Lesson 3Transient Response of Transmission Lines

楊尚達 Shang-Da YangInstitute of Photonics TechnologiesDepartment of Electrical EngineeringNational Tsing Hua University, Taiwan

Introduction

When there is an interface between different materials (ε, μ), discontinuity exists, ⇒ partial reflection & transmission, ⇒ total v, i are determined by superposition (why?).Goal: Transient response of a terminated transmission line or cascaded lines excited by a step-like voltage source.

(1) :

Example 3-1: A finite line with resistive load-1

;otherwise ,0

if ,),( 1

1 ⎪⎩

⎪⎨⎧ <

=+

+ tvzVtzv p0>t 0

0

01 V

RZZ

VS+

=+

As if there were no load.

Example 3-1: A finite line with resistive load-2

(2) :dtt >⎪⎩

⎪⎨⎧ −−>

=−

−

otherwise ,0

)( if ,),( 1

1dp ttvlzV

tzv

),(1 tzv −

,)()(

0 LL

L RZtitv

≠=

Reflected wave must arise, otherwise,

in violation with the BC.

Meanwhile, acts as if it had kept on going to the right.),(1 tzv+

Example 3-1: A finite line with resistive load-3

),(),()( 11 tlvtlvtvL−+ +=

0

1

0

1 ),(),()(Z

tlvZ

tlvtiL

−+

−=

)()()( tRtitv LLL ⋅=BC:

⎥⎦

⎤⎢⎣

⎡−=+⇒ +

−

+

−

),(),(1

),(),(1

1

1

01

1

tlvtlv

ZR

tlvtlv L

[ ]),(),(),(),( 110

11 tlvtlvZRtlvtlv L −+−+ −=+⇒

Example 3-1: A finite line with resistive load-4

( )LL

L ZR

Γ−=Γ+⇒ 110

0

0

1

1

),(),(

ZRZR

tlvtlv

L

LL +

−=≡Γ +

−Load voltage reflection coefficient

can be used to calculate given is known.

LΓ−

1V+

1V

Example 3-1: A finite line with resistive load-5

(3) :dtt 2>⎪⎩

⎪⎨⎧ −<

=+

+

otherwise ,0

)2( if ,),( 2

2dp ttvzV

tzv

0

0

1

2

),(),(

ZRZR

tlvtlv

S

SS +

−=≡Γ −

+ Source voltage reflection coefficient

can be used to calculate given

is known.

SΓ+

2V−

1V

Example 3-1: A finite line with resistive load-6

At :dtt 2=

( )SLLdddddS tvtvtvtvtv ΓΓ+Γ+=++= ++−+ 1)2,0()2,0()2,0()2,0()2( 1211

( )SLLdddd

dS Ztv

Ztv

Ztv

Ztvti ΓΓ+Γ−=+−=

++−+

1)2,0()2,0()2,0()2,0()2(0

1

0

2

0

1

0

1

Example 3-1: A finite line with resistive load-7

( )...1),0()(lim 232221 +ΓΓ+ΓΓ+ΓΓ+ΓΓ+Γ+⋅∞→= +

→∞ SLSLSLSLLSttvtv

( ) ( ) ( )[ ]...111 221 +Γ+ΓΓ+Γ+ΓΓ+Γ+⋅= +

LSLLSLLVSL

LVΓΓ−Γ+

= +

11

1

At :∞→t

00

01 V

RZZ

VS+

=+

0)(lim VRR

RtvLS

LSt +

=⇒∞→

Steady state response is as if there were no line.

What is the usefulness of bounce diagram?

Bounce diagram is a convenient tool to solve for transient response of a TX line driven by a step voltage and terminated by a resistor.

How to draw a bounce diagram?

Draw a 2D window with 0< z/l <1, t/td >0. Denote the two reflection coefficients ΓS , ΓL .Mark the points (0,2n)and (1,2n+1) for n = 0, 1, 2, …Connecting the points by zigzag lines.Mark the voltage amplitude of each component wave.

0

0

ZRZR

L

LL +

−=Γ

0

0

ZRZR

S

SS +

−=Γ

Determine the spatial voltage distribution at some time instant-1

The spatial distribution must be of the form:

⎪⎩

⎪⎨⎧

<<

<<=

lzzV

zzVtzv

rightt

left

0

00 for ,

0for ,),(

The problem is how to find the three key parameters: z0, Vleft, and Vright, respectively.

Determine the spatial voltage distribution at some time instant-2

Draw a horizontal line t = t0 (i.e. t/td = t0/td), intersecting with some zigzag line at point P0with coordinate (z0,t0).

Determine the spatial voltage distribution at some time instant-3

The determination of Vleft and Vright has two cases.If P0 falls on a line with positive slope:

+= 1VVleft

( )L

right

V

VVV

Γ+=

+=+

−+

11

11

Vright has one more component.

Determine the spatial voltage distribution at some time instant-4

If P0 falls on a line with negative slope:

( )SLL

left

V

VVVV

ΓΓ+Γ+=

++=+

+−+

11

211

( )L

right

V

VVV

Γ+=

+=+

−+

11

11

Vleft has one more component.

Determine the temporal voltage distribution at some position

must have constant voltages switched at a series of time instants.Draw a vertical line z = za, intersecting with the zigzag lines at

),( tzv a

.... ,, , 211+−+= tttt

diiii ttttt 211 =−=− +−+

+++

Example 3-2: Single TX line with open termination-1

025.0 ZRs = ∞=LR

10

0 =+−

=ΓZRZR

L

LL

6.00

0 −=+−

=ΓZRZR

S

SS

Example 3-2: Single TX line with open termination-2

(1) :0=z ,01 =+t ,221 dttt == +− ,...432 dttt == +−

000

01 8.0 VV

RZZV

S

=+

=+

0)48.08.08.0( V−+

0)288.048.012.1( V+−

Example 3-2: Single TX line with open termination-3

(2) :lz = ,11 dttt == −+ ,...)12( dkk tktt −== −+

0)8.08.0( V+

0)48.048.06.1( V−−

Example 3-3: Single TX line with matched termination-1

04ZRs =0ZRL =

6.00

0 =+−

=ΓZRZR

S

SS

00

0 =+−

=ΓZRZR

L

LL

Example 3-3: Single TX line with matched termination-2

000

01 2.0 VV

RZZV

S

=+

=+No overshooting & ringing (good!)

Example 3-4: Cascaded TX line-1

Example 3-4: Cascaded TX line-2

(1) At :1 , dj ttlz ==

),(),(),( 111111 djBdjAdjA tlvtlvtlv +−+ =+BC:

Example 3-4: Cascaded TX line-3

…Transmission coefficient

,),(),(

),(),(

111

11

11

11

djA

djB

djA

djA

tlvtlv

tlvtlv

+

+

+

−

=+⇒

),(),(),( 111111 djBdjAdjA tlvtlvtlv +−+ =+

ABAB T=Γ+⇒1

Example 3-4: Cascaded TX line-4

00

0 =+−

=ΓZRZR

S

SS

31

−=ΓAB

31

=ΓBA

32

=ABT

34

=BAT

6.00

0

=+−

=ΓZRZR

L

LL

Example 3-4: Cascaded TX line-5

V 75.05.0 0

00

01

==+

=+

V

VRZ

ZVSA

AA

( ) ABB TV ⋅=+ V 75.01

( ) LBV Γ⋅=− V 5.01

( ) BAA TV ⋅=+ V 3.02

+AV1 −+ + AA VV 11

−−+ ++ AAA VVV 211

Example 3-4: Cascaded TX line-6