Problems Three

Question 5

Calculate the range of frequencies over which only the lowest order waveguide mode will propagate in rectangular waveguides having the following cross sectional dimensions.

     
        20cm by 15cm  2cm by 1.5cm   2mm by 1.5mm
                      2cm by 1cm
                      2cm by 0.8cm
                      2cm by 2cm
                      2cm by 4cm
        

• The cutoff wavelength for the TE10 mode is twice the longest cross section dimension. Remembering 30 = freq(GHz) times wavelength we have cutoff for the 2cm guides as 30/4 = 7.5GHz. For the 20 cm guide it is 750MHz and for the 2mm guide it is 75 GHz.
• The next mode is either the TE20 mode at twice the cutoff frequency of the TE10 mode, or the TE01 mode if this is lower.
  • Thus, 20cm by 15cm has 750MHz to 1GHz

    2cm by 1.5cm, 7.5GHz to 10GHz, and 2mm by 1.5mm, 75GHz to 100GHz.

    2cm by 1cm and 2cm by 0.8cm both have 7.5GHz to 15GHz.

    2cm by 2cm has two degenerate modes, TE10 and TE01 both with cutoff 7.5GHz. There is thus no band of frequency over which only one mode will propagate.

    2cm by 4cm has long dimension 4cm so the range is 3.75GHz to 7.5GHz.

For the 2cm by 1cm waveguide, determine the lowest frequency of the lowest mode which will propagate. Calculate the guide group and phase velocities, and the propagation angle alpha, for frequencies 100MHz 300MHz and 1GHz higher than the cutoff frequency.

• From the geometry of TE10 mode propagation, refer to the book of your choice, if the free space wavelength is lambda, the guide width a, the propagation angle alpha, and the guide wavelength lambdaG we can write

  • sin(alpha) = lambda/(2a)

  • cos(alpha) = lambda/lambdaG

  • tan(alpha) = lambdaG/(2a)

• The cutoff of the 2cm by 1cm guide is 7500MHz so we want the values above at 7600, 7800, and 8500MHz.

• The values of lambda respectively are 3.9474cm, 3.8462cm, and 3.5294cm.

• The values of alpha are 80.70degrees, 74.06degrees, and 61.93degrees.

• The values of lambdaG are 24.41cm, 14.0cm, and 7.50cm.

• The group velocities are c cos(alpha) = 0.4848, 0.8239, and 1.412, all in units of 10^8 metres per second.

• The phase velocities are c/cos(alpha) = 18.56, 10.92, and 6.376, all in units of 10^8 metres per second.

Question 6.

Calculate the scattering matrix of a 25 cm length of waveguide of internal rectangular cross section 5cm by 2.2cm at a frequency of 4.0 GHz. Hint, you will need to find out the guide wavelength.

• The cutoff frequency is 3GHz. The guide wavelength is 11.3389cm. There are 2.2048 guide wavelengths in the 25cm length of guide.
• The residual phase shift after allowing whole cycles is -73.73 degrees.
• The s parameters are s11=s22=0, s12=s21 = 1 angle -73.73 degrees.

The wave impedance of a waveguide mode is defined as the ratio of the transverse component of electric field strength to the transverse component of magnetic field strength. Show that the wave impedance depends on the angle alpha and the guide wavelength, and derive expressions for the dependencies.

• For a TE mode, the electric field is wholly transverse and the same as it is in a free space wave. The magnetic field is turned by cos(alpha) from being transverse, so is smaller than the free space wave value by cos(alpha). Thus the wave impedance is 120 pi/cos(alpha) or 120 pi lambdaG/lambda.
• For a TM mode the electric field is rotates through alpha but the magnetic field is entirely transverse. The wave impedance is thus 120 pi cos(alpha) or 120 pi lambda/lambdaG.

Calculate the wave impedance for the conditions of the first part of this question. Hint, you need to know that in a free space wave the wave impedance is 120 pi ohms.

• Applying the results above, the impedance at 4GHz is 570 ohms to the nearest ohm.