Electromagnetics - Plane wave reflections/transmissions

[satchmo05]

Homework Statement

A 1 GHz plane wave is free space is normally incident on a finite-width, lossless dielectric slab (relative permeability, u_R of 2, relative permittivity, ε_R of 18, slab of length l).

Determine the narrowest slab length for which 57.14% of power is reflected and the reflected signal has a phase angle of -160.89 degrees with respect to incident signal.

Homework Equations

- n_i = sqrt(u_R*ε_R)
- _wavelength = λ_{i =} c/(f*n_i)
- intrinsic impedance = ζ = ζ_O*n_i

The Attempt at a Solution

- Knowing the dielectric values for air and the slab, I know the index of refraction for air is simply 1. The dielectric exhibits an index of 6.

- I also know the intrinsic impedances of the two dielectrics as well - ζ_O = 377[Ω], ζ_SLAB = 125.67[Ω]

-Wavelength of free space is 0.30[m], whereas the wavelength within the slab is 0.05[m].

- I also know that the reflection coefficient is equal to 0.5714.

From these small steps I have completed, I hope I have gotten closer to the answer. Thank you for all help in advance! It is most appreciated.

 

[KataKoniK]

Thank you for your post. Your approach is on the right track, but there are a few things that need to be clarified and some additional steps that need to be taken to reach the solution.

Firstly, the index of refraction for a medium is not just determined by the relative permeability and permittivity, but also by the frequency of the incident wave. In this case, the frequency is given as 1 GHz, so we can use the equation ni = sqrt(uR*εR) to find the index of refraction for the slab. This gives us ni = sqrt(2*18) = 4.24.

Next, we need to calculate the wavelength within the slab using the formula λi = c/(f*ni), where c is the speed of light in a vacuum (3x10^8 m/s). This gives us a wavelength of 0.058 meters.

Now, we can calculate the intrinsic impedance of the slab using the equation ζ = ζO*ni, where ζO is the intrinsic impedance of free space (377Ω). This gives us ζSLAB = 1599.48Ω.

To determine the narrowest slab length for which 57.14% of power is reflected and the reflected signal has a phase angle of -160.89 degrees, we need to use the reflection coefficient formula: Γ = (ζSLAB - ζO)/(ζSLAB + ζO). Rearranging this equation, we get ζSLAB = (Γ + 1)/(1 - Γ)*ζO. Substituting the given values for Γ and ζO, we get ζSLAB = 1.4*ζO.

Finally, we can use the equation ζ = ζO*ni to solve for the slab length l, where ζ is the intrinsic impedance of the slab at a particular length l. Substituting the calculated value for ζSLAB, we get 1.4*ζO = ζO*ni*l. Rearranging, we get l = 1.4/ni = 0.33 meters.

Therefore, the narrowest slab length for which 57.14% of power is reflected and the reflected signal has a phase angle of -160.89 degrees is 0.33 meters. I hope this helps. Let me