Question

The amplitude and phase of a plane wave are both constant over any plane that is normal to the direction of propagation. In a nonplanar wave surfaces of constant amplitude and surfaces of constant phase are not the same. Show that the following nonplanar wave satisfies the phasor form of Maxwell's equations (current-free region): \begin{equation} \hat{\mathbf{E}} = E_0 e^{-\alpha y} e^{-j\beta z} \mathbf{a}_x \end{equation} \begin{equation} \hat{\mathbf{H}} = \frac{\beta}{\omega \mu} E_0 e^{-\alpha y} e^{-j\beta z} \mathbf{a}_y + j\frac{\alpha}{\omega \mu} E_0 e^{-\alpha y} e^{-j\beta z} \mathbf{a}_z \end{equation} where $\alpha > 0$ (otherwise, a planar wave) and $\alpha^2 - \beta^2 = -\omega^2 \mu \epsilon$.

          The amplitude and phase of a plane wave are both constant over any plane that is normal to the direction of
propagation. In a nonplanar wave surfaces of constant amplitude and surfaces of constant phase are not the
same. Show that the following nonplanar wave satisfies the phasor form of Maxwell's equations (current-free
region):
\begin{equation*}
\hat{\mathbf{E}} = E_0 e^{-\alpha y} e^{-j\beta z} \mathbf{a}_x
\end{equation*}
\begin{equation*}
\hat{\mathbf{H}} = \frac{\beta}{\omega \mu} E_0 e^{-\alpha y} e^{-j\beta z} \mathbf{a}_y + j\frac{\alpha}{\omega \mu} E_0 e^{-\alpha y} e^{-j\beta z} \mathbf{a}_z
\end{equation*}
where $\alpha > 0$ (otherwise, a planar wave) and $\alpha^2 - \beta^2 = -\omega^2 \mu \epsilon$.

The amplitude and phase of a plane wave are both constant over any plane that is normal to the direction of
propagation. In a nonplanar wave surfaces of constant amplitude and surfaces of constant phase are not the
same. Show that the following nonplanar wave satisfies the phasor form of Maxwell's equations (current-free
region):

𝐄̂ = E0 e^-α y e^-jβ z𝐚x

𝐇̂ = (β)/(ωμ) E0 e^-α y e^-jβ z𝐚y + j(α)/(ωμ) E0 e^-α y e^-jβ z𝐚z

where α > 0 (otherwise, a planar wave) and α^2 - β^2 = -ω^2 μϵ.

Added by Mark W.

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