Both the electric voltage and the current in a double conductor satisfy the telegraph equation

fxx′′− aftt′′− bft′− cf = 0,

(1)

where x is distance, t is time and a, b, c are non-negative constants. The equation is a generalised form of the wave equation.

If the initial conditions are f(x, 0) = f_t′(x, 0) = 0 and the boundary conditions f(0, t) = g(t), f(∞, t) = 0, then the Laplace transform of the solution function f(x, t) is

F(x, s) = G(s)e−x.

(2)

In the special case b² − 4ac = 0, the solution is

f(x, t) = e− g(t − x)H(t − x).

(3)

Justification of (2). Transforming the differential equation (1) gives

which due to the initial conditions simplifies to

The solution of this Ordinary Differential Equation is

Using the latter boundary condition, we see that

whence C₁ = 0. Thus the former boundary condition implies

So we obtain the equation (2).

Justification of (3). When the discriminant of the quadratic equation as²+bs+c = 0 vanishes, the roots coincide to s = −, and as²+bs+c = a(s + )². Therefore (2) reads

According to the delay theorem, we have

wnere H is Heaviside step function. Thus we obtain for ℒ⁻¹{F(x, s)} the expression of (3).