ECE3025C_Homework3_F2005_solution

ECE3025C_Homework3_F2005_solution - + + Find parameters VIN...

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l , the wave that is generated into the line immediately after the switch is closed at 0. 0 case Reflection Components Method id n F t t V + = < 0 case t l l l l Find parameters , , , , , , , , and . Here is before the switch is closed and , are after the switch is closed at 0. Where is the first wave launched in GS S SS IN SS SS SS SS G L G S SS IN VVIVV V I VI t V ++ + ΓΓΓ Γ Γ= ± ± l to the line when the line was initially charged, 75/(75 75) 45 v. For 0 : 90 v ; 0 a ; 45 volts, since ; 0; and 1. 100 75 25 0: 0.1429, 175 IN SS SS SS SS SS SS G L G V tV I V V V V V t + +− =+ = <= = = = = + Γ = Γ = ≥Γ = = = + l 1, 0 v and 0 a. SS L = = ± 0 45 IN V + = 0 SS IN VV −= 0.1429 Γ = 45 SS V = l () 0.1429 45 6.429 G SS V = Net voltage ( 0) = 45 45 90 volts t <+ = Net voltage ( 0) = 45 6.429 51.429 volts t ≥+ = l 1 = Net voltage ( 0) Net voltage ( 0) = 38.571 volts tt V + ≥− < The first wave after the closure of the switch is negative, thus it discharges the line, and the start of the plot would look like this for the discharge cycle: zt 0 90 V = l 1 = 38.571 V + 1 51.429 V = Z t 45 SS V =
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1 Problem 1A (Continued) Using this method, we can find the voltage at the generator side of the Check An line im swer with Thevenin Eqivalent Circu mediately after the switch is close it d M at et hod 0a V t = 0 nd valid for the time range 0 2 . The Thevenin resistance is and the voltage is 2 90 volts. The Thevenin Equivalent circuit is show below. Note that the voltage at the generator side load resi SS tT ZV ≤< = stor, after the switch is thrown to position B, is 51.429 volts, i.e., the same voltage as shown in the diagram on the bottom of the previous page.
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l + 1 + 1 = 38.571 = 0.514 V I ± 00 90, 0 VI = = 1 1 51.429 0.514 V I = =− Z t 0.1429 Γ= 1 L l 2 + 2 = 38.571 = 0 4 - .51 V I ± l + 3 + 3 = 5.512 0.074 V I ± 2 2 12.858 0 V I = = 3 3 7.346 0.074 V I = l 4 + 4 = 5.51 - 2 = 0.074 V I ± l + 5 + 5 = 0.788 = 0.011 V I ± 5 5 1.046 0.011 V I = T 3T 5T l 6 + 6 = 0.78 - 8 = 0.011 V I ± 6T 4T 2T l + 7 + 7 = 0.113 = 0.002 V I ± 6 6 0.258 0 V I = = 7 7 0.145 0.002 V I = Problem 1A (Continued) Refection Diagram for 0 and 1 L t = 0 L 4 4 1.834 0 V I = =
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l 1 , the wave that is generated into the line immediately after the switch is closed at 0. 0 case Reflection Components M ethod F ind t t V + = < 0 case t l l l Find parameters , , , , , , , , and . Here is before the switch is closed and is after the switch is closed at 0. Where is the first wave launched into the line whe GS S SS IN SS SS SS SS G L G G IN VVIVV V I t V ++ + ΓΓΓ Γ Γ= ± 0 n the line was initially charged, and 75/(75 75) 45 v. 50 For 0 : 90 36 v, / 36 / 50 0.72 a 75 50 50 75 18 45 volts; from Eq. (5.4) of TEM notes for le 25 0 IN SS SS SS L SS L SS L V tV I V R VRZ V R + + =+ = ⎛⎞ <= = == = ⎜⎟ + ⎝⎠ + + = l 0 cture 5 50 75 18 9 volts; from Eq. (5.5) of TEM notes for lecture 5 0 As a check note that 36 45 9 36 volts 50 75 25 0; and 0.2 50 75 125 1 For 0:
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ECE3025C_Homework3_F2005_solution - + + Find parameters VIN...

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