# Effective bandwidth of a electromagnetic waveguide

My homework problem says:

The effective bandwidth of a waveguide is the interval between the cutoff frequencies of the lower two modes (only the fundamental mode propagates). a) Show that the maximum bandwidth is for $b \leq a/2$. b) What's the value of $b$ for a maximum power transmitted?

I think it's wrong, because the maximum is for $b= a/2$.

More information: $a \times b$ is the size of the waveguide: I suppose $a \geq b$ and I take $a = \lambda b$, $b$ arbitrary but fixed, and $\lambda \geq 1$. So, from the theory $$f^c_{mn} =\dfrac{c}{2} \sqrt{\left(\frac{m}{a}\right)^2 +\left( \frac{n}{b}\right)^2} = \dfrac{c}{2b}\sqrt{\left(\frac{m}{\lambda}\right)^2+n^2}$$ where $c$ is the speed of light in the medium (suppose vacuum) and $m,n \in \mathbb{N}\cup \{0\}$ for TE$_{mn}$ and $m,n\in \mathbb{N}$ for TM$_{mn}$. For example, for TE$_{mn}$ the modes and cutoff frequencies are $$\begin{array}{|c|c|c|} \hline m & n & \frac{2b}{c}f^c_{mn} \\ \hline 1 & 0 & 1/\lambda \\ \hline 0 & 1 & 1 \\ \hline 1 & 1 & \sqrt{1/\lambda^2+1}\\ \hline 2 & 0 & 2/\lambda \\ \hline 0 & 2 & 2\\ \hline \end{array}$$

This is the graphic of the cutoff frequency (proportional to it) as a function of $\lambda$ (size of the waveguide). Dark blue $(m,n)=(1,0)$; green $(2,0)$; pink $(0,1)$; yellow $(1,1)$; light blue $(0,2)$. • What do a and b represent? – Crimson May 19 '17 at 12:00
Your calculations are completely correct. However, you found the maximum for $f_c(\lambda)$ while keeping $b$ constant, while the question wants to know the maximum for $f_c(b)$ while keeping $a$ constant.
If we plot $f_c(b)$ (a=1, c=2): And plot $f_c(\lambda)$: it is clear that the bandwidth is maximal for $b< a/2$ or $\lambda > 2$
• And the maximum power trasmitted occurs when $b=a/2$? $P_T= \dfrac{1}{4\eta} |E_0|^2 ab (1-\omega_c^2/\omega^2)^{1/2}$, then $P_T(1,0) \propto 1/\lambda,$ and the maximum of this function for $\lambda \geq 2$ is when $\lambda =2$, isn't it? (fixing the frequency $\omega =\omega_0$) – Clare Francis May 19 '17 at 15:39