ece 598 js lecture 06 multiconductorsjsa.ece.illinois.edu/ece598js/lect_06.pdf · 2013-03-23 ·...

Copyright © by Jose E. Schutt‐Aine , All Rights ReservedECE 598‐JS, Spring 2012 1

ECE 598 JS Lecture ‐ 06Multiconductors

Spring 2012

Jose E. Schutt-AineElectrical & Computer Engineering

University of [email protected]


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


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


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∂ ∂

=∂ ∂


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− −= = Λ


MODAL ANALYSIS

Second-order differential equation in modal space

2 22

2 2m m

mV Vz t

∂ ∂= Λ

∂ ∂

2 22

2 2m m

mI Iz t

∂ ∂= Λ

∂ ∂

Vm = EV Im = HI

V3(z)

V2(z)

V1(z)

...

L, C

z=0 z=l


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


EIGEN ANALYSIS

* The eigenvalues λ of A satisfy the relation |A - λI| = 0 where Iis the unit matrix.

* A scalar λ is an eigenvalue of a matrix A if there exists a vector X such that AX = λX. (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.


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.


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.


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.


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− = Λ

1 2mHCLH − = Λ


1

2

3

1 0 0

10 0

10 0

m

mm

m

v

v

v

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

Λ = ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

1

2

3

1 0 0

10 0

10 0

m

mm

m

v

v

v

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

Λ = ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

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− = Λgives

1 2mHCLH − = Λ

gives


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

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


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

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦


Line variables are recovered using

V = E−1Vm

I = H−1Im

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


MODAL AND LINE IMPEDANCE MATRICES

Zm = Λm−1ELH−1

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 = E−1ZmH = E−1Λm−1EL


1

2( )

m

m

mn

j uv

j uv

j uv

e

eW u

e

ω

ω

ω

−

−

−

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥•⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦

1

2

1

1

1

m

mm

mn

v

v

v

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥Λ = ⎢ ⎥

•⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦


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


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− −= Λ

1 1 1c m mZ E Z H E EL− − −= = Λ

Modal Variables

General solution in modal space is:

Modal impedance matrix

Line impedance matrix


N-Line Network

Zs : Source impedance matrix

ZL : Load impedance matrix

Vs : Source vector


Reflection Coefficients

Source reflection coefficient matrix

Γs = [1n + EZsL−1E−1Λm ]−1[1n − EZsL−1E−1Λm ]

Load reflection coefficient matrix

ΓL = [1n + EZLL−1E−1Λm ]−1[1n − EZLL−1E−1Λm ]

* 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


TERMINATION NETWORK

yp11

yp22

yp12

V1

V2

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)ij

1Also, z ijy

≠


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


1) Get L and C matrices and calculate LC product

2) Get square root of eigenvalues and eigenvectors of LC matrix → Λm

3) Arrange eigenvectors into the voltage eigenvector matrix E

4) Get square root of eigenvalues and eigenvectors of CL matrix →Λm

5) Arrange eigenvectors into the current eigenvector matrix H

6) Invert matrices E, H, Λm.

7) Calculate the line impedance matrix Zc → Zc = E-1Λm-1EL

Procedure for Multiconductor Solution


8) Construct source and load impedance matrices Zs(t) and ZL(t)

9) Construct source and load reflection coefficient matrices Γ1(t) and Γ2(t).

10) Construct source and load transmission coefficient matrices T1(t), T2(t)

11) Calculate modal voltage sources g1(t) and g2(t)

12) Calculate modal voltage waves:

a1(t) = T1(t)g1(t) + Γ1(t)a2(t - τm)

a2(t) = T2(t)g2(t) + Γ2(t)a1(t - τm)

b1(t) = a2(t - τm)

b2(t) = a1(t - τm)

Procedure for Multiconductor Solution


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−•−−

=−

−−

−−

−−

)(

)()(

a

mod

22mod

11mod

mi

mnnei

mei

mei

ta

tata

t

τ

ττ

τ )(

τmi is the delay associated with mode i. τmi = 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


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.


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


2001000-1

0

1

2Sense Line at Near End

Time (ns)2001000

-1

0

1

2Sense Line at Far End

Time (ns)

This image cannot currently be displayed.

Sense Line


Multiconductor Simulation


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


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


11 13( )V IL Lz tα α∂ ∂

∂ ∂− = −

11 13( )I VC Cz tα α∂ ∂

∂ ∂− = −

Subtract (1c) from (1a) and (2c) from (2a), we get

This defines the Alpha mode with:

1 3V V Vα = − 1 3I I Iα = −and

The wave impedance of that mode is:

11 13

11 13

L LZC Cα

−=

−

and the velocity is( )( )11 13 11 13

1uL L C C

α =− −

Three‐Line – Alpha Mode


In order to determine the next mode, assume that

1 2 3V V V Vβ β= + +

1 2 3I I I Iβ β= + +

( ) ( ) ( ) 31 211 21 31 12 22 32 13 23 33

V II IL L L L L L L L Lz t t tβ∂ ∂∂ ∂β β β

∂ ∂ ∂ ∂− = + + + + + + + +

( ) ( ) ( ) 31 211 21 31 12 22 32 13 23 33

I VV VC C C C C C C C Cz t t tβ∂ ∂∂ ∂β β β

∂ ∂ ∂ ∂− = + + + + + + + +

By reciprocity L32 = L23, L21 = L12, L13 = L31

By symmetry, L12 = L23

Also by approximation, L22 ≈ L11, L11+L13 ≈ L11

Three‐Line – Modal Decomposition


( ) ( )31 211 12 13 12 112

V II IL L L L Lz t t tβ∂ ∂∂ ∂β β

∂ ∂ ∂ ∂⎛ ⎞− = + + + + +⎜ ⎟⎝ ⎠

In order to balance the right-hand side into Iβ, we need to have

( ) ( ) ( )12 11 2 11 12 13 2 11 12 22L L I L L L I L L Iβ β β β β+ = + + ≈ +

212 122L Lβ=

or 2β = ±

Therefore the other two modes are defined as

The Beta mode with

Three‐Line – Modal Decomposition


The Beta mode with

1 2 32V V V Vβ = + +

1 2 32I I I Iβ = + +

The characteristic impedance of the Beta mode is:

11 12 13

11 12 13

22

L L LZC C Cβ

+ +=

+ +

and propagation velocity of the Beta mode is

( )( )11 12 13 11 12 13

1

2 2u

L L L C C Cβ =

+ + + +

Three‐Line – Beta Mode


The Delta mode is defined such that

1 2 32V V V Vδ = − +

1 2 32I I I Iδ = − +

The characteristic impedance of the Delta mode is

11 12 13

11 12 13

22

L L LZC C Cδ

− +=

− +

The propagation velocity of the Delta mode is:

( )( )11 12 13 11 12 13

1

2 2u

L L L C C Cδ =

− + − +

Three‐Line – Delta Mode


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


1 2 3

εr = 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


73 0 0( ) 0 48 0

0 0 94mZ

⎛ ⎞⎜ ⎟Ω = ⎜ ⎟⎜ ⎟⎝ ⎠

56 23 8( ) 22 119 22

8 23 56cZ

⎛ ⎞⎜ ⎟Ω = ⎜ ⎟⎜ ⎟⎝ ⎠

0.15 0 0( / ) 0 0.17 0

0 0 0.18pv m ns

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

1 2 3

εr = 4.3

Coplanar Waveguide


WS' S'WS

εrh

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

πε

=+

1/ 22

1cpr

v cε

⎛ ⎞= ⎜ ⎟+⎝ ⎠

Coplanar Waveguide


S WW

εrh

120 '( ) (ohm)( )1

2

ocsr

K kZK k

πε

=+

Coplanar Strips


Characteristic Microstrip Coplanar Wguide Coplanar strips

εeff* ~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 εr=10 and h=0.025 inch

Qualitative Comparison


3-Line Matching Network

• Extend matching network synthesis to multiconductor transmission line (MTL) systems, specifically coupled microstrip

• Characterize MTL systems by their “normal mode parameters”– Mode configurations– Propagation constants– Characteristic impedance matrices


Longitudinal Immittance Matrix Functions (LIMFs)

• Given terminations, the immittance matrices are expressed as “input”longitudinal functions looking toward the load

m,c m,cin,out

m,c m,cin,out in,out

, ,

,

ΓS

Y Z


Calibration Methods1st order calibration: 6 distinct TRL calibrations were required

to deembed the fan-in and fan-out sections 2nd order calibration: 1 multimode TRL calibration required to

deembed all fan-ins and fan-outs

Multiconductor transmission line matching network device under test


Bias Networks and Multiline/Multimode TRL

• Self-biasing for DC power done using air bridges

• Regulator circuit may be built on board with surface mount/chip components

• Calibration must include all modes and allow for the 6-port discontinuity S-parameter measurement

Proposed calibration fixture for transistor amplifier device discontinuity with self-biasing network


3-line Transistor Amplifier

The general amplifier structure with generic matching networks

Note that a transistor seating hole may be present for convenience, but is not necessary

Photograph of a three-line transistor amplifier autobiased w/o matching networks


Transistor Amplifier Matching

S-parameter comparison between unmatched and matched three-line amplifier. Dotted line - unmatched; dashed line -matched. Design frequency: 3.1 GHz.

Matching• 2 - 0.8 pF chip cap.• 2 - 0.4 pF chip cap.

Gains• 3.38 dB @ 3.1 GHz

m=5.14 / um= 1.76• 6.22 dB @ 3.22 GHz

m=6.54 / um= 0.32


2p

h

w

α

dielectric

ground

signal strip

V Transmission Line


0

50

100

150

200

250

0 0.1 0.2 0.3 0.4 0.5 0.6w/h

CHARACTERISTIC IMPEDANCE

30º37.5º45º52.5º

60ºMicrostrip

0.7 0.8 0.9 1

Zo (o

hms)

Calculated values of the characteristic impedance for a single-line v-strip structure as a function of width-to-height ratio w/h. The relative dielectric constant is εr = 2.55.

V‐Line Characteristic Impedance


1.8

1.85

1.9

1.95

2

2.05

2.1

2.15

2.2

0 0.1 0.2 0.3 0.4 0.5 0.6

w/h

EFFECTIVE PERMITTIVITY

30º

37.5º

45º

52.5º

60º

Microstrip

0.7 0.8 0.9 1

Effe

ctiv

e R

elat

ive

Die

lect

ric C

onst

ant

Calculated values of the effective relative dielectric constant for a single-line v-strip structure as a function of width-to-height ratio w/h. The relative dielectric constant is εr = 2.55

V‐Line – Effective Permittivity


h

y

x

Region 2

Region 3

Region 4

Region 6

Region 1

Region 5

2pw

dielectricground surface

signal strip

s

Three‐Line – V Transmission Line


74.00 -0.97 -0.23C (pF/m) = -0.97 74.07 -0.97

-0.23 -0.97 74.00

425.84 20.35 7.00L (nH/m) = 20.36 422.85 20.36

7.00 20.35 425.84

52.49 -5.90 -0.57C (pF/m) = -5.90 53.27 -5.90

-0.57 -5.90 52.49

609.74 113.46 41.79L (nH/m) = 113.54 607.67 113.54

41.79 113.46 609.74

Microstrip V-line

Comparison of the inductance and capacitance matricesbetween a three-line v-line and microstrip structures. The parameters are p/h = 0.8 mils, w/h = 0.6 and εr = 4.0.

h

y

x

Region 2

Region 3

Region 4

Region 6

Region 1

Region 5

2pw

dielectricground surface

signal strip

s1 2 3

V‐Line and Microstrip


0

50

100

150

200

250

300

350

0 0.2 0.4 0.6 0.8 1 1.2s/h

Mutual Inductance

V-linemicrostrip

Lm

(nH

/m)

0

5

10

15

20

25

30

35

40

0 0.2 0.4 0.6 0.8 1 1.2s/h

Mutual Capacitance

V-linemicrostrip

C m(p

F/m

)

Plot of mutual inductance (top) and mutual capacitance (bottom) versus spacing-to-height ratiofor v-line and microstrip configurations. The parameters are w/h = 0.24, εr = 4.0.

V‐Line: Coupling Coefficients


0

0.1

0.2

0.3

0.4

0.5

0.6

0 0.2 0.4 0.6 0.8 1 1.2s/h

Coupling Coefficient

V-linemicrostrip

Kv

Plot of the coupling coefficient versus spacing-to-height ratio for v-line and microstrip configurations. The parameters are w/h = 0.24, εr = 4.0.

V‐Line vs Microstrip: Coupling Coefficients


0.5

0.6

0.7

0.8

0.9

1

1 3 5 7 9 11 13 15 17

microstrip

V-60

V-30

Frequency (GHz)

Linea

r Atte

nua ti

on

Propagation Function

2p

h

w

α

dielectric

ground

signal strip

Advantages of V Line* Higher bandwidth

* Lower crosstalk

* Better transition

ground reference

ground reference

dielectric

substrate

signal strip

Advantages of V‐Line


-8

-7

-6

-5

-4

-3

-2

-1

0

0 5 10 15 20 25

Insertion Loss

VlineMicrostrip20

*log1

0*|S

21|

Frequency (GHz)

V‐Line vs Microstrip: Insertion Loss


• J. E. Schutt-Aine and R. Mittra, "Transient analysis of coupled lossy transmission lines with nonlinear terminations," IEEE Trans. Circuit Syst., vol. CAS-36, pp. 959-967, July 1989.

• J. E. Schutt-Aine and D. B. Kuznetsov, "Efficient transient simulation of distributed interconnects," Int. Journ. Computation Math. Electr. Eng. (COMPEL), vol. 13, no. 4, pp. 795-806, Dec. 1994.

• D. B. Kuznetsov and J. E. Schutt-Ainé, "The optimal transient simulation of transmission lines," IEEE Trans. Circuits Syst.-I., vol. cas-43, pp. 110-121, February 1996.

• W. T. Beyene and J. E. Schutt-Ainé, "Efficient Transient Simulation of High-Speed Interconnects Characterized by Sampled Data," IEEE Trans. Comp., Hybrids, Manufacturing Tech. vol. 21, pp. 105-114, February 1998.

References

ece 598 js lecture 06 multiconductorsjsa.ece.illinois.edu/ece598js/lect_06.pdf · 2013-03-23 ·...

Documents