Lecture 24
14
1 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Lecture 24 Time Domain Analysis of Transmission Lines In this lecture you will learn: • Time domain analysis of transmission lines • Transients in transmission lines ECE 303 – Fall 2007 – Farhan Rana – Cornell University o Z 0 = z () t V s l − = z Time Domain Analysis - Basics Question: How does one handle transmission lines for signals that are NOT time harmonic and when one is NOT dealing with the sinusoidal steady state? a) First thing to realize is that the notion of complex impedance has meaning only for the sinusoidal steady state b) For an arbitrary source voltage V s (t ), one needs to work in the time domain and start from the basic time-domain equations: s R L R ( ) ( ) t t z I L z t z V ∂ ∂ − = ∂ ∂ , , ( ) ( ) t t z V C z t z I ∂ ∂ − = ∂ ∂ , , ( ) ( ) 2 2 2 2 2 , 1 , t t z V v z t z V ∂ ∂ = ∂ ∂ ( ) ( ) 2 2 2 2 2 , 1 , t t z I v z t z I ∂ ∂ = ∂ ∂ LC v 1 =
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Lecture 24
Transcript of Lecture 24
1
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Lecture 24
Time Domain Analysis of Transmission Lines
In this lecture you will learn:
• Time domain analysis of transmission lines
• Transients in transmission lines
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
oZ
0=z
( )tVs
l−=z
Time Domain Analysis - Basics
Question: How does one handle transmission lines for signals that are NOT time harmonic and when one is NOT dealing with the sinusoidal steady state?
a) First thing to realize is that the notion of complex impedance has meaning only for the sinusoidal steady state
b) For an arbitrary source voltage Vs(t ), one needs to work in the time domain and start from the basic time-domain equations:
sRLR
( ) ( )t
tzILz
tzV∂
∂−=
∂∂ ,,
( ) ( )t
tzVCz
tzI∂
∂−=
∂∂ ,,
( ) ( )2
2
22
2 ,1,t
tzVvz
tzV∂
∂=
∂∂
( ) ( )2
2
22
2 ,1,t
tzIvz
tzI∂
∂=
∂∂
LCv 1=
2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Time Domain Analysis - Basics
oZ
The equation: ( ) ( )2
2
22
2 ,1,t
tzVvz
tzV∂
∂=
∂∂
Has forward moving solutions of the form: ( ) ( )vtzVtzV −= +,
And backward moving solutions of the form: ( ) ( )vtzVtzV += −,
Examples:
( )vtzV −+
z
v
( )vtzV +−
z
v
LCv 1=
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Voltages and Currents
oZ
( )vtzV −+
z
vVoltage:
( )vtzI −+
z
vCorresponding Current:
( ) ( )t
tzILz
tzV∂
∂−=
∂∂ ,,The current is related to the voltage and satisfies:
And this: ( ) ( )t
tzVCz
tzI∂
∂−=
∂∂ ,,
( ) ( ) ( ) ( )oo Z
vtzVvtzIZ
vtzVvtzI +−=+
−=− −
−+
+ andThe solution is:
++ ++ ++
-- -- --
Current is proportional to voltage since higher voltage means more surface charges and more surface charges mean more current flow
v
3
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Visualizing Propagation
oZ
0=zl−=z
t
1
Forward moving solutions are of the form: ( )vtzV −+
And backward moving solutions are of the form: ( )vtzV +−
So suppose somebody tells you that at z = -ℓ : ( ) =−+ tV ,l
Then what is the forward moving voltage on the line at t = ℓ / 2v ? T
0=zl−=z
( )vtzV 2, l=+
vt
2l
=
vT
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Load End Boundary Condition
oZ
0=z
( )tVs
l−=z
sR
LR
Load end boundary condition:
( ) ( ) LRtzItzV ,0,0 ===
( ) ( ) ( )tzVtzVtzV ,0,0,0 =+=== −+
( ) ( ) ( )( ) ( )
oo ZtzV
ZtzV
tzItzItzI,0,0
,0,0,0=
−=
=
=+===
−+
−+
( ) ( ) LtzVtzV Γ===⇒ +− ,0,011
+−
=ΓoL
oLL ZR
ZR
+
-( )tzV ,0=
( )tzI ,0=
For all time t we must have:
Substitute these in this to get:
4
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Source End Boundary Condition
oZ
0=z
( )tVs
l−=z
sR
LR+
-
Source end boundary condition:
( ) ( ) ( )tzVRtzItV ss ,, ll −=+−==
( ) ( ) ( )tzVtzVtzV ,,, lll −=+−==−= −+
( ) ( ) ( )( ) ( )
oo ZtzV
ZtzV
tzItzItzI,,
,,,ll
lll
−=−
−==
−=+−==−=
−+
−+
( ) ( ) ( )os
oss ZR
ZtVtzVtzV+
+Γ−==−=⇒ −+ ,, ll11
+−
=Γos
oss ZR
ZR
( )tzV ,l−=( )tzI ,l−=
For all time t we must have:
Substitute these in this to get:
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-On Transient - I
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR Ω= 150LR
Suppose the source voltage is a step function:( ) ( )tutVs 4=
t
4
21
=ΓL0=Γs
( ) ( )
( )
( ) 20,
,,
=>−=⇒
++
Γ−==−=
+
−+
tzV
ZRZtV
tzVtzV
os
os
s
l
ll
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
0=t
5
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-On Transient - II
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR Ω= 150LR
Suppose the source voltage is a step function:( ) ( )tutVs 4=
t
4
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
2
2
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
2
2
vt
2l
=v
t23l
=
21
=ΓL
1
3
0=Γs
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR Ω= 150LR
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
2
3
vt l2≥
21
=ΓL
1
• The wave reflected back from the load end does not suffer a reflection at the source end since the source impedance is matched to the line impedance
• After a time greater than 2ℓ/v the line voltage is at constant 3 Volts
Step Voltage Source: Turn-On Transient - III
0=Γs
( ) ( )
( )os
os
s
ZRZtV
tzVtzV
++
Γ−==−= −+
,, ll
6
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-Off Transient - I
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR Ω= 150LR
Suppose the source voltage has been at 4 Volts for a long long time – but is shut off at time t = 0
( ) ( )[ ]tutVs −= 14t
4
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
2
vt
2l
=
21
=ΓL
1
3
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
2
3
0=t
1
0=Γs
1
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-Off Transient - II
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR Ω= 150LR
t
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
vt l2≥
21
=ΓL
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
vt
23l
=
1
0=Γs
1
• After a time greater than 2ℓ/v the line voltage is at constant 0 Volts
7
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-On Transient – General Case
oZ
0=z
( )tVs
l−=z
sR LR
LΓsΓ
Suppose the source voltage is a step function:( ) ( )tuVtV os =
t
oV
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
vt
2l
=v
t23l
=
os
oo ZR
ZV+
os
oo ZR
ZV+
Los
oo ZR
ZV Γ+
os
oo ZR
ZV+
( )Los
oo ZR
ZV Γ++
1os
oo ZR
ZV+
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-On Transient – General Case
oZ( )tVssR LR
LΓsΓ
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
vt
25l
=
os
oo ZR
ZV+
Los
oo ZR
ZV Γ+
( )Lsos
oo ZR
ZV ΓΓ++
1
( )⎥⎦
⎤⎢⎣
⎡Γ+
ΓΓ++ L
Ls
os
oo ZR
ZV1
( )Los
oo ZR
ZV Γ++
1
• The waves will keep bouncing forever
• But the net voltage on the line will slowly converge to the value one would expect in DC operation
( ) ( ) ( )[ ]
( ) ⎥⎦
⎤⎢⎣
⎡ΓΓ−+
=∞
+ΓΓ+ΓΓ+ΓΓ++
=∞
+
+
Lsos
oo
LsLsLsos
oo
ZRZVzV
ZRZVzV
11,
........1, 32
( ) ( ) ( )[ ]
( ) ⎥⎦
⎤⎢⎣
⎡ΓΓ−
Γ+
=∞
+ΓΓ+ΓΓ+ΓΓ+Γ+
=∞
−
−
Ls
L
os
oo
LsLsLsLos
oo
ZRZVzV
ZRZVzV
1,
........1, 32
( ) ( ) ( )
Ls
Lo
Ls
L
os
oo
RRRV
ZRZV
zVzVzV
+=
ΓΓ−Γ+
+=
∞+∞=∞ −+
11
,,,
8
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-On Transient
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 150sR Ω= 150LR
21
=ΓL21
=Γs
Source voltage is a step function: ( ) ( )tutVs 1=
t
1
( )tzV ,
0=zl−=z
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-On Transient – Capacitive Load - I
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR0=Γs
LC
Source voltage is a step function: ( ) ( )tutVs 1=
t
1
( )tzV ,
0=zl−=z
9
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-On Transient in a RC Circuit
( )tVsR C
+
-( )tVC
Source voltage is a step function:
( ) ( )tutVs 1=
t
1
( )tVCSolution for is:
t
1( )tVC
time constant CR== τ
( )tVs
( ) ( )tuetVt
C ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−=
−τ1
At short times the capacitor acts like a shortAt long times the capacitor acts like an open
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-On Transient – Capacitive Load - II
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR0=Γs
LC
Source voltage is a step function: ( ) ( )tutVs 1=
t
1
How does one solve this problem?
Transmission line is a linear system
So make a Thevenin equivalent circuit looking in from the load end
( )tVththR
LC
oth ZR =This holds even if the source impedance were not matched to the line impedance (no such thing as impedance transformations in time domain)This is because in transients situations it does not really matter what is on the other side of the line
+
-( )tzV ,0=
10
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-On Transient – Capacitive Load - II
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR0=Γs LC
t
1
A forward voltage wave of 0.5 Volts from the source will reach the load at time
vt l=
vt
vll 3
≤≤
To find Vth(t ) remove the load capacitor and look at the open circuit voltage:
+
-( )tVth
vt l3=
?????
( )tVth
vt l=
A reflected voltage wave of 0.5 Volts will be generated at the same time v
t l=
( ) 1=⇒ tVth for
( )tVththR
LC
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-On Transient – Capacitive Load - III
( )tVththR
LC
So the Thevenin circuit for times is:v
tv
ll 3≤≤
t
1
vt l=
vt l3=
( )tVth
oth ZR = +
-
( )tzV ,0=Solution for is:
t
1
vt l=
vt l3=
( )tzV ,0=
( )tzV ,0= time constant Lth CR== τ
11
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-On Transient – Capacitive Load - IV
t
1
vt l=
vt l3=
( )tzV ,0=
Since ( ) ( ) ( )tzVtzVtzV ,0,0,0 =+=== −+And is:( )tzV ,0=+
t
5.0
vt l=
vt l3=
( )tzV ,0=+
Therefore must be:( )tzV ,0=−
t
5.0
vt l=
vt l3=
( )tzV ,0=−
5.0−
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR0=Γs
LCΩ= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR0=Γs
LC
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR0=Γs
LC
Step Voltage Source: Turn-On Transient – Capacitive Load - V
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
0.5
0.5
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
0.5
vt
2l
=v
t23l
=
0.5
- 0.5
0.5 1
12
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR0=Γs
LC
Step Voltage Source: Turn-On Transient – Capacitive Load - VI
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
0.5
τ52+>
vt l
0.5
1
At sufficiently long times the capacitor is charged to 1 Volt
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR0=Γs
LC
Step Voltage Source: Turn-On Transient – Another Perspective
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
0.5
0.5
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
0.5
vt
2l
=v
t23l
=
0.5
- 0.5
0.5 1
At short times the capacitor acts like a short (ΓL = -1)At long times the capacitor acts like an open (ΓL = +1)CAREFUL: these kind of approximate arguments do not tell you what are the involved time constants
13
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Step Voltage Source: Turn-On Transient in a LR Circuit
( )tVsR L
+
-( )tVL
Source voltage is a step function:
( ) ( )tutVs 1=
t
1
( )tVLSolution for is:
t
1( ) ( )tuetV
t
L τ−
=
time constantRL
== τ
( )tVs
1
( )tVL
At short times the inductor acts like an openAt long times the inductor acts like a short
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR0=Γs
Step Voltage Source: Turn-On Transient –Inductive Load - I
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
0.5
0.5
0=zl−=z
( )tzV ,+
( )tzV ,−
( )tzV ,
0.5
vt
2l
=v
t23l
=
- 0.5
0.5
0.5 0.0
Source voltage is a step function
L
At short times the inductor acts like an open (ΓL = +1)At long times the inductor acts like a short (ΓL = -1)
14
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Ω= 50oZ
0=z
( )tVs
l−=z
Ω= 50sR0=Γs
L
Step Voltage Source: Turn-On Transient – Inductive Load - II
Source voltage is a step function: ( ) ( )tutVs 1=
t
1
( )tzV ,
