Provide answer of each part and provide explanation also2. Consider the mode shown in Figure 1,
yb
FEE A EE = an
ow or x dA x x x fA x 4 ox
AX + x fd x A ox hf dk ox box Some Relevant Information:

Question

Provide answer of each part and provide explanation also2. Consider the mode shown in Figure 1,
yb
FEE A EE = an
ow or x dA x x x fA x 4 ox
AX + x fd x A ox hf dk ox box Some Relevant Information:
x fd dox or dx oe ow fron
EE EE TS CE Nt Since the TE modes have I, — 0, we onside the s-component ofthe magnetic fed as
Tox + x +d x ox Ak oxo xf Hy = He) = XY (es
VR Weuse., uy 71 spchcaly ote the venue ope wove popsgtn i te
# + 4+ x f Unbound mem. 1: mst ty he Hemhotz equation
hi [Fa
Figure 1: Electric fields are purple. Magnetic fields are red. & N = N - +h 0! XY Ge =0
7 Exe Er a]
_— a ED x same
LENE LE)
(2) The mode indices and the type of mode being shown (TE or TM) Justify. Xo a tv ap tO
(b) The expression forthe longitudinal field component of this mode. Sepang he tems is
(¢) The expressions for the transverse field components (E,, E,, H, and H,). Link these to (a). exe, Lave,
(d) The expression fo the Poynting vector for this mode. Xa a + oa te
J ———
Fa
" “ian on inom, a4
en 2 ] pln vim] = 5 A
sx@a] =o, yar] = yp]  Be) = hep) + Bn) Ba) = [Aos,0) + Bints)
Likewise for the TM modes. [Ceon(Ag) + Danial] x[Cen(8) + Dein(3u)]
Puy = gE AY) = 75 (EH; BH) = 73 (H

VIPSolutions · Accepted Answer

Let's answer each part step-by-step, using the figures and the reference sheet provided.

---

## (a) The mode indices and the type of mode being shown (TE or TM). Justify.

### **Explanation:**
- **Color code:** The electric fields are shown in **purple**, and the magnetic fields are in **red**.
- **Field orientation:** The pattern shows that the electric field vectors have only **transverse (in-plane)** components (no longitudinal component visible). This suggests that $ E_z = 0 $.
- **TE vs TM:** For **TE (Transverse Electric) modes**, $ E_z = 0 $ and $ H_z \neq 0 $. For **TM (Transverse Magnetic) modes**, $ H_z = 0 $ and $ E_z \neq 0 $.

### **Mode indices:**
- The field pattern shows **one half-wave variation along the $ x $-axis** and **one half-wave along the $ y $-axis**.
- This corresponds to indices $ m = 1 $, $ n = 1 $.

### **Answer:**
- **Mode:** TE$_{11}$
- **Justification:** The electric field has no longitudinal component ($ E_z = 0 $), and the field varies as one half-wave in both $ x $ and $ y $ directions.

---

## (b) The expression for the longitudinal field component of this mode.

### **Explanation:**
- For TE$_{mn}$ modes, the only longitudinal component is $ H_z $.
- From the reference ("TE Modes - III" and "Problem 1 - Compute the Fields"):

$$
H_z(x, y) = H_0 \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{-j\beta z}
$$

For $ m = n = 1 $:

$$
H_z(x, y) = H_0 \sin\left(\frac{\pi x}{a}\right) \sin\left(\frac{\pi y}{b}\right) e^{-j\beta z}
$$

### **Answer:**
$$
H_z(x, y) = H_0 \sin\left(\frac{\pi x}{a}\right) \sin\left(\frac{\pi y}{b}\right) e^{-j\beta z}
$$

---

## (c) The expressions for the transverse field components ($E_x, E_y, H_x, H_y$). Link these to (a).

### **Explanation:**
- Use the expressions for TE$_{mn}$ modes (from reference sheet):

$$
\begin{align*}
E_x(x, y) &= -j \frac{\omega \mu}{k_c^2} \frac{\partial H_z}{\partial x} \
E_y(x, y) &= -j \frac{\omega \mu}{k_c^2} \frac{\partial H_z}{\partial y} \
H_x(x, y) &= -j \frac{\beta}{k_c^2} \frac{\partial H_z}{\partial x} \
H_y(x, y) &= -j \frac{\beta}{k_c^2} \frac{\partial H_z}{\partial y}
\end{align*}
$$
Where $ k_c^2 = \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 $.

For $ m = n = 1 $:

$$
\begin{align*}
E_x(x, y) &= -j \frac{\omega \mu}{k_c^2} H_0 \cos\left(\frac{\pi x}{a}\right) \sin\left(\frac{\pi y}{b}\right) \left(\frac{\pi}{a}\right) e^{-j\beta z} \
E_y(x, y) &= -j \frac{\omega \mu}{k_c^2} H_0 \sin\left(\frac{\pi x}{a}\right) \cos\left(\frac{\pi y}{b}\right) \left(\frac{\pi}{b}\right) e^{-j\beta z} \
H_x(x, y) &= -j \frac{\beta}{k_c^2} H_0 \cos\left(\frac{\pi x}{a}\right) \sin\left(\frac{\pi y}{b}\right) \left(\frac{\pi}{a}\right) e^{-j\beta z} \
H_y(x, y) &= -j \frac{\beta}{k_c^2} H_0 \sin\left(\frac{\pi x}{a}\right) \cos\left(\frac{\pi y}{b}\right) \left(\frac{\pi}{b}\right) e^{-j\beta z}
\end{align*}
$$

### **Link to (a):**
- These field configurations match the TE$_{11}$ mode, where both $ m $ and $ n $ are 1.

---

## (d) The expression for the Poynting vector for this mode.

### **Explanation:**
- The Poynting vector (average power) for TE modes is:

$$
P_{avg, TE} = \frac{1}{2} \text{Re} \left\{ E_x H_y^* - E_y H_x^* \right\}
$$
- Or integrated over the guide cross-section:

$$
\langle S_z \rangle = \frac{|E_{z}|^2}{2 Z_{TE}}
$$
But for TE modes, $ E_z = 0 $, so use:

$$
\langle S_z \rangle = \frac{|H_{z}|^2 Z_{TE}}{2}
$$

Where $ Z_{TE} = \frac{\omega \mu}{\beta} $.

### **Answer:**
$$
\langle S_z \rangle = \frac{|H_{z}|^2 Z_{TE}}{2}
$$
$$
Z_{TE} = \frac{\omega \mu}{\beta}
$$
So for your specific mode:
$$
\langle S_z \rangle = \frac{|H_0|^2 \sin^2\left(\frac{\pi x}{a}\right) \sin^2\left(\frac{\pi y}{b}\right) Z_{TE}}{2}
$$

---

### **Summary Table**

| Part | Answer |
|------|--------|
| (a) | Mode: TE$_{11}$, because $ E_z = 0 $ and field pattern shows one half-wave in both $ x $, $ y $. |
| (b) | $ H_z(x, y) = H_0 \sin\left(\frac{\pi x}{a}\right) \sin\left(\frac{\pi y}{b}\right) e^{-j\beta z} $ |
| (c) | $ E_x, E_y, H_x, H_y $ as derived above (see equations). |
| (d) | $ \langle S_z \rangle = \frac{|H_{z}|^2 Z_{TE}}{2} $, $ Z_{TE} = \frac{\omega \mu}{\beta} $ |

If you need further derivations or clarifications on any part, let me know!

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Study Help, Fast Answers

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Related Answers

Study Help, Fast Answers