V(z, t) t < L v. z = 0 z = vt z = L. I(z, t) z = L

Transcription

1 W.C.Chew ECE 35 Lecure Noes 12. Transiens on a Transmission Line. When we do no have a ime harmonic signal on a ransmission line, we have o use ransien analysis o undersand he waves on a ransmission line. A pulse waveform is an example of a ransien waveform. We have shown previously ha if we have a forward going wave for a volage on a ransmission line, he volage is V ( ) =f( ; v): (1) The corresponding curren can be derived via he elegrapher's equaion I( ) = 1 Z f( ; v): (2) If insead, we have a wave going in he negaive direcion, hen he curren from he elegrapher's equaions, is Hence, in general, if V ( ) =g( + v) (3) I( ) =; 1 Z g( + v): (4) V ( ) =V + ( )+V;( ) (5) I( ) =Y V + ( ) ; V;( ) (6) where Y = 1, and he subscrip + indicaes a posiive going wave, while Z he subscrip ; indicaes a negaive going wave. 1

2 (a) Reecion of a Transien Signal from a Shored Terminaion + V Z, v = If we swich on he volage of he above nework a =,he volage a =has he form V ( = )=V U(): (7) The volage on he ransmission line is ero iniially, he disurbance a = will creae a wave fron propagaing o he righ as increases. V(, ) < L v V V+ v = = v I(, ) V Y I+ v = = v When he wave reaches he righ end erminaion, he volage and he curren wave frons will be reeced. However, he shor a requires ha V ( ) = always. Hence he reeced volage wave, which is negaive going, has an ampliude of ;V. The corresponding curren can be derived from (4) and is as shown. 2

3 V + (, ) V + V > L v = V (, ) > L v V V V V(, ) = V + + V > L v (, ) Y V = I (, ) I Y V 2Y V I(, ) Y V = 3

4 When he signal reaches he source end, i is being reeced again. A volage source looks like a shor circui because he reeced volage mus cancel he inciden volage in order for he volage across he volage source remains unchanged. Hence he negaive going volage and curren are again reeced like ashor. Hence, if one is o measure he volage a =,iwill always be V. However, he curren a = will increase indeniely wih ime as shown. I( =, ) 7 Y V 3 Y V 3 Y V 5 Y V Y V = 2L/v = 4L/v = 6L/v The curren will evenually become inniely large because he ransmission line will become like a shor circui o he D.C. volage source. Therefore, he curren becomes innie. (b) Open-Circuied Terminaion If we have an open-circuied erminaion a, hen he curren has o be ero always. In his case, he reeced curren is negaive ha of he inciden curren such ha I( ) =always. For example, if he source waveform looks like as shown below, he reeced waveform will behave as shown. 4

5 V S () V V V(, ) 1 V + < L/v = = v( 1 ) = v V Y I(, ) < L/v = = v( 1 ) = v I(, ) Y V > L/v I V(, ) 2 V > L/v V V + I 5

6 (c) Resisive Terminaion We can hink of ransien signals as superposiions of ime harmonic signals. This is a consequence of Fourier analysis. We see ha he volage reecion coecien is;1 for a shored erminaion for all frequencies. Hence, he volage reecion coecien is ;1 for a ransien signal. By a similar argumen, he volage reecion coecien for an open-circuied erminaion is +1. When he erminaion is resisive on a lossless ransmission line, we recall ha he volage reecion coecien is v = Z L ; Z Z L + Z = R L ; Z R L + Z : (8) Hence, he reecion coecien is frequency independen. All frequency componens in a ransien signal will experience he same reecion. Hence, v is also he reecion coecien for a volage pulse. + V R A Z, v B Zin = R Consider, for example, a ransmission line being driven via a source resisance R and a load erminaion R. If R = 1 2 Z, le us see wha happens when we urn on he swich. For < L, he ransmission line appears o be inniely long o he V source. Hence, Z in looks like Z o he source. Hence, V A = Z V Z+R = 2 V 3 for R = 1 Z 2. Hence, we have awavefron of 2 V 3 propagaing o he righ for < L V. 6

7 2 V /3 2 V /9 2 Y V /3 V(, ) I(, ) 2L/v > > L/v V + 4 V /9 V = 2 V /9 8 Y V /9 I = 2 V Y /9 For > L, a reeced volage wave is generaed a he erminaion and V is ampliude is 2 3 vv. v = ; 1 for his erminaion. 3 2 V /3 2 V /9 2 Y V /3 V(, ) I(, ) I = V+ 2 V V = 9 2 V Y /9 2L/v > > L/v 4 V 9 8 Y V /9 For >2 L V,avolage source looks like a shor o he ransien signal. The reecion from he lef is again ; 1 for he volage and + 1 for he curren

8 2 V /3 V(, ) I(, ) 14 V 27 V2+ = 2 V/27 V = 2 V/9 V1+ = 2 3 V 26 Y V 8 27 Y V 9 I 2+ = 2 Y V /27 I = 2 Y V /9 4 V 9 I1+ = 2 YV /3 When! 1, he volage and curren on he line will sele down o a seady sae. In ha case, we have only DC signal on he line, and we need only o use DC circui analysis o nd he seady sae soluion. A DC,he ransmission line becomes rs wo pieces of wires, V A = V B = R V 2R = 1 V 2. The curren hrough he circui is V. If one is o measure V A as a funcion Z of ime, i will look like V A () 2 V 2 V /3 14 V /27 3 V /2 2 3 V Y V Y I A () 2L/v 4L/v 6L/v 26 V Y /27 2L/v 4L/v 6L/v Transien analysis has imporan applicaion o compuer circuiry. We noe ha when we swich on a circui wih a delay line, we do no immediaely arrive a he desired seady sae value when we have a ransmission line or a delay line. The seling ime depends on he lengh of he line involved. 8