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# Transmission Line Equation

Module by: Bill Wilson. E-mail the author

Summary: This document continues derivation of Transmission Line equations. It also introduces characteristic impedance.

We need to solve the telegrapher's equations,

Vxt x =(LVxt t ) x V x t L t V x t
(1)
Ixt x =(CVxt t ) x I x t C t V x t
(2)
The way we will proceed to a solution, and the way you always proceed when confronted with a pair of equations such as these, is to take a spatial derivative of one equation, and then substitute the second equation in for the spatial derivative in the first and you end up with...well, let's try it and see.

Taking a derivative with respect to xx of Equation 1

2 Vxt x 2 2 =(L2Ixt t x ) x 2 2 V x t L t x I x t
(3)
Now we substitute in for Ixt x x I x t from Equation 2
2 Vxt x 2 2 =LC 2 Vxt t 2 2 x 2 2 V x t L C t 2 2 V x t
(4)
It should be very easy for you to derive
2 Ixt x 2 2 =LC 2 Ixt t 2 2 x 2 2 I x t L C t 2 2 I x t
(5)
Oh, I know you all love differential equations! Well, let's take a look at these and just think for a minute. For either Vxt V x t or Ixt I x t , we need to find a function that has some rather stringent requirements. First of all, the function must be of the form such that no matter whether we take its second derivative in space (xx) or in time (tt), it must end up differing in the way it behaves in xx or tt by no more than just a constant ( LC L C ).

In fact, we can be more specific than that. First Vxt V x t must have the same functional form for both its xx and tt variation. At most, the two derivatives must differ only by a constant. Let's try a "lucky" guess and let:

Vxt= V 0 fxvt V x t V 0 f x v t
(6)
where V 0 V 0 is the amplitude of the voltage, and ff is some function, of a form yet undetermined. Well
fxvt t =(vdfxvtd) t f x v t v f x v t
(7)
and
((vfxvt)) t =v2d 2 fxvtd t 2 t v f x v t v 2 t 2 f x v t
(8)
Note also, that
2 fxvt x 2 2 =d 2 dt 2 fxvt x 2 2 f x v t t 2 f x v t
(9)
Thus, we have
2 Vxt x 2 2 = V 0 d 2 fxvtd t 2 =1LC 2 Vxt t 2 2 =v2LC V 0 d 2 fxvtd t 2 x 2 2 V x t V 0 t 2 f x v t 1 L C t 2 2 V x t v 2 L C V 0 t 2 f x v t
(10)
which works as a solution as long as
v=1LC v 1 L C
(11)
So, what is this fxvt f x v t ? We don't know yet what its actual functional form will be, but suppose at some point in time, t 1 t 1 , the function looks like Figure 1. At point x 1 x 1 , the function takes on the value V 1 V 1 . Now, let's advance to time t 2 t 2 . We look at the function and we see Figure 2. If tt increases from t 1 t 1 to t 2 t 2 then xx will have to increase from x 1 x 1 to x 2 x 2 in order for the argument of ff to have the same value, V 1 V 1 . Thus we find
x 1 v t 1 = x 2 v t 2 x 1 v t 1 x 2 v t 2
(12)
which can be re-written as
x 2 x 1 t 2 t 1 =ΔxΔt v p =1LC x 2 x 1 t 2 t 1 Δ x Δ t v p 1 L C
(13)
where v p v p is the velocity with which the function is moving along the x-axis! (We use the subscript "p" to indicated that what we have here is what is called the phase velocity. We will encounter another velocity called the group velocity a little later in the course.)

If we had "guessed" an fx+vt f x v t for our function, it should be pretty easy to see that this would have given us a signal moving in the minus xx direction, instead of the plus xx direction. Thus we shall denote

V plus = V + fx1LCt V plus V + f x 1 L C t
(14)
the positive going voltage function and
V minus = V - fx+1LCt V minus V - f x 1 L C t
(15)
which is the negative going voltage function. Notice that since we are taking the second derivative of ff with respect to tt, we are free to choose either a 1LC 1 L C or a 1LC 1 L C in front of the time argument inside ff. Also note that these are our only choices for a solution. As we know from Differential Equations, a second order equation has, at most, two independent solutions.

Since Ixt I x t has the same differential equation describing its behavior, the solutions for II must also be of the exact same form. Thus we can let

I plus = I + fx1LCt I plus I + f x 1 L C t
(16)
represent the current function which goes in the positive xx direction, and
I minus = I - fx+1LCt I minus I - f x 1 L C t
(17)
represent the negative going current function.

Now, let's take Equation 16 and Equation 14 and substitute them into Equation 1:

V + LCfx1LCt=L I + fx1LCt V + L C f x 1 L C t L I + f x 1 L C t
(18)
This can be solved for V + V + in terms of I + I + .
V + =LC I + Z 0 I + V + L C I + Z 0 I +
(19)
where Z 0 =LC Z 0 L C is called the characteristic impedance of the transmission line. We will leave it as an exercise to the reader to ensure that indeed LC L C has units of Ohms. For practice, and understanding about just how these equations work, the reader should ensure him/her self that
V - =(LC I - )( Z 0 I - ) V - L C I - Z 0 I -
(20)
Note the "subtle" difference here, with a "-" sign in front of the RHS of the equation!

We've been through lots of equations recently, so it is probably worth our while to summarize what we know so far.

1. The telegrapher's equations allow two solutions for the voltage and current on a transmission line. One moves in the x x direction and the other moves in the x x direction.
2. Both signals move at a constant velocity v p v p given by Equation 21.
3. The voltage and current signals are related to one another by the characteristic impedance Z 0 Z 0 , with Equation 22
v p =1LC v p 1 L C
(21)
Z 0 =LC Z 0 L C
(22)
V + I + = Z 0 V + I + Z 0 V - I - = Z 0 V - I - Z 0

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