Question

Zez 10082 Ri=5022 M V-IV ZR=10012 NAT Zei=502 Given the amplitude of a stop-Fundian wou vtiv, find the reflected voltage varv, transmitter! voltage Vt-TV", incident power pois reflected power po transmitted. Power pt, and powers Pi and Pz dissipated in R, and R2

Answer 1

Load impedance as seen from C1

ZI

= 50 $\small \Omega$ + 100 $\small \Omega$ || 100 $\small \Omega$

= 50 Ω + 100 * 100 100 + 100 Ω

100 και 100 = 50 Ω + Ω 200

50 Ω + 50 Ω

$\small = 100\: \Omega$

Therefore, reflection coefficient

$\small \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \Gamma = \frac{Z_L - Z_{c1}}{Z_L + Z_{c1}}\\\\ = \frac{100 - 50}{100 + 50}\\\\ = \frac{50}{150}\\\\ =\frac{1}{3}$

$\small \Rightarrow$ Reflected voltage $\small V^-$

+ 11 =

$\small =\frac{1}{3}*1\: \text{V}$

$\small =\frac{1}{3}\: \text{V}$

$\small \approx 0.33\: \text{V}$

Transmitted voltage $\small V^t$

= Incident voltage - reflected voltage

= 1 V - 0.33 V

= 0.67 V

Incident power $\small P^{+}$

$\small = \frac{(V^{+})^2}{Z_{c1}}$

$\small = \frac{(1\: V)^2}{50\: \Omega}$

= 0.02 W

= 20 mW

Reflected power $\small P^{-}$

$\small = \frac{(V^{-})^2}{Z_{c1}}$

$\small = \frac{(1/3\: V)^2}{50\: \Omega}$

= 0.00222 W

= 2.22 mW

Transmitted power $\small P^{t}$

$\small = \frac{(V^{t})^2}{Z_{c2}}$

$\small = \frac{(2/3\: V)^2}{100\: \Omega}$

= 0.00444 W

= 4.44 mW

Total power dissipated across $\small R_1$ and $\small R_2$

= 20 mW - (2.22 + 4.44) mW

= 20 mW - 6.66 mW

= 13.34 mW