z l 0 4 Z e Z Z Z = G = Z Z Z 0, e S e o e o 3 Z o RW RW Z =, G o = R Z G + G G G = = 0, S = = 0 G e = 0, G o = 0 Z = Z, R = Z 0 W 0 W 0 0 For Z 0 =50W Z =70.7W, R W =00W

48 Evaluation of S Exiting port with ports and 3 losed with Z 0, there is no urrent through R W, so it an be disarded. l 0 4 Z 0 Z in Z Z l 4 0 Z 0 Being Z = Z0 also S =0. Z in Z Z Z = G = = Z Z Z 0, in S Evaluation of S = S 3 Power entering port is not refleted and there is there is no dissipation in R W. So the power is all transferred to ports e 3; for the symmetry, the power exiting at eah port is the half of Pin: S = S3 = 0.5 The phase for both parameters is -90.

49 Frequeny response 0-5 Divisore di Wilkinson Trasmissione Adattamento DB( S(,) ) Ideale DB( S(,) ) Ideale DB( S(3,) ) Ideale Frequeny (GHz) The bandwidth for RL=0 db is about 40% of f 0 Transmission ( S = S 3 ) is independent on frequeny Dissipation in R W is zero provided that the load at ports and 3 is the same

50 Mirostrip implementation The very small size of Rw (pseudo lumped omponent) must be aounted for. The two output lines must be enough lose to allow the onnetion of Rw. R W On the other hand is not advisable to have the output lines too lose eah other beause an unwanted oupling may arise. So diverging lines are often used.

51 Example of derivation of S parameters from the eigenvalues of S matrix f f Z jb f Z G e Z jb jb/ Even eigenetwork f Y j B G e = GBexp = exp Y + j B ( j f) ( j f) ( j f) ( j f) G = G exp = exp o G o Z Odd eigenetwork G e + G o Y exp ( j B jb S = S = = j f) = exp ( j f) Y + j B Y + jb Ge Go Y exp ( j B Y ) exp ( S = S = = j f + = j f) Y + j B Y + jb