hw2_solutions

# hw2_solutions - 2 TRANSIENT RESPONSE OF TRANSMISSION LINES...

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rL=o 2 TRANSIENT RESPONSE OF TRANSMISSION LINES . 2·2. Resistive loads. Since Its = 0, rs = -1. (a) For RL = 250, we have r _ Rt. - Zo _ 25 - 50 _ 1 L - Rt. + Zo - 25 + 50 - -3" At t = 0, an incident voltage of amplitude given by + Zo 50 'V I = Yo = --(3V) = 3 V Its+Zo 0+50 is launched at the source end of the line. Using these values, the bounce diagram and the load Fig. 2.2. Figure for Problem 2·2a. Rt. = 25n. (b) For RL = 500, rL = O. The bounce diagram and the sketch of the load voltage for this case are as shown. . z 3 I'VL=O O.Sns 2 I I 'V L =3V ,------- - - -- t(ns) 2 3 Fig. 2.3. Figure for Problem 2-2a. Rt. = son. voltage 'VL(t) versus t are as shown. z I 'VL=o 3 O.Sns I 2 Ins r L =-113 I 'V L=3-1=2V I l.Sns I 2ns I 'V L=2+1-1I3=813V I 2.Sns I 3ns I 'V L =813+ 113-1/9=26/9V -2.89V -2.61V I I I. I t(ns) 2 3
TRANSIENT RESPONSE OF TRANSMISSION LINES Ins 2ns r L =+I13 .z I 'VL=O O.5ns I 2.5ns I 2 Fig. 1.4. Figure for Problem z,,2c. Rt. = lOOn. (c) For RL = 100.0, the load reflection coefficient is Again, the bounce diagram and °VUt) versus t are as shown. 3 -2.67V -3.llV t(ns) 2 :3 2·3. Ringing. Using the line parameters L = 4.5 nH-cm- 1 and C = 0.8 pF-cm- 1 , the characteristic impedance Zo and the phase velocity vp of the line can be calculated as Zo=f5= 4.5 X 10- 9 = 75.0 0.8 x 10- 12 1 1 v = --= ~ 16.7 cm-ns- I p J LC v' (4.5 x 10- 9 )(0.8 x 10- 12 ) The one-way time delay td from one to the other end of the line is td = ljv p ~(30 cm)/(16.7 cm-ns- 1 )=1.8 ns. The reflection coefficients at the two ends of the line are :r _ Rs - Zo _ 15

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hw2_solutions - 2 TRANSIENT RESPONSE OF TRANSMISSION LINES...

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