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Both the electric voltage and the current in a double conductor satisfy the telegraph equation
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(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 $f(0,\,t) = g(t)$ , $f(\infty,\,t) = 0$ , then the Laplace transform of the solution function $f(x,\,t)$ is
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(2) |
In the special case $b^2-4ac = 0$ , the solution is
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(3) |
Justification of (2). Transforming the partial differential equation (1) ($x$ may be regarded as a parametre) gives $$F_{xx}''(x,\,s)-a[s^2F(x,\,s)-sf(x,\,0)-f_t'(x,\,0)]-b[sF(x,\,s)-f(x,\,0)]-cF(x,\,s) = 0,$$ which due to the initial conditions simplifies to $$F_{xx}''(x,\,s) = (\underbrace{as^2+bs+c}_{K^2})F(x,\,s).$$ The solution of this ordinary differential equation is $$F(x,\,s) = C_1e^{Kx}+C_2e^{-Kx}.$$ Using the latter boundary condition, we see that $$F(\infty,\,s) = \int_0^\infty e^{-st}f(\infty,\,t)\,dt \equiv 0,$$ whence $C_1 = 0$ .
Thus the former boundary condition implies $$C_2 = F(0,\,s) = \mathcal{L}\{g(t)\} = G(s).$$ So we obtain the equation (2).
Justification of (3). When the discriminant of the quadratic equation $as^2\!+\!bs\!+\!c = 0$ vanishes, the roots coincide to $s = -\frac{b}{2a}$ , and $as^2\!+\!bs\!+\!c = a(s+\frac{b}{2a})^2$ . Therefore (2) reads $$F(x,\,s) = G(s)a^{-x\sqrt{a}(s+\frac{b}{2a})} = e^{-\frac{bx}{2\sqrt{a}}}e^{-x\sqrt{a}s}G(s).$$ According to the delay theorem, we have $$\mathcal{L}^{-1}\{e^{-ks}G(s)\} = g(t-k)H(t-k).$$ Thus we obtain for $\mathcal{L}^{-1}\{F(x,\,s)\}$ the expression of (3).
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