Ch13 Antennas
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In order to have EM wave, we need **Time Varying Source**:
* **加速電荷** or **AC current**

**Antenna**: 承載這些Source的就是天線
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分析上使用球面座標：$E(r, \theta, \phi, t) , H(r, \theta, \phi, t)$
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# Hertzian Dipole antenna
由電流匯入的位置，分成兩種

Monopole
: 從天線的**末端**輸入電流

Dipole
: 從天線的**中間**輸入電流
> 為什麼會有Dipole？ 實際上是輸入了cos(wt)交流電，等效成了一個Dipole
> 求Ｅ時，看成electric dipole
> 求Ｈ時，看成current element
### 步驟一（求Ａ）
因為$H={1\over\mu}\nabla \times{\overrightarrow{A}}$ (curl A除以$\mu$)
* 要求磁場，就要先求$\overrightarrow{A}$(vector potential)
考慮到波傳遞的時間，修正後的Retarded Vector Potential $\overrightarrow{A}(r,t)$:$${\mu I(t-r/u)dl\over4\pi r(t-r/u)}a_z$$
> Retarded Current:
$$I_0 cos(\omega t-\beta r)$$ (phasor: $Re[I_0 e^{j(\omega t-\beta r)}]$ )

* 若為**Stationary antenna**:$$A_s = {\mu I_0dl\over4\pi r}e^{-j\beta r}(cos\theta a_r-sin\theta a_{\theta})$$

### 步驟二（求Ｈ）：
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磁場只有$\phi$分量 $H_{\phi s}$
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### 步驟三（求Ｅ）：
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電場有$r, \theta$分量 $E_{rs}, E_{\theta s}$
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## Far Field:
條件：當$r\ge {e^{-j\beta r}\over r}$ , d為天線最長的長度

### Far Field Characteristics:
* $E_{\theta s}$ and $H_{\phi s}$ ⇨ **In phase** , **orthgonal**. (像平面波)
* 單位面積之**Time-average power density**:$$P_{ave}={\eta\beta^2I_0^2dl^2\over 32\pi^2r^2 }sin^2\theta$$
* 整個球面之**Time-average rediated power**: $$P_{rad}=\int P_{ave}\cdot dS$$
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化簡過程：$$=\int_{\phi =0}^{2\pi}\int_{\theta =0}^{\pi}{\eta\beta^2I_0^2dl^2\over 32\pi^2r^2 }sin^2\theta\cdot r^2sin\theta \ d\theta d\phi$$ $$={\eta\beta^2I_0^2dl^2\over 16\pi}\cdot {4\over 3} $$
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In free space: $P_{rad}=40\pi^2I_0^2[{dl\over\lambda}]^2$
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* 等效輻射Ｒ：$R_{rad}={2P_{rad}\over I_0^2}$
## Antenna Characteristic
* ### Antenna & Radiation Pattern
Pattern就是把far field的**E,H**圖像化
* **Hertzian Dipole**的Radiation pattern: $H_{\phi s}, E_{\theta s}$ 皆約等於 $sin\theta$
Normalized pattern function:$$f(\theta)=|sin\theta|$$
* **General** Radiation pattern:
Normalized pattern function:$$f(\theta ,\phi)$$
通常只討論兩個特定平面
* E-plane: $f(\theta,0)$
* H-plane: $f({\pi \over 2}, \phi)$
* **Power pattern**:$$f^2(\theta)=sin^2\theta$$

* ### Radiation Intensity $U(\theta, \phi)$
定義$U(\theta, \phi)=r^2 P_{ave} \propto f^2(\theta, \phi)$
在求整顆球power($P_{rad}$)時，可寫成：$$\int_{\phi =0}^{2\pi}\int_{\theta =0}^{\pi}U(\theta,\phi) d\Omega$$
==**Average radiation intensity**: $U_{ave}={P_{ave} \over 4\pi}$==

* ### Directive Gain($G_d$) & Directivity(D)
Directive Gain: 該天線在每個角度的Gain為多少$$G_d(\theta,\phi)={U(\theta,\phi)\over U_{ave}} = {4\pi f^2(\theta,\phi)\over \int f^2(\theta, \phi)d\Omega} = ({P_{ave}(\theta,\phi)\over P_{rad}})4\pi r^2$$

Directivity:天線Gain最強的方向$$D={U_{max}\over U_{ave}} = {4\pi U_{max}\over P_{rad}}$$
Power Gain:$$G_p(\theta,\phi) = \eta_r G_d(\theta,\phi)$$
$\eta_r$:Radiation Efficiency$$\eta_r = {P_{rad}\over P_{in}}$$
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轉dB⇨$10log_{10}$(D or G)
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* ### Beam solid angle($\Omega_A$)
將radiation pattern等效成一圓錐，得到的**立體角**：
$$\Omega_A = {4\pi\over D}$$

# Pratical Antenna
記兩個場：$$H_{\phi s} = {jI_0e^{-j\beta r}cos({\pi \over 2}cos\theta)\over 2\pi rsin\theta}$$
## Half-Wave Dipole Antennas
解法：
1. 從$H_0$推$I_0$
2. $R_{rad}(_{\lambda\over 2})$固定為73$\Omega$
3. 利用以下公式得$P_{rad}$ $$P_{rad} = {1\over 2}I_0^2 R_{rad}$$

## Quarter-Wave Monopole Antenna
基本和半波天線一樣，只是只取一半。
因此解法上，只要注意$R_{rad} = 36.5\Omega$
Power也就要$* {1\over 2}$

## Small-loop Antenna
* ### Duality:
Hertzian -> Loop
$dl a_\phi$ -> $j\beta S a_\theta = j{2\pi S\over \lambda}a_\theta$
$$S(面積) = N\pi\rho_0^2,\ \rho_0 = 半徑, N=圈數$$
# Antenna Arrays

# Effective Area and the Friis Equation
* ### Effective Area: $${A_E \over D} = {\lambda^2\over 4\pi}$$

* ### Friis Transmission Formula:$${P_r\over P_t} = G_{dt}G_{dr}[{\lambda\over4\pi r}]^2$$
# Radar Equation

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# 解題要背的公式：
### $R_{rad}$
* Hertzian:$$R_{rad} = 80\pi^2({dl\over\lambda})^2$$
* Half-wave: $$R_{rad} = 73\ \Omega$$
* Quarter-wave: $$R_{rad} = 36.5\ \Omega$$
### Far field:
$H_s=H_0{e^{-j\beta r}\over r}a_H$,
* Hertzian :$$|H_s| = H_0 = {j\beta I_0 dl sin\theta\over4\pi} \ \ , dl = Hertzian dipole的長度$$

* Half-wave, Quarter-wave : $$H_0 = {jI_0 cos({\pi \over 2}cos\theta)\over2\pi sin\theta}$$
* Loop :
$$S(面積) = N\pi\rho_0^2,\ \rho_0 = 半徑, N=圈數$$
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