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ECE 546 – Jose Schutt-Aine 1 ECE 546 Lecture - 06 Multiconductors Spring 2014 Jose E. Schutt-Aine Electrical & Computer Engineering University of Illinois jesa@illinois.edu
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ECE 546 – Jose Schutt-Aine 2 TELGRAPHER’S EQUATION FOR N COUPLED TRANSMISSION LINES V and I are the line voltage and line current VECTORS respectively (dimension n).
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ECE 546 – Jose Schutt-Aine 3 N-LINE SYSTEM L and C are the inductance and capacitance MATRICES respectively
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ECE 546 – Jose Schutt-Aine 4 N-LINE ANALYSIS 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.
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ECE 546 – Jose Schutt-Aine 5 MODAL ANALYSIS LC and CL are adjoint matrices. Find matrices E and H such that is the diagonal eigenvalue matrix whose entries are the inverses of the modal velocities squared.
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ECE 546 – Jose Schutt-Aine 6 MODAL ANALYSIS Second-order differential equation in modal space
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ECE 546 – Jose Schutt-Aine 7 V m and I m are the modal voltage and current vectors respectively. EIGENVECTORS
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ECE 546 – Jose Schutt-Aine 8 EIGEN ANALYSIS *The eigenvalues of A satisfy the relation |A - I| = 0 where I is the unit matrix. *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.
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ECE 546 – Jose Schutt-Aine 9 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.
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ECE 546 – Jose Schutt-Aine 10 EIGEN ANALYSIS For an n-line system, it can be shown that * Each eigenvalue is associated with a mode; the propagation velocity of that mode is the inverse of the eigenvalue. * 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.
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ECE 546 – Jose Schutt-Aine 11 EIGEN ANALYSIS * Each normalized eigenvector represents the relative line excitation required to excite the associated mode. * Each eigenvector is associated to an eigenvalue and therefore to a particular mode.
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ECE 546 – Jose Schutt-Aine 12 E : Voltage eigenvector matrix H : Current eigenvector matrix EIGENVECTORS
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ECE 546 – Jose Schutt-Aine 13 Eigenvalues and Eigenvectors gives
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ECE 546 – Jose Schutt-Aine 14 Modal Voltage Excitation Voltage Eigenvector Matrix
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ECE 546 – Jose Schutt-Aine 15 Modal Current Excitation Current Eigenvector Matrix
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ECE 546 – Jose Schutt-Aine 16 Line variables are recovered using V E 1 V m I H 1 I m EIGENVECTORS 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] E : voltage eigenvector matrix H : current eigenvector matrix E and H depend on both L and C
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ECE 546 – Jose Schutt-Aine 17 MODAL AND LINE IMPEDANCE MATRICES By requiring V m = Z m I m, we arrive at 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
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ECE 546 – Jose Schutt-Aine 18
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ECE 546 – Jose Schutt-Aine 19 MATRIX CONSTRUCTION
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ECE 546 – Jose Schutt-Aine 20 Modal Variables General solution in modal space is: Modal impedance matrix Line impedance matrix
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ECE 546 – Jose Schutt-Aine 21 N-Line Network Z s : Source impedance matrix Z L : Load impedance matrix V s : Source vector
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ECE 546 – Jose Schutt-Aine 22 Reflection Coefficients Source reflection coefficient matrix s [1 n EZEZ s L 1 E 1 m ] 1 [1 n EZEZ s L 1 E 1 m ] Load reflection coefficient matrix L [1 n EZEZ L L 1 E 1 m ] 1 [1 n EZEZ 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
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ECE 546 – Jose Schutt-Aine 23 TERMINATION NETWORK I 1 = y p11 V 1 +y p12 (V 1 -V 2 ) I 2 = y p22 V 2 +y p12 (V 2 -V 1 ) In general for a multiline system Note : y ii y pii, y ij = -y ji, (i≠j)
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ECE 546 – Jose Schutt-Aine 24 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
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ECE 546 – Jose Schutt-Aine 25 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 m -1 EL Procedure for Multiconductor Solution
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ECE 546 – Jose Schutt-Aine 26 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 ) Procedure for Multiconductor Solution
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ECE 546 – Jose Schutt-Aine 27 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
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ECE 546 – Jose Schutt-Aine 28 7-Line Coupled-Microstrip System L s = 312 nH/m; C s = 100 pF/m; L m = 85 nH/m; C m = 12 pF/m.
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ECE 546 – Jose Schutt-Aine 29 Drive Line 3
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ECE 546 – Jose Schutt-Aine 30 Sense Line
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ECE 546 – Jose Schutt-Aine 31 Multiconductor Simulation
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ECE 546 – Jose Schutt-Aine 32 LOSSY COUPLED TRANSMISSION LINES Solution is best found using a numerical approach (See References)
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ECE 546 – Jose Schutt-Aine 33 Three-Line Microstrip
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ECE 546 – Jose Schutt-Aine 34 Subtract (1c) from (1a) and (2c) from (2a), we get This defines the Alpha mode with: and The wave impedance of that mode is: and the velocity is Three-Line – Alpha Mode
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ECE 546 – Jose Schutt-Aine 35 In order to determine the next mode, assume that 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 Three-Line – Modal Decomposition
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ECE 546 – Jose Schutt-Aine 36 In order to balance the right-hand side into I , we need to have or Therefore the other two modes are defined as The Beta mode with Three-Line – Modal Decomposition
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ECE 546 – Jose Schutt-Aine 37 The Beta mode with The characteristic impedance of the Beta mode is: and propagation velocity of the Beta mode is Three-Line – Beta Mode
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ECE 546 – Jose Schutt-Aine 38 The Delta mode is defined such that The characteristic impedance of the Delta mode is The propagation velocity of the Delta mode is: Three-Line – Delta Mode
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ECE 546 – Jose Schutt-Aine 39 Alpha mode Beta mode* Delta mode* Symmetric 3-Line Microstrip In summary: we have 3 modes for the 3-line system *neglecting coupling between nonadjacent lines
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ECE 546 – Jose Schutt-Aine 40 Coplanar Waveguide
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ECE 546 – Jose Schutt-Aine 41 Coplanar Waveguide
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ECE 546 – Jose Schutt-Aine 42 Coplanar Waveguide
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ECE 546 – Jose Schutt-Aine 43 Coplanar Strips
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ECE 546 – Jose Schutt-Aine 44 CharacteristicMicrostripCoplanar WguideCoplanar strips e eff *~6.5 ~5 ~5 Power handlingHigh Medium Medium Radiation lossLow Medium Medium Unloaded QHigh Medium Low or High DispersionSmall Medium Medium Mounting (shunt)Hard Easy Easy Mounting (series)Easy Easy Easy * Assuming r =10 and h=0.025 inch Qualitative Comparison
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