Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) PDF Download

 Rectangular Waveguide 

In the last lecture, we have coonsidered the case of guided wave between a pair of infinite conducting planes. In this lecture, we consdier a rectangular wave guide which consists of a hollow pipe of infinite extent but of rectangular cross section of dimension α x b. The long direction will be taken to be the z direction.

Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Unlike in the previous case  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) is not zero in this case. However, as the propagation direction is along the z direction, we have  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) We can write the Maxwell’s curlequations as

                                 Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The second set of equations are obtained from the Faraday’s law, (these can be written down from above by  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

                            Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

As in the case of parallel plate waveguides, we can express all the field quantities in terms of the derivatives of Eand Hz . For instance, we have

                        Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

which gives,

                     Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

so that

                        Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

where

                        Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The other components can be similarly written down

                  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

 As before, we will look into the TE mode in detail. Since Ez = 0, we need to solve for Hz from the Helmholtz equation,

                 Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Remember that the complete solution is obtained by multiplying with  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) We solve equation (5) using the technique of separation of variables which we came across earlier. Let

                Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Substituting this in (5) and dividing by XY throughout, we get

           Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

where Ky is a constant. We further define  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

We now have two second order equations,

                                                                 Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The solutions of these equations are well known

                                                           Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

This gives,

                                                    Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The boundary conditions that must be satisfied to determine the constants is the vanishing of the tangential component of the electric field on the plates. In this case, we have   two pairs of plates. The tangential direction on the plates at x= 0 and x = α is the y direction, so that the y component of the electric field

                                                          Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Likewise, on the plates at y = 0 and y = b,

                                                         Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

We need first to evaluate Ex and Ey using equations (1) and (2) and then substitute the boundary conditions. Since Ez = 0, we can write eqn. (1) and (2) as

                                                               Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

                                                          Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

                                                     Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Since Ey = 0 at x = 0, we must have C2 = 0 and then we get,

                                                       Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Further, since Ex = 0 at y=0, we have C= 0. Combining these, we get, on defining a constant C = C1C3

                                                 Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

And

                                              Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

We still have the boundary conditions, Ex= 0 at y = b and Ey = 0 at x = α to be satisfied. The former gives  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) while the latter gives  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) where m and n are integers. Thus we have,

                                       Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

 

and            

                                      Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

However  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) so that,

                                   Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

For propagation to take place, γ must be imaginary, so that the cutoff frequency below which propagation does not take place is given by

                                Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The minimum cutoff is for TE1,0 (or TE0,1 ) mode which are known as dominant mode. For these modes Ex (or Ey ) is zero.

TM Modes

We will not be deriving the equations for the TM modes for which Hz = 0 . In this case, the solution for Ez , becomes,

                               Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

As the solution is in terms of product of sine functions, neither m nor n can be zero in this case. This is why the lowest TE mode is the dominant mode

For propagating solutions, we have,

                              Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

where,

                            Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

We have,  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Differentiating both sides, we have,

                                  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The group velocity of the wave is given by

                 Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

which is less than the speed of light. The phase velocity, however, is given by

                 Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

It may be noted that  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) which in vacuum equal the square velocity of light.

For propagating TE mode, we have, from (1) and (4) using Ez = 0

                    Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

where ηTE is the characteristic impedance for the TE mode. It is seen that the characteristic impedance is resistive. Likewise

                              Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Power Transmission

We have seen that in the propagating mode, the intrinsic impedance is resistive, indicating that there will be average flow of power. The Poynting vector for TE mode is given by

                               Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Substituting the expressions for the field components, the average power flow through the x y plane is given by

                         Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

                             Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Impossibility of TEM mode in Rectangular waveguides

We have seen that in a parallel plate waveguide, a TEM mode for which both the electric and magnetic fields are perpendicular to the direction of propagation, exists. This, however is not true of rectangular wave guide, or for that matter for any hollow conductor wave guide without an inner conductor.

We know that lines of H are closed loops. Since there is no z component of the magnetic field, such loops must lie in the x-y plane. However, a loop in the x-y plane, according to Ampere’s law, implies an axial current. If there is no inner conductor, there cannot be a real current. The only other possibility then is a displacement current. However, an axial displacement current requires an axial component of the electric field, which is zero for the TEM mode. Thus TEM mode cannot exist in a hollow conductor. (for the parallel plate waveguides, this restriction does not apply as the field lines close at infinity.)

Resonating Cavity

In a rectangular waveguide we had a hollow tube with four sides closed and a propagation direction which was infinitely long. We wil now close the third side and consider electromagnetic wave trapped inside a rectangular parallopiped of dimension α × b × d with walls being made of perfector conductor. (The third dimension is taken as d so as not to confuse with the speed of light c).

Resonanting cavities are useful for storing electromagnetic energy just as LC circuit does but the former has an advantage in being less lossy and having a frequency range much higher, above 100 MHz.

The Helmholtz equation for any of the components Eα of the electric field can be written as

                                  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

We write this in Cartesian and use the technique of separation of variables,

                          Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Introducing this into the equation and dividing by Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) , we get,

                     Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Since each of the three terms on the right is a function of an independent variable, while the right hand side is a constant, we must have each of the three terms equaling a constant such that the three constants add up to the constant on the right. Let

                           Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

                          Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

such that  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

As each of the equations has a solution in terms of linear combination of sine and cosine functions, we write the solution for the electric field as

       Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Tha tangential component of above must vanish at the metal boundary. This implies,

  1. At x= 0 and at x = α , Ey and Ez = 0 for all values of y,z
  2. At y = 0 and y = b, Ex and Ez = 0 for all values of x,z
  3. At z = 0 and z=d, Ex and Ey = 0 for all values of x, y

Let us consider Ex which must be zero at Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) . This is posible if

        Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

with  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Here m and n are non-zero integers.

In a similar way, we have,

          Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

with  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) with ℓ being non-zerointeger.

We now use

          Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

This relation must be satisfied for all values of x, y, z within the cavity. This requires

          Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Let us choose some special points and try to satisfy this equation. Let x = 0, y, z arbitrary,

This requires Bx = 0. Likewise, takingy = 0, x, z arbitrary, we require Dy = 0 and finally, z = 0,

x, yarbitrary gives Gz = 0.

With these our solutions for the components of the electric field becomes,

                           Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

where we have redefined our constants Ex0, Ey0 and Ez0.

We also have,

                        Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

the integers ℓ, m, n cannot be simultaneously zero for then the field will identically vanish. Note further that since Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) must be satisfied everywhere within the cavity, we must have, calculating the divergence explicitly from the above expressions

                   Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

This requires that  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) is perpendicular to  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

One can see that the modes within a cavity can exist only with prescribed “resonant” frequency corresponding to the set of integers ℓ, m, n, 

            Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Though there is nothing like a propagation direction here, one can take the longest dimension (say d) to be the propagation direction.

We can have modes like Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) corresponding to the set l, m, n for which Ez = 0. For this set

           Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

with Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) The magnetic field components can be obtained from the Faraday’s law,

          Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Tutorial Assignment 

  1. A rectangular waveguide has dimensions 8 cm x 4 cm. Determine all the modes that can propagate when the operating frequency is (a) 1 GHz, (b) 3 GHz and (c) 8 GHz.
  2. Two signals, one of frequency 10 GHz and the other of 12 GHz are simultaneously launched in an air filled rectangular waveguide of dimension 2 cm x 1 cm. Find the time interval between the arrival of the signals at a distance of 10 m from the common place of their launch.
  3. What should be the third dimension of a cavity having a length of 1 cm x 1 cm which can operate in a TE103 mode at 24 GHZ?

Solutions to Tutorial Assignments

1. The cutoff frequency for (m,n) mode is  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) . Some of the lowest cutoffs are as under (in GHz) :

mnCutoff frequency (GHz)
013.75
027.5
0311.25
101.875
114.192
127.731
1311.405
203.75
215.30
228.39
305.625
316.76
329.375
407.5
418.38
509.375

At 1 GHz, no propagation takes place. At 3 GHz only TE10 mode propagates (recall no TM mode is possible when either of the indices is zero.) At 8 GHz, we have TE01, TE02, TE10, TE11, TE12, TE20, TE21, TE30, TE31, TE40, TM11, TM12, TM21 and TM31, i.e. a total of 14 modes propagating.

2. The cutoff frequency is given by

                        Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

     The propagation of the wave is given by the group velocity  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) which is 1.98x 10and 2.07x10m/s respectively for f=10 GHz and 12 GHz respectively.           Thus the difference in speed is 9x 106 m/s, resulting in a time difference of approximately 10-5 s in travelling 10m.

3. The operating frequency is given by

                                                 Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

Substitutingl = 1, m = 0, n = 3, we get d = 2.4 cm

Self Assessment Questions

  1. A rectangular, air filled waveguide has dimension 2 cm x 1 cm. For what range of frequencies, there is a “single mode” operation in the guide?
  2. The cutoff frequency for a TE10 mode in an air filled waveguide is 1.875 GHz. What would be the cutoff frequency of this mode if the guide were to be filled with a dielectric of permittivity 9∈0?
  3. In an air filled waveguide of dimension 2 cm x 2 cm, the x component of the electric field is given by

                                    Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

            Identify the propagating mode, determine the frequency of operation and find Hz and Ez .

Solutions to Self Assessment Questions

  1. The cutoff frequency for (m,n) mode is   Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Substituting values, the cutoff for (1,0) mode is 7.5 GHz and for (2,0) is 15 GHz. The frequency for (0,1) is also 15 GHz. All other modes have higher cut off. Thus in order that only one mode propagates, the operating frequency should be in the range 7.5 < ν < 15 GHz. In this range, only TE10 mode operates. Recall that there is no TM mode with either of the indices being zero.
  2. The cutoff frequency is proportional to  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) Thus the frequency would be reduced by a factor of 3 making it 625 MHz.
  3. Since one of the standing wave factor is missing, one of the mode indices is zero. Thus it is a TE mode. The electric field of TEmn mode is given by (with a=b)

                                                                    Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

          Thus the mode is TE02 mode. From the relation  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)           the operating                   frequency is 15.74 GHz.

           Since kx = 0, Ey = 0 . Further, by definition for TE mode Ez= 0.

           Here we have  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) The ratio  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

           so that  Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)                                    

 

The document Wave Guides - 2 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Electromagnetic Fields Theory (EMFT).
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FAQs on Wave Guides - 2 - Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

1. What is a waveguide?
Ans. A waveguide is a structure that is used to guide electromagnetic waves, such as radio waves, microwaves, and light waves. It consists of a hollow metallic or dielectric tube or channel that allows the waves to propagate without significant loss or distortion.
2. What are the advantages of using waveguides?
Ans. Waveguides offer several advantages, including low transmission loss, high power handling capability, and the ability to confine and control the direction of the electromagnetic waves. They are also more efficient than traditional transmission lines and can support a wider range of frequencies.
3. How do waveguides work?
Ans. Waveguides work by confining and guiding electromagnetic waves within their structure. The waves bounce off the walls of the waveguide, allowing them to propagate along a specific path without significant loss. The dimensions and shape of the waveguide determine the frequency range and mode of operation.
4. What are the different types of waveguides?
Ans. There are several types of waveguides, including rectangular waveguides, circular waveguides, coaxial waveguides, and optical waveguides. Each type is designed to work with specific frequencies and applications. Rectangular waveguides are commonly used for microwave applications, while optical waveguides are used for transmitting light signals.
5. What are some common applications of waveguides?
Ans. Waveguides are widely used in various applications, such as satellite communications, radar systems, microwave ovens, medical imaging, fiber optic communication systems, and laser technology. They are essential components in many electronic devices and systems that require the efficient transmission of electromagnetic waves.
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