Sebuah gelombang bidang seragam berfrekuensi 20 GHz merambat ke arah $+z$ (forward $z$ direction) dalam medium lossless dengan karakteristik $\epsilon_r = \mu_r = 3$. Carilah (a) Kecepatan fase ($v_p$), (b) Konstanta fase ($\beta$), (c) Panjang gelombang ($\lambda$), (d) Bentuk phasor medan listrik ($\mathbf{E}_s$), (e) Bentuk phasor medan magnet ($\mathbf{H}_s$), (f) Rata-rata vektor Poynting ($\langle \mathbf{S} \rangle$).
(a) Kecepatan Fase ($v_p$)
$$v_p = \frac{c}{\sqrt{\mu_r \epsilon_r}}$$
$$v_p = \frac{3 \times 10^8}{\sqrt{3 \cdot 3}}$$
$$v_p = \frac{3 \times 10^8}{3}$$
$$\mathbf{v_p = 10^8 \text{ m/s}}$$
(b) Konstanta Fase ($\beta$)
$$\beta = \frac{\omega}{v_p}$$
$$\beta = \frac{40\pi \times 10^9}{10^8}$$
$$\beta = \mathbf{40 \cdot 10 \cdot 3,14}$$
$$\mathbf{\beta = 1256 \text{ rad/m}}$$
(c) Panjang Gelombang ($\lambda$)
$$\lambda = \frac{v_p}{f}$$
$$\lambda = \frac{10^8}{20 \times 10^9}$$
$$\mathbf{\lambda = \mathbf{0.005 \text{ m}}}$$
(d) Phasor Medan Listrik ($\mathbf{E}_s$)
Karena merambat ke $+z$ dan terpolarisasi-$x$ dengan amplitudo $E_0$:
$$E_s = E_0 e^{-j\beta z} \text{a}_x$$
$$\mathbf{E_s = E_0 e^{-j400\pi z} \mathbf{a}_x \text{ V/m}}$$
(e) Phasor Medan Magnet ($\mathbf{H}_s$)
Kita cari impedansi intrinsik ($\eta$):
$$\eta = \sqrt{\frac{\mu}{\epsilon}}$$
$$\eta = \eta_0 \sqrt{\frac{\mu_r}{\epsilon_r}}$$
$$\eta = 120\pi \sqrt{\frac{3}{3}}$$
$$\eta = 120 \cdot 3,14 \cdot \sqrt{1}$$
$$\mathbf{\eta = 377 \ \Omega}$$
Medan magnet tegak lurus terhadap $\mathbf{E}$ dan arah rambat ($\mathbf{a}_H = \mathbf{a}_k \times \mathbf{a}_E = \mathbf{a}_z \times \mathbf{a}_x = \mathbf{a}_y$):
$$H_s = \frac{E_0}{\eta} e^{-j\beta z} \text{a}_y$$
$$\mathbf{H_s = \frac{E_0}{120\pi} e^{-j400\pi z} \mathbf{a}_y \text{ A/m}}$$
(f) Rata-rata Vektor Poynting ($\langle \mathbf{S} \rangle$)
Kita cari konjugat dari $H_s$
$$H_s^* = \frac{E_0}{120\pi} e^{j400\pi z} \mathbf{a}_y$$
Untuk medium lossless, rumusnya adalah:
$$\langle \text{S} \rangle = \frac{1}{2} \text{Re} [\mathbf{E}_s \times \mathbf{H}_s^*]$$
$$\langle \text{S} \rangle = \frac{1}{2} \text{Re} [E_0 e^{-j400\pi z} \mathbf{a}_x \times \frac{E_0}{120\pi} e^{j400\pi z} \mathbf{a}_y]$$
$$\langle \text{S} \rangle = \frac{1}{2} \text{Re} \left[\left(\frac{E_0 \times E_0}{120\pi}\right) \cdot (e^{-j400\pi z} \times e^{j400\pi z}) \cdot (\text{a}_x \times \text{a}_y) \right]$$
$$\langle \text{S} \rangle = \frac{1}{2} \text{Re} \left[\left(\frac{E_0^2}{120\pi}\right) \mathbf{a}_z \right]$$
$$\langle \text{S} \rangle = \frac{1}{2} \times \left(\frac{E_0^2}{120\pi}\right) \mathbf{a}_z$$
$$\mathbf{\langle \text{S} \rangle = \frac{E_0^2}{240\pi} \mathbf{a}_z \text{ W/m}}$$
