Finding attenuation, phase constant, and velocity

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SUMMARY

The discussion focuses on calculating the attenuation constant (alpha), phase constant (beta), and phase velocity (v) for a material with given parameters: conductivity σ=ωε, relative permeability μr=1, relative permittivity εr=2.5, and wavelength in free space λ₀=30 cm. The calculated value for alpha is 9.53 nepers/m. The phase constant beta is derived using the equation β=sqrt((με/2)(1+sqrt(1+(σ/ωε)^2))) and is found to be approximately 22.976. The relationship between alpha and beta is clarified, emphasizing that alpha is the real part and beta is the imaginary part of the complex propagation constant Γ.

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  • Knowledge of material parameters: conductivity, permittivity, and permeability
  • Ability to manipulate equations involving wave equations and constants
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DODGEVIPER13
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Homework Statement


Find the attenuation constant alpha, phase constant β, and phase velocity v if the conductivity of the material is σ=ωε the material parameters are μr=1, εr =2.5, and the wavelength in free space is λ naught = 30cm


Homework Equations





The Attempt at a Solution


So using some big hairy equations I found alpha to be equal to 9.53 nepers/m. The part I am having trouble with is the beta part? I start out using this equation β=sqrt((με/2)(1+sqrt(1+(σ/ωε)^2))). I have tried a ton of different ways to arrive at this equation β=(ω/c)(sqrt(εr))(sqrt((1+sqrt(2))/(2))) what do I do?
 
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I can't reproduce your equations, sorry.

If we assume an E wave polarized in the y direction (propagation along x direction), the y component of E is

Ey = Emexp(jωt +/-Γx)
where Em = constant and
Γ is a (very!) complex number including ω, μ, ε, and σ. I leave it to you to obtain or derive this relationship. It will be in your textbook somewhere I'm sure.

Then, Γ = α + jβ so the answer to your problem is the imaginary part of Γ.
 
I went through it using the gamma equation I found in my book gamma= alpha+jbeta and then jomega(sqrt(mu(epsilon)))(1-j(sigma/((omega)(epsilon)))) once I had gamma I uses gamm=alpha+jbeta and found for beta using alpha I got 22.976
 
I used mu=4pix10^-7 and epsilon=8.85e-12
 
DODGEVIPER13 said:
I went through it using the gamma equation I found in my book gamma= alpha+jbeta and then jomega(sqrt(mu(epsilon)))(1-j(sigma/((omega)(epsilon)))) once I had gamma I uses gamm=alpha+jbeta and found for beta using alpha I got 22.976

You have the right equation for Gamma.

You don't need alpha to get beta. Alpha is the real part of Gamma and beta is the imaginary part.

I did not check your numbers. What did you wind up with for alpha and beta in terms of omega, epsion, mu, sigma?
 

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