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Frequency domain analysis is extremely revealing in the case of the wave equation in particular. Considering the case of a string defined over the spatial domain
, a Fourier transform (in space) and a two-sided Laplace transform (in time) may be used. See Chapter 5 for more details. As a short-cut, one may analyze the behaviour of the test solution
![$\displaystyle u(x,y) = e^{st+j\beta x}$](img412.png) |
(6.6) |
where
is interpreted as a complex frequency variable,
, and
is a real wavenumber. It is easy to see that if
, such a test solution is, in fact a traveling wave of temporal frequency
, and wavelength
. Substituting such a solution into (6.2) gives
![$\displaystyle (s^2+\gamma^2\beta^2)u = 0$](img418.png) |
(6.7) |
Clearly the test solution
is non-zero everywhere, leading the characteristic equation
![$\displaystyle s^2+\gamma^2\beta^2 = 0$](img419.png) |
(6.8) |
which has roots
![$\displaystyle s = \pm j\gamma\beta\qquad \Longrightarrow\qquad \omega = \pm \gamma\beta$](img420.png) |
(6.9) |
This is known as the dispersion relation for the system (6.2); the frequency and wavenumber of a plane-wave solution are not independent. The fact that there are two solutions results from the fact that the wave equation is second order in time--the solutions correspond to left-going and right-going wave solutions.
Next: Phase and Group Velocity
Up: Definition
Previous: Initial Conditions
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Stefan Bilbao
2006-11-15