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: (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: (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 the products and since these do not commute in general. A general solution is obtained by either solving (9) or(10). We will consider the solution of (10). (9) (10)

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: (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 The eigenvalues of and are referred to as propagation constants and are used to form the solution: Note that 1) the form of the solution is the same as for two-conductor lines; 2) the constants Equations (16) and (17) may be cast as: where (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 actual currents are found from: The voltage solution is obtained from: Observe that, since, then Hence (22) may be rewritten (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: Let us now see how we can solve for the unknown constants. We need to add some additional constraints. For this purpose we look at the transmission line as a four-port circuit: Three-conductor line This allows us to write: Figure 1 (26) and (27) (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: Evaluating the voltage, given by (22), at yields: When (29) and (30) are combined with (26) we obtain: Similarly, evaluating the voltage at yields: When (32) and (33) are combined with (27) we obtain: (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 matrix equation that determines After the previous equation is solved current and voltage along the line are computed using equations (21) and (22). Looking at the line as a four-port network may be particularly useful if we are interested only in the values of voltages and currents at the end points of the line. For this purpose we need the chain parameter matrix (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 expressions are: Finally, another representation, obtained substituting (26)-(27) into (36), for the terminal currents is: (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 solution that is obtained by making the following assumptions: 1) three-conductor line; 2) lossless; 3) homogenous medium Lossless implies: Homogeneous implies: from which And we also observe that is diagonal so that. The propagation constant matrix is: (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: When (45)-(48) are introduced into the representation (41) for the terminal current we obtain: (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: (50) When (49) and (50) are solved for the terminal voltages due to crosstalk among the wires one finds: (51) (52)

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:

University of Illinois at Chicago ECE 423, Dr. D. Erricolo, Lecture 19 University of Illinois at Chicago ECE 423, Dr. D. Erricolo, Lecture 19

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. are the characteristic impedances of each circuit in the presence of the other circuit. are the time constants of the circuit. are the voltage and current of the generator circuit for DC excitation, respectively. The remaining terms give the ratio of the termination resistance to the characteristic 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.