Derivation of Transmission Equations

As a current progresses down the line, a certain amount is lost through parallel admittance y consisting of the shunt conductance and the capacitance of each element. This current loss at each element is Ey, and thus

Figure 3. A transmission line consists of distributed resistance, inductance, capacitance, and conductance.

It is now necessary to differentiate these two equations to eliminate E and I. Thus, when equation 2 is differentiated with respect to l it becomes

similarly, equation 3 becomes

The undesired variables can now be eliminated by substituting in equations 4 and 5 the values from equations 3 and 2 respectively. Then,

and

These two equations are simple linear differential equations of the second order, and a knowledge of differential equations is required for their solution. Without presenting the details, the general solution for equations of this form is

and

The four constants of integration A₁, B₁, A₂, and B₂ must now be found. To do this, equations 8 and 9 are first differentiated with respect to l. They then become

and

It will be noted that these two expressions are equal to equations 2 and 3, respectively. It can therefore be written that at the sending end, where l = 0,

and

These values are preceded by a negative sign because the current / and the voltage E in any part of the line are less than the sending-end values I_s and E_s in the infinite line (having no reflection) under consideration. If l is also assumed equal to zero in equations 8 and 9, then E becomes the sending-end voltage E_s, and

Similarly,

When these values of A₁ and B₁ are substituted in equations 12 and 13_?

Also,

Furthermore, when these values of A₁, B₁, A₂, and B₂ are substituted in equations 8 and 9,

and

Although it would be possible to use these equations in this form_; they can be simplified by the use of hyperbolic functions. In textbooks on this subject it is shown that

where cosh x and sinh x are the hyperbolic cosine and sine, respectively. Accordingly, equations 18 and 19 become

and

These two equations give the voltage E or the current I (in either maximum or effective values) at any point l miles from the sending end of a line having a series impedance of z = R-jX (where X = 2πfL) ohms per loop mile, and a parallel admittance of y = G + jB (where B = 2πfC) mhos per loop mile (page 194).

If the values in equations 12 and 13 had been preceded by a positive instead of a negative sign and if E_r and I_r had been used, equations for voltage and current at any point in the line in terms of the receiving-end voltage and current and the distance l from the receiving end would have been obtained. These equations are

and

Equations 21, 22, 23, and 24 are general transmission equations applying, for steady-state conditions, to telephone lines and cables. Other methods of deriving these equations are also available (references 2 to 7 inclusive).