Question

A propagating electric field is given by 100e -0.1z COS E (z, t) V/m T x 10' tTZ : (a) Determine the attenuation constant, the wave frequency, the wavelength, the propagation velocity, and the phase shift. (b) How far must the wave travel before its amplitude is reduced to 1.0 V/m?

Answer 1

If you compare the equation with a general travelling wave equation travelling in the direction z, then you can find the parameters as required in this question

$\vec{E}\left ( z,t \right )=\vec{E}_{0}\cos\left ( \omega t+kz-\delta \right )=Ae^{-bz}\cos\left ( \omega t+kz-\delta \right )$

where, $\inline \omega ,\ k,\ \delta ,\ A_{0}, E_{0}$ are angular frequency, wave vector, phase, initial amplitude of Electric field, electric field with damping term.

(a) The attenuation is proportional to the intensity of the EM wave or the square of electric field. Hence the attenuation term will be 2b according to the above equation. If you compare it with the given equation you will obtain the attenuation constant as 2 x 0.1 or 0.2 /m.

The wave frequency will be

$Frequency=\frac{\omega }{2\pi }=\frac{\pi \times 10^{7}}{2\pi }=5\times 10^{6}\ Hz= 5\ MHz$

$Wavelength=\frac{2\pi }{k }=\frac{2\pi}{\pi }=2\ m$

$Propagation\ velocity=frequency\times Wavelength=5\times 10^{6}\times 2=10^{7}\ m/s$

The phase shift is $\inline \delta =\frac{\pi }{3}$

(b) the electric field equation is

$\left |\vec{E} \right |=100e^{-0.1z}$

let us consider the amplitude reduced to 1 V/m when the propagation distance will be z. then from the above equation

$z=\frac{1}{0.1}\ln \left (100 \right )=46.05\ m$