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Consider a wave guide with a rectangular cross section of sides \(a\) and \(b\) in the \(x\) and the \(y\) directions, respectively. (a) Show that the separated DEs have the following solutions: $$\begin{array}{ll}X_{n}(x)=\sin \left(\frac{n \pi}{a} x\right), & \lambda_{n}=\left(\frac{n \pi}{a}\right)^{2} \text { for } n=1,2, \ldots, \\\Y_{m}(y)=\sin \left(\frac{m \pi}{b} y\right), & \mu_{m}=\left(\frac{m \pi}{b}\right)^{2} \quad \text { for } m=1,2, \ldots, \end{array}$$ with \(\gamma_{m n}^{2}=\lambda_{n}+\mu_{m}\). (b) Using the fact that the wave number must be real, show that there is a cutoff frequency given by $$\omega_{m n}=c \sqrt{\left(\frac{n \pi}{a}\right)^{2}+\left(\frac{m \pi}{b}\right)^{2}} \text { for } m, n \geq 1$$ (c) Show that the most general solution for \(E_{z}\) is therefore $$E_{z}=\sum_{m, n=1}^{\infty} A_{m n} \sin \left(\frac{n \pi}{a} x\right) \sin \left(\frac{m \pi}{b} y\right) e^{i\left(\omega t \pm k_{m n} z\right)}$$

Short Answer

Expert verified
The solutions to the wave equation in a rectangular waveguide show how the field varies in x, y, z directions. Using these, the cutoff frequencies were found. The general solution is a superposition of individual solutions, depends on the \(E_z\) field direction and shows how it varies in propagation direction and with time.

Step by step solution

01

Applying Separation of Variables

Start by trusting that the wave equation solution in a rectangular waveguide can be written in separated variables form \(E_z=X(x)Y(y)Z(z)\). Substitute this expression into the Helmholtz equation to get the separated differential equations.
02

Solving Separated Differential Equations

Next, solve these separated differential equations for \(X(x)\), \(Y(y)\) and \(Z(z)\) independently. This can be achieved by setting each component to equal a constant. Solve \(X''/\lambda = X\) to get \(X_n(x)=\sin(n\pi x/a)\) and similar for \(Y\). Solving gives us the solutions as provided in the problem given above.
03

Applying Boundary Conditions

Enforce the boundary conditions that the field is zero at the walls of the waveguide, i.e., \(E_z=0\) at \(x=0, a\) and \(y=0, b\). The boundary conditions determine the solutions of differential equations.
04

Determining Cutoff Frequency

Now that we have the expressions for \(lambda_n\), \(mu_m\), and \(gamma_{mn}^2\), we can substitute these into the relation for the phase constant \(\gamma\) to get the dispersion relation. This can be used to express the cutoff frequency \(\omega_{mn}\).
05

Writing Most General Solution for \(E_z\)

The most general solution would include the solution to all modes of the waveguide. Hence, summing over all \(m\) and \(n\), we get the general solution for \(E_z\). Note that the signs in the exponent account for the direction of wave propagation.

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Key Concepts

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

Separation of Variables
Separation of variables is a crucial technique when solving differential equations, especially in waveguide theory. It allows us to break a complex equation into simpler parts. For a waveguide, we assume that the electric field can be expressed as a product of functions, each depending only on one coordinate. For example, in our case, the expression is given as \(E_z = X(x)Y(y)Z(z)\). This simplification helps in transforming the Helmholtz equation into separate ordinary differential equations for each coordinate.
\(X(x)\) and \(Y(y)\) functions depend on the shape and boundary conditions of the waveguide, while \(Z(z)\) describes the propagation along the waveguide. Once separated, the equations can be solved independently, making the computation easier. It's like solving small puzzles to find the overall picture.
Helmholtz Equation
The Helmholtz equation is a fundamental part of waveguide analysis, often written as \(abla^2 E + k^2E = 0\), where \(E\) represents the electric field, and \(k\) is the wave number. In a rectangular waveguide, this equation is adapted to three spatial coordinates. By separating variables as seen in the previous section, we obtain simpler equations. These equations are similar to harmonic oscillators and have solutions expressed in sine and cosine terms.
This harmonic nature is dictated by boundary conditions and the geometry of the waveguide. Solving the Helmholtz equation ultimately gives us the modes of the waveguide, which describe how electromagnetic waves propagate through it.
Cutoff Frequency
Cutoff frequency is a pivotal concept when discussing wave propagation in waveguides. It's the minimum frequency at which a mode can propagate down the waveguide without attenuation. Below this frequency, the wave cannot travel, as it is reflected back.
The cutoff frequency \(\omega_{mn}\) is derived using the expressions for \(\lambda_n\) and \(\mu_m\), as \(\omega_{mn} = c \sqrt{(n\pi/a)^2 + (m\pi/b)^2}\). Each mode \((m,n)\) within the waveguide has its cutoff frequency, based on the waveguide dimensions \(a\) and \(b\). Knowing this helps engineers design waveguides for specific frequencies and ensure efficient transmission.
Rectangular Waveguide
Rectangular waveguides are a specific type of waveguide with a cross-section shaped like a rectangle, characterized by its width \(a\) in the \(x\)-direction and height \(b\) in the \(y\)-direction. They are widely used in microwave engineering, given their ability to confine and guide electromagnetic waves effectively.
The geometry of a rectangular waveguide allows certain modes to propagate, determined by its dimensions and the frequency of the waves. These modes are the solutions to the Helmholtz equation, considering the geometry. Practically, rectangular waveguides are advantageous because they can handle high power with low loss and are easier to manufacture.
Boundary Conditions
Boundary conditions play a significant role in waveguide theory because they define how electromagnetic fields behave at the edges of the waveguide. For rectangular waveguides, the fields must satisfy conditions at the walls to ensure there is no electric field outside the boundary.
In mathematical terms, boundary conditions lead to the requirement that certain functions, like \(X(x)\) and \(Y(y)\), be zero at the walls \(x=0, a\) and \(y=0, b\). These conditions influence the form of solutions, ensuring they fit within the waveguide boundaries. Without satisfying these, the solutions wouldn't accurately describe the physical reality of the waveguide.

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Most popular questions from this chapter

The circumference of a flat circular plate of radius \(a\), lying in the \(x y\) plane, is held at \(T=0\). Find the temperature distribution for all time if the temperature distribution at \(t=0\) is given-in Cartesian coordinates \(-\) by (a) \(\frac{T_{0}}{a} y\). (b) \(\frac{T_{0}}{a} x\). (c) \(\frac{T_{0}}{a^{2}} x y\). (d) \(T_{0}\).

The lateral faces of a cube are grounded, and its top and bottom faces are held at potentials \(f_{1}(x, y)\) and \(f_{2}(x, y)\), respectively. (a) Find a general expression for the potential inside the cube. (b) Find the potential if the top is held at \(V_{0}\) volts and the bottom at \(-V_{0}\) volts.

Consider a rectangular heat-conducting plate with sides of lengths \(a\) and \(b\). Three of the sides are held at \(T=0\), and the fourth side has a temperature variation \(T=f(x)\) (see Fig. 19.4). The flat faces are insulated, so they cannot lose heat to the surroundings. Assume a steady-state heat transfer. (a) Show that the separation of variables yields $$X_{n}(x)=\sin \left(\frac{n \pi}{a} x\right), \quad Y(y)=\sinh \left(\frac{n \pi}{a} y\right) \quad \text { for } n=1,2, \ldots$$ (b) Show that the most general solution is $$T(x, y)=X(x) Y(y)=\sum_{n=1}^{\infty} B_{n} \sin \left(\frac{n \pi}{a} x\right) \sinh \left(\frac{n \pi}{a} y\right)$$ with $$B_{n}=\frac{2}{a \sinh (n \pi b / a)} \int_{0}^{a} \sin \left(\frac{n \pi}{a} x\right) f(x) d x .$$ (c) Show that if the fourth side is held at the constant temperature \(T_{0}\), then we obtain $$T(x, y)=\frac{4 T_{0}}{\pi} \sum_{k=0}^{\infty} \frac{1}{2 k+1} \frac{\sin [(2 k+1) \pi x / a] \sinh [(2 k+1) \pi y / a]}{\sinh [(2 k+1) \pi b / a]} .$$$(19.47)\( (d) If the temperature variation of the fourth side is of the form \)f(x)=\( \)T_{0} \sin (\pi x / a)$, then $$T(x, y)=T_{0} \frac{\sin (\pi x / a) \sinh (\pi y / a)}{\sinh (\pi b / a)}$$

Find the steady-state temperature distribution \(T(\rho, \varphi, z)\) in a semiinfinite solid cylinder of radius \(a\) if the temperature distribution of the base is \(f(\rho, \varphi)\) and the lateral surface is held at \(T=0\).

All sides of the plate of Sect. \(19.5 .2\) are held at \(T=0\). Find the temperature distribution for all time if the initial temperature distribution is given by (a) \(T(x, y, 0)=\left\\{\begin{array}{ll}T_{0} & \text { if } \frac{1}{4} a \leq x \leq \frac{3}{4} a \text { and } \frac{1}{4} b \leq y \leq \frac{3}{4} b, \\ 0 & \text { otherwise. }\end{array}\right.\) (b) \(T(x, y, 0)=\frac{T_{0}}{a b} x y, \quad\) where \(0 \leq x

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