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
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