Corrections to Chapter 2 and Exercise 2

4 files changed, 16 insertions, 11 deletions

diff --git a/chapter02/content_ch02.tex b/chapter02/content_ch02.tex
index 8ae492c..d596c93 100644
--- a/chapter02/content_ch02.tex
+++ b/chapter02/content_ch02.tex
@@ -325,11 +325,11 @@ Now, you can prove that the cosine and sine functions are orthogonal to each oth
 
 Furthermore, the sine and cosine functions with \underline{different} indices are orthogonal to each other.
 \begin{equation}
-	\int\limits_{-\frac{T_0}{2}}^{\frac{T_0}{2}} \cos\left(n \omega_0 t\right) \cos\left(p \omega_0 t\right) \, \mathrm{d} t = \frac{\pi}{\omega_0} \cdot \delta_{np} \qquad \forall \; n, p \in \mathbb{N}
+	\int\limits_{-\frac{T_0}{2}}^{\frac{T_0}{2}} \cos\left(n \omega_0 t\right) \cos\left(p \omega_0 t\right) \, \mathrm{d} t = \frac{\pi}{\omega_0} \cdot \delta_{np} \qquad \forall \; n, p \in \mathbb{N} \backslash \{0\}
 	\label{eq:ch02:orth_rel_cos}
 \end{equation}
 \begin{equation}
-	\int\limits_{-\frac{T_0}{2}}^{\frac{T_0}{2}} \sin\left(m \omega_0 t\right) \sin\left(q \omega_0 t\right) \, \mathrm{d} t = \frac{\pi}{\omega_0} \cdot \delta_{mq} \qquad \forall \; m, q \in \mathbb{N}
+	\int\limits_{-\frac{T_0}{2}}^{\frac{T_0}{2}} \sin\left(m \omega_0 t\right) \sin\left(q \omega_0 t\right) \, \mathrm{d} t = \frac{\pi}{\omega_0} \cdot \delta_{mq} \qquad \forall \; m, q \in \mathbb{N} \backslash \{0\}
 	\label{eq:ch02:orth_rel_sin}
 \end{equation}
 with the Kronecker delta
@@ -362,10 +362,13 @@ The orthogonality relations are useful to extract the coefficients $a_n$ and $b_
 Using the orthogonality relations, the coefficients $\tilde{a}_n$ and $\tilde{b}_n$ can be obtained by:
 \begin{subequations}
 	\begin{align}
-		\tilde{a}_n &= \frac{\omega_0}{\pi} \int\limits_{-\frac{T_0}{2}}^{\frac{T_0}{2}} \tilde{x}_p(t) \cdot \cos\left(n \omega_0 t\right) \, \mathrm{d} t \label{eq_ch02_fourier_series_coeff_an} \\
-		\tilde{b}_m &= \frac{\omega_0}{\pi} \int\limits_{-\frac{T_0}{2}}^{\frac{T_0}{2}} \tilde{x}_p(t) \cdot \sin\left(m \omega_0 t\right) \, \mathrm{d} t \label{eq_ch02_fourier_series_coeff_bm}
+		\tilde{a}_n &= \frac{\omega_0}{\pi} \int\limits_{-\frac{T_0}{2}}^{\frac{T_0}{2}} \tilde{x}_p(t) \cdot \cos\left(n \omega_0 t\right) \, \mathrm{d} t \label{eq_ch02_fourier_series_coeff_an} \qquad \forall \; n > 0 \\
+		\tilde{b}_m &= \frac{\omega_0}{\pi} \int\limits_{-\frac{T_0}{2}}^{\frac{T_0}{2}} \tilde{x}_p(t) \cdot \sin\left(m \omega_0 t\right) \, \mathrm{d} t \label{eq_ch02_fourier_series_coeff_bm} \qquad \forall \; m > 0 \\
+		\tilde{a}_0 &= \frac{\omega_0}{2 \pi} \int\limits_{-\frac{T_0}{2}}^{\frac{T_0}{2}} \tilde{x}_p(t) \, \mathrm{d} t \\
+		\tilde{b}_0 &= 0
 	\end{align}
 \end{subequations}
+\textit{Remark: } $a_0$ and $b_0$ need a special treatment, because of slightly changed orthogonality relations.
 
 \begin{proof}{Parameter Extraction for $\tilde{a}_n$}
 	Given is a periodic function $\tilde{x}_p(t)$, which can be decomposed into:
@@ -441,6 +444,8 @@ It is based on the orthogonality relation:
 	\label{eq:ch02:orth_rel_exp}
 \end{equation}
 
+\vspace*{1em}
+
 \begin{definition}{Complex-Valued Fourier series}
 	A complex-valued, periodic signal $\underline{x_p}(t)$ can be decomposed into a series complex-valued mono-chromatic signals \eqref{eq:ch02:fourier_series_cmplx} -- the \index{Fourier series!complex-valued} \textbf{complex-valued Fourier series}.
 	\begin{equation*}
diff --git a/exercise02/exercise02.tex b/exercise02/exercise02.tex
index 3c5e412..e561555 100644
--- a/exercise02/exercise02.tex
+++ b/exercise02/exercise02.tex
@@ -139,10 +139,10 @@
 				 &= \frac{1}{\pi n} \left(
 				 	\SI{0.8}{V} \cdot \left(
 				 		\sin\left(\frac{\pi}{2}n\right)
-				 		- \underbrace{\sin\left(-\frac{\pi}{2}n\right)}_{= \sin\left(\frac{\pi}{2}n\right)}
+				 		- \underbrace{\sin\left(-\frac{\pi}{2}n\right)}_{= -\sin\left(\frac{\pi}{2}n\right)}
 				 	\right)
 				 	- \SI{0.2}{V} \cdot \left(
-				 		\underbrace{\sin\left(\frac{3\pi}{2}n\right)}_{= \sin\left(\frac{\pi}{2}n\right)}
+				 		\underbrace{\sin\left(\frac{3\pi}{2}n\right)}_{= -\sin\left(\frac{\pi}{2}n\right)}
 				 		- \sin\left(\frac{\pi}{2}n\right)
 				 	\right)
 				 \right) \\
@@ -806,7 +806,7 @@
 		\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)
+			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.
 	\end{tasks}
diff --git a/main/chapter02.tex b/main/chapter02.tex
index 91b8ed9..b2bb59f 100644
--- a/main/chapter02.tex
+++ b/main/chapter02.tex
@@ -13,8 +13,8 @@
 % Configuration
 \def\thekindofdocument{Lecture Notes}
 \def\thesubtitle{Chapter 2: Time-Continuous Signals and Systems}
-\def\therevision{1}
-\def\therevisiondate{2020-05-11}
+\def\therevision{2}
+\def\therevisiondate{2020-05-29}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 % Header
diff --git a/main/exercise02.tex b/main/exercise02.tex
index d7bae02..f6f7963 100644
--- a/main/exercise02.tex
+++ b/main/exercise02.tex
@@ -13,8 +13,8 @@
 % Configuration
 \def\thekindofdocument{Exercise}
 \def\thesubtitle{Chapter 2: Time-Continuous Signals and Systems}
-\def\therevision{1}
-\def\therevisiondate{2020-05-18}
+\def\therevision{2}
+\def\therevisiondate{2020-05-29}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 % Header