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Figure Q5 shows the cross-section of the typical layout of two tracks (A and B) on a printed circuit board with a ground plane C. The tracks are terminated as shown in the figure, with track A carrying digital signals. You can assume that the lines are electrically short. Draw the equivalent circuit that models the crosstalk and write down a relation between the voltage applied to track A and the voltage coupled on to track B at low frequencies, VNE-. (a) (b) If the total mutual inductance between the two tracks L =9 nH, and the total mutual capacitance between the two tracks is C = 2 pF, calculate: (i) the amplitude of the near-end pulse VNE- (ii) the amplitude of the near-end pulse VNE if the driven line is terminated in a short circuit instead. Comment on the crosstalk. A B C 50 Ω 50 2 A V(t) 50 2 VNE VE 3 50 N V{t) 5 V 50 ns 50 ns 1 us

Figure Q5 shows the cross-section of the typical layout of two tracks (A and B) on a printed circuit board with a ground plane C. The tracks are terminated as shown in the figure, with track A carrying digital signals. You can assume that the lines are electrically short. Draw the equivalent circuit that models the crosstalk and write down a relation between the voltage applied to track A and the voltage coupled on to track B at low frequencies, VNE-. (a) (b) If the total mutual inductance between the two tracks L =9 nH, and the total mutual capacitance between the two tracks is C = 2 pF, calculate: (i) the amplitude of the near-end pulse VNE- (ii) the amplitude of the near-end pulse VNE if the driven line is terminated in a short circuit instead. Comment on the crosstalk. A B C 50 Ω 50 2 A V(t) 50 2 VNE VE 3 50 N V{t) 5 V 50 ns 50 ns 1 us

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Q5 Figure Q5 shows the cross-section of the typical layout of two tracks (A and B) on a printed circuit board with a ground plane C. The tracks are terminated as shown in the figure, with track A carrying digital signals. You can assume that the lines are electrically short. Draw the equivalent circuit that models the crosstalk and write down a relation between the voltage applied to track A and the voltage coupled on to track B at low frequencies, VNE- (a) (b) If the total mutual inductance between the two tracks L =9 nH, and the total mutual capacitance between the two tracks is C =2 pF, calculate: (i) the amplitude of the near-end pulse VNE- (ii) the amplitude of the near-end pulse VNE if the driven line is terminated in a short circuit instead. Comment on the crosstalk. A B 50 2 50 N A B 50 Ω V. VE 50 Ω Vs(t) 5 V 50 ns 50 ns 1 μs Figure Q5
Transcribed Image Text:Q5 Figure Q5 shows the cross-section of the typical layout of two tracks (A and B) on a printed circuit board with a ground plane C. The tracks are terminated as shown in the figure, with track A carrying digital signals. You can assume that the lines are electrically short. Draw the equivalent circuit that models the crosstalk and write down a relation between the voltage applied to track A and the voltage coupled on to track B at low frequencies, VNE- (a) (b) If the total mutual inductance between the two tracks L =9 nH, and the total mutual capacitance between the two tracks is C =2 pF, calculate: (i) the amplitude of the near-end pulse VNE- (ii) the amplitude of the near-end pulse VNE if the driven line is terminated in a short circuit instead. Comment on the crosstalk. A B 50 2 50 N A B 50 Ω V. VE 50 Ω Vs(t) 5 V 50 ns 50 ns 1 μs Figure Q5
(a) E = 10e¬4xt ( mv /m we know, intrinsic impedance is given as, E = Ni H. SO, 10* 103 188.5 H. H = 53.05 * 10-° A/m = ax aE * aH 父* aH Å * ŷ = 2 hence, aH = ŷ and H = 53.05e-J4nz ý µA/m
Transcribed Image Text:(a) E = 10e¬4xt ( mv /m we know, intrinsic impedance is given as, E = Ni H. SO, 10* 103 188.5 H. H = 53.05 * 10-° A/m = ax aE * aH 父* aH Å * ŷ = 2 hence, aH = ŷ and H = 53.05e-J4nz ý µA/m
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