STEM Problem Detail

Problem Statement

Prompt

provide solution of given question from problem image

Solution Rendering...

**Problem 8.6**

Design a three-cavity quarter-wave coupled Chebyshev filter with the following
specifications:

- Waveguide width: $$a = 0.9''$$
- Band edge frequency: $$f_1 = 10{,}400$$~MHz
- Passband tolerance: $$a_m^2 = 0.0233$$
- Inductive diaphragms with circular holes are used

Determine the hole radii and diaphragm spacings.

---

**Given:**

\[
\begin{aligned}
n &= 3 \quad \text{(number of cavities)} \\
a &= 0.9'' = 2.286~\text{cm} \\
f_1 &= 10.4~\text{GHz} \\
a_m^2 &= 0.0233
\end{aligned}
\]

---

**Step 1: Chebyshev Ripple Factor**

For a Chebyshev response, the ripple factor $$\epsilon$$ is

\[
\epsilon = \sqrt{a_m^2} = \sqrt{0.0233} = 0.1527
\]

---

**Step 2: Normalized Low-Pass Prototype Constants**

For a three-section Chebyshev filter with $$\epsilon = 0.1527$$, the low-pass
prototype values are:

\[
\begin{aligned}
g_0 &= 1 \\
g_1 &= 1.5963 \\
g_2 &= 1.0967 \\
g_3 &= 1.5963 \\
g_4 &= 1
\end{aligned}
\]

---

**Step 3: Fractional Bandwidth**

The cutoff frequency of the waveguide is

\[
f_c = \frac{c}{2a}
\]

Substituting $$a = 2.286$$~cm:

\[
f_c = \frac{3 \times 10^{10}}{2 \times 2.286} = 6.56~\text{GHz}
\]

The centre frequency is approximated as

\[
f_0 \approx f_1 = 10.4~\text{GHz}
\]

The fractional bandwidth is

\[
\Delta = \frac{f_1 - f_c}{f_0}
\]

\[
\Delta = \frac{10.4 - 6.56}{10.4} = 0.369
\]

---

**Step 4: External Quality Factors**

The external quality factors are given by

\[
Q_{e1} = \frac{g_0 g_1}{\Delta}, \quad
Q_{e3} = \frac{g_3 g_4}{\Delta}
\]

\[
Q_{e1} = Q_{e3} = \frac{1 \times 1.5963}{0.369} = 4.33
\]

---

**Step 5: Coupling Coefficients**

The coupling coefficients between adjacent cavities are

\[
k_{12} = \frac{\Delta}{\sqrt{g_1 g_2}}, \quad
k_{23} = \frac{\Delta}{\sqrt{g_2 g_3}}
\]

\[
k_{12} = k_{23} = \frac{0.369}{\sqrt{1.5963 \times 1.0967}} = 0.28
\]

---

**Step 6: Diaphragm Hole Radii**

For inductive diaphragms with circular apertures, the coupling coefficient is
related to the hole radius $$r$$ by

\[
k \propto \left( \frac{r}{a} \right)^3
\]

Thus,

\[
\frac{r}{a} = k^{1/3}
\]

\[
\frac{r}{2.286} = (0.28)^{1/3}
\]

\[
r = 1.50~\text{cm}
\]

---

**Step 7: Diaphragm Spacing**

For quarter-wave coupled cavities, the spacing between diaphragms is approximately

\[
l = \frac{\lambda_g}{4}
\]

The guided wavelength is

\[
\lambda_g = \frac{\lambda_0}{\sqrt{1 - (f_c/f_0)^2}}
\]

\[
\lambda_0 = \frac{c}{f_0} = \frac{3 \times 10^{10}}{10.4 \times 10^9}
= 2.885~\text{cm}
\]

\[
\lambda_g = \frac{2.885}{\sqrt{1 - (6.56/10.4)^2}} = 3.72~\text{cm}
\]

\[
l = \frac{3.72}{4} = 0.93~\text{cm}
\]

---

**Final Results**

\[
\boxed{
\begin{aligned}
\text{Hole radius} &\approx 1.5~\text{cm} \\
\text{Diaphragm spacing} &\approx 0.93~\text{cm}
\end{aligned}
}
\]

Loading solution...

Problem #72