  Question:
Published on: 25 September, 2022

Find out the expressions for the different components of electric and magnetic fields inside a rectangular waveguide for TE mode of propagation. Hence find out the expression for the guide wavelength and propagation constant.

In order to study transmission lines electromagnetically, one should know the laws governing time varying electromagnetic fields. The interaction between charges and their mutual energy is expressed in terms of four vectors E and B, D and H. The first pair defines the factor on a charge density ρ moving with vector v given by;

F=ρ[E+(v+B)]

The vectors E and B satisfy the field equation and we have the Faraday’s Law

$$\nabla\times E=-\frac{\partial B}{\partial t}$$ … (1)

and $$\nabla.B=0$$ … (2)

The second pair of field vectors are determined by the charges and current present and satisfy the equation, we have the Ampere’s Law

$$\nabla\times H=J+\frac{\partial D}{\partial t}$$ … (3)

And Gauss Law

$$\nabla.D=\ \rho$$ … (4)

Here J is the current density

The interrelations between the four vectors E, B, D, and H depend on the medium in which the field exists

B=μ H … (5)

D=∈ E … (6)

and J=σ E … (7)

Here μ is the magnetic permeability and ∈ is electric permittivity of the medium.

For the free space equation (5) and (6) are modified as B=μ0H, D=∈0E

Where μ0=1.257×10-6 henry/meter and

Wave equation:

For a perfect dielectric media containing no charges and no conduction current, then we can write ρ=0 and J=0, thus the equation (1) and (3) can be written as

$$\nabla\times E=-\frac{\partial(\mu H)}{\partial t}=-\mu\frac{\partial H}{\partial t}$$ … (8)

and $$\nabla\times H=\frac{\partial(\in E)}{\partial t}=\in\frac{\partial E}{\partial t}$$ … (9)

now $$\nabla\times\nabla\times E=\nabla\times\left(-\mu\frac{\partial H}{\partial t}\right)=-\mu.\ \frac{\partial}{\partial t}\left(\nabla\times H\right)$$

$$=-\mu.\ \in\frac{\partial^2E}{\partial t^2}$$

Using vector identity,

$$\nabla\times\nabla\times E=\nabla.\nabla.E-\nabla^2E$$

$$\nabla.\nabla.E-\nabla^2E=\ =-\mu.\ \in\frac{\partial^2E}{\partial t^2}$$ … (10)

We know that ∇.D= ρ and D=∈ E for perfect dielectric media ρ=0

Hence, ∇.E=0, therefore ∇.∇.E=0

The equation (10) becomes

$$\nabla^2E=\ =\mu.\ \in\frac{\partial^2E}{\partial t^2}$$ … (11)

The equation (11) is the wave equation of electric vector E

Similarly for the magnetic vector H, we can derive the same type expression

For a perfect dielectric media containing no charges and no conduction current, then we can write ρ=0 and J=0, thus the equation (1) and (3) can be written as

$$\nabla\times E=-\frac{\partial(\mu H)}{\partial t}=-\mu\frac{\partial H}{\partial t}$$ … (12)

and $$\nabla\times H=\frac{\partial(\in E)}{\partial t}=\in\frac{\partial E}{\partial t}$$ … (13)

now $$\nabla\times\nabla\times H=\nabla\times\left(\in\frac{\partial E}{\partial t}\right)=\ \in.\ \frac{\partial}{\partial t}\left(\nabla\times E\right)$$

$$\nabla\times\nabla\times H=\ -\mu\in.\ \frac{\partial}{\partial t}\left(\frac{\partial H}{\partial t}\right)$$

$$=-\mu.\ \in\frac{\partial^2H}{\partial t^2}$$

Using vector identity,

$$\nabla\times\nabla\times H=\nabla.\nabla.H-\nabla^2H$$

$$\nabla.\nabla.H-\nabla^2H=\ =-\mu.\ \in\frac{\partial^2H}{\partial t^2}$$ … (14)

We know that J=σ E, ∇.D= ρ and D= ∈ E for perfect dielectric media ρ=0

Hence, ∇.H=0, therefore ∇.∇.H=0

The equation (14) becomes

$$\nabla^2H=\ =\mu.\ \in\frac{\partial^2H}{\partial t^2}$$ … (15)

The equation (15) is the wave equation of magnetic vector H

For fields that vary harmonically with time Maxwell’s equation takes a simple form.

Let $$E=E_0e^{j\omega t} and H=H_0e^{j\omega t}$$

Differentiating w. r. t. time we get

$$\frac{\partial E}{\partial t}=j\omega E_0e^{j\omega t}=j\omega E$$

$$\frac{\partial^2E}{\partial t^2}=j\omega\frac{\partial E}{\partial t}=-\omega^2E$$ … (16)

and

$$\frac{\partial H}{\partial t}=j\omega H_0e^{j\omega t}=\ j\omega H$$

$$\frac{\partial^2H}{\partial t^2}=\ j\omega\frac{\partial H}{\partial t}==-\omega^2H$$ … (17)

Using equation (11) and (15), the equations (16) and (17) can be written as

$$\nabla^2E=-\omega^2\mu\epsilon E$$

and $$\nabla^2H=-\omega^2\mu\epsilon H$$

$$\binom{\nabla^2E=-\omega^2\mu\epsilon E}{\nabla^2H=-\omega^2\mu\epsilon H}$$ .... (18)

The solution of equation (18) depends on the appropriate boundary conditions at the waveguide wall, leads to the transmission characteristics of the wave guide.

The propagation constant γg in the guide differs from the intrinsic propagation constant γ of the dielectric. Considering propagation of wave along Z- direction, the

$${\gamma^2}_g=\gamma^2+{K^2}_x+{K^2}_y=\gamma^2+{K^2}_c$$

$$K_c=\sqrt{{K^2}_x+{K^2}_y}$$ is called cut-off wave number.

We know that propagation constant

$$\gamma=\alpha\pm j\beta$$

Now for loss less dielectric α=0, then from the definition of propagation function

$$\gamma^2=j\omega\mu.\left(j\omega\epsilon\right)=-\omega^2\mu\epsilon$$

Or

$$\gamma_g=j\omega\sqrt{\mu\epsilon}$$,

So, phase constant $$\beta=j\omega\sqrt{\mu\epsilon}$$

The guide wavelength $$\lambda_g=\frac{\lambda}{\sqrt{1-(\frac{f_c}{f})}}$$

Where λ, f and fc are the free space wavelength, operating frequency and cut – off frequency of the waveguide respectively.

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