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Chapter 9: Problem 28

Consider a rectangular wave guide with dimensions \(2.28 \mathrm{~cm} \times 1.01 \mathrm{~cm}\). What TE modes will propagate in this wave guide, if the driving frequency is \(1.70 \times 10^{10} \mathrm{~Hz}\) ? Suppose you wanted to excite only one TE mode; what range of frequencies could you use? What are the corresponding wavelengths (in open spacc)?

Short Answer

Expert verified
TE modes TE\(_{01}\) and TE\(_{10}\) will propagate. Use frequencies between 6.58 GHz and 14.9 GHz for one TE mode. Corresponding wavelengths: 4.56 cm to 2.01 cm.

Step by step solution

01

Understanding Waveguide Dimensions

The waveguide has dimensions of \(a = 2.28 \text{ cm}\) and \(b = 1.01 \text{ cm}\). These represent the larger and smaller sides, respectively, of the rectangular cross-section of the waveguide.
02

Calculating Cut-off Frequency

The cutoff frequency for a TE\(_{mn}\) mode in a rectangular waveguide is given by \(f_{c_{mn}} = \frac{1}{2\pi \sqrt{\mu \varepsilon}} \sqrt{\left(\frac{m \pi}{a}\right)^2 + \left(\frac{n \pi}{b}\right)^2}\), where \(\mu\) and \(\varepsilon\) are the permeability and permittivity of free space. For free space, this simplifies to \(f_{c_{mn}} = \frac{c}{2} \sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}\), where \(c = 3 \times 10^8 \text{ m/s}\) is the speed of light.
03

Determining Propagation Conditions

For a TE mode to propagate, the frequency \(f\) must be greater than the cutoff frequency \(f_{c_{mn}}\). Begin by calculating \(f_{c_{01}}\) and \(f_{c_{10}}\), which are the lowest cutoff frequencies. Compare these to the given frequency \(1.70 \times 10^{10} \text{ Hz}\).
04

Calculating\(f_{c_{01}}\) and \(f_{c_{10}}\)

Calculate \(f_{c_{01}} = \frac{c}{2b} = \frac{3 \times 10^8}{2 \times 1.01 \times 10^{-2}} \approx 1.49 \times 10^{10} \text{ Hz}\) and \(f_{c_{10}} = \frac{c}{2a} = \frac{3 \times 10^8}{2 \times 2.28 \times 10^{-2}} \approx 6.58 \times 10^{9} \text{ Hz}\).
05

Identifying Propagating TE Modes

The given frequency \(1.70 \times 10^{10} \text{ Hz}\) is greater than both \(f_{c_{01}}\) and \(f_{c_{10}}\), so TE modes TE\(_{01}\) and TE\(_{10}\) can propagate. Check higher order modes, such as TE\(_{11}\), TE\(_{20}\), etc., to ensure they have higher cutoff frequencies.
06

Range for Single Mode Propagation

To excite only one TE mode, typically the TE\(_{10}\) mode, operate just above \(f_{c_{10}}\) and below \(f_{c_{01}}\). Hence, the frequency range is \(6.58 \times 10^{9} \text{ Hz} < f < 1.49 \times 10^{10} \text{ Hz}\).
07

Calculating Corresponding Wavelengths

Use the relation \(\lambda = \frac{c}{f}\) for frequencies at the extremes. For \(f = 6.58 \times 10^{9} \text{ Hz}\), \(\lambda = \frac{3 \times 10^8}{6.58 \times 10^{9}} \approx 4.56 \text{ cm}\). For \(f = 1.49 \times 10^{10} \text{ Hz}\), \(\lambda = \frac{3 \times 10^8}{1.49 \times 10^{10}} \approx 2.01 \text{ cm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

TE Modes
In a rectangular waveguide, Transverse Electric (TE) modes are a type of electromagnetic wave pattern where the electric field is entirely transverse to the direction of wave propagation, meaning it has no component in the direction of travel (the axis of the waveguide). These modes are denoted as TE\(_{mn}\), where \(m\) and \(n\) are integers determining the number of half-wave variations in the electric field across the waveguide's broader (\(a\)) and narrower (\(b\)) sides, respectively. TE modes become crucial in determining the range of frequencies over which a waveguide will effectively transmit signals. The selection of modes affects the efficiency and bandwidth of the waveguide, ensuring that only specific frequency components pass through efficiently while others attenuate rapidly.
Cutoff Frequency
Cutoff frequency is a critical concept in waveguide operations, marking the minimum frequency at which a particular mode can propagate through the waveguide. If the operating frequency of electromagnetic waves is below this threshold, the waves will not propagate and will be attenuated. In mathematical terms, for a TE\(_{mn}\) mode, the cutoff frequency \(f_{c_{mn}}\) is calculated using the formula:
  • \(f_{c_{mn}} = \frac{c}{2} \sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}\)
where \(c\) is the speed of light. This equation highlights that waveguide dimensions \(a\) and \(b\) and mode indices \(m\) and \(n\) influence the cutoff frequency. Ensuring the operating frequency exceeds the cutoff frequency of all desired modes is essential for efficient waveguide operation.
Rectangular Waveguide
A rectangular waveguide is a two-dimensional conduit, typically made of metal, through which electromagnetic waves travel. It features a rectangular cross section, defined by two dimensions: the width \(a\) and the height \(b\). These structures guide waves by reflecting them off the walls of the waveguide, keeping them contained within.
  • The rectangular waveguide's ability to transmit specific electromagnetic modes, such as TE\(_{mn}\), depends on its dimensions and the operating frequency.
  • Advantages of rectangular waveguides include high power handling capabilities and low signal loss.
These features make them ideal for applications like radar and satellite communications, where precision and efficiency are paramount.
Propagation Conditions
Propagation conditions in waveguides determine whether electromagnetic waves can travel through the guide. A key condition for propagation is the operating frequency's relationship to the cutoff frequency. For a given waveguide mode, the frequency \(f\) must be greater than the cutoff frequency \(f_{c_{mn}}\).
  • This ensures that the wave 'fits' within the dimensions of the guide and can travel without significant loss.
  • For practical applications, especially in communications, maintaining frequencies above the lowest cutoff frequency helps avoid signal attenuation.
Understanding these conditions allows engineers to design waveguides that maximize transmission efficiency by selecting appropriate frequencies and dimensions.
Electromagnetic Waves
Electromagnetic waves are a form of energy transmission through the oscillation of electric and magnetic fields. These waves include radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays, covering a vast spectrum of frequencies and wavelengths. In the context of waveguides, electromagnetic waves are harnessed and guided to travel long distances or to specific areas with minimal loss.
  • Waveguides confine these waves, enabling efficient transmission of information or energy.
  • The relationship between wavelength and frequency is given by \(\lambda = \frac{c}{f}\), where \(\lambda\) is the wavelength, \(c\) is the speed of light, and \(f\) is the frequency.
This relationship is vital in matching the wave characteristics to the waveguide dimensions for optimal transmission.

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Most popular questions from this chapter

Write down the (real) electric and magnetic fields for a monochromatic plane wave of amplitude \(E_{0}\), frequency \(\omega\), and phase angle zero that is (a) traveling in the negative \(x\) direction and polarized in the \(z\) direction; (b) traveling in the direction from the origin to the point \((1,1,1)\), with polarization parallel to the \(x z\) plane. In each case, sketch the wave, and give the explicit Cartesian components of \(\mathrm{k}\) and \(\hat{\mathrm{n}}\).

Suppose \(A e^{i a x}+B e^{i b x}=C e^{i c x}\), for some nonzero constants \(A, B, C, a, b .\) \(c\), and for all \(x\). Prove that \(a=b=c\) and \(A+B=C\).

Show that the mode \(\mathrm{TE}_{00}\) cannot occur in a rectangular wave guide. [Hint: In this case \(\omega / c=k\), so Eqs. \(9.180\) arc indetcrminate, and you must go back to \(9.179\). Show that \(B_{z}\) is a constant, and hence- applying Faraday's law in integral form to a cross scction-that \(B_{z}=0\), so this would be a TEM mode.]

Light of (angular) frequency \(\omega\) passes from medium 1, through a slab (thickness d) of medium 2, and into medium 3 (for instance, from water through glass into air, as in Fig. 9.27). Show that the transmission cocfficient for normal incidence is given by $$ T^{-1}=\frac{1}{4 n_{1} n_{3}}\left[\left(n_{1}+n_{3}\right)^{2}+\frac{\left(n_{1}^{2}-n_{2}^{2}\right)\left(n_{3}^{2}-n_{2}^{2}\right)}{n_{2}^{2}} \sin ^{2}\left(\frac{n_{2} \omega d}{c}\right)\right] $$ [Hint: To the left, there is an incident wave and a reflected wave; to the right, there is a transmitted wave; inside the slab there is a wave going to the right and a wave going to the left. Express each of these in terms of its complex amplitude, and relate the amplitudes by imposing suitable boundary conditions at the two interfaces. All three media are linear and homogeneous; assume \(\left.\mu_{1}=\mu_{2}=\mu_{3}=\mu_{0} .\right]\)

The intensity of sunlight hitting the earth is about \(1300 \mathrm{~W} / \mathrm{m}^{2}\). If sunlight strikes a perfect absorber, what pressure does it exert? How about a perfect reflector? What fraction of atmospheric pressure does this amount to?
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