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diff --git a/exercise03/exercise03.tex b/exercise03/exercise03.tex new file mode 100644 index 0000000..be192ce --- /dev/null +++ b/exercise03/exercise03.tex @@ -0,0 +1,521 @@ +% SPDX-License-Identifier: CC-BY-SA-4.0 +% +% Copyright (c) 2020 Philipp Le +% +% Except where otherwise noted, this work is licensed under a +% Creative Commons Attribution-ShareAlike 4.0 License. +% +% Please find the full copy of the licence at: +% https://creativecommons.org/licenses/by-sa/4.0/legalcode + +\phantomsection +\addcontentsline{toc}{section}{Exercise 3} +\section*{Exercise 3} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{question}[subtitle={Stochastic Process}] + A normally distributed random process produces the sequences $x_1(t)$, $x_2(t)$ and $x_3(t)$. + \begin{table}[H] + \centering + \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|} + \hline + $t$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ + \hline + \hline + $x_1(t)$ & 4.99 & 4.37 & 8.57 & 4.01 & 3.77 & 3.35 & 3.87 & 8.39 & 6.89 & 1.96 \\ + \hline + $x_2(t)$ & 3.95 & 5.35 & 2.94 & 6.38 & 4.78 & 7.62 & 5.25 & 6.81 & 5.65 & 5.29 \\ + \hline + $x_3(t)$ & 7.01 & 4.40 & 4.26 & 6.54 & 4.53 & 6.85 & 4.46 & 5.81 & 6.49 & 4.11 \\ + \hline + \end{tabular} + \end{table} + \begin{tasks} + \task + Calculate the stochastic mean for each time instance! + + \task + Calculate the temporal mean for each sequence! + + \task + The process is ergodic with $\mu_x = 5.00$. However, why is the condition $\E\left\{\vect{x}\right\} = \overline{x} = \mu_x$ not fulfilled? + \end{tasks} +\end{question} + +\begin{solution} + \begin{tasks} + \task + \begin{table}[H] + \centering + \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|} + \hline + $t$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ + \hline + \hline + $\E\left\{\vect{x}(t)\right\}$ & 5.32 & 4.71 & 5.26 & 5.64 & 4.36 & 5.94 & 4.53 & 7.00 & 6.34 & 3.79 \\ + \hline + \end{tabular} + \end{table} + + \task + \begin{itemize} + \item $\overline{x_1} = 5.01$ + \item $\overline{x_1} = 5.40$ + \item $\overline{x_1} = 5.45$ + \end{itemize} + + \task + \begin{itemize} + \item There are only $N = 3$ sequences drawn from the random process. $\E\left\{\vect{x}\right\}$ will converge to $5.00$ for $N \rightarrow \infty$. $N = 3$ is too short. + \item Each sequence is only $L = 10$ samples long. $\overline{x}$ will converge to $5.00$ for $L \rightarrow \infty$. $L = 10$ is too short. + \end{itemize} + \end{tasks} +\end{solution} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{question}[subtitle={Cross-Correlation and Autocorrelation}] + Two signals are given $f_1(t)$ and $f_2(t)$. The signals are value- and time-continuous. Ten samples are given. Both signals are zero for $t < -5$ and $t > 5$. + \begin{table}[H] + \centering + \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|r|} + \hline + $t$ & -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ + \hline + \hline + $f_1(t)$ & 0 & 1 & 2 & 3 & 4 & 5 & 4 & 3 & 2 & 1 & 0 \\ + $f_2(t)$ & -1.34 & 0.30 & 14.54 & -1.54 & -3.03 & -1.72 & 14.16 & 2.70 & 1.17 & -2.44 & -4.66 \\ + \hline + \end{tabular} + \end{table} + \begin{tasks} + \task + Calculate the cross-correlation + \begin{equation*} + \mathrm{R}_{f_1 f_2}(\tau) \approx \left(f_1 \star f_2\right)(\tau) = \sum\limits_{t=-5}^{5} f_1(t) f_2(t + \tau) + \end{equation*} + for $t = -2$, $t = -1$ and $t = 0$. + + \task + Signal $f_2(t)$ contains signal $f_1(t)$ which is superimposed by noise. Using the values calculated in a), how much is the time $\Delta t$ lag between $f_1(t)$ and $f_2(t)$? + + \task + Calculate the autocorrelation + \begin{equation*} + \mathrm{R}_{f_1 f_1}(\tau) \approx \left(f_1 \star f_1\right)(\tau) = \sum\limits_{t=-5}^{5} f_1(t) f_1(t + \tau) + \end{equation*} + for $t = -2$, $t = -1$, $t = 0$, $t = 1$ and $t = 2$. + + \task + How much is the signal energy $E$ of $f_1(t)$? + \begin{equation*} + E \approx \frac{1}{T} \sum\limits_{t=-5}^{5} \left|f_1(t)\right|^2 + \end{equation*} + with $T = 11$? + \end{tasks} + + \textit{Remark:} Only samples of the functions with a spacing of $1$ are given. Therefore, the indefinite integrals of both cross-correlation and autocorrelation can be approximated using the sums. +\end{question} + +\begin{solution} + \begin{tasks} + \task + Full cross-correlation between $-10 \geq t \geq 10$. The cross-correlation is zero everywhere outside of the interval. + \begin{table}[H] + \centering + \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|} + \hline + $\tau$ & -10 & -9 & -8 & -7 & -6 & -5 & -4 & -3 & -2 \\ + \hline + $\mathrm{R}_{f_1 f_2}(\tau)$ & 0.0 & -4.66 & -11.76 & -17.69 & -20.92 & -9.99 & 8.54 & 28.92 & 45.42 \\ + \hline + \hline + $\tau$ & -1 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ + \hline + $\mathrm{R}_{f_1 f_2}(\tau)$ & 71.06 & 68.68 & 63.74 & 62.42 & 65.35 & 41.9 & 32.01 & 23.08 & 11.12 \\ + \hline + \hline + $\tau$ & 8 & 9 & 10 & & & & & & \\ + \hline + $\mathrm{R}_{f_1 f_2}(\tau)$ & -2.38 & -1.34 & 0.0 & & & & & & \\ + \hline + \end{tabular} + \end{table} + \begin{itemize} + \item $\left(f_1 \star f_2\right)(-2) = 45.42$ + \item $\left(f_1 \star f_2\right)(-1) = 71.06$ + \item $\left(f_1 \star f_2\right)(0) = 68.68$ + \end{itemize} + + \task + \begin{itemize} + \item The maximum value is at $\tau = -1$. + \item $f_2(t)$ is advanced by $\Delta t = 1$ in relation to $f_1(t)$. + \end{itemize} + + \task + Full autocorrelation between $-10 \geq t \geq 10$. The autocorrelation is zero everywhere outside of the interval. + \begin{table}[H] + \centering + \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|r|} + \hline + $\tau$ & -10 & -9 & -8 & -7 & -6 & -5 & -4 & -3 & -2 & -1 & 0 \\ + \hline + $\mathrm{R}_{f_1 f_1}(\tau)$ & 0 & 0 & 1 & 4 & 10 & 20 & 3. & 52 & 68 & 80 & 85 \\ + \hline + \hline + $\tau$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \\ + \hline + $\mathrm{R}_{f_1 f_1}(\tau)$ & 80 & 68 & 52 & 35 & 20 & 10 & 4 & 1 & 0 & 0 & \\ + \hline + \end{tabular} + \end{table} + Only three values must be calculated: + \begin{itemize} + \item $\mathrm{R}_{f_1 f_1}(\tau)(-2) = 68$ + \item $\mathrm{R}_{f_1 f_1}(\tau)(-1) = 80$ + \item $\mathrm{R}_{f_1 f_1}(\tau)(0) = 85$ + \end{itemize} + The other two values can be deducted from the symmetry rules: + \begin{itemize} + \item $\mathrm{R}_{f_1 f_1}(\tau)(1) = \mathrm{R}_{f_1 f_1}(\tau)(-1) = 80$ + \item $\mathrm{R}_{f_1 f_1}(\tau)(2) = \mathrm{R}_{f_1 f_1}(\tau)(-2) = 68$ + \end{itemize} + + \task + \begin{itemize} + \item It is not necessary to solve the sum again. + \item The sum is $\mathrm{R}_{f_1 f_1}(\tau)(0)$. + \item The energy is + \begin{equation*} + E = \frac{1}{11} \mathrm{R}_{f_1 f_1}(\tau)(0) = 7.23 + \end{equation*} + \end{itemize} + \end{tasks} +\end{solution} + + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{question}[subtitle={Power Spectral Density}] + Are the following statements about the power spectral density (PSD) true or false? If false, correct the statement! + \begin{tasks} + \task + The PSD is measure for the fraction of power which is located at a certain frequency. + + \task + The unit of the PSD is the same as the signal. + + \task + The PSD is even, if the signal is purely real-valued. + + \task + The PSD is odd, if the signal is complex-valued. + + \task + The PSD is always real-valued. + + \task + The PSD can be used to determine the output signal of an LTI system in amplitude and phase. + + \end{tasks} +\end{question} + +\begin{solution} + \begin{tasks} + \task + True. The signal power can be obtained by integration of the PSD over all frequencies. + \begin{equation*} + P = \int\limits_{-\infty}^{\infty} \mathrm{S}_{xx}(\omega) \; \mathrm{d} \omega + \end{equation*} + + \task + False. + \begin{equation*} + \left(\text{Unit of PSD}\right) = \frac{\left(\text{Unit of signal}\right)^2}{\left(\text{Unit of frequency}\right)} + \end{equation*} + + For example \si{W/Hz}. Never \si{W} only, because it is the density of the power across the frequency. + + \task + True + + \task + False. There are no symmetry rules for the PSD, if the signal is complex-valued. + + \task + True. + \begin{itemize} + \item The autocorrelation function is always Hermitian. + \item The Fourier transform of a Hermitian function is always real-valued. + \item Therefore, the PSD is real-valued. + \end{itemize} + + \task + False. + \begin{itemize} + \item Only the amplitude can be calculated. + \item The phase information of the system is lost by squaring the absolute value of the transfer funtion. + \item Furthermore, the PSD does not contain phase information of the signal. + \end{itemize} + \begin{equation*} + \mathrm{S}_{yy}(\omega) = \left|\underline{H}\left(j \omega\right)\right|^2 \cdot \mathrm{S}_{xx}(\omega) + \end{equation*} + \end{tasks} +\end{solution} + + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{question}[subtitle={Decibel}] + \begin{tasks} + \task + Convert to logarithmic scale or vice versa! + \begin{enumerate} + \item \SI{2}{mW}, $P_0 = \SI{1}{mW}$ + \item \SI{2}{V}, $U_0 = \SI{1}{V}$ + \item \num{400000} + \item \SI{5}{GHz}, $f_0 = \SI{1}{Hz}$ + \item \SI{8}{fW/MHz}, $P_0 = \SI{1}{mW}$ + \item \SI{5}{dB\micro\volt} + \item \SI{-10}{dBm} + \end{enumerate} + + \task + Convert to logarithmic scale or vice versa! + \begin{enumerate} + \item $1$ + \item $2$ + \item $4$ + \item $0.5$ + \item $0.25$ + \item $10$ + \item $100$ + \item $0.1$ + \item $0.01$ + \item $500$ + \item $0.04$ + \end{enumerate} + + \task + A signal is transmitted with a power of \SI{13}{dBm}. The signal is attenuated by \SI{86}{dB} on its way to the receiver. Assume that a ohmic resistance of \SI{50}{\ohm} is connected to the antenna. What voltage can be measured at this resistance? Give your result in linear and logarithmic scale! + \end{tasks} +\end{question} + +\begin{solution} + \begin{tasks} + \task + \begin{enumerate} + \item $\SI{10}{dBm} \cdot \log_{10} \left(\frac{\SI{2}{mW}}{\SI{1}{mW}}\right) = \SI{3}{dBm}$ + \item $\SI{20}{dBV} \cdot \log_{10} \left(\frac{\SI{2}{V}}{\SI{1}{V}}\right) = \SI{6}{dBV}$ (20 instead of 10 for voltages and currents) + \item $\SI{10}{dB} \cdot \log_{10} \left(400000\right) = \SI{53}{dB}$ + \item $\SI{10}{dBHz} \cdot \log_{10} \left(\frac{\SI{5}{GHz}}{\SI{1}{Hz}}\right) = \SI{97}{dbHz}$ + \item $\SI{10}{dBm/Hz} \cdot \log_{10} \left(\frac{\SI{8}{fW}}{\SI{1}{mW}}\right) = \SI{-111}{dB/Hz}$ + \item $\SI{1}{\micro\volt} \cdot 10^{\SI{5}{dB\micro\volt} / 20} = \SI{1.8}{\micro\volt}$ + \item $\SI{1}{mW} \cdot 10^{\SI{-10}{dBm} / 10} = \SI{100}{\micro.W}$ + \end{enumerate} + + \task + \begin{enumerate} + \item $\SI{0}{dB}$ + \item $\SI{3}{dB}$ + \item $\SI{6}{dB}$, Hint: $4 = 2 \cdot 2 \equiv \SI{3}{dB} + \SI{3}{dB} = \SI{6}{dB}$ + \item $\SI{-3}{dB}$ + \item $\SI{-6}{dB}$, Hint: $0.25 = 0.5 \cdot 0.5 \equiv \SI{-3}{dB} + \SI{-3}{dB} = \SI{-6}{dB}$ + \item $\SI{10}{dB}$ + \item $\SI{20}{dB}$ + \item $\SI{-10}{dB}$ + \item $\SI{-20}{dB}$ + \item $\SI{27}{dB}$, Hint: $50 = 10 \cdot 10 \cdot 10 \cdot 0.5 \equiv \SI{10}{dB} + \SI{10}{dB} + \SI{10}{dB} - \SI{3}{dB} = \SI{27}{dB}$ + \item $\SI{-14}{dB}$, Hint: $0.04 = 0.1 \cdot 0.1 \cdot 2 \cdot 2 \equiv \SI{-10}{dB} + \SI{-10}{dB} + \SI{3}{dB} + \SI{3}{dB} = \SI{-14}{dB}$ + \end{enumerate} + Remember these rules as an RF engineer! You will need them often. ;) + + \task + Power at the receiver antenna: + \begin{equation*} + \SI{13}{dBm} \underbrace{- \SI{86}{dB}}_{\text{Atennuation}} = \SI{-73}{dBm} + \end{equation*} + + \begin{itemize} + \item 1st way: Convert to linear scale, then back to logarithmic scale + \begin{equation*} + \begin{split} + \SI{-73}{dBm} &\equiv \SI{50.1}{pW} \\ + U = \sqrt{P \cdot R} &= \sqrt{\SI{50.1}{pW} \cdot \SI{50}{\ohm}} = \SI{50}{\micro\volt} \\ + \SI{50}{\micro\volt} &\equiv \SI{17}{dB\micro\volt} + \end{split} + \end{equation*} + \item 2nd way: Convert everything to logarithmic scale. The multiplication of power and resistance must happen with ``pure'' SI units. + \begin{equation*} + \begin{split} + \SI{50}{\ohm} &\equiv \SI{10}{dB\ohm} \cdot \log_{10} \left(\frac{\SI{50}{\ohm}}{\SI{1}{\ohm}}\right) = \SI{17}{dB\ohm} \\ + \SI{-73}{dBm} \underbrace{- \SI{30}{dB}}_{\text{\si{mW} to \si{W}}} &= \SI{-103}{dBW} \quad \text{SI unit!} \\ + \SI{-103}{dBW} + \SI{17}{dB\ohm} &= \SI{-86}{dBV} \quad \text{SI unit!} \\ + \SI{-86}{dBV} \underbrace{+ \SI{120}{dB}}_{\text{\si{V} to \si{\micro\volt}}} &= \SI{34}{dB\micro\volt} \\ + \SI{34}{dB\micro\volt} &\equiv \SI{50}{\micro\volt} + \end{split} + \end{equation*} + \end{itemize} + \end{tasks} +\end{solution} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{question}[subtitle={Noise}] + The following system is given: + \begin{figure}[H] + \centering + \begin{adjustbox}{scale=0.7} + \begin{tikzpicture} + \node[draw, block] (BPF){Band-pass filter\\ $L_{G,1} = \SI{-3}{dB}$\\ $L_{F,1} = \SI{3}{dB}$}; + \node[draw, block, right=of BPF] (Mix){Mixer\\ $L_{G,2} = \SI{3}{dB}$\\ $L_{F,2} = \SI{20}{dB}$}; + \node[draw, block, right=of Mix] (Demod){Demodulator\\ $L_{G,3} = \SI{6}{dB}$\\ $L_{F,3} = \SI{10}{dB}$}; + \node[draw, block, right=of Demod] (Amp){Amplifier\\ $L_{G,4} = \SI{50}{dB}$\\ $L_{F,4} = \SI{16}{dB}$}; + + \draw[o->] ([xshift=-1cm]BPF.west) node[left,align=right]{Input $u(t)$} -- (BPF.west); + \draw[->] (BPF.east) -- (Mix.west); + \draw[->] (Mix.east) -- (Demod.west); + \draw[->] (Demod.east) -- (Amp.west); + \draw[->] (Amp.east) -- ([xshift=1cm]Amp.east) node[right,align=left]{Output}; + \end{tikzpicture} + \end{adjustbox} + \end{figure} + \begin{itemize} + \item The band-pass filter has an input impedance of $\SI{50}{\ohm}$. + \item The band-pass filter has a bandwidth of $\SI{2}{MHz}$. + \item All system components are at room temperature: \SI{25}{\degreeCelsius}. + \end{itemize} + The input signal is: + \begin{equation*} + u(t) = \SI{70.71}{\micro\volt} \cdot \cos\left(\omega_0 t\right) + \end{equation*} + + Give all results in the logarithmic scale! + \begin{tasks} + \task + What is the signal power, the thermal noise power and the SNR at the input? + + \task + What is the overall noise figure and overall gain of the chain? What is the signal power, the thermal noise power and the SNR at the output of the chain? + + \task + Now a low noise amplifier is placed in front of the mixer. The gain of the last amplifier is reduced, so that the overall gain does not change. + \begin{figure}[H] + \centering + \begin{adjustbox}{scale=0.5} + \begin{tikzpicture} + \node[draw, block] (Amp){Low noise amplifier\\ $L_{G,0} = \SI{30}{dB}$\\ $L_{F,0} = \SI{6}{dB}$}; + \node[draw, block, right=of Amp] (BPF){Band-pass filter\\ $L_{G,1} = \SI{-3}{dB}$\\ $L_{F,1} = \SI{3}{dB}$}; + \node[draw, block, right=of BPF] (Mix){Mixer\\ $L_{G,2} = \SI{3}{dB}$\\ $L_{F,2} = \SI{20}{dB}$}; + \node[draw, block, right=of Mix] (Demod){Demodulator\\ $L_{G,3} = \SI{6}{dB}$\\ $L_{F,3} = \SI{10}{dB}$}; + \node[draw, block, right=of Demod] (Amp2){Amplifier\\ $L_{G,4} = \SI{20}{dB}$\\ $L_{F,4} = \SI{16}{dB}$}; + + \draw[o->] ([xshift=-1cm]Amp.west) node[left,align=right]{Input $u(t)$} -- (Amp.west); + \draw[->] (Amp.east) -- (BPF.west); + \draw[->] (BPF.east) -- (Mix.west); + \draw[->] (Mix.east) -- (Demod.west); + \draw[->] (Demod.east) -- (Amp2.west); + \draw[->] (Amp2.east) -- ([xshift=1cm]Amp2.east) node[right,align=left]{Output}; + \end{tikzpicture} + \end{adjustbox} + \end{figure} + For this new constellation: What is the overall noise figure and overall gain of the chain? What is the signal power, the thermal noise power and the SNR at the output of the chain? + \end{tasks} +\end{question} + +\begin{solution} + \begin{tasks} + \task + \begin{itemize} + \item The RMS value of the signals is: $\frac{\SI{70.71}{\micro\volt}}{\sqrt{2}} = \SI{50}{\micro\volt}$ + \item At $\SI{50}{\ohm}$, this yields a power of: $P = \frac{U^2}{R} = \SI{50}{pW} \equiv L_{P,S} = \SI{-73}{dBm}$ + \item The noise floor at room temperature is: $k_B T \equiv \SI{-174}{dBm/Hz}$ + \item The filter's bandwidth is: $\SI{63}{dBHz}$ + \item The noise power is: $k_B T \Delta f \equiv L_{P,N} = \SI{-174}{dBm/Hz} + \SI{63}{dBHz} = \SI{-111}{dBm}$ + \item The SNR is: $L_{\mathrm{SNR}} = L_{P,S} - L_{P,N} = \SI{-73}{dBm} - (\SI{-111}{dBm}) = \SI{38}{dBm}$ + \end{itemize} + + \task + Convert to linear scale: + \begin{itemize} + \item $G_1 = 0.5$ + \item $F_1 = 2$ + \item $G_2 = 2$ + \item $F_2 = 100$ + \item $G_3 = 4$ + \item $F_3 = 10$ + \item $G_4 = 100000$ + \item $F_4 = 40$ + \end{itemize} + + Using the Friis formula: + \begin{equation*} + \begin{split} + F_{\text{total}} &= F_1 + \frac{F_2 - 1}{G_1} + \frac{F_3 - 1}{G_1 G_2} + \frac{F_4 - 1}{G_1 G_2 G_3} \\ + &= 218 \\ + L_{F,\text{total}} &= \SI{23.4}{dB} + \end{split} + \end{equation*} + Overall gain: + \begin{equation*} + L_{G,\text{total}} = \SI{-3}{dB} + \SI{3}{dB} + \SI{6}{dB} + \SI{50}{dB} = \SI{56}{dB} + \end{equation*} + \begin{itemize} + \item Output signal power: $\SI{-73}{dBm} + L_{G,\text{total}} = \SI{-17}{dBm}$ + \item Output noise power: $\SI{-111}{dBm} + L_{G,\text{total}} + L_{F,\text{total}} = \SI{-31.6}{dBm}$ + \item Output SNR: $\SI{38}{dBm} - L_{F,\text{total}} = \SI{14.6}{dB}$ + \end{itemize} + + \task + Convert to linear scale: + \begin{itemize} + \item $G_0 = 1000$ + \item $F_0 = 4$ + \item $G_1 = 0.5$ + \item $F_1 = 2$ + \item $G_2 = 2$ + \item $F_2 = 100$ + \item $G_3 = 4$ + \item $F_3 = 10$ + \item $G_4 = 10$ + \item $F_4 = 40$ + \end{itemize} + + Using the Friis formula: + \begin{equation*} + \begin{split} + F_{\text{total}} &= F_0 + \frac{F_1 - 1}{G_0} + \frac{F_2 - 1}{G_0 G_1} + \frac{F_3 - 1}{G_0 G_1 G_2} + \frac{F_4 - 1}{G_0 G_1 G_2 G_3} \\ + &= 4.22 \\ + L_{F,\text{total}} &= \SI{6.3}{dB} + \end{split} + \end{equation*} + Overall gain: + \begin{equation*} + L_{G,\text{total}} = \SI{50}{dB} + \SI{-3}{dB} + \SI{3}{dB} + \SI{6}{dB} = \SI{56}{dB} + \end{equation*} + \begin{itemize} + \item Output signal power: $\SI{-73}{dBm} + L_{G,\text{total}} = \SI{-17}{dBm}$ + \item Output noise power: $\SI{-111}{dBm} + L_{G,\text{total}} + L_{F,\text{total}} = \SI{-48.7}{dBm}$ + \item Output SNR: $\SI{38}{dBm} - L_{F,\text{total}} = \SI{31.7}{dB}$ + \end{itemize} + + \textbf{Conclusion:} + \begin{itemize} + \item Placing the low noise amplifier in front of the chain, gives a much better ($\SI{16.1}{dB}$) SNR. The gain is equal. + \item The overall noise figure is dominated by the first component in the chain. + \item In this constellation, first component is a low noise amplifier with high gain and low noise figure. + \item The high gain scales down the noise contribution of the components following in the chain. + \item Therefore, it is always feasible to place a low noise amplifier (high gain, low noise) to the front of the chain. + \end{itemize} + \end{tasks} +\end{solution} + + + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%\begin{question}[subtitle={Decibel}] +% \begin{tasks} +% \end{tasks} +%\end{question} +% +%\begin{solution} +% \begin{tasks} +% \end{tasks} +%\end{solution} |