Why does the attenuation constant not affect the wavelength?

Question

Reference: https://en.wikipedia.org/wiki/Propagation_constant#Phase_constant

Let's say I have a propagation constant ($\gamma$), where $jk_z = \gamma = \alpha +j \beta$. Why does the attenuation constant have no role in determining the wavelength, especially in a waveguide where TE/TM modes exists peacefully?

Background

1. According to the reference we can write wavelength as $\lambda = \frac{2 \pi}{Im\{ \gamma \}}$ (see reference)
2. Constant equation in a waveguide is $\gamma ^2 = h_{cut}^2 - k^2$

No Problem in TEM Mode because $h_{cut} = 0$, hence, $\gamma = jk$ and $im \{ \gamma \} = im\{jk\} = k$. Conclusion: $\lambda = \frac{2 \pi}{k}$

Problem in TE/TM Mode because $h_{cut} \ne 0$, hence, $\gamma = \sqrt{ h_{cut}^2 - k^2}$ and $im\{\gamma \} = Im\{ \sqrt{ h_{cut}^2 - k^2} \}$. Conclusion: $\lambda = \frac{2 \pi}{ Im\{ \sqrt{ h_{cut}^2 - k^2} \} }$ and sources I am finding define the phase constant ($\beta$) as $\beta = Im\{ \sqrt{ h_{cut}^2 - k^2} \}$

So why is the attenuation constant being ignored, even in wave guides?

Answer 1

I think your question should be asked in a different way, and the answer to that simpler question will also answer your question implicitly, namely: what is the frequency of a damped oscillation, such as $f(t)=e^{-\alpha t } sin(\omega_0 t)$?

First of all, this $f(t)$ is an exponentially unbounded signal and has no Fourier transform, so one cannot really analyze it over the full time axis but it can be analyzed for $\alpha >0 $ as a transient from $t=0$ to $t\to \infty$,

$$F(\omega) = \mathcal F [f] = \int_0^{\infty}e^{-\mathfrak j \omega t}e^{-\alpha t } sin(\omega_0 t)dt\\ =\frac{\omega_0}{(\alpha +\mathfrak j \omega)^2+\omega_0^2}$$ This $|F(\omega)|$ has frequency content everywhere but where is its peak? Obviously the peak is where the denominator's modulus $|(\alpha +\mathfrak j \omega)^2+\omega_0^2|$ is the smallest, that is we have to find the minimum of $(\alpha^2 + \omega_0^2 - \omega^2)^2+2\alpha^2\omega^2$. Simple differentiation shows that happens at the frequency $\omega = \hat \omega$ where $$\hat \omega = \sqrt{\omega_0^2 - \alpha^2}.$$ If $\omega_0 <\alpha$ then the oscillation is so heavily damped that there is no clear frequency at which we can meaningfully say that it oscillates. If $\omega_0 >> \alpha$ then $\hat \omega \approx \omega _0\left(1-\frac{\alpha}{2\omega_0}\right)$, and you see that indeed the peak is dependent on the attenuation, but its relative importance decreases with their ratio and in a waveguide with metallic walls this is the normal propagating situation.