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ECE 546 – Jose Schutt‐Aine 1
ECE 546Lecture ‐ 07
MulticonductorsSpring 2018
Jose E. Schutt-AineElectrical & Computer Engineering
University of [email protected]
ECE 546 – Jose Schutt‐Aine 2
TELGRAPHER’S EQUATION FOR NCOUPLED TRANSMISSION LINES
V and I are the line voltage and line current VECTORSrespectively (dimension n).
V ILz t
I VCz t
V3(z)
V2(z)
V1(z)
...
L, C
z=0 z=l
ECE 546 – Jose Schutt‐Aine 3
N-LINE SYSTEM
L and C are the inductance and capacitance MATRICES respectively
1
2
n
II
I
I
1
2
n
VV
V
V
11 12
21 22
nn
L LL L
L
L
11 12
21 22
nn
C CC C
C
C
V3(z)
V2(z)
V1(z)
...
L, C
z=0 z=l
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 atransformation on the variables V and I.
Diagonalize LC and CL is equivalent to finding the eigenvalues ofLC and CL.
Since LC and CL are adjoint, they must have the sameeigenvalues.
2 2
2 2
V VLCz t
2 2
2 2
I ICLz t
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 ofthe modal velocities squared.
2 21
2 2
EV EVELCEz t
2 21
2 2
HI HIHCLHz t
1 1 2mELCE HCLH
ECE 546 – Jose Schutt‐Aine 6
MODAL ANALYSIS
Second-order differential equation in modal space
2 22
2 2m m
mV Vz t
2 22
2 2m m
mI Iz t
Vm = EV Im = HI
V3(z)
V2(z)
V1(z)
...
L, C
z=0 z=l
ECE 546 – Jose Schutt‐Aine 7
Vm and Im are the modal voltage and current vectors respectively.
EIGENVECTORS
1
2
m
m
m
mn
VV
V
V
1
2
m
m
m
mn
II
I
I
Im = HI
Vm = EV
ECE 546 – Jose Schutt‐Aine 8
EIGEN ANALYSIS
* The eigenvalues of A satisfy the relation |A - I| = 0 where Iis 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 isan eigenvector of A.
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.
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 propagationvelocity of that mode is the inverse of the eigenvalue.
* LC can be transformed into a diagonal matrix whose entriesare the eigenvalues.
* LC possesses n distinct eigenvalues (possibly degenerate).
* There exist n eigenvectors which are linearly independent.
ECE 546 – Jose Schutt‐Aine 11
EIGEN ANALYSIS
* Each normalized eigenvector represents the relative lineexcitation required to excite the associated mode.
* Each eigenvector is associated to an eigenvalue and thereforeto a particular mode.
ECE 546 – Jose Schutt‐Aine 12
E : Voltage eigenvector matrix
H : Current eigenvector matrix
EIGENVECTORS
11 12 13
21 22 23
31 32 33
e e eE e e e
e e e
11 12 13
21 22 23
31 32 33
h h hH h h h
h h h
1 2mELCE
1 2mHCLH
ECE 546 – Jose Schutt‐Aine 13
1
2
3
1 0 0
10 0
10 0
m
mm
m
v
v
v
1
2
3
1 0 0
10 0
10 0
m
mm
m
v
v
v
11 12 13
21 22 23
31 32 33
h h hH h h h
h h h
11 12 13
21 22 23
31 32 33
e e eE e e e
e e e
Eigenvalues and Eigenvectors1 2
mELCE gives
1 2mHCLH
gives
ECE 546 – Jose Schutt‐Aine 14
MATCHINGNETWORK
+++
---e31
e32
e33 MODE C
MATCHINGNETWORK
+++
---e11
e12
e13 MODE A
MATCHINGNETWORK
+++
---e21
e22
e23 MODE B
Modal Voltage Excitation
Voltage EigenvectorMatrix
11 12 13
21 22 23
31 32 33
e e eE e e e
e e e
ECE 546 – Jose Schutt‐Aine 15
Modal Current Excitation
Current EigenvectorMatrix
MATCHINGNETWORK
h11
h12
h13 MODE A
MATCHINGNETWORK
h21
h22
h23 MODE B
MATCHINGNETWORK
h31
h32
h33 MODE C
11 12 13
21 22 23
31 32 33
h h hH h h h
h h h
ECE 546 – Jose Schutt‐Aine 16
Line variables are recovered using
V E1Vm
I H1Im
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
ECE 546 – Jose Schutt‐Aine 17
MODAL AND LINE IMPEDANCE MATRICES
Zm m1ELH1
By requiring Vm = ZmIm, we arrive at
Zm is a diagonal impedance matrix which relates modal voltagesto modal currents
By requiring V = ZcI, one can define a line impedance matrix Zc.Zc relates line voltages to line currents
Zm is diagonal. Zc contains nonzero off-diagonal elements. We can show that
Zc E1ZmH E1m1EL
ECE 546 – Jose Schutt‐Aine 18
1
2( )
m
m
mn
j uv
j uv
j uv
e
eW u
e
1
2
1
1
1
m
mm
mn
v
v
v
ECE 546 – Jose Schutt‐Aine 19
MATRIX CONSTRUCTION
siiL L ( )m
ijijL L
( )
1
[ ]n
mii s ij
j
C C C
( )[ ] mij ijC C
[ ]ii sR R [ ] mij ijR R
( )
1
[ ]n
mii s ij
j
G G G
( )[ ] mij ijG G
ECE 546 – Jose Schutt‐Aine 20
Vm = EV Im = HI
Vm (z) = W(z)A +W(- z)B
Im (z) = Zm- 1 W(z)A - W(- z)B[ ]
1 1m mZ ELH
1 1 1c m mZ E Z H E EL
Modal Variables
General solution in modal space is:
Modal impedance matrix
Line impedance matrix
ECE 546 – Jose Schutt‐Aine 21
N-Line Network
Zs : Source impedance matrix
ZL : Load impedance matrix
Vs : Source vector
ECE 546 – Jose Schutt‐Aine 22
Reflection Coefficients
Source reflection coefficient matrix
s [1n EZsL1E1m ]1[1n EZsL1E1m ]
Load reflection coefficient matrix
L [1n EZLL1E1m ]1[1n EZLL1E1m ]
* The reflection and transmission coefficient are n x nmatrices with off-diagonal elements. The off-diagonal elements account for the coupling between modes.
* Zs and ZL can be chosen so that the reflection coefficient matrices are zero
ECE 546 – Jose Schutt‐Aine 23
TERMINATION NETWORK
yp11
yp22
yp12
V1
V2
I1 = yp11V1 +yp12(V1-V2)
I2 = yp22V2 +yp12(V2-V1)
In general for a multiline system
I YVV ZI Z Y1
Note : yii ypii, yij = -yji , (i≠j)ij
1Also, z ijy
ECE 546 – Jose Schutt‐Aine 24
Termination Network Construction
To get impedance matrix Z• Get physical impedance values ypij• Calculate yij's from ypij's• Construct Y matrix• Invert Y matrix to obtain Z matrix
Remark: If ypij = 0 for all i≠j, then Y = Z-1 and zii = 1/yii
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 Zc Zc = E-1m-1EL
Procedure for Multiconductor Solution
ECE 546 – Jose Schutt‐Aine 26
8) Construct source and load impedance matrices Zs(t) and ZL(t)
9) Construct source and load reflection coefficient matrices 1(t) and 2(t).
10) Construct source and load transmission coefficient matrices T(t), T2(t)
11) Calculate modal voltage sources g1(t) and g2(t)
12) Calculate modal voltage waves:
a1(t) = T1(t)g1(t) + 1(t)a2(t - m)
a2(t) = T2(t)g2(t) + 2(t)a1(t - m)
b1(t) = a2(t - m)
b2(t) = a1(t - m)
Procedure for Multiconductor Solution
ECE 546 – Jose Schutt‐Aine 27
)(
)()(
a
mod
22mod
11mod
mi
mnnei
mei
mei
ta
tata
t
)(
mi is the delay associated with mode i. mi = length/velocity of mode i. The modal volage wave vectors a1(t) and a2(t) need to be stored since they contain the history of the system.
13) Calculate total modal voltage vectors:
Vm1(t) = a1(t) + b1(t)Vm2(t) = a2(t) + b2(t)
14) Calculate line voltage vectors:
V1(t) = E-1Vm1(t)V2(t) = E-1Vm2(t)
Wave‐Shifting Solution
ECE 546 – Jose Schutt‐Aine 28
ALS04 ALS240Drive Line 1
Drive Line 2
z=0 z=l
Drive Line 3
Sense Line 4
Drive Line 5
Drive Line 6
Drive Line 7
ALS04
ALS04
ALS04
ALS04
ALS04
ALS240
ALS240
ALS240
ALS240
ALS240
7-Line Coupled-Microstrip System
Ls = 312 nH/m; Cs = 100 pF/m;
Lm = 85 nH/m; Cm = 12 pF/m.
ECE 546 – Jose Schutt‐Aine 29
20010000
1
2
3
4
5Drive line 3 at Near End
Time (ns)2001000
-1
0
1
2
3
4
5Drive Line 3 at Far End
Time (ns)
Drive Line 3
ECE 546 – Jose Schutt‐Aine 30
2001000-1
0
1
2Sense Line at Near End
Time (ns)2001000
-1
0
1
2Sense Line at Far End
Time (ns)
The picture can't be displayed.
Sense Line
ECE 546 – Jose Schutt‐Aine 31
Multiconductor Simulation
ECE 546 – Jose Schutt‐Aine 32
LOSSY COUPLED TRANSMISSION LINES
Solution is best found using a numerical approach(See References)
V IRI Lz t
I VGV Cz t
V3(z)
V2(z)
V1(z)
...
L, C
z=0 z=l
ECE 546 – Jose Schutt‐Aine 33
1 2 3
31 1 211 12 13
VI V VC C Cz t t t
32 1 221 22 23
VI V VC C Cz t t t
3 31 231 32 33
I VV VC C Cz t t t
31 1 211 12 13
IV I IL L Lz t t t
32 1 221 22 23
IV I IL L Lz t t t
3 31 231 32 33
V II IL L Lz t t t
Three‐Line Microstrip
ECE 546 – Jose Schutt‐Aine 34
11 13( )V IL Lz t
11 13( )I VC Cz t
Subtract (1c) from (1a) and (2c) from (2a), we get
This defines the Alpha mode with:
1 3V V V 1 3I I I and
The wave impedance of that mode is:
11 13
11 13
L LZC C
and the velocity is 11 13 11 13
1uL L C C
Three‐Line – Alpha Mode
ECE 546 – Jose Schutt‐Aine 35
In order to determine the next mode, assume that
1 2 3V V V V
1 2 3I I I I
31 211 21 31 12 22 32 13 23 33
V II IL L L L L L L L Lz t t t
31 211 21 31 12 22 32 13 23 33
I VV VC C C C C C C C Cz t t t
By reciprocity L32 = L23, L21 = L12, L13 = L31
By symmetry, L12 = L23
Also by approximation, L22 L11, L11+L13 L11
Three‐Line – Modal Decomposition
ECE 546 – Jose Schutt‐Aine 36
31 211 12 13 12 112
V II IL L L L Lz t t t
In order to balance the right-hand side into I, we need to have
12 11 2 11 12 13 2 11 12 22L L I L L L I L L I
212 122L L
or 2
Therefore the other two modes are defined as
The Beta mode with
Three‐Line – Modal Decomposition
ECE 546 – Jose Schutt‐Aine 37
The Beta mode with
1 2 32V V V V
1 2 32I I I I
The characteristic impedance of the Beta mode is:
11 12 13
11 12 13
22
L L LZC C C
and propagation velocity of the Beta mode is
11 12 13 11 12 13
1
2 2u
L L L C C C
Three‐Line – Beta Mode
ECE 546 – Jose Schutt‐Aine 38
The Delta mode is defined such that
1 2 32V V V V
1 2 32I I I I
The characteristic impedance of the Delta mode is
11 12 13
11 12 13
22
L L LZC C C
The propagation velocity of the Delta mode is:
11 12 13 11 12 13
1
2 2u
L L L C C C
Three‐Line – Delta Mode
ECE 546 – Jose Schutt‐Aine 39
1 0 1
1 2 1
1 2 1
E
1 2 3
Alpha modeBeta mode*
Delta mode*
Symmetric 3‐Line Microstrip
In summary: we have 3 modes for the 3-line system
*neglecting coupling between nonadjacent lines
ECE 546 – Jose Schutt‐Aine 40
1 2 3
r = 4.3
113 17 5( / ) 16 53 16
5 17 113C pF m
346 162 67( / ) 152 683 152
67 162 346L nH m
0.45 0.12 0.450.5 0 0.50.45 0.87 0.45
E
0.44 0.49 0.440.5 0 0.50.10 0.88 0.10
H
Coplanar Waveguide
ECE 546 – Jose Schutt‐Aine 41
73 0 0( ) 0 48 0
0 0 94mZ
56 23 8( ) 22 119 22
8 23 56cZ
0.15 0 0( / ) 0 0.17 0
0 0 0.18pv m ns
1 2 3
r = 4.3
Coplanar Waveguide
ECE 546 – Jose Schutt‐Aine 42
WS' S'WS
rh
2Sk
S W
( ) : Complete Elliptic Integral of the first kindK k
'( ) ( ')K k K k
2 1/ 2' (1 )k k
30 '( ) (ohm)( )1
2
ocpr
K kZK k
1/ 22
1cpr
v c
Coplanar Waveguide
ECE 546 – Jose Schutt‐Aine 43
S WW
rh
120 '( ) (ohm)( )1
2
ocsr
K kZK k
Coplanar Strips
ECE 546 – Jose Schutt‐Aine 44
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
Qualitative Comparison