Waveguides:

Waveguides are utilized for same objectives as transmission lines i.e. to transmit electromagnetic energy. Though, waveguides are utilized at microwave frequencies and they are not as lossy as transmission lines that are utilized at lower frequencies. There are two kinds of waveguides, the metal tubes of any cross section and dielectric rods. Wave travels on inside of the first type and on outside of the other.

Rectangular waveguide:

Rectangular waveguide is hollow infinite rectangular pipe of highly conducting material. Rectangular waveguide is composed of two pairs of planes. For sake of simplicity a pair of planes will be considered first.

Transmission of waves in a pair of parallel conducting:

Planes:

Let separation between pair of planes be distance, d, assuming planes are perfectly conducting and of infinite length, features of electric and magnetic fields which can travel down between planes have to be determined. It is essential for fields to obey Maxwell's equations in free space in between plates. The following are boundary conditions on planes.

(i) Parallel or tangential component of electric field and perpendicular or normal component of magnetic field should be zero at every point on planes. This is because there is no electromagnetic field inside perfect conductors.

(ii) Perpendicular or normal component of electric field and parallel or tangential component of magnetic field need not equal zero. This is due to there can be charges on surfaces of plane, and surface currents in them.

Obeying Maxwell's equations implies satisfaction of electric field wave equations and magnetic field wave equations i.e.

∇²E = ε₀μ₀d²E/dt²

∇²B = ε₀μ₀d²B/dt²

Linearly polarized plane wave whose electric and magnetic fields, respectively are

E = exE₀exp[jw(t-kz/w)]

B = -eyE₀/cexp[jw(t - kz/w)]

It is not only solution to equations but also satisfies boundary conditions (i) & (ii) stated above.

Linearly polarized wave whose electric and magnetic fields are expressed above is transverse electric and magnetic (TEM) wave. This is due to both fields are not only perpendicular to each other but are also transverse to direction of propagation such that three are mutually perpendicular i.e. orthogonal. Although TEM waves can travel in pair of planes just as they could in parallel wire and coaxial cable transmission lines, they can't, though, be propagated in wave guide which is a hollow conductor. This is because perfectly conducting walls of waveguide will short circuit the electric field, electric potential not being able to exist across perfect conductors.

Superposition of incident and reflected waves whose electric fields and magnetic fields are represented by E₁ and B₁, and E₂ and B₂ respectively in Figure is also solution to equations.

If E₁ oscillates along y - direction, its expression becomes

E₁ = u_yE₀exp[jw(t - k₁r/w)]

=uyE₀exp[jw(t - r/w)]

Where v₁ = -u_xsinΘ/V + u_zsinΘ/V

Such that 1/V₁.r = -xsinΘ/v + zcosΘ/v

And equation becomes,

E₁ = u_yE₀exp[jw(t - zcosΘ/v + xsinΘ/v)]

On reflection wave has velocity

1/V₂ = u_xsinΘ/v + u_zcosΘ/v

Electric field, E₂, of reflected wave is given by equation

E₂ =-u_yE₀exp[jw(wt - zcosΘ/v - xsinΘ/v)]

Superposing E₁ and E₂ to obtain resultant field E provides

E = u_y2jsin((wx/v) sinΘ)exp[jw(t - zcosΘ/v)]

Planes are placed at x = 0 and x = a as by our boundary conditions E = 0 at walls i.e. first at x = 0; substituting x = 0 in equation, E automatically becomes zero showing that boundary condition is satisfied. If Θ is chosen such that sin(wx/v sinΘ) = 0 then E = 0 at x = a. This satisfies boundary conditions on second plane. Condition that

sin((wx/v)sinΘ) = 0

Implies that wx/vsinΘ = nπ or sinΘ = nπv/wx

Putting sinΘ≤1 and x = a we have, nvπ/wa≤1

Suppose vπ/wa ≥ 1 when n =1, λ/2a≥1 or f≤c/2a

f is cut- off frequency when n = 1. For n>1, f_n = nc/2a. For instance when pair of planes is separated by 2cm, from equation i.e. sinΘ = nvπ/wx

CosΘ = [1 - (nvπ/wx)²]1/2

Substituting sin θ and cos θ into equation provides

E = u_y2jE₀sin(nπx/a)exp[jw(t-z/Vg)]

Where 1/Vg = (1/V² - n²π²/w²b²)^1/2

Using fact that 1/V = k/w

1/Vg = kg/w and kg = (k² - n²π²/a)^1/2

Electric field components for TE waves are

E_y = 2jE₀sin(nπx/a)exp[jw(t-z/vg)]

E_x = 0 and E_z = 0

By differential form of Faraday's law i.e. ∇xE = dB/dt we can get components of magnetic fields by using

We have

ux(-dE_y/dz) + u_z(dE_y/dx) = -dB/dt or

dB/dt = jw/v_g.2jE₀sin(nπx/a)exp[jw(t-z/vg)]u_x - (nπ/a).2jE₀cos(nπx/a)exp[jw(t-z/vg)]u_z

Integrating provides

B_x = (2j/vg)E₀sin(nπx/a)exp[jw(t - z/v_g)]

B_y = 0

B_z = -(2nπ/wa)E₀cos(nπx/a)exp[jw(t-z/v_g)]

= - 2(1/v² - 1/v²_g)1/2E₀cos(nπx/a)exp[jw(t-z/vg)] where 1/v² - 1/v²g = n²π²/w²a²

We can also add two waves having their magnetic field in y-direction only such that

 B₁ = e_yB₀exp[jw(t - z/vg)] and B₂ = -e_yB₀exp[jw(t-z/vg)]

Superposing two fields gives the set of equations in which resultant B has only y-component, B_y; B_x and B_z being equal to zero. Electric field, E will then have component in z-direction. Such a wave is known as transverse magnetic (TM) wave.

Waveguide:

Two pairs of parallel conducting planes form a rectangular waveguide. There two types of waveguides- the metal tubes of any cross section and the dielectric rods. Wave travels on the inside of the first type and on the outside of the other. The second type is outside the scope of this course. Also, we limit our discussion to rectangular waveguides.

Adding another pair of perfectly conducting parallel planes to the pair already treated such that these new planes are located at y = 0 and y = b, we obtain the waveguide shown in Figure 2. The waveguide is a hollow rectangular metal pipe. Only the TE and TM waves can propagate in the waveguide. By the presence of the new pair of parallel plates separated along y - axis, the electric field is expected to also vary in the y - direction in addition to its variation in the x - direction such that,

E ∝ Kf(y)sin(nπx/a)exp[jw(t-z/v)]

Or E = Kf(y)sin(nπx/a)exp[t - z/v]

For wave to be able to propagate in waveguide, equation should satisfy equation electric field wave equation and Maxwell's equations. Taking 1st Maxwell's equation i.e. ∇.E = 0 or

dE_x/dx + dE_y/dy + dE_z/dz = 0

E_z = 0 as wave can't have component in direction of propagation. Consequently dE_z/dz = 0 therefore equation becomes

dE_x/dx + dE_y/dy = 0 or dE_x/dx = -dE_y/dy

dE_x/dx = -K(df(y)/dy)sin(mπx/a)exp[jw(t-z/v)]

By integration Ex becomes

E_x = (ca/mπ)(df(y)/dy)cos(mπx/a)exp[jw(t-z/v)]

We can find out f (y) by making E_x satisfy boundary conditions that it be zero at walls y = 0 and y = b and satisfy wave equation and ensuring that wave equation is satisfied by E_x.

For E_x = 0 at y = 0, (a/mπ)(df(y)/dy)=sinky = sinnπ or df(y)/dy = (mπ/a)sinky = (mπ/a)sinmπ

Putting y = b, nπ = kb or k = nπ/b, and, df(y)/dy = (mπ/a)sin(nπy/b)

Integrating df(y)/dy we have

f(y) = -(mπ/a)cos(nπy/b)

For pair of parallel plates we had,

k_g² = k² - (n²π²/b² + (m²π²/a²))

or k² - k_g² = (n²π²/b² + (m²π²/a²)) which is waveguide equation.

Guide wave number, k_g, from which cut-off wave length is determined for particular mode is obtained from dimensions (width and length) of wave guide and free space wave number. We can write equation as

k²_c = k² - k_g² where

k_c² = m²π²/a² + n²π²/b² or 4π²/λ_c² = m²π²/a² + n²π²/b²

Therefore equation can be written as:

4π²/λ_c² = 4π²/λ² - 4π²/λ_g² or 1/λ_c² = 1/λ² - 1/λ_g²

Substituting for f(y) in equations  we obtained components of electric field as shown below.

E_x = (knπ/b)sin(nπy/b)cos(mπx/a)exp[jw(t-z/v)]

E_x = (kmπ/b)cos(nπy/b)sin(mπx/a)exp[jw(t-z/v)]

E_z = 0

That E_z = 0 is really the reason the wave is referred to as TE as it is transverse to direction of propagation i.e. z-direction having no component in that direction. Expressions for magnetic fields are obtained from differential form of Faraday's law i.e. ∇xE = -dB/dt as follows:

B_x = -(k_gKmπ/wa)cos(nπy/b)sin(mπx/a)exp[jw(t-z/v)]

B_x = -(k_gKnπ/wb)sin(nπy/b)cos(mπx/a)exp[jw(t-z/v)]

B_z = jw(1/v² - 1/v²_g)cos(nπy/b)cos(mπx/a)exp[jw(t-z/v)]

Direction of propagation is z-direction. Therefore energy of wave is in z-direction. Rate of energy flow across unit area is referred to as poynting vector, N. it can be illustrated that N is vector product of electric field and magnetic field of wave i.e.

N = ExB

Expanding equation (4), we have,

Thus,

N_x = E_yB_z - E_zB_y = E_yB_z since E_z = 0.

N_y = E_zB_x - E_xB_z = -E_xB_z since E_z = 0, and

N_z = E_xB_y - E_yB_x

As direction of energy is z-direction as mentioned above, N_x = N_y = 0 i.e. E_yB_z = E_xB_z = 0 implying that E_y and B_z are out of phase and E_x and B_z are out of phase. Energy produced by the products only flows in and out of the planes. Cut-off frequency of TE waves is obtained from equation and f_c = c/λ_c to give

f_c = c/2√(m/a)² + (n/b)²

Only waves of frequency greater than f_c can be propagated in guide of given a×b. Therefore wave guide behaves like high-pass filter. In other words only modes whose cut-off frequencies are lower than radiation frequency can be propagated.

Transverse Magnetic (TM) waves in which there is no magnetic field in direction of propagation are also propagated in guide. They should, though, satisfy boundary conditions of guide and wave equation. For these waves electric field has component in direction of propagation. TM modes would have cut-off frequencies at same values as TE waves given by equation.

We should though note that unlike TE there cannot be TM_on or TM_mo. This is due to field lines of magnetic field in waveguide is two-dimensional there being no magnetic charges or mono poles and as such magnetic field has to differ with both coordinates other than in direction of propagation. Therefore lowest TM modes which can exist is TM₁₁ mode. Electric field, on other side is one dimensional in guide. As such its lines of force go from one plane to another. Therefore lowest TE mode is either TE₀₁ or TE₁₀ depending on which of width, a, or height, b, is longer. For example cut-off frequencies for TE₀₁ i.e. f₀₁ = c/2√(1/b)² = c/2b

While that of TE₀₁ i.e. f₀₁ = c/2√(1/a)² = c/2a

Lowest TE mode is fundamental and significant mode in rectangular waveguide. Take for instance waveguide of dimensions a = 3 cm and b= 1.5cm in which wave whose wavelength in free space is 5cm corresponding to the frequency of 6 × 10⁹Hz i.e. 6GHz, cut-off frequencies of given modes are determined and compared. It is only f₁₀ that is lower than frequency of propagated wave. As such it is only f₁₀ that is permitted in guide. Note that f₁₀ is fundamental and most significant mode. Velocity with which wave crests and/or troughs travel through medium is called as phase velocity, v_p. From fact that velocity is quotient of angular frequency and wave number i.e. v = w/k, phase velocity in a wave guide is v_p = w/k_g (k_g being guide wave number). k_g can be determined from combination of equations from which we can get v_p. In example above v_p = 5.5×10⁸ms^-1 and v_p > c (velocity of light)

Group velocity, v_g is velocity of group of waves whereas phase velocity is velocity of individual waves within group. Waves within group are observed to travel faster than group itself, therefore v_p > v_g. It is, though, group velocity which is usually estimated as wave velocities are estimated by arrival of disturbances and it is velocity by which energy of wave is propagated. v_g < c and as such there is no contradiction of Einstein's. The product of phase velocity and group velocity equals c². Therefore v_g = c/v_p = 1.64ms^-1.

The poynting vector of guide in direction of propagation i.e. N_z for TE₁₀ mode = E_yH_x = (k_g/wμ₀)E²_ysin²(πx/a) by combining expressions of E_y and B_x above and noting that B = H_x/μ₀. For TE₁₀ mode, m = 1 while n = 0. Substituting for m and n in expressions of E_x and B_y above shows E_x = B_y = 0 such that N_z = E_yB_x implying that first term of left hand side of equation is zero. Average power of operation of the guide for TE₁₀ mode is therefore provided by (1/2)(kg/μ₀w)E₁₀²(ab/2), k_g, a b and w are obtainable from wavelength and guide dimension known; electric field, E, would be given and average power can then be obtained. Integrating expression over area of the guide, we get

(1/2)(k_g/wμ₀)E_oy² sin²(nπ/a)

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