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Transcript of University of Illinois at Chicago ECE 423, Dr. D. Erricolo, Lecture 19 University of Illinois at...
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
Frequency analysis of three-conductor lines
We have already found that the transmission line equations for a three-conductor system are, in time domain:
For a sinusoidal steady-state excitation the above equations become:
t
tzILtzIR
z
tzV
),(),(
),(
t
tzVCtzVG
z
tzI
),(),(
),(
)(ˆˆ)(ˆ
)(ˆˆ)(ˆ
zVYdz
zId
zIZdz
zVd
(1)
(2)
(3)
(4)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
where we have introduced the voltage and current phasors
and the per-unit length impedance and admittance matrices:
)(ˆ
)(ˆ)(ˆ
)(ˆ
)(ˆ)(ˆ
zI
zIzI
zV
zVzV
R
G
R
G
CjGY
LjRZ
ˆ
ˆ
(5)
(6)
(7)
(8)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
Equations (3) and (4) are coupled first-order differential equations.In order to obtain uncoupled equations we can differentiate each equation with respect to z and substitute into the other to obtain
When (9) and (10) are obtained, it is important to keep the order of theproducts and since these do not commute in general.
A general solution is obtained by either solving (9) or(10). We willconsider the solution of (10).
YZ ˆˆ ZY ˆˆ
(9)
(10))(ˆˆˆ)(ˆˆ)(ˆ
)(ˆˆˆ)(ˆˆ)(ˆ
2
2
2
2
2
2
2
2
zIZYdz
zVdY
dz
zId
zVYZdz
zIdZ
dz
zVd
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
First we introduce a transformation into nodal currents by setting
And substituting into (10), which yields
If the transformation can diagonalize we obtain
And the modal equations (12) become uncoupled:
mITI ˆˆˆ
T
ZY ˆˆT
)(ˆˆˆˆˆ)(ˆ 1
2
2
zITZYTdz
zIdm
m
)(ˆ)(ˆ
)(ˆ)(ˆ
22
2
22
2
zIdz
zId
zIdz
zId
mRRmR
mGGmG
2
221
ˆ0
0ˆˆˆˆˆˆ
R
GTZYT
(11)
(12)
(13)
(14)
(15)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
2Gˆ ,Zˆ YThe eigenvalues of and are referred to as propagation
constants and are used to form the solution:
2ˆR
mRz
mRz
mR
mGz
mGz
mG
IeIezI
IeIezI
RR
GG
ˆˆ)(ˆ
ˆˆ)(ˆ
ˆˆ
ˆˆ
Note that 1) the form of the solution is the same as for two-conductor lines; 2) the constants
mRmRmGmG IIII ˆ,ˆ,ˆ,ˆ
Equations (16) and (17) may be cast as: mG
zmG
zm IeIezI GG ˆˆ)(ˆ ˆˆ
where
z
zz
R
G
e
ee
ˆ
ˆˆ
0
0
mR
mGm
I
II
ˆ
ˆˆ
(16)
(17)
(18)
(19) (20)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
Once a solution is found in terms of the nodal currents, the actualcurrents are found from:
mz
mz
m IeIeTITI GG ˆˆˆˆˆˆ ˆˆ
The voltage solution is obtained from:
m
zm
z IeIeTYdz
zIdYzV GG ˆˆˆˆˆ)(ˆˆ)(ˆ ˆˆ11
Observe that, since , then21ˆˆˆˆˆ
TZYT
ˆˆˆˆˆˆ 11 TYTZ
Hence (22) may be rewritten
m
zm
z IeIeTTTZzV GG ˆˆˆˆˆˆˆ)(ˆ ˆˆ11
(21)
(22)
(23)
(24)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
The terms inside square brackets in the previous equation is the characteristic impedance matrix:
11 ˆˆˆˆˆ TTZZ c Let us now see how we can solve for the unknown constants . Weneed to add some additional constraints. For this purpose we look at the transmission line as a four-port circuit:
Three-conductor line
SR
NER NER VV ˆ)0(ˆ +
-
+
-
)0(GV)0(ˆ
GI)0(ˆ
RI
SV
FER VLV ˆ)(ˆ +
- - -
FER
FER
+
)(ˆ LVG
)(ˆ LIG)(ˆ LI R
z0zLz
This allows us to write:
Figure 1
)0(ˆˆˆ)0(ˆ IZVV ss )(ˆˆ)(ˆ LIZLV L(26) and (27)
mI
(25)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
The previous quantities correspond to:
FE
LL
NE
ss
ss R
RZ
R
RZ
VV
0
0ˆ,0
0ˆ,0
ˆˆ
Evaluating the voltage, given by (22), at yields:0z
mm
mmc
IITI
IITZV
ˆˆˆ)0(ˆ
ˆˆˆˆ)0(ˆ
When (29) and (30) are combined with (26) we obtain:
mmcmmss IITZIITZV ˆˆˆˆˆˆˆˆˆ
Similarly, evaluating the voltage at yields:Lz
mL
mL
mL
mL
c
IeIeTLI
IeIeTZLV
ˆˆˆ)(ˆ
ˆˆˆˆ)(ˆ
ˆˆ
ˆˆ
When (32) and (33) are combined with (27) we obtain:
mL
mL
cmL
mL
L IeIeTZIeIeTZ ˆˆˆˆˆˆˆˆ
(28)
(29)
(30)
(31)
(32)
(33)
(34)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
Equations (31) and (34) are now combined to give an algebraic matrixequation that determines
mI
0
ˆ
ˆ
ˆ
ˆˆˆˆˆˆ
ˆˆˆˆˆˆ
ˆˆ
s
m
m
LLc
LLc
scsc V
I
I
eTZZeTZZ
TZZTZZ
After the previous equation is solved current and voltage along the lineare computed using equations (21) and (22).
Looking at the line as a four-port network may be particularly useful ifwe are interested only in the values of voltages and currents at the endpoints of the line. For this purpose we need the chain parameter matrix
)0(ˆ
)0(ˆ
ˆˆ
ˆˆ
)(ˆ
)(ˆ
2221
1211
I
V
LI
LV
(35)
(36)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
The parameters are computed from (29)-(32), and their expressionsare:
ij
1ˆˆ
22
11ˆˆ
21
1ˆˆ1
12
1ˆˆ1
11
ˆˆ2
1ˆ
ˆˆˆˆ2
1ˆ
ˆˆˆˆ2
1ˆ
ˆˆˆˆ2
1ˆ
TeeT
YTeeT
TeeTY
YTeeTY
LL
LL
LL
LL
Finally, another representation, obtained substituting (26)-(27) into (36), for the terminal currents is:
)0(ˆˆˆˆˆˆ)(ˆ
ˆˆˆˆ)0(ˆˆˆˆˆˆˆˆˆ
212221
112121221112
IZVLI
VZIZZZZ
ss
sLsLLs
(37)
(38)
(39)
(40)
(41)
(42)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
An exact solution (lossless line immersed in homogeneous medium)
The solution previously found does not provide insights into the mechanism of crosstalk; hence here we consider an analytical solutionthat is obtained by making the following assumptions: 1) three-conductor line; 2) lossless; 3) homogenous medium
Lossless implies: CjYLjZGR ˆ,ˆ,0,0
Homogeneous implies: ICL from which
IZY 2ˆˆ
And we also observe that is diagonal so that . The propagation constant matrix is:
ZY ˆˆ IT ˆˆ
IvjIj ˆ
(43)
(44)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
The elements of the chain parameter matrix reduce to:
IL
CL
LLjCLjv
LL
LLjLLjv
IL
cosˆ
sinsinˆ
sinsinˆ
cosˆ
22
21
12
11
When (45)-(48) are introduced into the representation (41) for theterminal current we obtain:
sL
sLLs
VLCjZL
LIL
ILjZLCjZL
LZZL
ˆˆsincos
)0(ˆˆˆsinˆˆcos
(45)
(46)
(47)
(48)
(49)
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
The expression for may be obtained upon substitution of (45)-(48)into (42) or by reciprocity. Applying reciprocity we obtain:
sLsLs VLILjZLCjZL
LZZL ˆ)(ˆˆˆsinˆˆcos
(50)
When (49) and (50) are solved for the terminal voltages due to crosstalkamong the wires one finds:
DC
DC
GLG
mFENE
FENE
GLGmFENE
NENER
VSk
LjCLcj
RR
RR
ISk
LjCLlj
RR
RSVV
ˆ1
1
2
ˆ1
2
Denˆ)0(ˆ
2
2
DCDC Gm
FENE
FENEGm
FENE
FEFER VLcj
RR
RRILlj
RR
RSVLV ˆˆ
Denˆ)(ˆ
(51)
(52)
)(ˆ LI
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
The various parameters that appear in (51) and (52) are:
)()1)(1(
)1)(1(1Den 2222
RGLGSGLRSR
SRLGLRSGRG CSjkSC
LC cos
L
LS
sin
1,))((
kcccc
c
ll
lk
mRmG
m
RG
m
Ls
LsmG
Ls
GG RR
RRLcc
RR
Ll
)(
FENE
FENEmR
FENE
RR RR
RRLcc
RR
Ll
)(
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
sLs
LG V
RR
RV
DC
ˆˆ
sLs
G VRR
IDC
ˆ1ˆ
21ˆ kvlcc
lZ G
mG
GCG
21ˆ kvlcc
lZ R
mR
RCR
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
University of Illinois at Chicago
ECE 423, Dr. D. Erricolo, Lecture 19
Among the previous parameters, k is referred to as the coupling coefficient between the generator and receptor circuits; and when the two circuits are weakly coupled.
1k1k
are the characteristic impedances of each circuit in the CRCG ZZ ˆandˆ
presence of the other circuit.
RG and are the time constants of the circuit.
DCDC GG IV ˆandˆ are the voltage and current of the generator circuit for DC
excitation, respectively.
The remaining terms give the ratio of the termination resistance to thecharacteristic impedance of the line:
When one of the previous ratios is smaller or greater than one, the corresponding termination impedance is referred to as being a low-impedance load or high-impedance load, respectively.
CRFELRCRNESRCGLLGCGSSG ZRZRZRZR /,/,/,/