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# ece 3025 ECE 3025 A

Georgia Tech

GPA 3.72

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This 21 page Class Notes was uploaded by kayden ton on Sunday February 7, 2016. The Class Notes belongs to ECE 3025 A at Georgia Institute of Technology taught by Prof. Alenka Zajic in Spring 2016. Since its upload, it has received 13 views. For similar materials see Electromagnetics in Electrical Engineering at Georgia Institute of Technology.

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Date Created: 02/07/16

− + + ▯ ▯ ▯ Find parametersVVIN ,SSV,SS, SS,SSS,VGΓ,G ISL ,SS and . Here G is before the switch is closed SndΓG,VS SSand ▯ are after the swittch is closed at0. WhereV IN is the first wave launched inthe line when the line was initially charged, and 7IN(75 75)v.45 For t< 0: SS ==90 v ; SS+= 0 a ; SS− SS 45 volts, sinceSS SS SS ;G 0; andL 1. ▯ 100 75 25 ▯ ▯ For 0:≥Γ G = = 0.1429, Γ=L1, V=I SS 0 v an=d SS 0 a. 100 75 175 Reflection ComponentsMethod + Fid V , the wave that is generated into the lineimediately after the switch is closed att = 0. t< 0 case t ≥ 0 case − VSS 45 V IN= 45 + + ▯ − VVSS IN 0 Γ=GVSS 0.1=29 () 6.429 Γ= 0 − VSS 45 Γ = 0.1429 Net voltage (t= +) = 45 45 90 volts Net voltage (t ≥ 0 +== 45 6.429 51.429 volts + V 1 Net voltage (t ≥t−0) Net vol<tage−( 0) = 38.571 volts The first wave after the closure of the switch is negative, thus it discharges the line, and the start of the zt− plot would look like this for the discharge cycle: Z V0= 90 + t V1= 51.429 V = 3 −8.571 1 Problem 1A (Continued) Check Answer with Thevenin Eqivalent Circu it Method Using this method, we can find the voltage V at th1 generator side of the line immediately after the switch is closed at t = 0a nd valid for the time range 0 ≤< tT2 . The Thevenin resistance is Z aVnd the voltage is 2 − 90 volts. 0 SS The Thevenin Equivalent circuit is show below. Note that the voltage at the generator side load resistor, 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. Problem 1A (Continued) Refection Diagram for t=>Γ0 and 1 L 0 L Z + V0 0 90, 0 = V 1 3 −8.571 V 1 51.429 ▯ + I =− 0.514 I1= 0−.514 1 - ΓL 1 V 2 3−8.571 T + I2= 0.514 Γ= 0.1429 + V =12.858 V 3 5−.512 2 2T + I =0 I3=− 0.074 2 V3= 7.346 I3=− 0.074 V = 5 −.51 2 4 3T ▯+ I4= 0.074 V =1.834 V = 0−.788 4 4T 5 I4= 0 ▯+ I5= 0−.011 V5=1.046 I =− 0.011 5 - V 6 0 −.78 8 5T + I6= 0.011 V6= 0.258 I = 0 6 6T V = 0 −.113 7 ▯+ I7 = −.002 V7= 0.145 t I7=− 0.002 Find parametersVV,I VV , , , SVΓ,Γ GI S ,▯ SSand ▯ .ΓHere is before IN SS SS SS SS G L G the switch is closed andΓG s after the switch is closed att0. Where V + is the first wave launched into the line n the line was initially charged, IN and 75/(75 75)v. 45 IN ⎛ ⎞50 For t V0: SS= 90=⎜ ⎟ VI 36 v, SS SS/ L 36/50 0.72 a ⎝ ⎠ 50 + VSS⎛ ⎞L+ 0 ⎛ ⎞ 75 0 VSS = =2 5 R 18 ⎜ ⎟ 45 volts; from Eq. (5.4) of TEM notes for leure 5 ⎝ ⎠ L ⎝ ⎠ − V − = = =⎛ ⎞L − 0 18⎛ ⎞0 75 9 volts; from Eq. (5.5)of TEM notes for lecture 5 0 SS 2 5 ⎟ R ⎝ ⎠ ⎝ ⎠ L As a check note thatV V == + =− +=− 45 9 36 volts SS SS SS 50 −− 25 anΓG L Γ= = 0.2− 50 +5 125 1 00 75 25 For 0≥=Γ ▯ G == 0.1429, =SSa=0d SS 0.2 L 100 75 175 Reflection Components M ethod + FindV▯1 , the wave that is generated into the lmediately after the switch is clostd at = 0. t < 0 case t ≥ 0 case − + V SS−9 VIN = 45 VSS=+ IN 0 Γ= 0 − V SS 9 ▯ − Γ = 0.1429 Γ=GVSS −0.1429 () 9 1.286 Net voltage (t=<−) = 45 9 36 volts Net voltage (t −0) −= 9 1.286 10.286 volts + V1= Net voltage (t ≥t0) Net vol<−g=e ( 0) 46.286 volts Problem 1B (Continued) The first wave after the closure of the switch is negative, thus it discharges the line, and the start of the zt−plot would look like this for the discharge cycle: Z V = 36 0 t V =− 10.286 ▯ + 1 V 1 4 −6.286 Check Answer with Thevenin Eqivalent Circuit Method Using this method, we can find the voltage V1 at the generator side of the line immediately after the switch is closed att =0 and valid for the time range 0 ≤<tT 2 . The Thevenin resistance is Z aVnd the voltage is 2 − =−18 volts. 0 SS The Thevenin Equivalent circuit is show below. Note that the voltage at the generator side load resistor, after the switch is thrown to position B, is -10.286 volts, i.e., the same voltage as shown in the diagram above. Problem 1B (Continued) Refection Diagram for − t= >Γ0 and L 0.2 0 L Z + V 0 036, 0.7=2 V 1 4 −6.286 V =− 10.286 1 I1= 0−.6171 I1= 0.1029 ▯ - Γ=L 0.2 V 2 9.257 T + I2 =1−0. 23 Γ= 0.1429 ▯+ V2= −1.029 2T V 3 1.323 + I2=− 0.0205 I3= 0.0176 V 3 0.294 I3=− 0.0029 - V 4 0 −.265 + 3T I4= 0.0035 + V4= 0.029 V 5 0 −.038 4T + I4= 0.0006 I5= 0−.0005 V5=1.046 I5= 0.0001 V 6 0.008 5T ▯ + I 6− 00 .0 1 V6= 0.258 I6= 0 6T ▯+ V 7 0.001 + I7∼ 0 V7= 0 t I7= 0 Problem 1B (Continued) Bergeron Plot (Graphical) Method Generator Resistor 75 Ω slope V0 0 36, 0.72 Load Resistor 50 Ω slope Transmission Line 75 Ω slope V SS= 0 I SS= 0 V3= 0.294 I =− 0.0029 3 V =− 1.029 2 I2=− 0.0205 V1= −10.286 I1= 0.1029 Discharge Resistor 100 Ω slope + For 0 <<tT V/2, INis computed by voltage dividing the 25v souce tweeZ G and 0, ⎛ ⎞ 75 VIN = 25⎜ ⎟ 12.5v ⎝ ⎠5+75 At tT= / 2, the switch is opened. This pro duces a zero voltage on the generator end of the line after the switch is opened, athe wave generated is equal to the voltage state after the switch is opene d minus the state before the wave. Therefore opening the switch sends a -12.5v wave into the left side of the transmission line. Γ changes at the generator end of the line due to closing and opening the switch, i.e., Γ= 1; < t 0 G ΓG 0; <<0 tT /2 ΓG 1; > tT /2 Finally, compute the reflection coefficient at the load the line using Γ= RL G = 150 =5 1 L RR − + 150 75 3 L G + The above values forΓG L N IndV 12.5v are used to construct the reflection diagram shown on the next page. Problem 2A (Continued) Refection Diagram for t >0 0 L Z 0 v + 12.5 v V1= 12.5v + V1= 1−2.5v T Γ=L1/3 Γ= 0; t< 0 v T 2 T V = 4.167v 16.67 v Γ= 1; t> 2 2 - V2= − .167v 2T 4.167 v + 8.33 v V3= 4.167v 0 v 4.167 v + V 3 4−.167v 3T 0 v V = 1.389v 5.56 v 4 ▯- V 4 1− .389v 4T 1.389v 2.8 v V = 1.389v 0 v 5 1.389 v V = 1−.389v 5 5T 0 v 1.85 v − V 6 0.463v 0.463 v - V 6 0− .463v 6T 0 v 0.93 v t Vola 0 z = V Lg zL= IN Problem 2A (Continued) V INd V ploLs V IN 15 12.5v 10 8.3v 5 2.8v 0 t 0 T 2T 3T 4T 5T 6T V L 25 20 16.7v 15 10 5.6v 5 1.9v 0 t 0 T 2T 3T 4T 5T 6T Problem 2B: For the case when the switch is ected at t =and disconnected at (/3 ) All voltages for the solution are the same, except the pulse wi()h is nnstead()f 1/2 , thus, can just change the times on the voltage time plots accordingly. V IN 15 12.5v 10 8.3v 5 2.8v 0 t 0 T 2T 3T 4T 5T 6T V L 25 20 16.7v 15 10 5.6v 5 1.9v 0 t 0 T 2T 3T 4T 5T 6T For 0 <<tT V/2, INis computed by voltage dividing the 10v souce tweenZ G and 0, + ⎛ ⎞ 150 VIN = =5⎜ 75+150 16.67v ⎝ ⎠ The peak voltage of the generator waveform is 10v, this corresponds to aIN of + ⎛ ⎞ 50 VIN = 10⎜ 75+50 4v ⎝ ⎠ ⎛ ⎞5− − ⎛ ⎞ 25 50 The reflection coefficents arΓG ⎜ ⎟ = ⎜ ⎟Γ=nd − 1/3 ⎝ ⎠5+ + ⎝ ⎠ 25 50 To construct the reflection diagram, use a 4v wave generated at= /8, this is the solid line of the reflection diagram shown on the next page. Next, stetch in wave lines beginning Tt t=0=and /4; these ar e the dashed lines shown on the reflection diagram. In the region between the leading dashed line and the trailing dashed line the voltage ramps up to the value at the crest midway through the region, and then back down to zero at the back edge of the wave, i.e., the second dashed line. Problem 3 Refection Diagram for 0 ≤≤ tT 6 0 L Z 4 v at peak 0 v 4 v at peak 1T 4 ΓL− 1/3 T 0 v T Γ=G 0.2 −1.333 v at peak 2.667 v at peak 5T 4 2T −1.6 v at peak 0 v −0.267 v at peak 9T 4 3T 0 v 0.089 v at peak −0.178 v at peak 13T 4 4T 0.107 v at peak 0.018 v at peak 0 v 17 T 4 5T 0 v −0.006 v at peak 0.012 v at peak Vola 0 z = IN 21T 4 6T −0.007 v at peak 25 T 4 t −0.001 v at peak V Lg z= Problem 3 (Continued): V INd V ploLs V IN 4 v 4 2 0.107 v 0 −1.6 v -2 t 0 T 2T 3T 4T 5T 6T V L 4 2.67 v 2 0.01 v 0 −0.18 v -2 t 0 T 2T 3T 4T 5T 6T + The above is the equivalent circuit foV 0 where 2.5 v. Adding voltages gives di ZRi i=+ 0 L dt + or di + = ⎞0 L i 2V dt ⎝ ⎠ L L L 220 x 10-6 thus τ = = = 2.2 sec ZR0 L 100 For t< = V L 0 ttd+ V V = L LL2 Γ= 5 vI; = L 1and 0 ( appears like as open cirucit) + t▯ V V L L L =2.5 v; I= 0 andL 0.05 A ( appears like a short cirucit) Now wecan sketch the plots using these limiting values. Since neither the transit time, nor the length and propagtion velocity of the lineare specified, then all we can do it sketch in the exponential functional solution forms and note time constan2.2 sec on the plots. Problem 4 (Continued): (VtIN − )tand ( L )plots V ()olts 25=+ IN L τ = = 2.2 μsec time constant Z0 + L 4 2 V = 2.5 v 0 0 T 2T 3T V ()olts v= + L τ = = L 2.2 μ sec time constant Z0 + L 4 2 + V = 2.5 v 0 0 T 2T 3T Problem 4 (Continued): (tL− L )atnd ( )plots V L()olts v= + L τ = = 2.2 μsec time constant 4 Z0+ L 2 V = 2.5 v 0 0 T 2T 3T I ()mps L IL= 0.05 A 0.05 L τ = = 2.2 μsec time constant Z0+ L 0 0 T 2T 3T The solution of the load current may be obtained by inspection of the current plot: ⎡ ⎤ − ⎜ ⎟ It() =0.05 1 ⎥ e− ⎝ t = ()= T ; where τ L 2.2μ sec L ⎢ ⎥ ZR + ⎣ ⎦ 0 L From the equivalent circuit below, t V=where + 12.5 v. dvv Adding currents gives: ii 1 2 +C , dt R Relating the total voltage and current to wave compos across the load gives, + − + − VV v=+VV i= −d , ZZ0 0 2Vv eliminatingV −bitween these equations gives = − ZZ0 0 d V 2⎛ Z0+ R L + Finally, summing currents in the circuit below produces: ⎜ ⎟ i dt ⎝ ⎠0RCL CZ0 -12 The time constant is the reciprocal of the term:=Z0RCL 7=ii0 5 x 10 150 sec 5 2Z0+ R L Now the following limiting values of the voltage and current can be used to sketch the plots. For t< VL = 0 t=+d V = Γ−= 0; = 1 and 0.167 (C appears like as short cirucit) L L L ⎛ ⎞50 tT▯ V= L L 2L⎜ ⎟ 10 v; I0.2 and 0.2 A ( C appears like an open cirucit) ⎝ ⎠ 50 Now we can sketch the plots using these limiting values. Since neither the transit time, nor the length and propagtion velocity of the line are specified, then all we can do it sketch in the exponential functional solution forms and note time consta150p sec on the plots Problem 5 (Continued): (VIN− −V)and ( L ) plots V IN()olts VL= 12.5 v VL= 10 v 10 Z0RL τ = = 150 pec Z0 + RL 0 0 T 2T 3T VL ()olts VL= 10 v 10 Z RC τ = =0 L 150 pec Z0+ R L 0 0 T 2T 3T The solution of the voltage across theload may be obtained by inspection: ⎛ ⎞ ⎡ ⎤ − ⎜τ⎟ Z RC t V L() p=10 1 ⎥ e − ⎠ () =T ; where τ 0 L 150 sec ⎢ ⎥ Z0+ R L ⎣ ⎦ 0o5= Ω i → di L = 220Hμ v= dt + v25 += - R = 50 Ω L 5o7=Ω i i2 2 V 2V + + C5= p L - RL=Ω 50

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