# 8.8 TM Modes in Rectangular Waveguide

In the preceding section, we developed expressions for E˜x, E˜y, H˜x, and H˜y in terms of E˜z and H˜z. Since H˜z=0 for the TM mode, our task reduces to obtaining a valid solution for E˜z. Our starting point is the homogeneous wave equation for E˜. For a lossless medium characterized by an unbounded-medium wavenumber k, the wave equation is given by Eq. (7.19) as

∇2 E˜+k 2 E˜=0. (8.92)

To satisfy Eq. (8.92), each of its x ˆ, y ˆ, and z ˆ components has to be satisfied independently. Its z ˆ component is given by

∂2 E˜z∂x 2+∂2 E˜z∂y 2+∂2 E˜z∂z 2+k 2 E˜z=0. (8.93)

By adopting the mathematical form given by Eq. (8.85), namely,

E˜z(x,y,z)=e˜z(x,y)e−j β z, (8.94)

Eq. (8.93) reduces to

∂2 e˜z∂x 2+∂2 e˜z∂y 2+k c 2 e˜z=0, (8.95)

where k c 2 is as defined by Eq. (8.90).

The form of the partial differential equation (separate, uncoupled derivatives with respect to x and y) allows us to assume a product solution of the form

e˜z(x,y)=X(x)Y(y). (8.96)

Substituting Eq. (8.96) into Eq. (8.95) and then dividing all terms by X(x)Y(y) leads to:

1 X d 2 X d x 2+1 Y d 2 Y d y 2+k c 2=0. (8.97)

To satisfy Eq. (8.97), each of the first two terms has to equal a constant. Hence, we define separation constants k x and k y such that

d 2 X d x 2+k x 2 X=0, (8.98a)
d 2 Y d y 2+k y 2 Y=0, (8.98b)

and

k c 2=k x 2+k y 2. (8.99)

Before proposing solutions for Eqs. (8.98a and b), we should consider the constraints that the solutions must meet. The electric field E˜z is parallel to all four walls of the waveguide. Since E=0 in the conducting walls, the boundary conditions require E˜z in the waveguide cavity to go to zero as x approaches 0 and a and as y approaches 0 and b (Fig. 8-22). To satisfy these boundary conditions, sinusoidal solutions are chosen for X(x) and Y(y) as:

e˜z=X(x)Y(y)=(A cos k x x+B sin k x x)(C cos k y y+D sin k y y). (8.100)

These forms for X(x) and Y(y) definitely satisfy the differential equations given by Eqs. (8.98a and b). The boundary conditions for e˜z are

e˜z=0 at x=0 and a, (8.101a)
e˜z=0 at y=0 and b. (8.101b)

Satisfying e˜z=0 at x=0 requires that we set A=0, and similarly, satisfying e˜z=0 at y=0 requires C=0. Satisfying e˜z=0 at x=a requires

k x=m π a,m=1,2,3,… (8.102a)

and similarly, satisfying e˜z=0 at y=b requires

k y=n π b,n=1,2,3,… (8.102b)

Consequently,

E˜z=e˜z e−j β z=E 0 sin m π x a sin n π y b e−j β z, (8.103)

where E 0=B D is the amplitude of the wave in the guide. Keeping in mind that H˜z=0 for the TM mode, the transverse components of E˜ and H˜ now can be obtained by applying Eq. (8.103) to Eqs. (8.89a–d),

E˜x=−j β k c 2 m π a E 0 cos m π x a sin n π y b e−j β z, (8.104a)
E˜y=−j β k c 2 n π b E 0 sin m π x a cos n π y b e−j β z, (8.104b)
H˜x=j ω ε k c 2 n π b E 0 sin m π x a cos n π y b e−j β z, (8.104c)
H˜y=−j ω ε k c 2 m π a E 0 cos m π x a sin n π y b e−j β z. (8.104d)

Each combination of the integers m and n represents a viable solution—or a mode—denoted T M m n. Associated with each m n mode are specific field distributions for the region inside the guide. Figure 8-23 depicts the E and H field lines for the T M 11 mode across two different cross sections of the guide.

# Figure 8-23 T M 11 electric and magnetic field lines across two cross-sectional planes.

(a) Cross-sectional planes

(b) Field lines for front view

(c) Field lines for side view

Figure 8 23 Full Alternative Text

According to Eq. (8.103), a rectangular waveguide with cross section (a×b) can support the propagation of waves with many different, but discrete, field configurations specified by the integers m and n. The only quantity in the fields’ expressions that we have yet to determine is the propagation constant β, which is contained in the exponential e−j β z. By combining Eqs. (8.90), (8.99), and (8.102), we obtain the following expression for β:

β=k 2−k c 2=ω 2 μ ε−m π a 2−n π b 2.(TE and TM) (8.105)

Even though the expression for β was derived for TM modes, it is equally applicable to TE modes.

The exponential e−j β z describes a wave traveling in the +z direction—provided that β is real—which corresponds to k>k c. If k<k c, β becomes imaginary: β=−j α with α real, in which case e−j β z=e−α z, yielding evanescent waves characterized by amplitudes that decay rapidly with z due to the attenuation function e−α z. Corresponding to each mode (m,n), there is a cutoff frequency f m n at which β=0. By setting β=0 in Eq. (8.105) and then solving for f, we have

where u p 0=1/μ ε is the phase velocity of a TEM wave in an unbounded medium with constitutive parameters ε and μ.

The mode with the lowest cutoff frequency is known as the dominant mode. The dominant mode is T M 11 among TM modes and T E 10 among TE modes. Whereas a value of zero for m or n is allowed for TE modes, it is not for TM modes [because if either m or n is zero, E˜z in Eq. (8.103) becomes zero and all other field components vanish as well].

By combining Eqs. (8.105) and (8.106), we can express β in terms of f m n,

The phase velocity of a TE or TM wave in a waveguide is

The transverse electric field consists of components E˜x and E˜y given by Eqs. (8.104a and b). For a wave traveling in the +z direction, the magnetic field associated with E˜x is H˜y [according to the right hand rule given by Eq. (7.39a)]. Similarly, the magnetic field associated with E˜y is −H˜x. The ratios, obtained by employing Eq. (8.104e), constitute the wave impedance in the guide,

Z TM=E˜x H˜y=−E˜y H˜x=β η k=η 1−f m n f 2, (8.109)

where η=μ/ε is the intrinsic impedance of the dielectric material filling the guide.

# Example 8-8: Mode Properties

A TM wave propagating in a dielectric-filled waveguide of unknown permittivity has a magnetic field with the y component given by

H y=6 cos(25 π x) sin(100 π y)× sin(1.5 π×10 10 t−109 π z)(mA/m).

If the guide dimensions are a=2 b=4 cm, determine (a) the mode numbers, (b) the relative permittivity of the material in the guide, and (c) the phase velocity, and (d) obtain an expression for E x.

Solution: (a) By comparison with the expression for H˜y given by Eq. (8.104d), we deduce that the argument of x is (m π/a) and the argument of y is (n π/b). Hence,

25 π=m π 4×10−2,100 π=n π 2×10−2,

which yield m=1 and n=2. Therefore, the mode is TM 12.

(b) The second sine function in the expression for H y represents sin(ω t−β z), which means that

ω=1.5 π×10 10 (rad/s) or f=7.5 GHz,β=109 π (rad/m).

By rewriting Eq. (8.105) to obtain an expression for ε r=ε/ε 0 in terms of the other quantities, we have

ε r=c 2 ω 2 β 2+m π a 2+n π b 2,

where c is the speed of light. Inserting the available values, we obtain

ε r=(3×10 8)2(1.5 π×10 10)2 ⋅(109 π)2+π 4×10−2 2+2 π 2×10−2 2=9.

(c)

u p=ω β=1.5 π×10 10 109 π=1.38×10 8 m/s,

which is slower than the speed of light. However, as explained later in Section 8-10, the phase velocity in a waveguide may exceed c, but the velocity with which energy is carried down the guide is the group velocity u g, which is never greater than c.

(d) From Eq. (8.109),

Z TM=η 1−(f 12/f)2.

Application of Eq. (8.106) yields f 12=5.15 GHz for the T M 12 mode. Using that in the expression for Z TM, in addition to f=7.5 GHz and

η=μ/ε=μ 0/ε 0 ε r=377 9=125.67 Ω,

gives

Z TM=91.3 Ω.

Hence,

E x=Z TM H y=91.3×6 cos(25 π x) sin(100 π y)× sin(1.5 π×10 10 t−109 π z)(m V/m)=0.55 cos(25 π x) sin(100 π y)× sin(1.5 π×10 10 t−109 π z)(V/m).

Concept Question 8-8: What are the primary limitations of coaxial cables at frequencies higher than 30 GHz?

Concept Question 8-9: Can a TE mode have a zero magnetic field along the direction of propagation?

Concept Question 8-10: What is the rationale for choosing a solution for e˜z that involves sine and cosine functions?

Concept Question 8-11: What is an evanescent wave?