ECE 546 Lecture 07 Multiconductors Spring 2018 Jose E. Schutt-Aine Electrical & Computer Engineering University of Illinois jesa@illinois.edu ECE 546 Jose Schutt Aine 1
TELGRAPHER S EQUATION FOR N COUPLED TRANSMISSION LINES V 1 (z) V 2 (z) z=0 z=l... V 3 (z) L, C V z L I t I z C V t V and I are the line voltage and line current VECTORS respectively (dimension n). ECE 546 Jose Schutt Aine 2
ECE 546 Jose Schutt Aine 3 N-LINE SYSTEM L and C are the inductance and capacitance MATRICES respectively 1 2 n I I I I 1 2 n V V V V 11 12 21 22 nn L L L L L L 11 12 21 22 nn C C C C C C V 3 (z) V 2 (z) V 1 (z)... L, C z=0 z=l
N-LINE ANALYSIS V z V LC t 2 2 I z 2 2 I CL t 2 2 2 2 In general LC CL and LC and CL are not symmetric matrices. GOAL: Diagonalize LC and CL which will result in a transformation on the variables V and I. Diagonalize LC and CL is equivalent to finding the eigenvalues of LC and CL. Since LC and CL are adjoint, they must have the same eigenvalues. ECE 546 Jose Schutt Aine 4
MODAL ANALYSIS EV z EV t 2 2 1 ELCE 2 2 HI z HI t 2 2 1 HCLH 2 2 LC and CL are adjoint matrices. Find matrices E and H such that ELCE HCLH 1 1 2 m is the diagonal eigenvalue matrix whose entries are the inverses of the modal velocities squared. ECE 546 Jose Schutt Aine 5
MODAL ANALYSIS V 1 (z) V 2 (z) z=0 z=l... V 3 (z) L, C I z 2 2 m 2 m 2 m 2 V z I t V t 2 2 m 2 m 2 m 2 V m = EV I m = HI Second-order differential equation in modal space ECE 546 Jose Schutt Aine 6
ECE 546 Jose Schutt Aine 7 V m and I m are the modal voltage and current vectors respectively. EIGENVECTORS 1 2 m m m mn V V V V 1 2 m m m mn I I I I I m = HI V m = EV
EIGEN ANALYSIS * A scalar is an eigenvalue of a matrix A if there exists a vector X such that AX = X. (i.e. for which multiplication by a matrix is equivalent to an elongation of the vector). * The vector X which satisfies the above requirement is an eigenvector of A. * The eigenvalues of A satisfy the relation A - I = 0 where I is the unit matrix. ECE 546 Jose Schutt Aine 8
EIGEN ANALYSIS Assume A is an n x n matrix with n distinct eigenvalues, then, * A has n linearly independent eigenvectors. * The n eigenvectors can be arranged into an n n matrix E; the eigenvector matrix. * Finding the eigenvalues of A is equivalent to diagonalizing A. * Diagonalization is achieved by finding the eigenvector matrix E such that EAE -1 is a diagonal matrix. ECE 546 Jose Schutt Aine 9
EIGEN ANALYSIS For an n-line system, it can be shown that * LC can be transformed into a diagonal matrix whose entries are the eigenvalues. * LC possesses n distinct eigenvalues (possibly degenerate). * There exist n eigenvectors which are linearly independent. * Each eigenvalue is associated with a mode; the propagation velocity of that mode is the inverse of the eigenvalue. ECE 546 Jose Schutt Aine 10
EIGEN ANALYSIS * Each eigenvector is associated to an eigenvalue and therefore to a particular mode. * Each normalized eigenvector represents the relative line excitation required to excite the associated mode. ECE 546 Jose Schutt Aine 11
EIGENVECTORS E : Voltage eigenvector matrix 1 2 ELCE m e e e E e e e e e e 11 12 13 21 22 23 31 32 33 H : Current eigenvector matrix 1 2 HCLH m h h h H h h h h h h 11 12 13 21 22 23 31 32 33 ECE 546 Jose Schutt Aine 12
1 2 ELCE m gives 1 2 HCLH m gives Eigenvalues and Eigenvectors e e e E e e e e e e 11 12 13 21 22 23 31 32 33 h h h H h h h h h h 11 12 13 21 22 23 31 32 33 1 0 0 v m1 1 m 0 0 v m2 1 0 0 v m3 1 0 0 v m1 1 m 0 0 v m2 1 0 0 v m3 ECE 546 Jose Schutt Aine 13
Modal Voltage Excitation Voltage Eigenvector Matrix e e e E e e e e e e 11 12 13 21 22 23 31 32 33 e 11 e 12 + + + e 13 - - - MODE A e 21 e 22 + + + e 23 - - - MODE B MATCHING NETWORK MATCHING NETWORK e 32 + + + e 31 e 33 - - MODE C - MATCHING NETWORK ECE 546 Jose Schutt Aine 14
Modal Current Excitation Current Eigenvector Matrix h 11 h 12 h 13 MODE A MATCHING NETWORK h h h H h h h h h h 11 12 13 21 22 23 31 32 33 h 21 h 22 h 23 MODE B MATCHING NETWORK MATCHING NETWORK h 31 h 32 h 33 MODE C ECE 546 Jose Schutt Aine 15
EIGENVECTORS E : voltage eigenvector matrix H : current eigenvector matrix E and H depend on both L and C Special case: For a two-line symmetric system, E and H are equal and independent of L and C. The eigenvectors are [1,1] and [1,-1] Line variables are recovered using V E 1 V m I H 1 I m ECE 546 Jose Schutt Aine 16
MODAL AND LINE IMPEDANCE MATRICES By requiring V m = Z m I m, we arrive at Z m m 1 ELH 1 Z m is a diagonal impedance matrix which relates modal voltages to modal currents By requiring V = Z c I, one can define a line impedance matrix Z c. Z c relates line voltages to line currents Z m is diagonal. Z c contains nonzero off-diagonal elements. We can show that Z c E 1 Z m H E 1 m 1 EL ECE 546 Jose Schutt Aine 17
ECE 546 Jose Schutt Aine 18 1 2 ( ) m m mn j u v j u v j u v e e Wu e 1 2 1 1 1 m m m mn v v v
MATRIX CONSTRUCTION L ii L ( m) s L L ij ij n ( m) ii s ij j1 [ C] C C [ C] ij C ( m) ij [ R] ii R s [ R] ij R m ij n ( m) ii s ij j1 [ G] G G ( m) [ G] ij G ij ECE 546 Jose Schutt Aine 19
Modal Variables V m = EV I m = HI General solution in modal space is: V m (z) = W(z)A +W(- z)b - I m (z) = Z 1 m [ W(z)A- W(- z)b] Modal impedance matrix Z m ELH 1 1 m Line impedance matrix Z E Z H E EL 1 1 1 c m m ECE 546 Jose Schutt Aine 20
N-Line Network Z s : Source impedance matrix Z L : Load impedance matrix V s : Source vector ECE 546 Jose Schutt Aine 21
Reflection Coefficients Source reflection coefficient matrix s [1 n EZ s L 1 E 1 m ] 1 [1n EZ s L 1 E 1 m ] Load reflection coefficient matrix L [1 n EZ L L 1 E 1 m ] 1 [1n EZ L L 1 E 1 m ] * The reflection and transmission coefficient are n x n matrices with off-diagonal elements. The off-diagonal elements account for the coupling between modes. * Z s and Z L can be chosen so that the reflection coefficient matrices are zero ECE 546 Jose Schutt Aine 22
TERMINATION NETWORK V 1 y p11 y p12 V 2 y p22 I1 = yp11v1 +yp12(v1-v2) I2 = yp22v2 +yp12(v2-v1) In general for a multiline system I YV V ZI Z Y 1 Note : yii ypii, yij = -yji, (i j) Also, z ij 1 y ij ECE 546 Jose Schutt Aine 23
Termination Network Construction To get impedance matrix Z Get physical impedance values y pij Calculate y ij 's from y pij 's Construct Y matrix Invert Y matrix to obtain Z matrix Remark: If y pij = 0 for all i j, then Y = Z -1 and z ii = 1/y ii ECE 546 Jose Schutt Aine 24
Procedure for Multiconductor Solution 1) Get L and C matrices and calculate LC product 2) Get square root of eigenvalues and eigenvectors of LC matrix m 3) Arrange eigenvectors into the voltage eigenvector matrix E 4) Get square root of eigenvalues and eigenvectors of CL matrix m 5) Arrange eigenvectors into the current eigenvector matrix H 6) Invert matrices E, H, m. 7) Calculate the line impedance matrix Z c Z c = E -1-1 m EL ECE 546 Jose Schutt Aine 25
Procedure for Multiconductor Solution 8) Construct source and load impedance matrices Z s (t) and Z L (t) 9) Construct source and load reflection coefficient matrices 1 (t) and 2 (t). 10) Construct source and load transmission coefficient matrices T (t), T 2 (t) 11) Calculate modal voltage sources g 1 (t) and g 2 (t) 12) Calculate modal voltage waves: a 1 (t) = T 1 (t)g 1 (t) + 1 (t)a 2 (t - m) a 2 (t) = T 2 (t)g 2 (t) + 2 (t)a 1 (t - m) b 1 (t) = a 2 (t - m ) b 2 (t) = a 1 (t - m ) ECE 546 Jose Schutt Aine 26
ai ai a i( t m ) ai mode1 mode2 ( t ( t moden ( t mi is the delay associated with mode i. mi = length/velocity of mode i. The modal volage wave vectors a 1 (t) and a 2 (t) need to be stored since they contain the history of the system. 13) Calculate total modal voltage vectors: V m1 (t) = a 1 (t) + b 1 (t) V m2 (t) = a 2 (t) + b 2 (t) 14) Calculate line voltage vectors: V 1 (t) = E -1 V m1 (t) V 2 (t) = E -1 V m2 (t) Wave Shifting Solution m1 m2 mn ) ) ) ECE 546 Jose Schutt Aine 27
7-Line Coupled-Microstrip System ALS04 Drive Line 1 ALS240 ALS04 Drive Line 2 ALS240 ALS04 Drive Line 3 ALS240 Sense Line 4 ALS04 Drive Line 5 ALS240 ALS04 Drive Line 6 ALS240 ALS04 Drive Line 7 ALS240 z=0 z=l L s = 312 nh/m; C s = 100 pf/m; L m = 85 nh/m; C m = 12 pf/m. ECE 546 Jose Schutt Aine 28
5 Drive line 3 at Near End Drive Line 3 5 Drive Line 3 at Far End 4 4 3 2 1 3 2 1 0 0 0 100 Time (ns) 200-1 0 100 Time (ns) 200 ECE 546 Jose Schutt Aine 29
The picture can't be displayed. Sense Line 2 Sense Line at Near End 2 Sense Line at Far End 1 1 0 0-1 0 100 Time (ns) 200-1 0 100 Time (ns) 200 ECE 546 Jose Schutt Aine 30
Multiconductor Simulation ECE 546 Jose Schutt Aine 31
LOSSY COUPLED TRANSMISSION LINES V 1 (z) V 2 (z) z=0 z=l... V 3 (z) L, C V RI L z I GV C z Solution is best found using a numerical approach (See References) I t V t ECE 546 Jose Schutt Aine 32
Three Line Microstrip 1 2 3 V1 I1 I2 I L11 L12 L13 z t t t 3 I1 V1 V2 V C11 C12 C13 z t t t 3 V2 I1 I2 I L21 L22 L23 z t t t 3 I2 V1 V2 V C21 C22 C23 z t t t 3 V I I I L L L z t t t 3 1 2 3 31 32 33 I V V V C C C z t t t 3 1 2 3 31 32 33 ECE 546 Jose Schutt Aine 33
Three Line Alpha Mode Subtract (1c) from (1a) and (2c) from (2a), we get V I ( L11 L13) z t I V ( C11 C13) z t This defines the Alpha mode with: and I V V1V I 1 I3 3 The wave impedance of that mode is: Z L C L C 11 13 11 13 and the velocity is u 1 L L C C 11 13 11 13 ECE 546 Jose Schutt Aine 34
Three Line Modal Decomposition In order to determine the next mode, assume that V V V V 1 2 3 I I I I 1 2 3 V I I I 11 21 31 12 22 32 13 23 33 z t t t 1 2 3 L L L L L L L L L I V V V 11 21 31 12 22 32 13 23 33 z t t t 1 2 3 C C C C C C C C C By reciprocity L 32 = L 23, L 21 = L 12, L 13 = L 31 By symmetry, L 12 = L 23 Also by approximation, L 22 L 11, L 11 +L 13 L 11 ECE 546 Jose Schutt Aine 35
Three Line Modal Decomposition V I I3 I L L L 2L L z t t t 1 2 11 12 13 12 11 In order to balance the right-hand side into I, we need to have 2L L I L L L I L L I 12 11 2 11 12 13 2 11 12 2 2L L 2 12 12 or 2 Therefore the other two modes are defined as The Beta mode with ECE 546 Jose Schutt Aine 36
The Beta mode with Three Line Beta Mode V V 2V V 1 2 3 I I 2I I 1 2 3 The characteristic impedance of the Beta mode is: Z L 2L L 11 12 13 C 2C C 11 12 13 and propagation velocity of the Beta mode is 1 u L11 2L12 L13 C11 2C12 C13 ECE 546 Jose Schutt Aine 37
Three Line Delta Mode The Delta mode is defined such that V V 2V V 1 2 3 I I 2I I 1 2 3 The characteristic impedance of the Delta mode is Z L 2L L 11 12 13 C 2C C 11 12 13 The propagation velocity of the Delta mode is: 1 u L11 2L12 L13 C11 2C12 C13 ECE 546 Jose Schutt Aine 38
Symmetric 3 Line Microstrip 1 2 3 In summary: we have 3 modes for the 3-line system E 1 0 1 1 2 1 1 2 1 Alpha mode Beta mode* Delta mode* *neglecting coupling between nonadjacent lines ECE 546 Jose Schutt Aine 39
Coplanar Waveguide 1 2 3 r = 4.3 346 162 67 LnH ( / m) 152 683 152 67 162 346 113 17 5 C( pf/ m) 16 53 16 5 17 113 E 0.45 0.12 0.45 0.5 0 0.5 0.45 0.87 0.45 H 0.44 0.49 0.44 0.5 0 0.5 0.10 0.88 0.10 ECE 546 Jose Schutt Aine 40
Coplanar Waveguide 1 2 3 r = 4.3 Z m 73 0 0 ( ) 0 48 0 0 0 94 Z c 56 23 8 ( ) 22 119 22 8 23 56 vp 0.15 0 0 ( m/ ns) 0 0.17 0 0 0 0.18 ECE 546 Jose Schutt Aine 41
Coplanar Waveguide S' W S W S' h r Kk ( ) : Complete Elliptic Integral of the first kind S k 30 K'( k) S 2W Zocp (ohm) ( ) r 1 Kk K'( k) K( k') 2 k' (1 k ) 2 1/2 v cp 2 r 1 1/2 c ECE 546 Jose Schutt Aine 42
Coplanar Strips W S W h r Z ocs 120 K'( k) (ohm) 1 Kk ( ) r 2 ECE 546 Jose Schutt Aine 43
Qualitative Comparison Characteristic Microstrip Coplanar Wguide Coplanar strips eff * ~6.5 ~5 ~5 Power handling High Medium Medium Radiation loss Low Medium Medium Unloaded Q High Medium Low or High Dispersion Small Medium Medium Mounting (shunt) Hard Easy Easy Mounting (series) Easy Easy Easy * Assuming r =10 and h=0.025 inch ECE 546 Jose Schutt Aine 44