Physics Library
 An open source physics library
Encyclopedia | Forums | Docs | Random | Template Test |  
Login
create new user
Username:
Password:
forget your password?
Main Menu
Sections

Talkback

Downloads

Information
[parent] telegraph equation (Topic)

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

$\displaystyle f_{xx}''-af_{tt}''-bf_t'-cf = 0,$ (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   $f(0,\,t) = g(t)$ $f(\infty,\,t) = 0$,  then the Laplace transform of the solution function  $f(x,\,t)$  is

$\displaystyle F(x,\,s) = G(s)e^{-x\sqrt{as^2+bs+c}}.$ (2)
In the special case  $b^2-4ac = 0$,  the solution is
$\displaystyle f(x,\,t) = e^{-\frac{bx}{2\sqrt{a}}}g(t-x\sqrt{a})H(t-x\sqrt{a}).$ (3)

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

$\displaystyle F_{xx}''(x,\,s)-a[s^2F(x,\,s)-sf(x,\,0)-f_t'(x,\,0)]-b[sF(x,\,s)-f(x,\,0)]-cF(x,\,s) = 0,$
which due to the initial conditions simplifies to

$\displaystyle F_{xx}''(x,\,s) = (\underbrace{as^2+bs+c}_{K^2})F(x,\,s).$
The solution of this ordinary differential equation is

$\displaystyle F(x,\,s) = C_1e^{Kx}+C_2e^{-Kx}.$
Using the latter boundary condition, we see that

$\displaystyle F(\infty,\,s) = \int_0^\infty e^{-st}f(\infty,\,t)\,dt \equiv 0,$
whence  $C_1 = 0$.  Thus the former boundary condition implies

$\displaystyle C_2 = F(0,\,s) = \mathcal{L}\{g(t)\} = G(s).$
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

$\displaystyle F(x,\,s) = G(s)a^{-x\sqrt{a}(s+\frac{b}{2a})} = e^{-\frac{bx}{2\sqrt{a}}}e^{-x\sqrt{a}s}G(s).$
According to the delay theorem, we have

$\displaystyle \mathcal{L}^{-1}\{e^{-ks}G(s)\} = g(t-k)H(t-k),$
wnere $H$ is Heaviside step function.  Thus we obtain for $\mathcal{L}^{-1}\{F(x,\,s)\}$ the expression of (3).



Anyone with an account can edit this entry. Please help improve it!

"telegraph equation" is owned by pahio.

View style:

See Also: wave equations

Also defines:  telegraph equation

This object's parent.

Cross-references: theorem, quadratic equation, ordinary differential equation, differential equation, function, Laplace transform, boundary, wave equation

This is version 2 of telegraph equation, born on 2008-05-13, modified 2008-05-24.
Object id is 280, canonical name is TelegraphEquation.
Accessed 1609 times total.

Classification:
Physics Classification02.30.Jr (Partial differential equations)
 41.20.-q (Applied classical electromagnetism)

Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:

No messages.

Testing some escape charachters for html category with a generator has an injective cogenerator" now escape ” with "