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Work out the theory of TM modes for a rectangular wave guide. In particular, find the longitudinal electric field, the cutoff frequencies, and the wave and group velocities. Find the ratio of the lowest TM cutoff frequency to the lowest TE cutoff frequency, for a given wave guide. [Caution: What is the lowest TM mode?]

Short Answer

Expert verified

The longitudinal electric field is Ez=E0sinmπxasinnπya, the cut-off frequency is

ωmn=cπma2+nb2, the wave velocity isV=C1-ωmnω2, and the ratio of the lowest TM cutoff frequency to the lowest TE cutoff frequency for a given waveguide is ω11ω10=1+ab2.

Step by step solution

01

Expression for the components of an electric and magnetic field along the z-axis in a rectangular wave:

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

[2x2+2y2+ωc2=k2]Ez=0[2x2+2y2+ωc2-k2]Bz=0

Here,Ez is the longitudinal component of electric field and Bzis the longitudinal component of a magnetic field, ωis the frequency of a wave, c is the speed of light, and k is the wavenumber.

02

Determine the longitudinal electric field:

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

Write the boundary conditions at the wall.

E=0,B=0

Let,Ez(x,y)=X(x)Y(y)

Here, X(x)=Asin(kxx)+Bcos(Kxx)

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

X=0B=0

Hence, it is known that:

kx=mπa;m=1,2,3ky=nπa;n=1,2,3.

So, the longitudinal electric field will be,

Ez=E0sinmπxasinnπya

03

Determine the cut off frequency and wave velocity:

Write the expression for wave number.

k=ωc2-π2ma2+nb2

Hence, the cut off frequency will be,

ωmn=cπma2+nb2

Write the expression for the wave velocity,

v=ωk …… (1)

Write the expression for the lowest cut-off frequency (ω11)for modeTM11.

ω11=cπ1a2+1b2 …… (2)

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

k=1cω2-ωmn2

Substitute ω11=cπ1a2+1b2andk=1cω2-ωmn2in equation (1).

v=cπ1a2+1b21cω2-ωmn2

V=C1-ωmnω2

04

Determine the group velocity:

Write the expression for the group velocity .

Vg=1dkdω

Substitute k=1cω2-ωmn2in the above expression.

vg=1ddω1cω2-ωmn2Vg=c1-ωmnω2

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

ω10=cπa …… (3)

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

ω11ω10=cπ1a2+1b2cπaω11ω10=1+ab2

Therefore, the longitudinal electric field is Ez=E0sinmπxasinnπya,

the cut-off frequency is ωmn=cπma2+nb2, the wave velocity is

V=C1-ωmnω2, the group velocity is Vg=c11-ωmnω2, and the ratio of the

lowest TM cutoff frequency to the lowest TE cutoff frequency for a given wave guide isω11ω10=1+ab2 .

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Most popular questions from this chapter

(a) Calculate the (time-averaged) energy density of an electromagnetic plane wave in a conducting medium (Eq. 9.138). Show that the magnetic contribution always dominates.

(b) Show that the intensity is(k2μω)E02e-2xz

Consider the resonant cavity produced by closing off the two ends of a rectangular wave guide, at z=0 and at z=d, making a perfectly conducting empty box. Show that the resonant frequencies for both TE and TM modes are given by

ωlmn=(ld)2+(ma)2+(nb)2 (9.204)

For integers l, m, and n. Find the associated electric and magneticfields.

Consider a rectangular wave guide with dimensions 2.28cm×1.01cm. What TE modes will propagate in this waveguide if the driving frequency is 1.70×1010Hz? Suppose you wanted to excite only one TE mode; what range of frequencies could you use? What are the corresponding wavelengths (in open space)?

Show that the mode TE00 cannot occur in a rectangular wave guide. [Hint: In this case role="math" localid="1657512848808" ωc=k, so Eqs. 9.180 are indeterminate, and you must go back to Eq. 9.179. Show thatrole="math" localid="1657512928835" Bz is a constant, and hence—applying Faraday’s law in integral form to a cross section—thatrole="math" localid="1657513040288" Bz=0 , so this would be a TEM mode.]

Question:According to Snell's law, when light passes from an optically dense medium into a less dense one the propagation vector bends away from the normal (Fig. 9.28). In particular, if the light is incident at the critical angle

θc=sin-(n2n1)

Then , and the transmitted ray just grazes the surface. If exceeds , there is no refracted ray at all, only a reflected one (this is the phenomenon of total internal reflection, on which light pipes and fiber optics are based). But the fields are not zero in medium ; what we get is a so-called evanescent wave, which is rapidly attenuated and transports no energy into medium 2.26

Figure 9.28

A quick way to construct the evanescent wave is simply to quote the results of Sect. 9.3.3, with and

kT=kTsinθTx^+cosθTz^

the only change is that

sinθT=n1n2sinθI

is now greater than, and

cosθT=1-sin2θT

is imaginary. (Obviously, can no longer be interpreted as an angle!)

(a) Show that

ET(r,t)=E0Te-kzeI(kx-ωt)

Where

kωc(n1sinθ1)2-n22

This is a wave propagating in the direction (parallel to the interface!), and attenuated in the direction.

(b) Noting that (Eq. 9.108) is now imaginary, use Eq. 9.109 to calculate theirreflection coefficient for polarization parallel to the plane of incidence. [Notice that you get reflection, which is better than at a conducting surface (see, for example, Prob. 9.22).]

(c) Do the same for polarization perpendicular to the plane of incidence (use the results of Prob. 9.17).

(d) In the case of polarization perpendicular to the plane of incidence, show that the (real) evanescent fields are

Er,t=E0e-kzcoskx-ωty^Br,t=E0ωe-kzksinkx-ωtx^+kcoskx-ωtz^

(e) Check that the fields in (d) satisfy all of Maxwell's equations (Eq. 9.67).

(f) For the fields in (d), construct the Poynting vector, and show that, on average, no energy is transmitted in the z direction.

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