91影视

Write the expression for the components of electric and magnetic fields along the z-axis in a rectangular wave.

$$\begin{gathered} \left[\frac{{鈭�}^2}{鈭�x^2}+\frac{{鈭�}^2}{鈭�y^2}+{\left(\frac{蝇}{c}\right)}^2=k^2\right]{\mathit{E}}_{\mathbf{z}}\mathbf{=}\mathbf{0} \\ \left[\frac{{鈭�}^2}{鈭�x^2}+\frac{{鈭�}^2}{鈭�y^2}+{\left(\frac{蝇}{c}\right)}^2-k^2\right]{\mathit{B}}_{\mathbf{z}}\mathbf{=}\mathbf{0} \end{gathered}$$

Here,$E_z$ is the longitudinal component of electric field and $B_z$is the longitudinal component of a magnetic field, $蝇$is the frequency of a wave, c is the speed of light, and k is the wavenumber.

For the TM wave, the value of the longitudinal component of the magnetic field is zero.

Write the boundary conditions at the wall.

$$\begin{gathered} E=0\text{,} \\ {B}^{鈯�}=0 \end{gathered}$$

Let,$E_z(x,y)=X\left(x\right)Y\left(y\right)$

Here, $X\left(x\right)=Asin鈦�\left(k_{x}x\right)+Bcos鈦�\left(K_{x}x\right)$

At walls $E_z=0$, then at $x=0$, the value of X and B will be,

$$\begin{gathered} X=0 \\ B=0 \end{gathered}$$

Hence, it is known that:

$$\begin{gathered} k_x=\frac{m蟺}{a};m=1,2,3鈥� \\ {k}_y=\frac{n蟺}{a};n=1,2,3鈥�. \end{gathered}$$

So, the longitudinal electric field will be,

$E_z=E_0\sin \left(\frac{m蟺x}{a}\right)\sin \left(\frac{n蟺y}{a}\right)$

Write the expression for wave number.

$k=\sqrt{{\left(\frac{蝇}{c}\right)}^2-{蟺}^2\left[{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2\right]}$

Hence, the cut off frequency will be,

${蝇}_{mn}=c蟺\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}$

Write the expression for the wave velocity,

$v=\frac{蝇}{k}$ 鈥︹�� (1)

Write the expression for the lowest cut-off frequency $\left({蝇}_{11}\right)$for mode$T{M}_{11}$.

${蝇}_{11}=c蟺\sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}$ 鈥︹�� (2)

Hence, the wavenumber in terms of the cut-off frequency will be,

$k=\frac{1}{c}\sqrt{{蝇}^2-{蝇}_{mn}^2}$

Substitute ${蝇}_{11}=c蟺\sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}\text{and}k=\frac{1}{c}\sqrt{{蝇}^2-{蝇}_{mn}^2}$in equation (1).

$v=\frac{c蟺\sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}}{\frac{1}{c}\sqrt{{蝇}^2-{蝇}_{mn}^2}}$

$V=\frac{C}{\sqrt{1-{\left(\frac{{蝇}_{mn}}{蝇}\right)}^2}}$

Write the expression for the group velocity .

$V_g=\frac{1}{\left(\frac{dk}{d蝇}\right)}$

Substitute $k=\frac{1}{c}\sqrt{{蝇}^2-{蝇}_{mn}^2}$in the above expression.

$$\begin{gathered} v_g=\frac{1}{\left(\frac{d}{d蝇}\left(\frac{1}{c}\sqrt{{蝇}^2-{蝇}_{mn}^2}\right)\right)} \\ {V}_g=c\sqrt{1-{\left(\frac{{蝇}_{mn}}{蝇}\right)}^2} \end{gathered}$$

Write the expression for the lowest cut-off frequency for Transverse electric mode (TE).

${蝇}_{10}=\frac{c蟺}{a}$ 鈥︹�� (3)

Take the ratio of equations (2) and (3).

$$\begin{gathered} \frac{{蝇}_{11}}{{蝇}_{10}}=\frac{c蟺\sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}}{\frac{c蟺}{a}} \\ \frac{{蝇}_{11}}{{蝇}_{10}}=\sqrt{1+{\left(\frac{a}{b}\right)}^2} \end{gathered}$$

Therefore, the longitudinal electric field is $E_z=E_0\sin \left(\frac{m蟺x}{a}\right)\sin \left(\frac{n蟺y}{a}\right),$

the cut-off frequency is ${蝇}_{mn}=c蟺\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}$, the wave velocity is

$V=\frac{C}{\sqrt{1-{\left(\frac{{蝇}_{mn}}{蝇}\right)}^2}}$, the group velocity is $V_g=c_1\sqrt{1-{\left(\frac{{蝇}_{mn}}{蝇}\right)}^2}$, and the ratio of the

lowest TM cutoff frequency to the lowest TE cutoff frequency for a given wave guide is$\frac{{蝇}_{11}}{{蝇}_{10}}=\sqrt{1+{\left(\frac{a}{b}\right)}^2}$ .