Exercise 11.1 Transmit power

a)

An answer:
From an electrical point of view, a resistor is the only passive component that converts electrical energy (or power) into energy (or power) in another domain (heat, light, etc.).

$$\begin{array}{rcll}& P_{\mathit{diss}}& =i_{\mathit{eff}}^2{R}_a& \\ & & =\frac{v_{\mathit{eff}}^2}{{\left|Z_{\mathit{load}}\right|}^2}{R}_a& \\ & & =\frac{v_{\mathit{eff}}^2}{R_a^2+X_a^2}{R}_a& \end{array}$$

b)

An answer:
As the antenna is specified to be a monopole, it is basically a conductor with a specific length, connected on one end. An antenna is tuned to the transmit/receive frequency, hence the ”quarter-wave”. At that frequency the antenna has the specified impedance. At $f=\omega =0$ the antenna (monopole or dipole) obeys KCL and KVL and hence is an open: $Z\to \infty \mathrm{\Omega }$ @ 0 Hz.

c)

An answer:

$$\begin{array}{llll}\hfill P_{\mathit{transmit}}& =\Re \left(i_{\mathit{antenna}}\cdot v_{\mathit{antenna}}\right)& \hfill & \\ \hfill & =i_{\mathit{eff}}^2\cdot \Re \left(Z_{\mathit{antenna}}\right)& \hfill & \\ \hfill & =\frac{v_{\mathit{eff}}^2}{{\left|Z_{\mathit{antenna}}\right|}^2}\cdot \Re \left(Z_{\mathit{antenna}}\right)& \hfill & \end{array}$$

d)

An answer:

$$\begin{array}{llll}\hfill P_{10+j1000\mathrm{\Omega }}& =\frac{{\widehat{V}}^2}{2\cdot 10^6}\cdot 10=\frac{1}{2}{\widehat{V}}^2\cdot 10^{-5}[\mathrm{\text{W}}]& \hfill & \\ \hfill 1)P_{10\mathrm{\Omega }}& =\frac{{\widehat{V}}^2}{2\cdot 10^2}\cdot 10=\frac{1}{2}{\widehat{V}}^2\cdot 10^{-1}[\mathrm{\text{W}}]\mathrm{\text{ratio }}10^{-4}\mathrm{\text{ or }}10^4& \hfill & \\ \hfill 2)P_{50\mathrm{\Omega }}& =\frac{{\widehat{V}}^2}{2\cdot 50^2}\cdot 50=\frac{1}{2}{\widehat{V}}^2\cdot 2\cdot 10^{-2}[\mathrm{\text{W}}]\mathrm{\text{ratio }}5\cdot 10^{-4}\mathrm{\text{ or }}2\cdot 10^3& \hfill & \\ \hfill & & \hfill \end{array}$$