From 0ab48e46d376c406647a161bb184e4e671a930e7 Mon Sep 17 00:00:00 2001 From: Philipp Le Date: Mon, 18 May 2020 23:33:06 +0200 Subject: Exercise 2 rev 1 --- exercise02/exercise02.tex | 69 ++++++++++++++++++++++++++++++++++++++++++----- 1 file changed, 63 insertions(+), 6 deletions(-) (limited to 'exercise02') diff --git a/exercise02/exercise02.tex b/exercise02/exercise02.tex index 51099f7..580ac67 100644 --- a/exercise02/exercise02.tex +++ b/exercise02/exercise02.tex @@ -37,7 +37,7 @@ \end{tasks} \end{solution} -\begin{question} +\begin{question}[subtitle={Fourier Series}] The following periodic signal is given. \begin{figure}[H] \centering @@ -315,7 +315,7 @@ \end{tasks} \end{solution} -\begin{question} +\begin{question}[subtitle={System Analysis}] The following circuit is given. \begin{figure}[H] \centering @@ -333,8 +333,8 @@ Find a differential equation which connects $u_i(t)$ and $u_o(t)$! \task Determine the transfer function $\underline{H} \left(j \omega\right)$! - \task - Calculate the impulse response! + %\task + %Calculate the impulse response! \task Is the system causal? Why? \task @@ -342,7 +342,11 @@ \end{tasks} \end{question} -\begin{question} +\begin{solution} + %TODO +\end{solution} + +\begin{question}[subtitle={Amplitude and Phase Response}] \begin{figure}[H] \centering \begin{circuitikz} @@ -377,7 +381,7 @@ \task The following signal is applied to the input of the system $u_i(t)$. \begin{equation} - u_i(t) = \SI{2}{V} \cos\left(2 \pi \cdot \SI{25}{kHz} \cdot t\right) + u_i(t) = \SI{2}{V} \cdot \cos\left(2 \pi \cdot \SI{2.5}{kHz} \cdot t\right) \end{equation} Calculate the output signal $u_o(t)$ as either a time domain function or a phasor! \end{tasks} @@ -407,6 +411,59 @@ \end{itemize} The system is stable because the real part of its pole is negative. + \task + \begin{equation*} + \begin{split} + \underline{H}\left(j \omega\right) &= \frac{j \omega RC}{j \omega RC + 1} \\ + &= \frac{j \omega RC \left(j \omega RC - 1\right)}{\left(j \omega RC + 1\right)\left(j \omega RC - 1\right)} \\ + &= \frac{-\left(\omega RC\right)^2 - j \omega RC}{- \left(j \omega RC\right)^2 - 1} \\ + &= \frac{\left(\omega RC\right)^2 + j \omega RC}{\left(j \omega RC\right)^2 + 1} \\ + \Re\left\{\underline{H}\left(j \omega\right)\right\} &= \frac{\left(\omega RC\right)^2}{\left(j \omega RC\right)^2 + 1} \\ + \Im\left\{\underline{H}\left(j \omega\right)\right\} &= \frac{\omega RC}{\left(j \omega RC\right)^2 + 1} + \end{split} + \end{equation*} + \begin{equation*} + \begin{split} + \left|\underline{H}\left(j \omega\right)\right| &= \sqrt{\frac{\left(\omega RC\right)^4 + \left(\omega RC\right)^2}{\left(\left(j \omega RC\right)^2 + 1\right)^2}} \\ + &= \sqrt{\frac{\left(\omega RC\right)^2 \left(\left(j \omega RC\right)^2 + 1\right)}{\left(\left(j \omega RC\right)^2 + 1\right)^2}} \\ + &= \sqrt{\frac{\left(\omega RC\right)^2}{\left(j \omega RC\right)^2 + 1}} + \end{split} + \end{equation*} + + \task + \begin{equation*} + \begin{split} + \arg\left(\underline{H}\left(j \omega\right)\right) &= \mathrm{atan2}\left(\Im\left\{\underline{H}\left(j \omega\right)\right\}, \Re\left\{\underline{H}\left(j \omega\right)\right\}\right) \\ + &= \begin{cases} + \arctan\left(\frac{\Im\left\{\underline{H}\left(j \omega\right)\right\}}{\Re\left\{\underline{H}\left(j \omega\right)\right\}}\right) &\quad \text{ if } \Re\left\{\underline{H}\left(j \omega\right)\right\} > 0, \\ + \arctan\left(\frac{\Im\left\{\underline{H}\left(j \omega\right)\right\}}{\Re\left\{\underline{H}\left(j \omega\right)\right\}}\right) + \pi &\quad \text{ if } \Re\left\{\underline{H}\left(j \omega\right)\right\} < 0 \text{ and } \Im\left\{\underline{H}\left(j \omega\right)\right\} \geq 0, \\ + \arctan\left(\frac{\Im\left\{\underline{H}\left(j \omega\right)\right\}}{\Re\left\{\underline{H}\left(j \omega\right)\right\}}\right) - \pi &\quad \text{ if } \Re\left\{\underline{H}\left(j \omega\right)\right\} < 0 \text{ and } \Im\left\{\underline{H}\left(j \omega\right)\right\} < 0, \\ + +\frac{\pi}{2} &\quad \text{ if } \Re\left\{\underline{H}\left(j \omega\right)\right\} = 0 \text{ and } \Im\left\{\underline{H}\left(j \omega\right)\right\} > 0, \\ + -\frac{\pi}{2} &\quad \text{ if } \Re\left\{\underline{H}\left(j \omega\right)\right\} = 0 \text{ and } \Im\left\{\underline{H}\left(j \omega\right)\right\} < 0, \\ + \text{undefined} &\quad \text{ if } \Re\left\{\underline{H}\left(j \omega\right)\right\} = 0 \text{ and } \Im\left\{\underline{H}\left(j \omega\right)\right\} = 0. \\ + \end{cases} \\ + &= \arctan\left(\frac{\Im\left\{\underline{H}\left(j \omega\right)\right\}}{\Re\left\{\underline{H}\left(j \omega\right)\right\}}\right) + \end{split} + \end{equation*} + + \task + \begin{equation*} + \underline{U}_i\left(j \omega\right) = \mathcal{F}\left\{u_i(t)\right\} = \SI{2}{V} \pi \left(\delta\left(\omega - 2 \pi \cdot \SI{2.5}{kHz}\right) + \delta\left(\omega + 2 \pi \cdot \SI{2.5}{kHz}\right)\right) + \end{equation*} + \begin{equation*} + \begin{split} + \underline{U}_i\left(j \omega\right) &= \underline{H}\left(j \omega\right) \underline{U}_i\left(j \omega\right) \\ + &= \SI{2}{V} \pi \left(\underline{H}\left(j \left(2 \pi \cdot \SI{2.5}{kHz}\right)\right) \delta\left(\omega - 2 \pi \cdot \SI{2.5}{kHz}\right) \right. + \\ &\qquad \left. \underline{H}\left(-j \left(2 \pi \cdot \SI{2.5}{kHz}\right)\right) \delta\left(\omega + 2 \pi \cdot \SI{2.5}{kHz}\right)\right) \\ + &= \SI{2}{V} \pi \left(0.594 \cdot e^{j \SI{53.6}{\degree}} \cdot \delta\left(\omega - 2 \pi \cdot \SI{2.5}{kHz}\right) \right. + \\ &\qquad \left. 0.594 \cdot e^{-j \SI{53.6}{\degree}} \cdot \delta\left(\omega + 2 \pi \cdot \SI{2.5}{kHz}\right)\right) \\ + &= \SI{1.19}{V} \pi \left(e^{j \SI{53.6}{\degree}} \cdot \delta\left(\omega - 2 \pi \cdot \SI{2.5}{kHz}\right) + e^{-j \SI{53.6}{\degree}} \cdot \delta\left(\omega + 2 \pi \cdot \SI{2.5}{kHz}\right)\right) + \end{split} + \end{equation*} + Using the time-shift theorem: + \begin{equation*} + u_o(t) = \mathcal{F}^{-1}\left\{\underline{U}_i\left(j \omega\right)\right\} = \SI{1.19}{V} \cdot \cos\left(2 \pi \cdot \SI{2.5}{kHz} \cdot t - \SI{53.6}{\degree}\right) + \end{equation*} + The signal has been attenuated and phase-shifted. + %TODO \end{tasks} \end{solution} -- cgit v1.1