Merge pull request #797 from defrenned3238/patch-2

Correction erreur dans ex2

src/q7/radiation-ELEC2795/exam/2019/Janvier/All/radiation-ELEC2795-exam-2019-Janvier-All.tex

@@ -90,13 +90,13 @@ A wireless horizontal link is established over a 100 meter distance at \SI{2.4}{
   \item $ P_T = \SI{-30}{dB} $
   \item The impedance matching factor is 
   \[
-   \abs{1-\Gamma}^2 = \abs{1-\frac{Z_L-Z_c}{Z_L+Z_c}}^2 = \abs{1-\frac{50 - 20 j}{150 - 20 j}}^2 = 0.437.
+   1-\abs{\Gamma}^2 = 1-\abs{\frac{Z_L-Z_c}{Z_L+Z_c}}^2 = 1-\abs{\frac{50 - 20 j}{150 - 20 j}}^2 = 0.873.
   \]
   The transmitter polarization is $ \Vect{i} = \frac{1}{\sqrt{2}} \left( 1 + \e^{j\pi\sin(\SI{20}{\degree}}\right) \Base{x} $ and the receiver polarization is $ \Base{p} = \frac{1}{\sqrt{2}} \left( \Base{x} + j \Base{y} \right)$. The polarization matching factor is
   \[
    \abs{\mathbf{\Vect{i}\cdot\Base{p}}}^2 = \abs{\frac{1}{\sqrt{2}} \left( 1 + \e^{j\pi\sin(\SI{20}{\degree}}\right)\frac{1}{\sqrt{2}}}^2 = 0.74.
   \]
-  $ L_R = \abs{1-\Gamma}^2\abs{\mathbf{\Base{i}\cdot\Base{p}}}^2 = 0.323 = \SI{-4.89}{dB}$.
+  $ L_R = (1-\abs{\Gamma}^2)\abs{\mathbf{\Base{i}\cdot\Base{p}}}^2 = 0.646 = \SI{-1.90}{dB}$.
   \item $L_T = 1$.
   \item Since $\theta_{\SI{3}{dB}} = \frac{\SI{160}{\degree}}{\sqrt{G_R(u_0)}}$, $G_R(u_0) = \left(\frac{\SI{160}{\degree}}{25}\right)^2 = \SI{16.12}{dB}$. The gain of 10 dB has to be added: $G_R = \SI{26.12}{dB}$.
   \item $G_T = \SI{2.15}{dB}$ for a dipole.
@@ -105,12 +105,12 @@ A wireless horizontal link is established over a 100 meter distance at \SI{2.4}{
 
  The received power is thus
  \[
-  P_R = -30 - 4.89 + 26.12 + 2.15 - 80.05 = \SI{-86.67}{dB}.
+  P_R = -30 - 1.9 + 26.12 + 2.15 - 80.05 = \SI{-83.67}{dB}.
  \]
  The noise power is $N = kBT = \SI{-130.97}{dB}$.
  The output SNR is
  \[
-  \mathrm{SNR}_o = \mathrm{SNR}_i / F = -86.67 + 130.97 -4 = \SI{40.3}{dB}.
+  \mathrm{SNR}_o = \mathrm{SNR}_i / F = -83.67 + 130.97 -4 = \SI{43.3}{dB}.
  \]
  \item The angle of propagation for the reflected wave is $\theta = \arctan(20/50) = \SI{21.8}{\degree}$. It will create a wave being the sum of the waves from the two dipole. Taking into account the initial phase difference from the dipoles, the total phase shift is $ \pi (\sin(\SI{20}{\degree}) - \sin(\SI{21.8}{\degree}))$. The reflected wave is thus proportional to
  \[