# 8.11 Cavity Resonators

A rectangular waveguide has metal walls on four sides. When the two remaining sides are terminated with conducting walls, the waveguide becomes a cavity. By designing cavities to resonate at specific frequencies, they can be used as circuit elements in microwave oscillators, amplifiers, and bandpass filters.

The rectangular cavity shown in Fig. 8-28(a) with dimensions (a×b×d) is connected to two coaxial cables that feed and extract signals into and from the cavity via input and output probes. As a bandpass filter, the function of a resonant cavity is to block all spectral components of the input signal except for those with frequencies that fall within a narrow band surrounding a specific center frequency f 0, which is the cavity’s resonant frequency. Comparison of the spectrum in Fig. 8-28(b), which describes the range of frequencies that might be contained in a typical input signal, with the narrow output spectrum in Fig. 8-28(c) demonstrates the filtering action imparted by the cavity.

# Figure 8-28 A resonant cavity supports a very narrow bandwidth around its resonant frequency f 0.

(a) Resonant cavity

Figure 8 28-a Full Alternative Text

(b) Input spectrum

(c) Output spectrum

In a rectangular waveguide, the fields constitute standing waves along the x and y directions and a propagating wave along z ˆ. The terms TE and TM were defined relative to the propagation direction; TE meant that E was entirely transverse to z ˆ, and TM meant that H had no component along z ˆ. In a cavity, there is no unique propagation direction, as no fields propagate. Instead, standing waves exist along all three directions. Hence, the terms TE and TM need to be modified by defining the fields relative to one of the three rectangular axes. For the sake of consistency, we will continue to define the transverse direction to be any direction contained in the plane whose normal is z ˆ.

The TE mode in the rectangular waveguide consists of a single propagating wave whose H˜z component is given by Eq. (8.110e) as

H˜z=H 0 cos m π x a cos n π y b e−j β z, (8.119)

where the phase factor e−j β z signifies propagation along +z ˆ. Because the cavity has conducting walls at both z=0 and z=d, it will contain two such waves: one with amplitude H 0 traveling along +z ˆ and another with amplitude H 0− traveling along −z ˆ. Hence,

H˜z=(H 0 e−j β z+H 0−e j β z) cos m π x a cos n π y b. (8.120)

Boundary conditions require the normal component of H˜ to be zero at a conducting boundary. Consequently, H˜z must be zero at z=0 and z=d. To satisfy these conditions, it is necessary that H 0−=−H 0 and β d=p π, with p=1,2,3,…. In this case, Eq. (8.120) becomes

H˜z=−2 j H 0 cos m π x a cos n π y b sin p π z d. (8.121)

Given that E˜z=0 for the TE modes, all of the other components of E˜ and H˜ can be derived readily through the application of the relationships given by Eq. (8.89). A similar procedure also can be used to characterize cavity modes for the TM case.

# 8.11.1 Resonant Frequency

The consequence of the quantization condition imposed on β, namely β=p π/d with p assuming only integer values, is that for any specific set of integer values of (m,n,p) the wave inside the cavity can exist at only a single resonant frequency, f m n p, whose value has to satisfy Eq. (8.105). The resulting expression for f m n p is

For TE, the indices m and n start at 0, but p starts at 1. The exact opposite applies to TM. By way of an example, the resonant frequency for a TE 101 mode in a hollow cavity with dimensions a=2 cm, b=3 cm, and d=4 cm is f 101=8.38 GHz.

# 8.11.2 Quality Factor

In the ideal case, if a group of frequencies is introduced into the cavity to excite a certain TE or TM mode, only the frequency component at exactly f m n p of that mode will survive, and all others will attenuate. If a probe is used to couple a sample of the resonant wave out of the cavity, the output signal will be a monochromatic sinusoidal wave at f m n p. In practice, the cavity exhibits a frequency response similar to that shown in Fig. 8-28(c), which is very narrow, but it is not a perfect spike. The bandwidth Δ f of the cavity is defined as the frequency range between the two frequencies (on either side of f m n p) at which the amplitude is 1/2 of the maximum amplitude (at f m n p). The normalized bandwidth, defined as Δ f/f m n p, is approximately equal to the reciprocal of the quality factor Q of the cavity:

Q≈f m n p Δ f. (8.123)

For an ideal cavity with perfectly conducting walls, no energy loss is incurred, as a result of which Q is infinite and Δ f≈0. Metals have very high (but not infinite) conductivities, so a real cavity with metal walls stores most of the energy coupled into it in its volume. However, it also loses some energy to heat conduction. A typical value for Q is on the order of 10,000, which is much higher than can be realized with lumped RLC circuits.

# Example 8-11: Q of a Resonant Cavity

The quality factor for a hollow resonant cavity operating in the TE 101 mode is

Q=1 δ s a b d(a 2+d 2)[a 3(d+2 b)+d 3(a+2 b)], (8.124)

where δ s=1/π f m n p μ 0 σ c is the skin depth and σ c is the conductivity of the conducting walls. Design a cubic cavity with a TE 101 resonant frequency of 12.6 GHz and evaluate its bandwidth. The cavity walls are made of copper.

Solution: For a=b=d, m=1, n=0, p=1, and u p 0=c=3×10 8 m/s, Eq. (8.122) simplifies to

f 101=3 2×10 8 2 a(Hz).

For f 101=12.6 GHz, this gives

a=1.68 cm.

At f 101=12.6 GHz, the skin depth for copper (with σ c=5.8×10 7 S/m) is

δ s=1[π f 101 μ 0 σ c]1/2=1[π×12.6×10 9×4 π×10−7×5.8×10 7]1/2=5.89×10−7 m.

Upon setting a=b=d in Eq. (8.124), the expression for Q of a cubic cavity becomes

Q=a 3 δ s=1.68×10−2 3×5.89×10−7≈9,500.

Hence, the cavity bandwidth is

Δ f≈f 101 Q≈12.6×10 9 9,500≈1.3 MHz.