Download Examples: Telegrapher`s Equation

Examples: Telegrapher’s Equation
Rev 28 Jan 2013
“Candygram for Mr. Mongo.” – Blazing Saddles
This equation describes an electrical signal traveling
along a transmission cable. First derived in the
horse and buggy days of the telegraph (hence the
name), it is still useful for describing long distance
power lines and cable TV systems.
Think telegraphy is extinct? Why does RS-232 still include such teletype codes as
ReadyToSend and RingIndicator? Why does ASCII still include characters for
EndOfTransmission and BEL? Why is “line feed” not the same as “carriage return”
and thus silly little programs like unix2dos are still hanging around? Despite continued
use of the radio distress call “Mayday Mayday Mayday” by boaters and pilots, why do
people still know that “SOS” means big trouble?
A little electromagnetism for background:
Electrical current (the flow of electrical charge) flows through a wire whenever there is a
difference in electrical potential energy from one end of the wire to the other. Such a
potential energy difference per unit of electric charge is known as voltage (V); current (I)
flows in accordance with Ohm’s Law, V = IR. In this expression, R is the wire’s
resistance (the degree to which the material of the wire decreases electrical energy). We
want our wires to be as low resistance as possible.
Current also sets up a magnetic field (see photo at right) in
proportion to the wire’s inductance (L). Changing current
generates a changing magnetic field – and that generates a
voltage (that process is how electricity is generated in the first
place). This “induced voltage” opposes the voltage that is
already present across the wire; as a result, the wire has a
measurable inductance per unit length.
Whenever two wires are close together (or a wire is close to a
slightly conductive material (like the ground), there is a small
electric field between the wires. The result is a measurable
capacitance per unit length.
Ideal transmission lines – no leakage to ground
It is assumed that we are talking about time varying signals, such as those that might
carry cable TV. We want to find functions for voltage and current in space and time:
V(x,t) and I(x,t).
We model the transmission line as
if it is a long string of inductors and
capacitors connected as shown.
L = inductance per unit length
C = capacitance per unit length
Ohm’s Law analysis of this circuit gives us two simultaneous equations:
V
I
 L
x
t
I
V
 C
x
t
We combine these by taking the appropriate derivatives:
 2V
2 I
 L 2
t x
t
2
 I
 2V
 C
x 2
xt
and because mixed partials are equal, we obtain the second order PDE
which is a wave equation with velocity v 
2 I
1 2 I

,
t 2 LC x 2
1
. Of course, all wave functions must
LC
obey their BCs.
Use a similar treatment to obtain the corresponding second order PDE for voltage.
Not so ideal – wires have resistance and cables leak
The above model did not include any internal resistance of the wires comprising the
transmission lines, nor did it include the leakage of current out of the wire to the outside
environment (collectively called ‘ground’). Leakage occurs due to a combination of
factors: conductance G through the wire’s insulation and capacitive coupling C between
wire and ground. We want the conductance (which is measured in units that are the
2
inverse of resistance) between the wire and the ground as low as possible – and we want
this coupling C to be fairly small as well.
The figure at right is a model of how these
factors fit together to explain the changing
voltage in a long wire. In the figure, all
these quantities are divided by length:
R = resistance (ohms) per unit length,
L = inductance (henries) per unit length,
C = capacitance (farads) per unit length,
G = conductance per unit length (mhos/m).
Ohm’s Law analysis of this circuit gives us two simultaneous equations:
V
I
 GV 
0
t
x
,
V
I
 IR  L  0
x
t
C
where V(x,t) is the solution function for the voltage and I(x,t) is the current in the
wire.
Show how these two equations combine into one second-order PDE for voltage:
 2V
 2V
V

LC
 ( RC  GL)
 GRV , or by combining constants
2
2
x
t
t
c2
 2V  2V
V
G
R
1
 2  ( a  b)
 abV , with a  , b  , c 2 
2
x
t
t
C
L
LC
We see the wave equation plus a term in Vt and a term in V. We therefore suspect that
there is a traveling wave solution, V(x,t) = ½ (f(x - ct) + f(x + ct)). The function f must
accommodate the loss of voltage due to both the cable’s internal resistance R and the
leakage due to G and C.
A perfectly insulated cable will have G = 0, leaving just the loss term R/L Vt. Just as in
the wave equation, at the end of such a transmission line, a reflection of the out-going
signal may be generated (depending on the BCs). For an animation, see
http://www.math.ubc.ca/~feldman/demos/demo8.html
Typical values for 1 km of coaxial cable: C = 70 nF L = 360 mH G = 2.9x10-8 mho
3
Do the following steps:
Set up separation of variables to obtain a solution for V(x,t). Use V(x,t) = u(x) w(t);
choose a negative separability constant, - 2.
You should obtain u ( x)  A cos

x  B sin

x , where the ratio
c
c
the wavenumber (2 divided by the wavelength).

c
 k and k represents
The time equation should become w ''(t )  (a  b) w '(t )  (ab   2 ) w(t )  0 , which can be
solved by forming the characteristic polynomial of the 2nd order ODE. Verify that if
(a  b)  (a  b) 2  4(ab   2 )
w(t )  e , we have r 
.
2
rt
Consider the case where the radical is zero: r  
V ( x, t )  ( A cos kx  B sin kx)e

a b
t
2
ab
and therefore
2
. What condition does this place on the value of ?
Using initial conditions V(0,t) = V(L,t) = 0, you will find A = 0 and eigenfunctions in
2
your solution. Send a pulse V ( x, t )  f ( x)  e  x down the line.
Another way – a system of PDEs
Our two first order equations can be represented in matrix form
 C

V
   G
I   
   0

  V  
0    0
t
 
L
1 V
    
R   x   R
Return to the equation c 2
1   I 
G   t 
 
I
0  
  x 

 2V  2V
V
G
R
1
 2  ( a  b)
 abV , with a  , b  , c 2 
.
2
x
t
t
C
L
LC
Let’s write it differently:
2
 2V
V
2  V

c
 ( a  b)
 abV and make some clever substitutions.
2
2
t
x
t
Let V1  V , V2 
V
V
, or as a system of three first-order equations:
, V3 
x
t
4
V1
 V2
x
V1
 V3
,
t
V3
V
 c 2 2  (a  b)V3  abV1
t
x
equivalent to our second-order telegraph equation.
 V1 
 V1 
 x 
 t 
V1 




V2 
V2 




and Vx 
We now let the vectors V  V2  , Vt 
 x 
 t 
V3 




 V3 
 V3 
 t 
 x 
and seek to rewrite the system in the form of a linear matrix equation
AVt + BVx + CV = 0, where A, B and C are 3x3 matrices.
0 0 0 
1 0
1 0 0  V   0 0

 t 
 0 0 1 
 0 c 2
0
0 
 0 1
0  Vx   0 0
1  V  0
 ab 0 a  b 
0 

kt
Note that if we let v( x, t )  e 2 w( x, t ) , we obtain another PDE:
2
1
2w
2  w
a

 (b  k 2 ) w , known as the Klein-Gordon Equation. That’s often seen in
2
2
4
t
x
1  2  2
mc
relativistic quantum mechanics in the guise of: 2 2  2  ( ) 2  0 , which is of
x
c t

 it  ikx
.
course solved by   e
5