diff --git a/contents/slides/aequivalente_isotrope_strahlungsleistung_eirp_2.md b/contents/slides/aequivalente_isotrope_strahlungsleistung_eirp_2.md
index cedddfb..30e5a52 100644
--- a/contents/slides/aequivalente_isotrope_strahlungsleistung_eirp_2.md
+++ b/contents/slides/aequivalente_isotrope_strahlungsleistung_eirp_2.md
@@ -61,7 +61,7 @@ Eine ortsfeste Amateurfunkanlage ist nach § 9 BEMFV bei der BNetzA anzuzeigen,
 * gesucht: $P_{\mathrm{EIRP}}$
 
 <fragment>
-$\begin{equation}\begin{split} \nonumber P_{\mathrm{EIRP}} &= P_{\mathrm{Sender}} \cdot 10^{\frac{g_i - a}{10\mathrm{dB}}}\\ &= 250mW \cdot 10^{\frac{26\mathrm{dB}}{10\mathrm{dB}}}\\ &= 250mW \cdot 398\\ &\approx 100W \end{split}\end{equation}$
+$\begin{split} P_{\mathrm{EIRP}} &= P_{\mathrm{Sender}} \cdot 10^{\frac{g_i - a}{10\mathrm{dB}}}\\ &= 250mW \cdot 10^{\frac{26\mathrm{dB}}{10\mathrm{dB}}}\\ &= 250mW \cdot 398\\ &\approx 100W \end{split}$
 </fragment>
 
 ---
@@ -81,7 +81,7 @@ $\begin{equation}\begin{split} \nonumber P_{\mathrm{EIRP}} &= P_{\mathrm{Sender}
 * gesucht: $P_{\mathrm{Sender}}$
 
 <fragment>
-$\begin{equation}\begin{split} \nonumber P_{\mathrm{EIRP}} &= P_{\mathrm{Sender}} \cdot 10^{\frac{g_i - a}{10\mathrm{dB}}}\\ \Rightarrow P_{\mathrm{Sender}} &= \dfrac{P_{\mathrm{EIRP}}}{10^{\frac{g_i - a}{10\mathrm{dB}}}}\\ &= \dfrac{10W}{10^{\frac{5,15\mathrm{dB}}{10\mathrm{dB}}}}\\ &\approx \frac{10W}{3,27} \approx 3W \end{split}\end{equation}$
+$\begin{split} P_{\mathrm{EIRP}} &= P_{\mathrm{Sender}} \cdot 10^{\frac{g_i - a}{10\mathrm{dB}}}\\ \Rightarrow P_{\mathrm{Sender}} &= \dfrac{P_{\mathrm{EIRP}}}{10^{\frac{g_i - a}{10\mathrm{dB}}}}\\ &= \dfrac{10W}{10^{\frac{5,15\mathrm{dB}}{10\mathrm{dB}}}}\\ &\approx \frac{10W}{3,27} \approx 3W \end{split}$
 </fragment>
 
 <note>
@@ -121,7 +121,7 @@ Gewinn: 10dB, also auch Faktor 10
 * gesucht: $P_{\mathrm{EIRP}}$
 
 <fragment>
-$\begin{equation}\begin{split} \nonumber P_{\mathrm{EIRP}} &= P_{\mathrm{Sender}} \cdot 10^{\frac{g_d - a + 2,15\mathrm{dB}}{10\mathrm{dB}}}\\ &= 5W \cdot 10^{\frac{5\mathrm{dB} - 2\mathrm{dB} + 2,15\mathrm{dB}}{10\mathrm{dB}}}\\ &= 5W \cdot 3,27\\ &\approx 16,4W \end{split}\end{equation}$
+$\begin{split} P_{\mathrm{EIRP}} &= P_{\mathrm{Sender}} \cdot 10^{\frac{g_d - a + 2,15\mathrm{dB}}{10\mathrm{dB}}}\\ &= 5W \cdot 10^{\frac{5\mathrm{dB} - 2\mathrm{dB} + 2,15\mathrm{dB}}{10\mathrm{dB}}}\\ &= 5W \cdot 3,27\\ &\approx 16,4W \end{split}$
 </fragment>
 
 ---
diff --git a/contents/slides/antenne_laenge_resonanz.md b/contents/slides/antenne_laenge_resonanz.md
index 7650992..0537dcc 100644
--- a/contents/slides/antenne_laenge_resonanz.md
+++ b/contents/slides/antenne_laenge_resonanz.md
@@ -11,7 +11,7 @@
 
 Anstatt direkt die ungefähre Wellenlänge des 10m-Bands zu verwenden, wird hier erst die angegebene Frequenz in die exakte Wellenlänge umgerechnet.
 
-$\begin{equation} \begin{split} \nonumber l &= \frac{5}{8}\lambda\\ &= \frac{5}{8} \cdot \dfrac{300}{28,5MHz}\\ &= \frac{5}{8} \cdot 10,53m\\ &= 6,58m\\ \end{split} \end{equation}$
+$\begin{split} l &= \frac{5}{8}\lambda\\ &= \frac{5}{8} \cdot \dfrac{300}{28,5MHz}\\ &= \frac{5}{8} \cdot 10,53m\\ &= 6,58m\\ \end{split}$
 
 ---
 
diff --git a/contents/slides/daempfungsglieder.md b/contents/slides/daempfungsglieder.md
index 9b0ce1f..d61b699 100644
--- a/contents/slides/daempfungsglieder.md
+++ b/contents/slides/daempfungsglieder.md
@@ -42,7 +42,7 @@ Der Name T-Konfiguration stammt vom Aufbau der Widerstände als T
 * gesucht: $\Delta P = P_2 - P_1$
 
 <fragment>
-$\begin{align}\nonumber a &= 10 \cdot \log_{10}{(\frac{P_1}{P_2})}dB\\ \nonumber \Rightarrow \frac{a}{10} &= \log_{10}{(\frac{P_1}{P_2})}dB\\ \nonumber \Rightarrow 10^{\frac{a}{10}} &= \frac{P_1}{P_2}\\ \nonumber \Rightarrow P_2 &= \frac{P_1}{10^{\frac{a}{10}}}\end{align}$
+$\begin{split} a &= 10 \cdot \log_{10}{(\frac{P_1}{P_2})}dB\\ \Rightarrow \frac{a}{10} &= \log_{10}{(\frac{P_1}{P_2})}dB\\ \Rightarrow 10^{\frac{a}{10}} &= \frac{P_1}{P_2}\\ \Rightarrow P_2 &= \frac{P_1}{10^{\frac{a}{10}}}\end{split}$
 </fragment>
 ---
 <fragment>
diff --git a/contents/slides/dezibel_2.md b/contents/slides/dezibel_2.md
index b487b74..c9c73df 100644
--- a/contents/slides/dezibel_2.md
+++ b/contents/slides/dezibel_2.md
@@ -68,7 +68,7 @@ Details zur Berechnung des Faktor 20 sind im Online-Kurs. Kurzfassung: Im Spannu
 * gesucht: $P$
 
 <fragment>
-$\begin{equation}\begin{align} \nonumber p &= 10\cdot \log_{10}\left(\frac{P}{1W}\right)\textrm{dBW}\\ \nonumber \Rightarrow P &= 10^{\frac{p}{10}} \cdot 1W = 10^{\frac{20\textrm{dBW}}{10}} \cdot 1W = 10^2W \end{align}\end{equation}$
+$\begin{split} p &= 10\cdot \log_{10}\left(\frac{P}{1W}\right)\textrm{dBW}\\ \Rightarrow P &= 10^{\frac{p}{10}} \cdot 1W = 10^{\frac{20\textrm{dBW}}{10}} \cdot 1W = 10^2W \end{split}$
 </fragment>
 ---
 [question:AA107]
@@ -99,7 +99,7 @@ Aus der Tabelle in der Formelsammlung
 * gesucht: $U$
 
 <fragment>
-$\begin{equation}\begin{align} \nonumber u &= 20\cdot \log_{10}\left(\frac{U}{1\textrm{µV}}\right)\textrm{\textrm{dBµV}}\\ \nonumber \Rightarrow U &= 10^{\frac{u}{20}} \cdot 1\textrm{µV} = 10^{\frac{120\textrm{dBµV}/m}{20}} \cdot 1\textrm{µV} = 1V/m \end{align}\end{equation}$
+$\begin{split} u &= 20\cdot \log_{10}\left(\frac{U}{1\textrm{µV}}\right)\textrm{\textrm{dBµV}}\\ \Rightarrow U &= 10^{\frac{u}{20}} \cdot 1\textrm{µV} = 10^{\frac{120\textrm{dBµV}/m}{20}} \cdot 1\textrm{µV} = 1V/m \end{split}$
 </fragment>
 <fragment>
 In der Literatur ist oft zu finden: 120dBµV = 1V
diff --git a/contents/slides/doppelueberlagerungsempfaenger_doppelsuper.md b/contents/slides/doppelueberlagerungsempfaenger_doppelsuper.md
index b0daf95..a784d71 100644
--- a/contents/slides/doppelueberlagerungsempfaenger_doppelsuper.md
+++ b/contents/slides/doppelueberlagerungsempfaenger_doppelsuper.md
@@ -68,8 +68,8 @@
 $f_{ZF} = |f_E − f_{OSZ}| \Rightarrow f_{OSZ} = f_{ZF} \pm f_{E}$
 </fragment>
 <fragment>
-1. Lösung: $\begin{equation}\begin{split}\nonumber f_{OSZ} &= f_{ZF} \, + \, f_{E}\\ &= 50MHz \, + \, 3\dots30MHz\\ &= 53\dots80MHz \end{split}\end{equation}$
-2. Lösung: $\begin{equation}\begin{split}\nonumber f_{OSZ} &= f_{ZF} \, - \, f_{E}\\ &= 50MHz \, - \, 3\dots30MHz\\ &= 47\dots20MHz \end{split}\end{equation}$
+1. Lösung: $\begin{split} f_{OSZ} &= f_{ZF} \, + \, f_{E}\\ &= 50MHz \, + \, 3\dots30MHz\\ &= 53\dots80MHz \end{split}$
+2. Lösung: $\begin{split} f_{OSZ} &= f_{ZF} \, - \, f_{E}\\ &= 50MHz \, - \, 3\dots30MHz\\ &= 47\dots20MHz \end{split}$
 </fragment>
 ---
 [question:AF120]
diff --git a/contents/slides/effektive_strahlungsleistung_erp_2.md b/contents/slides/effektive_strahlungsleistung_erp_2.md
index cb7218c..2a31ca9 100644
--- a/contents/slides/effektive_strahlungsleistung_erp_2.md
+++ b/contents/slides/effektive_strahlungsleistung_erp_2.md
@@ -53,5 +53,5 @@ Diese Tabelle ist in der Formelsammlung enthalten und steht während der Prüfun
 * gesucht: $P_{\textrm{ERP}}$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber P_{\textrm{ERP}} &= P_S \cdot 10^{\frac{g_d - a}{10dB}}\\ &= 50W \cdot 10^{\frac{-20dBd - 0W}{10dB}}\\ &= 50W \cdot 10^{-2} = 0,5W$
+$\begin{split} P_{\textrm{ERP}} &= P_S \cdot 10^{\frac{g_d - a}{10dB}}\\ &= 50W \cdot 10^{\frac{-20dBd - 0W}{10dB}}\\ &= 50W \cdot 10^{-2} = 0,5W\end{split}$
 </fragment>
diff --git a/contents/slides/fm_3.md b/contents/slides/fm_3.md
index a51ba64..ca200fa 100644
--- a/contents/slides/fm_3.md
+++ b/contents/slides/fm_3.md
@@ -67,7 +67,7 @@ Etwa 99% der Sendeleistung befinden sich in der Bandbreite
 * gesucht: $B$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber B &\approx 2 \cdot (\Delta f_{\textrm{T}} + f_{\textrm{mod max}})\\ &= 2 \cdot (1,8kHz + 2kHz) = 7,6kHz \end{split}\end{equation}$
+$\begin{split} B &\approx 2 \cdot (\Delta f_{\textrm{T}} + f_{\textrm{mod max}})\\ &= 2 \cdot (1,8kHz + 2kHz) = 7,6kHz \end{split}$
 </fragment>
 ---
 [question:AE308]
@@ -78,7 +78,7 @@ $\begin{equation}\begin{split}\nonumber B &\approx 2 \cdot (\Delta f_{\textrm{T}
 * gesucht: $B$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber B &\approx 2 \cdot (\Delta f_{\textrm{T}} + f_{\textrm{mod max}})\\ &= 2 \cdot (2,5kHz + 2,7kHz) = 10,4kHz \end{split}\end{equation}$
+$\begin{split} B &\approx 2 \cdot (\Delta f_{\textrm{T}} + f_{\textrm{mod max}})\\ &= 2 \cdot (2,5kHz + 2,7kHz) = 10,4kHz \end{split}$
 </fragment>
 ---
 [question:AE311]
@@ -89,7 +89,7 @@ $\begin{equation}\begin{split}\nonumber B &\approx 2 \cdot (\Delta f_{\textrm{T}
 * gesucht: $f_{\textrm{mod max}}$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber B &\approx 2 \cdot (\Delta f_{\textrm{T}} + f_{\textrm{mod max}})\\ \Rightarrow f_{mod max} &= \frac{B}{2} - \Delta f_T\\ &= \frac{10kHz}{2} - 2,5kHz = 2,5kHz \end{split}\end{equation}$
+$\begin{split} B &\approx 2 \cdot (\Delta f_{\textrm{T}} + f_{\textrm{mod max}})\\ \Rightarrow f_{mod max} &= \frac{B}{2} - \Delta f_T\\ &= \frac{10kHz}{2} - 2,5kHz = 2,5kHz \end{split}$
 </fragment>
 ---
 [question:AE312]
@@ -100,7 +100,7 @@ $\begin{equation}\begin{split}\nonumber B &\approx 2 \cdot (\Delta f_{\textrm{T}
 * gesucht: $\Delta f_{\textrm{T}}$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber B &\approx 2 \cdot (\Delta f_{\textrm{T}} + f_{\textrm{mod max}})\\ \Rightarrow \Delta f_T &= \frac{B}{2} - f_{mod max}\\ &= \frac{10kHz}{2} - 2,7kHz = 2,3kHz \end{split}\end{equation}$
+$\begin{split} B &\approx 2 \cdot (\Delta f_{\textrm{T}} + f_{\textrm{mod max}})\\ \Rightarrow \Delta f_T &= \frac{B}{2} - f_{mod max}\\ &= \frac{10kHz}{2} - 2,7kHz = 2,3kHz \end{split}$
 </fragment>
 ---
 [question:AE310]
\ No newline at end of file
diff --git a/contents/slides/frequenzgenauigkeit.md b/contents/slides/frequenzgenauigkeit.md
index 34bf613..e2c045d 100644
--- a/contents/slides/frequenzgenauigkeit.md
+++ b/contents/slides/frequenzgenauigkeit.md
@@ -31,7 +31,7 @@ $435MHz \cdot \frac{1}{10^6} = \frac{435\cdot \cancel{10^6}Hz}{\cancel{10^6}} =
 * gesucht: $f_{min}, f_{max}$
 
 <fragment>
-$\begin{aligned}f_{min} &= f\,-\,f \cdot \frac{10}{10^6}\\ &= 14,2MHz\,-\,\frac{14,2\cdot \cancel{10^6}Hz\cdot 10}{\cancel{10^6}}\\ &= 14,2MHz\,-\,142Hz\\ &= 14,199858MHz\end{aligned}$
+$\begin{split}f_{min} &= f\,-\,f \cdot \frac{10}{10^6}\\ &= 14,2MHz\,-\,\frac{14,2\cdot \cancel{10^6}Hz\cdot 10}{\cancel{10^6}}\\ &= 14,2MHz\,-\,142Hz\\ &= 14,199858MHz\end{split}$
 </fragment>
 ---
 #### Lösungsweg Teil 2
@@ -40,7 +40,7 @@ $\begin{aligned}f_{min} &= f\,-\,f \cdot \frac{10}{10^6}\\ &= 14,2MHz\,-\,\frac{
 * gesucht: $f_{min}, f_{max}$
 
 <fragment>
-$\begin{aligned}f_{max} &= f\,+\,f \cdot \frac{10}{10^6}\\ &= 14,2MHz\,+\,\frac{14,2\cdot \cancel{10^6}Hz\cdot 10}{\cancel{10^6}}\\ &= 14,2MHz\,+\,142Hz\\ &= 14,200142MHz\end{aligned}$
+$\begin{split}f_{max} &= f\,+\,f \cdot \frac{10}{10^6}\\ &= 14,2MHz\,+\,\frac{14,2\cdot \cancel{10^6}Hz\cdot 10}{\cancel{10^6}}\\ &= 14,2MHz\,+\,142Hz\\ &= 14,200142MHz\end{split}$
 </fragment>
 ---
 [question:AI506]
@@ -51,7 +51,7 @@ $\begin{aligned}f_{max} &= f\,+\,f \cdot \frac{10}{10^6}\\ &= 14,2MHz\,+\,\frac{
 * gesucht: $\Delta f$
 
 <fragment>
-$\begin{aligned}\Delta f &= 29MHz \cdot 0,01\%\\ &= 29\cdot \cancel{10^6}Hz \cdot 100\cdot \cancel{10^{-6}}\\ &= 2900Hz\end{aligned}$
+$\begin{split}\Delta f &= 29MHz \cdot 0,01\%\\ &= 29\cdot \cancel{10^6}Hz \cdot 100\cdot \cancel{10^{-6}}\\ &= 2900Hz\end{split}$
 </fragment>
 ---
 [question:AI507]
@@ -62,7 +62,7 @@ $\begin{aligned}\Delta f &= 29MHz \cdot 0,01\%\\ &= 29\cdot \cancel{10^6}Hz \cdo
 * gesucht: $\Delta f$
 
 <fragment>
-$\begin{aligned}\Delta f &= 14100kHz \cdot 0,00001\%\\ &= 14,1\cdot \cancel{10^6}Hz \cdot 0,1\cdot \cancel{10^{-6}}\\ &= 1,41Hz\end{aligned}$
+$\begin{split}\Delta f &= 14100kHz \cdot 0,00001\%\\ &= 14,1\cdot \cancel{10^6}Hz \cdot 0,1\cdot \cancel{10^{-6}}\\ &= 1,41Hz\end{split}$
 </fragment>
 ---
 [question:AI508]
@@ -73,7 +73,7 @@ $\begin{aligned}\Delta f &= 14100kHz \cdot 0,00001\%\\ &= 14,1\cdot \cancel{10^6
 * gesucht: $\Delta f$
 
 <fragment>
-$\begin{aligned}\Delta f &= 100MHz \cdot \frac{1}{10^6}\\ &= \frac{100\cdot \cancel{10^6}Hz}{\cancel{10^6}}\\ &= 100Hz\end{aligned}$
+$\begin{split}\Delta f &= 100MHz \cdot \frac{1}{10^6}\\ &= \frac{100\cdot \cancel{10^6}Hz}{\cancel{10^6}}\\ &= 100Hz\end{split}$
 </fragment>
 ---
 [question:AI509]
@@ -84,14 +84,14 @@ $\begin{aligned}\Delta f &= 100MHz \cdot \frac{1}{10^6}\\ &= \frac{100\cdot \can
 * gesucht: $f_{min},f_{max}$
 
 <fragment>
-$\begin{aligned}\Delta f &= 145MHz \cdot \frac{10}{10^6}\\ &= \frac{145\cdot \cancel{10^6}Hz \cdot 10}{\cancel{10^6}}\\ &= 1450Hz\end{aligned}$
+$\begin{split}\Delta f &= 145MHz \cdot \frac{10}{10^6}\\ &= \frac{145\cdot \cancel{10^6}Hz \cdot 10}{\cancel{10^6}}\\ &= 1450Hz\end{split}$
 </fragment>
 <fragment>
 <left>
-$\begin{aligned}f_{min} &= f\,-\,\Delta f\\ &= 145MHz\,-\,1450Hz\\ &= 144,99855MHz\end{aligned}$
+$\begin{split}f_{min} &= f\,-\,\Delta f\\ &= 145MHz\,-\,1450Hz\\ &= 144,99855MHz\end{split}$
 </left>
 <right>
-$\begin{aligned}f_{max} &= f\,+\,\Delta f\\ &= 145MHz\,+\,1450Hz\\ &= 145,00145MHz\end{aligned}$
+$\begin{split}f_{max} &= f\,+\,\Delta f\\ &= 145MHz\,+\,1450Hz\\ &= 145,00145MHz\end{split}$
 </right>
 </fragment>
 ---
@@ -105,11 +105,11 @@ $\begin{aligned}f_{max} &= f\,+\,\Delta f\\ &= 145MHz\,+\,1450Hz\\ &= 145,00145M
 
 <fragment>
 <left>
-$\begin{aligned}\Delta f &= 144,4MHz \cdot \frac{1}{10^6}\\ &= \frac{144,4\cdot \cancel{10^6}Hz}{\cancel{10^6}}\\ &= 144,4Hz\end{aligned}$
+$\begin{split}\Delta f &= 144,4MHz \cdot \frac{1}{10^6}\\ &= \frac{144,4\cdot \cancel{10^6}Hz}{\cancel{10^6}}\\ &= 144,4Hz\end{split}$
 </left>
 </fragment>
 <fragment>
 <right>
-$\begin{aligned}f_{B,max,Abw} &= f_{B,max} + \Delta f\\ &= 2,7kHz + 144,4Hz\\ &= 2,8444kHz\end{aligned}$
+$\begin{split}f_{B,max,Abw} &= f_{B,max} + \Delta f\\ &= 2,7kHz + 144,4Hz\\ &= 2,8444kHz\end{split}$
 </right>
 </fragment>
diff --git a/contents/slides/impedanztransformation.md b/contents/slides/impedanztransformation.md
index ab07e82..1548933 100644
--- a/contents/slides/impedanztransformation.md
+++ b/contents/slides/impedanztransformation.md
@@ -45,7 +45,7 @@
 * gesucht: $Z$
 
 <fragment>
-$\begin{aligned}Z &= \sqrt{Z_E \cdot Z_A}\\ &= \sqrt{240Ω \cdot 60Ω}\\ &= 120Ω\end{aligned}$ 
+$\begin{split}Z &= \sqrt{Z_E \cdot Z_A}\\ &= \sqrt{240Ω \cdot 60Ω}\\ &= 120Ω\end{split}$ 
 </fragment>
 ---
 [question:AG418]
@@ -56,7 +56,7 @@ $\begin{aligned}Z &= \sqrt{Z_E \cdot Z_A}\\ &= \sqrt{240Ω \cdot 60Ω}\\ &= 120
 * gesucht: $Z$
 
 <fragment>
-$\begin{aligned}Z &= \sqrt{Z_E \cdot Z_A}\\ &= \sqrt{600Ω \cdot 240Ω}\\ &= 380Ω\end{aligned}$ 
+$\begin{split}Z &= \sqrt{Z_E \cdot Z_A}\\ &= \sqrt{600Ω \cdot 240Ω}\\ &= 380Ω\end{split}$ 
 </fragment>
 ---
 
diff --git a/contents/slides/kondensator_2.md b/contents/slides/kondensator_2.md
index 40accd6..05aa6a4 100644
--- a/contents/slides/kondensator_2.md
+++ b/contents/slides/kondensator_2.md
@@ -64,7 +64,7 @@ Ein VNA misst die Veränderung des Blindwiderstandes $X_C$ in Abhängigkeit der
 * gesucht: $X_{\textrm{C}}$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber X_{\textrm{C}} &= \frac{1}{\omega \cdot C} = \frac{1}{2\pi \cdot f \cdot C}\\ &= \frac{1}{2\pi \cdot 100MHz \cdot 10pF}\\ &\approx 159\Omega \end{split}\end{equation}$
+$\begin{split} X_{\textrm{C}} &= \frac{1}{\omega \cdot C} = \frac{1}{2\pi \cdot f \cdot C}\\ &= \frac{1}{2\pi \cdot 100MHz \cdot 10pF}\\ &\approx 159\Omega \end{split}$
 </fragment>
 
 ---
@@ -76,7 +76,7 @@ $\begin{equation}\begin{split}\nonumber X_{\textrm{C}} &= \frac{1}{\omega \cdot
 * gesucht: $X_{\textrm{C}}$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber X_{\textrm{C}} &= \frac{1}{\omega \cdot C} = \frac{1}{2\pi \cdot f \cdot C}\\ &= \frac{1}{2\pi \cdot 145MHz \cdot 50pF}\\ &\approx 22\Omega \end{split}\end{equation}$
+$\begin{split} X_{\textrm{C}} &= \frac{1}{\omega \cdot C} = \frac{1}{2\pi \cdot f \cdot C}\\ &= \frac{1}{2\pi \cdot 145MHz \cdot 50pF}\\ &\approx 22\Omega \end{split}$
 </fragment>
 ---
 [question:AC106]
@@ -87,7 +87,7 @@ $\begin{equation}\begin{split}\nonumber X_{\textrm{C}} &= \frac{1}{\omega \cdot
 * gesucht: $X_{\textrm{C}}$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber X_{\textrm{C}} &= \frac{1}{\omega \cdot C} = \frac{1}{2\pi \cdot f \cdot C}\\ &= \frac{1}{2\pi \cdot 100MHz \cdot 100pF}\\ &\approx 15,9\Omega \end{split}\end{equation}$
+$\begin{split} X_{\textrm{C}} &= \frac{1}{\omega \cdot C} = \frac{1}{2\pi \cdot f \cdot C}\\ &= \frac{1}{2\pi \cdot 100MHz \cdot 100pF}\\ &\approx 15,9\Omega \end{split}$
 </fragment>
 
 ---
@@ -99,7 +99,7 @@ $\begin{equation}\begin{split}\nonumber X_{\textrm{C}} &= \frac{1}{\omega \cdot
 * gesucht: $X_{\textrm{C}}$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber X_{\textrm{C}} &= \frac{1}{\omega \cdot C} = \frac{1}{2\pi \cdot f \cdot C}\\ &= \frac{1}{2\pi \cdot 435MHz \cdot 100pF}\\ &\approx 3,7\Omega \end{split}\end{equation}$
+$\begin{split} X_{\textrm{C}} &= \frac{1}{\omega \cdot C} = \frac{1}{2\pi \cdot f \cdot C}\\ &= \frac{1}{2\pi \cdot 435MHz \cdot 100pF}\\ &\approx 3,7\Omega \end{split}$
 </fragment>
 
 ---
@@ -120,7 +120,7 @@ $X_{\textrm{C}} = \frac{U}{I} = \frac{16V}{32mA} = 500\Omega$
 </fragment>
 
 <fragment>
-$\begin{equation}\begin{align}\nonumber X_{\textrm{C}} &= \frac{1}{\omega \cdot C} \\ \nonumber \Rightarrow C &= \frac{1}{\omega \cdot X_{\textrm{C}}} = \frac{1}{2\pi \cdot f \cdot X_{\textrm{C}}}\\ \nonumber &= \frac{1}{2\pi \cdot 50Hz \cdot 500\Omega}\\ \nonumber &\approx 6,37\mu F\end{align}\end{equation}$
+$\begin{split} X_{\textrm{C}} &= \frac{1}{\omega \cdot C} \\ \Rightarrow C &= \frac{1}{\omega \cdot X_{\textrm{C}}} = \frac{1}{2\pi \cdot f \cdot X_{\textrm{C}}}\\ &= \frac{1}{2\pi \cdot 50Hz \cdot 500\Omega}\\ &\approx 6,37\mu F\end{split}$
 </fragment>
 
 ---
diff --git a/contents/slides/ladung_energie.md b/contents/slides/ladung_energie.md
index 10cb366..bfe607e 100644
--- a/contents/slides/ladung_energie.md
+++ b/contents/slides/ladung_energie.md
@@ -26,7 +26,7 @@ in Joule (J) auf Sekunde bezogen oder Wattstunden (Wh)
 * gesucht: $W$
 
 <fragment>
-$\begin{equation} \nonumber W = P\cdot t = U\cdot I\cdot t = 230V\cdot 0,63A\cdot 7h = 1,01kWh \end{equation}$
+$$W = P\cdot t = U\cdot I\cdot t = 230V\cdot 0,63A\cdot 7h = 1,01kWh$$
 </fragment>
 
 ---
@@ -39,5 +39,5 @@ $\begin{equation} \nonumber W = P\cdot t = U\cdot I\cdot t = 230V\cdot 0,63A\cdo
 * gesucht: $W$
 
 <fragment>
-$\begin{equation}\begin{split} \nonumber W = P\cdot t = \frac{U^2}{R} \cdot t = \frac{(10V)^2}{100\Omega}\cdot 1h &= 1Wh \cdot 3600\frac{s}{h}\\ &= 3600J\end{split}\end{equation}$
+$\begin{split} W = P\cdot t = \frac{U^2}{R} \cdot t = \frac{(10V)^2}{100\Omega}\cdot 1h &= 1Wh \cdot 3600\frac{s}{h}\\ &= 3600J\end{split}$
 </fragment>
\ No newline at end of file
diff --git a/contents/slides/leistungsvertaerker.md b/contents/slides/leistungsvertaerker.md
index 4ea63f5..0331920 100644
--- a/contents/slides/leistungsvertaerker.md
+++ b/contents/slides/leistungsvertaerker.md
@@ -176,12 +176,12 @@
 
 <left>
 <fragment>
-$\begin{aligned}R_E &= \frac{(R_3+R_6) \cdot R_4}{(R_3 + R_6) + R_4}\\ &= \frac{220Ω + 150Ω) \cdot 6,8kΩ}{220Ω + 150Ω + 6,8kΩ}\\ &= \frac{2,516MΩ^2}{7170Ω}\\ &= 351Ω\end{aligned}$
+$\begin{split}R_E &= \frac{(R_3+R_6) \cdot R_4}{(R_3 + R_6) + R_4}\\ &= \frac{220Ω + 150Ω) \cdot 6,8kΩ}{220Ω + 150Ω + 6,8kΩ}\\ &= \frac{2,516MΩ^2}{7170Ω}\\ &= 351Ω\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}\frac{U_Z}{U_{GS}} &= \frac{R_2 + R_E}{R_E}\\ \Rightarrow \frac{6,2V}{U_{GS}} &= \frac{270Ω+351Ω}{351Ω}\\ &= 1,77\\ \Rightarrow U_{GS} &= \frac{6,2V}{1,77}\\ &= 3,50V\end{aligned}$
+$\begin{split}\frac{U_Z}{U_{GS}} &= \frac{R_2 + R_E}{R_E}\\ \Rightarrow \frac{6,2V}{U_{GS}} &= \frac{270Ω+351Ω}{351Ω}\\ &= 1,77\\ \Rightarrow U_{GS} &= \frac{6,2V}{1,77}\\ &= 3,50V\end{split}$
 </fragment>
 </right>
 
@@ -243,11 +243,11 @@ $\begin{aligned}\frac{U_Z}{U_{GS}} &= \frac{R_2 + R_E}{R_E}\\ \Rightarrow \frac{
 
 <left>
 <fragment>
-$\begin{aligned}g &= P_2 - P_1\\ &= 43dBm - (-5dBm)\\ &= 43dBm + 5dBm\\ &= 48dB\end{aligned}$
+$\begin{split}g &= P_2 - P_1\\ &= 43dBm - (-5dBm)\\ &= 43dBm + 5dBm\\ &= 48dB\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}g &= 10 \cdot \log_{10}{(\frac{P_2}{P_1})}dB\\ &= 10 \cdot \log_{10}{(\frac{20W}{0,3mW})}dB \\ &\approx 48dB\end{aligned}$
+$\begin{split}g &= 10 \cdot \log_{10}{(\frac{P_2}{P_1})}dB\\ &= 10 \cdot \log_{10}{(\frac{20W}{0,3mW})}dB \\ &\approx 48dB\end{split}$
 </fragment>
 </right>
\ No newline at end of file
diff --git a/contents/slides/leiterwiderstand.md b/contents/slides/leiterwiderstand.md
index 23eed60..d719c46 100644
--- a/contents/slides/leiterwiderstand.md
+++ b/contents/slides/leiterwiderstand.md
@@ -43,10 +43,10 @@ Damit lässt sich der ohmsche Widerstand eines Drahtes bei bekanntem Material, L
 * gesucht: $R$
 
 <fragment>
-$\begin{equation} \nonumber A_{\textrm{Dr}} = \frac{d^2\cdot \pi}{4} = \frac{(0,2mm)^2 \cdot \pi}{4} = \frac{\pi}{100}mm^2 = 0,0314mm^2 \end{equation}$
+$$A_{\textrm{Dr}} = \frac{d^2\cdot \pi}{4} = \frac{(0,2mm)^2 \cdot \pi}{4} = \frac{\pi}{100}mm^2 = 0,0314mm^2$$
 </fragment>
 <fragment>
-$\begin{equation} \nonumber R = \frac{\rho\cdot l}{A_{\textrm{Dr}}} = \frac{0,018 \frac{\Omega mm^2}{m} \cdot 1,8m}{0,0314mm^2} \approx 1,02\Omega \end{equation}$
+$$R = \frac{\rho\cdot l}{A_{\textrm{Dr}}} = \frac{0,018 \frac{\Omega mm^2}{m} \cdot 1,8m}{0,0314mm^2} \approx 1,02\Omega$$
 </fragment>
 ---
 [question:AB102]
@@ -58,7 +58,7 @@ $\begin{equation} \nonumber R = \frac{\rho\cdot l}{A_{\textrm{Dr}}} = \frac{0,01
 * gesucht: $l$
 
 <fragment>
-$\begin{equation}\begin{align} \nonumber R &= \frac{\rho\cdot l}{A_{\textrm{Dr}}}\\ \nonumber \Rightarrow l &= \frac{R\cdot A_{\textrm{Dr}}}{\rho} = \frac{1,5\Omega \cdot 0,5mm^2}{0,018 \frac{\Omega mm^2}{m}} \approx 41,7m \end{align}\end{equation}$
+$\begin{split} R &= \frac{\rho\cdot l}{A_{\textrm{Dr}}}\\ \Rightarrow l &= \frac{R\cdot A_{\textrm{Dr}}}{\rho} = \frac{1,5\Omega \cdot 0,5mm^2}{0,018 \frac{\Omega mm^2}{m}} \approx 41,7m \end{split}$
 </fragment>
 
 ---
diff --git a/contents/slides/mischer.md b/contents/slides/mischer.md
index 3f538a6..6a8b557 100644
--- a/contents/slides/mischer.md
+++ b/contents/slides/mischer.md
@@ -1,8 +1,8 @@
 <left>
 * Beim Mischen von zwei Eingangs-Frequenzen entstehen immer zwei Ausgangs-Frequenzen
 
-$\begin{equation}f_\text{A1} = f_\text{E1} + f_\text{E2}\end{equation}$
-$\begin{equation}f_\text{A2} = |f_\text{E1} - f_\text{E2}|\end{equation}$
+$$f_\text{A1} = f_\text{E1} + f_\text{E2}$$
+$$f_\text{A2} = |f_\text{E1} - f_\text{E2}|$$
 </left>
 <right>
 [picture:102:e_mischer:Mischer mit zwei Eingangsfrequenzen]
diff --git a/contents/slides/modulatoren.md b/contents/slides/modulatoren.md
index 00995a3..544fb6d 100644
--- a/contents/slides/modulatoren.md
+++ b/contents/slides/modulatoren.md
@@ -137,7 +137,7 @@
 * gesucht: $f_{USB}$
 
 <fragment>
-$\begin{aligned}f_{USB} &= f_Q - (f_{LSB} - f_Q)\\ &= 9MHz - (9,0015MHz - 9MHz)\\ &= 9MHz - 0,0015MHz\\ &=8,9985MHz\end{aligned}$ 
+$\begin{split}f_{USB} &= f_Q - (f_{LSB} - f_Q)\\ &= 9MHz - (9,0015MHz - 9MHz)\\ &= 9MHz - 0,0015MHz\\ &=8,9985MHz\end{split}$ 
 </fragment>
 
 ---
diff --git a/contents/slides/muf_luf_2.md b/contents/slides/muf_luf_2.md
index 2d87552..25f3d7d 100644
--- a/contents/slides/muf_luf_2.md
+++ b/contents/slides/muf_luf_2.md
@@ -57,12 +57,12 @@ Kritische Frequenz wird auch als $f_k$ oder $f_\mathrm{krit}$ angegeben
 
 <left>
 <fragment>
-$\begin{equation}\begin{split} \nonumber \mathrm{MUF} & \approx \frac{f_c}{\sin(\alpha)}\\&\approx \frac{\qty{3}{\mega\hertz}}{0,71}\\&\approx \qty{4,2}{\mega\hertz}\end{split}\end{equation}$
+$\begin{split} \mathrm{MUF} & \approx \frac{f_c}{\sin(\alpha)}\\&\approx \frac{3MHz}{0,71}\\&\approx 4,2MHz\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{equation}\begin{split} \nonumber f_{\mathrm{opt}} &= \mathrm{MUF}\cdot 0,85\\ &= \qty{4,2}{\mega\hertz} \cdot 0,85\\ &= \qty{3,6}{\mega\hertz} \end{split}\end{equation}$
+$\begin{split} f_{\mathrm{opt}} &= \mathrm{MUF}\cdot 0,85\\ &= 4,2MHz \cdot 0,85\\ &= 3,6MHz \end{split}$
 </fragment>
 </right>
 
diff --git a/contents/slides/naeherungsformel_1.md b/contents/slides/naeherungsformel_1.md
index d237e23..91c3c1b 100644
--- a/contents/slides/naeherungsformel_1.md
+++ b/contents/slides/naeherungsformel_1.md
@@ -6,7 +6,7 @@
 * Gilt nur im Freiraum <br/> ($d > \frac{\lambda}{2\pi}$)
 </left>
 <right>
-$\begin{equation}\begin{split}\nonumber E &= \dfrac{\sqrt{30\Omega \cdot P_A \cdot G_i}}{d}\\ &= \dfrac{\sqrt{30\Omega \cdot P_{\textrm{EIRP}}}}{d} \end{split}\end{equation}$
+$\begin{split} E &= \dfrac{\sqrt{30\Omega \cdot P_A \cdot G_i}}{d}\\ &= \dfrac{\sqrt{30\Omega \cdot P_{\textrm{EIRP}}}}{d} \end{split}$
 </right>
 
 ---
@@ -16,7 +16,7 @@ $\begin{equation}\begin{split}\nonumber E &= \dfrac{\sqrt{30\Omega \cdot P_A \cd
 * Umstellen nach $d$
 </left>
 <right>
-$\begin{equation}\begin{split}\nonumber d &= \dfrac{\sqrt{30\Omega \cdot P_A \cdot G_i}}{E}\\ &= \dfrac{\sqrt{30\Omega \cdot P_{\textrm{EIRP}}}}{E} \end{split}\end{equation}$
+$\begin{split} d &= \dfrac{\sqrt{30\Omega \cdot P_A \cdot G_i}}{E}\\ &= \dfrac{\sqrt{30\Omega \cdot P_{\textrm{EIRP}}}}{E} \end{split}$
 </right>
 
 ---
@@ -36,12 +36,12 @@ $\begin{equation}\begin{split}\nonumber d &= \dfrac{\sqrt{30\Omega \cdot P_A \cd
 
 <left>
 <fragment>
-$\begin{equation}\begin{split} \nonumber P_{\textrm{EIRP}} &= P_S \cdot 10^{\frac{g_d - a + 2,15dB}{10dB}}\\ &= 100W \cdot 10^{\frac{7,5dB - 1,5dB + 2,15dB}{10dB}}\\ &\approx 100W \cdot 6,5\\ &= 650W \end{split}\end{equation}$
+$\begin{split} P_{\textrm{EIRP}} &= P_S \cdot 10^{\frac{g_d - a + 2,15dB}{10dB}}\\ &= 100W \cdot 10^{\frac{7,5dB - 1,5dB + 2,15dB}{10dB}}\\ &\approx 100W \cdot 6,5\\ &= 650W \end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{equation}\begin{split} \nonumber d &= \dfrac{\sqrt{30\Omega \cdot P_{\textrm{EIRP}}}}{E}\\ &= \dfrac{\sqrt{30\Omega \cdot 650W}}{28\frac{V}{m}}\\ &\approx 5m \end{split}\end{equation}$
+$\begin{split} d &= \dfrac{\sqrt{30\Omega \cdot P_{\textrm{EIRP}}}}{E}\\ &= \dfrac{\sqrt{30\Omega \cdot 650W}}{28\frac{V}{m}}\\ &\approx 5m \end{split}$
 </fragment>
 </right>
 
@@ -51,7 +51,7 @@ $\begin{equation}\begin{split} \nonumber d &= \dfrac{\sqrt{30\Omega \cdot P_{\te
 Liegen die errechneten 5m nicht im Nahfeld für das 10m-Band aus der Frage?
 
 <fragment>
-$\begin{equation}\nonumber \begin{split} \nonumber d &> \frac{\lambda}{2\pi}\\ \nonumber 5m &> \frac{10m}{2\pi}\\ \nonumber 5m &\gtrapprox 1,6m \end{split}\end{equation}$
+$\begin{split} d &> \frac{\lambda}{2\pi}\\ 5m &> \frac{10m}{2\pi}\\ 5m &\gtrapprox 1,6m \end{split}$
 </fragment>
 
 ---
diff --git a/contents/slides/naeherungsformel_2.md b/contents/slides/naeherungsformel_2.md
index 69f1d1c..ee9fbb2 100644
--- a/contents/slides/naeherungsformel_2.md
+++ b/contents/slides/naeherungsformel_2.md
@@ -34,7 +34,7 @@ DIN EN 50413, für Amateurfunk relevant nur ATV mit 0,38 und SATV mit 0,54
 </right>
 
 <fragment>
-$\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_A \cdot G_i}}{d}\\ \Rightarrow d &= \frac{\sqrt{30Ω \cdot P_A \cdot G_i}}{E}\\ &= \frac{\sqrt{30Ω \cdot 100W \cdot 1,64}}{28\frac{V}{m}}\\ &= 2,5m\end{aligned}$
+$\begin{split}E &= \frac{\sqrt{30Ω \cdot P_A \cdot G_i}}{d}\\ \Rightarrow d &= \frac{\sqrt{30Ω \cdot P_A \cdot G_i}}{E}\\ &= \frac{\sqrt{30Ω \cdot 100W \cdot 1,64}}{28\frac{V}{m}}\\ &= 2,5m\end{split}$
 </fragment>
 
 ---
@@ -63,12 +63,12 @@ $P_\text{EIRP} = P_S\cdot10^{\frac{g_d - a + 2,15}{10}}$
 
 <fragment>
 <left>
-$\begin{aligned}P_{EIRP} &= P_S \cdot 10^{\frac{g_d -a + 2,15dB}{10dB}}\\ &= 300W \cdot 10^{\frac{0dBd - 0,5dB + 2,15dB}{10dB}}\\ &= 438,7W\end{aligned}$
+$\begin{split}P_{EIRP} &= P_S \cdot 10^{\frac{g_d -a + 2,15dB}{10dB}}\\ &= 300W \cdot 10^{\frac{0dBd - 0,5dB + 2,15dB}{10dB}}\\ &= 438,7W\end{split}$
 </left>
 </fragment>
 <fragment>
 <right>
-$\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{E}\\ &= \frac{\sqrt{30Ω \cdot 438,7W}}{28\frac{V}{m}}\\ &= 4,10m\end{aligned}$
+$\begin{split}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{E}\\ &= \frac{\sqrt{30Ω \cdot 438,7W}}{28\frac{V}{m}}\\ &= 4,10m\end{split}$
 </right>
 </fragment>
 
@@ -90,12 +90,12 @@ $\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \fr
 
 <fragment>
 <left>
-$\begin{aligned}P_{EIRP} &= P_S \cdot 10^{\frac{g_d -a + 2,15dB}{10dB}}\\ &= 700W \cdot 10^{\frac{0dBd - 0,5dB + 2,15dB}{10dB}}\\ &= 1023,5W\end{aligned}$
+$\begin{split}P_{EIRP} &= P_S \cdot 10^{\frac{g_d -a + 2,15dB}{10dB}}\\ &= 700W \cdot 10^{\frac{0dBd - 0,5dB + 2,15dB}{10dB}}\\ &= 1023,5W\end{split}$
 </left>
 </fragment>
 <fragment>
 <right>
-$\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{E}\\ &= \frac{\sqrt{30Ω \cdot 1023,5W}}{28\frac{V}{m}}\\ &= 6,26m\end{aligned}$
+$\begin{split}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{E}\\ &= \frac{\sqrt{30Ω \cdot 1023,5W}}{28\frac{V}{m}}\\ &= 6,26m\end{split}$
 </right>
 </fragment>
 
@@ -117,12 +117,12 @@ $\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \fr
 
 <fragment>
 <left>
-$\begin{aligned}P_{EIRP} &= P_S \cdot 10^{\frac{g_d -a + 2,15dB}{10dB}}\\ &= 75W \cdot 10^{\frac{11,5dBd - 1,5dB + 2,15dB}{10dB}}\\ &= 1230,4W\end{aligned}$
+$\begin{split}P_{EIRP} &= P_S \cdot 10^{\frac{g_d -a + 2,15dB}{10dB}}\\ &= 75W \cdot 10^{\frac{11,5dBd - 1,5dB + 2,15dB}{10dB}}\\ &= 1230,4W\end{split}$
 </left>
 </fragment>
 <fragment>
 <right>
-$\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{E}\\ &= \frac{\sqrt{30Ω \cdot 1230,4W}}{28\frac{V}{m}}\\ &= 6,86m\end{aligned}$
+$\begin{split}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{E}\\ &= \frac{\sqrt{30Ω \cdot 1230,4W}}{28\frac{V}{m}}\\ &= 6,86m\end{split}$
 </right>
 </fragment>
 
@@ -144,12 +144,12 @@ $\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \fr
 
 <fragment>
 <left>
-$\begin{aligned}P_{EIRP} &= P_S \cdot 10^{\frac{g_d -a + 2,15dB}{10dB}}\\ &= 100W \cdot 10^{\frac{10,5dBd - 1,5dB + 2,15dB}{10dB}}\\ &= 1303,2W\end{aligned}$
+$\begin{split}P_{EIRP} &= P_S \cdot 10^{\frac{g_d -a + 2,15dB}{10dB}}\\ &= 100W \cdot 10^{\frac{10,5dBd - 1,5dB + 2,15dB}{10dB}}\\ &= 1303,2W\end{split}$
 </left>
 </fragment>
 <fragment>
 <right>
-$\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{E}\\ &= \frac{\sqrt{30Ω \cdot 1303,2W}}{28\frac{V}{m}}\\ &= 7,1m\end{aligned}$
+$\begin{split}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{E}\\ &= \frac{\sqrt{30Ω \cdot 1303,2W}}{28\frac{V}{m}}\\ &= 7,1m\end{split}$
 </right>
 </fragment>
 
@@ -171,11 +171,11 @@ $\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \fr
 
 <fragment>
 <left>
-$\begin{aligned}P_{EIRP} &= P_S \cdot 10^{\frac{g_d -a + 2,15dB}{10dB}}\\ &= 40W \cdot 10^{\frac{18dBd - 2dB + 2,15dB}{10dB}}\\ &= 2612,5W\end{aligned}$
+$\begin{split}P_{EIRP} &= P_S \cdot 10^{\frac{g_d -a + 2,15dB}{10dB}}\\ &= 40W \cdot 10^{\frac{18dBd - 2dB + 2,15dB}{10dB}}\\ &= 2612,5W\end{split}$
 </left>
 </fragment>
 <fragment>
 <right>
-$\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{E}\\ &= \frac{\sqrt{30Ω \cdot 2612,5W}}{61\frac{V}{m}}\\ &= 4,6m\end{aligned}$
+$\begin{split}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow d &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{E}\\ &= \frac{\sqrt{30Ω \cdot 2612,5W}}{61\frac{V}{m}}\\ &= 4,6m\end{split}$
 </right>
 </fragment>
diff --git a/contents/slides/nicht_sinus_signale.md b/contents/slides/nicht_sinus_signale.md
index 8d2117f..5435874 100644
--- a/contents/slides/nicht_sinus_signale.md
+++ b/contents/slides/nicht_sinus_signale.md
@@ -110,5 +110,5 @@ $3 \cdot f = 3 \cdot 7,050MHz = 21,150MHz$
 * gesucht: mehrere Harmonische
 
 <fragment>
-$\begin{aligned}2 \cdot 144,300MHz &= 288,600MHz\\ 3 \cdot 144,300MHz &= \bold{432,900MHz}\\ &\vdots\\ 9 \cdot 144,300MHz &= \bold{1298,700MHz}\end{aligned}$
+$\begin{split}2 \cdot 144,300MHz &= 288,600MHz\\ 3 \cdot 144,300MHz &= \bold{432,900MHz}\\ &\vdots\\ 9 \cdot 144,300MHz &= \bold{1298,700MHz}\end{split}$
 </fragment>
diff --git a/contents/slides/oszilloskop_2.md b/contents/slides/oszilloskop_2.md
index a000b24..372e685 100644
--- a/contents/slides/oszilloskop_2.md
+++ b/contents/slides/oszilloskop_2.md
@@ -58,7 +58,7 @@ Definition: Die Pulsbreite liegt bei 50% des Spitzenwerts
 * gesucht: $P_{\textrm{PEP}}$
 
 <fragment>
-$\begin{equation}\begin{split} \nonumber P_{\textrm{PEP}} &= \frac{U_{\textrm{eff}}^2}{R} = \frac{(\frac{100V}{\sqrt{2}})^2}{50\Omega}\\ &=\frac{\frac{(100V)^2}{2}}{50\Omega} = \frac{5000V^2}{50\Omega} = 100W \end{split}\end{equation}$
+$\begin{split} P_{\textrm{PEP}} &= \frac{U_{\textrm{eff}}^2}{R} = \frac{(\frac{100V}{\sqrt{2}})^2}{50\Omega}\\ &=\frac{\frac{(100V)^2}{2}}{50\Omega} = \frac{5000V^2}{50\Omega} = 100W \end{split}$
 </fragment>
 ---
 [question:AI306]
@@ -69,7 +69,7 @@ $\begin{equation}\begin{split} \nonumber P_{\textrm{PEP}} &= \frac{U_{\textrm{ef
 * gesucht: $P_{\textrm{PEP}}$
 
 <fragment>
-$\begin{equation}\begin{split} \nonumber P_{\textrm{PEP}} &= \frac{U_{\textrm{eff}}^2}{R} = \frac{(\frac{6V\cdot 10}{\sqrt{2}})^2}{50\Omega}\\ &=\frac{\frac{(60V)^2}{2}}{50\Omega} = 36W \end{split}\end{equation}$
+$\begin{split} P_{\textrm{PEP}} &= \frac{U_{\textrm{eff}}^2}{R} = \frac{(\frac{6V\cdot 10}{\sqrt{2}})^2}{50\Omega}\\ &=\frac{\frac{(60V)^2}{2}}{50\Omega} = 36W \end{split}$
 </fragment>
 
 ---
diff --git a/contents/slides/parbolspiegel_2.md b/contents/slides/parbolspiegel_2.md
index 45b38a9..1226ff9 100644
--- a/contents/slides/parbolspiegel_2.md
+++ b/contents/slides/parbolspiegel_2.md
@@ -51,7 +51,7 @@ $g_i = 10 \cdot \log_{10}{(\frac{\pi \cdot d}{\lambda})^2} \cdot \eta\,dB$
 $\lambda = \frac{c}{f} = \frac{3\cdot 10^8\frac{m}{s}}{5,7GHz} = 0,053m$
 </fragment>
 <fragment>
-$\begin{aligned}g_i &= 10 \cdot \log_{10}{(\frac{\pi \cdot d}{\lambda})^2} \cdot \eta\,dB\\ &= 10 \cdot \log_{10}{(\frac{\pi \cdot 0,3m}{0,053m})^2} \cdot 1dB\\ &= 25,1dBi\end{aligned}$
+$\begin{split}g_i &= 10 \cdot \log_{10}{(\frac{\pi \cdot d}{\lambda})^2} \cdot \eta\,dB\\ &= 10 \cdot \log_{10}{(\frac{\pi \cdot 0,3m}{0,053m})^2} \cdot 1dB\\ &= 25,1dBi\end{split}$
 </fragment>
 ---
 [question:AG227]
@@ -70,7 +70,7 @@ $\begin{aligned}g_i &= 10 \cdot \log_{10}{(\frac{\pi \cdot d}{\lambda})^2} \cdot
 $\lambda = \frac{c}{f} = \frac{3\cdot 10^8\frac{m}{s}}{5,7GHz} = 0,053m$
 </fragment>
 <fragment>
-$\begin{aligned}g_i &= 10 \cdot \log_{10}{(\frac{\pi \cdot d}{\lambda})^2} \cdot \eta\,dB\\ &= 10 \cdot \log_{10}{(\frac{\pi \cdot 0,8m}{0,053m})^2} \cdot 1dB\\ &= 33,6dBi\end{aligned}$
+$\begin{split}g_i &= 10 \cdot \log_{10}{(\frac{\pi \cdot d}{\lambda})^2} \cdot \eta\,dB\\ &= 10 \cdot \log_{10}{(\frac{\pi \cdot 0,8m}{0,053m})^2} \cdot 1dB\\ &= 33,6dBi\end{split}$
 </fragment>
 ---
 [question:AG228]
@@ -89,7 +89,7 @@ $\begin{aligned}g_i &= 10 \cdot \log_{10}{(\frac{\pi \cdot d}{\lambda})^2} \cdot
 $\lambda = \frac{c}{f} = \frac{3\cdot 10^8\frac{m}{s}}{10,4GHz} = 0,029m$
 </fragment>
 <fragment>
-$\begin{aligned}g_i &= 10 \cdot \log_{10}{(\frac{\pi \cdot d}{\lambda})^2} \cdot \eta\,dB\\ &= 10 \cdot \log_{10}{(\frac{\pi \cdot 0,8m}{0,029m})^2} \cdot 1dB\\ &= 38,8dBi\end{aligned}$
+$\begin{split}g_i &= 10 \cdot \log_{10}{(\frac{\pi \cdot d}{\lambda})^2} \cdot \eta\,dB\\ &= 10 \cdot \log_{10}{(\frac{\pi \cdot 0,8m}{0,029m})^2} \cdot 1dB\\ &= 38,8dBi\end{split}$
 </fragment>
 ---
 [question:AG229]
@@ -108,5 +108,5 @@ $\begin{aligned}g_i &= 10 \cdot \log_{10}{(\frac{\pi \cdot d}{\lambda})^2} \cdot
 $\lambda = \frac{c}{f} = \frac{3\cdot 10^8\frac{m}{s}}{10,4GHz} = 0,029m$
 </fragment>
 <fragment>
-$\begin{aligned}g_i &= 10 \cdot \log_{10}{(\frac{\pi \cdot d}{\lambda})^2} \cdot \eta\,dB\\ &= 10 \cdot \log_{10}{(\frac{\pi \cdot 1,2m}{0,029m})^2} \cdot 1dB\\ &= 42,3dBi\end{aligned}$
+$\begin{split}g_i &= 10 \cdot \log_{10}{(\frac{\pi \cdot d}{\lambda})^2} \cdot \eta\,dB\\ &= 10 \cdot \log_{10}{(\frac{\pi \cdot 1,2m}{0,029m})^2} \cdot 1dB\\ &= 42,3dBi\end{split}$
 </fragment>
\ No newline at end of file
diff --git a/contents/slides/personenschutzabstand_3.md b/contents/slides/personenschutzabstand_3.md
index bd998da..351e0e4 100644
--- a/contents/slides/personenschutzabstand_3.md
+++ b/contents/slides/personenschutzabstand_3.md
@@ -49,12 +49,12 @@ $d > \dfrac{\lambda}{2\pi}$
 
 <fragment>
 <left>
-$\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow P_{EIRP} &= \frac{(E \cdot d)^2}{30Ω}\\ &= \frac{(28\frac{V}{m} \cdot 5m)^2}{30Ω}\\ &= 653W\end{aligned}$
+$\begin{split}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ \Rightarrow P_{EIRP} &= \frac{(E \cdot d)^2}{30Ω}\\ &= \frac{(28\frac{V}{m} \cdot 5m)^2}{30Ω}\\ &= 653W\end{split}$
 </left>
 </fragment>
 <fragment>
 <right>
-$\begin{aligned}P_{EIRP} &= P_S \cdot 10^{\frac{g_d - a + 2,15dB}{10dB}}\\ \Rightarrow P_S &= \frac{P_{EIRP}}{10^{\frac{g_d - a + 2,15dB}{10dB}}}\\ &= \frac{653W}{10^{\frac{6dBd - 0 + 2,15dB}{10dB}}}\\ &= \frac{653W}{6,53}\\ &\approx 100W\end{aligned}$
+$\begin{split}P_{EIRP} &= P_S \cdot 10^{\frac{g_d - a + 2,15dB}{10dB}}\\ \Rightarrow P_S &= \frac{P_{EIRP}}{10^{\frac{g_d - a + 2,15dB}{10dB}}}\\ &= \frac{653W}{10^{\frac{6dBd - 0 + 2,15dB}{10dB}}}\\ &= \frac{653W}{6,53}\\ &\approx 100W\end{split}$
 </right>
 </fragment>
 
@@ -75,12 +75,12 @@ $\begin{aligned}P_{EIRP} &= P_S \cdot 10^{\frac{g_d - a + 2,15dB}{10dB}}\\ \Righ
 
 <fragment>
 <left>
-$\begin{aligned}G_i &= 10^{\frac{g_i}{10dB}}\\ &= 10^{\frac{12,15dBi}{10dB}}\\ &= 16,4\end{aligned}$
+$\begin{split}G_i &= 10^{\frac{g_i}{10dB}}\\ &= 10^{\frac{12,15dBi}{10dB}}\\ &= 16,4\end{split}$
 </left>
 </fragment>
 <fragment>
 <right>
-$\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_A \cdot G_i}}{d}\\ &= \frac{\sqrt{30Ω \cdot 250W \cdot 16,4}}{30m}\\ &= \frac{350V}{30m}\\ &\approx 11,7\frac{V}{m}\end{aligned}$
+$\begin{split}E &= \frac{\sqrt{30Ω \cdot P_A \cdot G_i}}{d}\\ &= \frac{\sqrt{30Ω \cdot 250W \cdot 16,4}}{30m}\\ &= \frac{350V}{30m}\\ &\approx 11,7\frac{V}{m}\end{split}$
 </right>
 </fragment>
 
@@ -96,12 +96,12 @@ $\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_A \cdot G_i}}{d}\\ &= \frac{\sqrt{
 
 <fragment>
 <left>
-$\begin{aligned}P_{EIRP} &= P_{ERP} \cdot 1,64\\ &= 100W \cdot 1,64\\ &= 164W\end{aligned}$
+$\begin{split}P_{EIRP} &= P_{ERP} \cdot 1,64\\ &= 100W \cdot 1,64\\ &= 164W\end{split}$
 </left>
 </fragment>
 <fragment>
 <right>
-$\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ &= \frac{\sqrt{30Ω \cdot 164W}}{100m}\\ &= 0,7\frac{V}{m}\end{aligned}$
+$\begin{split}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ &= \frac{\sqrt{30Ω \cdot 164W}}{100m}\\ &= 0,7\frac{V}{m}\end{split}$
 </right>
 </fragment>
 
@@ -117,11 +117,11 @@ $\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ &= \frac{\sqrt{30Ω
 
 <fragment>
 <left>
-$\begin{aligned}P_{EIRP} &= P_{ERP} \cdot 1,64\\ &= 100W \cdot 1,64\\ &= 164W\end{aligned}$
+$\begin{split}P_{EIRP} &= P_{ERP} \cdot 1,64\\ &= 100W \cdot 1,64\\ &= 164W\end{split}$
 </left>
 </fragment>
 <fragment>
 <right>
-$\begin{aligned}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ &= \frac{\sqrt{30Ω \cdot 164W}}{100m}\\ &= 0,7\frac{V}{m}\end{aligned}$
+$\begin{split}E &= \frac{\sqrt{30Ω \cdot P_{EIRP}}}{d}\\ &= \frac{\sqrt{30Ω \cdot 164W}}{100m}\\ &= 0,7\frac{V}{m}\end{split}$
 </right>
 </fragment>
diff --git a/contents/slides/reihe_parallel_gemischt.md b/contents/slides/reihe_parallel_gemischt.md
index afc2291..2b8e49b 100644
--- a/contents/slides/reihe_parallel_gemischt.md
+++ b/contents/slides/reihe_parallel_gemischt.md
@@ -18,7 +18,7 @@ Eigenkapazität einer Spule
 * gesucht: $C_{\textrm{ges}}$
 
 <fragment>
-$\begin{equation}\begin{align}\nonumber \tfrac{1}{C_{\textrm{ges}}} &= \tfrac{1}{C_1} + \tfrac{1}{C_2} + \tfrac{1}{C_3} = \tfrac{1}{0,10nF} + \tfrac{1}{47pF} + \tfrac{1}{22pF}\\ \nonumber &= 7,67\cdot 10^{10} \tfrac{1}{F}\\ \nonumber \Rightarrow C_{\textrm{ges}} &= \frac{1}{7,67\cdot 10^{10} \frac{1}{F}} \approx 13,0pF \end{align}\end{equation}$
+$\begin{split} \tfrac{1}{C_{\textrm{ges}}} &= \tfrac{1}{C_1} + \tfrac{1}{C_2} + \tfrac{1}{C_3} = \tfrac{1}{0,10nF} + \tfrac{1}{47pF} + \tfrac{1}{22pF}\\ &= 7,67\cdot 10^{10} \tfrac{1}{F}\\ \Rightarrow C_{\textrm{ges}} &= \frac{1}{7,67\cdot 10^{10} \frac{1}{F}} \approx 13,0pF \end{split}$
 </fragment>
 ---
 [question:AD103]
@@ -32,7 +32,7 @@ $\begin{equation}\begin{align}\nonumber \tfrac{1}{C_{\textrm{ges}}} &= \tfrac{1}
 * gesucht: $C_{\textrm{ges}}$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber C_{\textrm{ges}} &= C_1 + C_2 + C_3 + C_{\textrm{L}}\\ &= 0,1nF + 1,5nF + 220pF + 1pF\\ &= 1821pF \end{split}\end{equation}$
+$\begin{split} C_{\textrm{ges}} &= C_1 + C_2 + C_3 + C_{\textrm{L}}\\ &= 0,1nF + 1,5nF + 220pF + 1pF\\ &= 1821pF \end{split}$
 </fragment>
 <note>
 Einheitenpräfixe beachten und mit Kehrwerten rechnen
@@ -49,7 +49,7 @@ Einheitenpräfixe beachten und mit Kehrwerten rechnen
 * gesucht: $Z$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber X_{\textrm{L}} &= \omega \cdot L = 2 \cdot \pi \cdot f \cdot L\\ &= 2 \cdot \pi \cdot 1MHz \cdot 100\mu H = 628\Omega\end{split}\end{equation}$
+$\begin{split} X_{\textrm{L}} &= \omega \cdot L = 2 \cdot \pi \cdot f \cdot L\\ &= 2 \cdot \pi \cdot 1MHz \cdot 100\mu H = 628\Omega\end{split}$
 </fragment>
 <fragment>
 $Z = \sqrt{R^2 + X^2} = \sqrt{(100\Omega)^2 + (628\Omega)^2} \approx 636\Omega$
@@ -66,7 +66,7 @@ $Z = \sqrt{R^2 + X^2} = \sqrt{(100\Omega)^2 + (628\Omega)^2} \approx 636\Omega$
 * gesucht: $Z$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber X_{\textrm{C}} &= \frac{1}{\omega \cdot C} = \frac{1}{2 \cdot \pi \cdot f \cdot C}\\ &= \frac{1}{2 \cdot \pi \cdot 1MHz \cdot 100nF} = 159\Omega\end{split}\end{equation}$
+$\begin{split} X_{\textrm{C}} &= \frac{1}{\omega \cdot C} = \frac{1}{2 \cdot \pi \cdot f \cdot C}\\ &= \frac{1}{2 \cdot \pi \cdot 1MHz \cdot 100nF} = 159\Omega\end{split}$
 </fragment>
 <fragment>
 $Z = \sqrt{R^2 + X^2} = \sqrt{(100\Omega)^2 + (159\Omega)^2} \approx 188\Omega$
diff --git a/contents/slides/reihe_parallel_widerstandsnetz_2.md b/contents/slides/reihe_parallel_widerstandsnetz_2.md
index 43f8e91..976b57c 100644
--- a/contents/slides/reihe_parallel_widerstandsnetz_2.md
+++ b/contents/slides/reihe_parallel_widerstandsnetz_2.md
@@ -87,7 +87,7 @@ $R_{ges} = 200Ω + \frac{100Ω \cdot (200Ω + 1kΩ)}{100Ω + 200Ω +1kΩ} \appro
 * gesucht: $R_{ges}$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber R_{ges} &= \frac{(R_1 + R_2) \cdot (R_3 + R_4)}{(R_1 + R_2) + (R_3 + R_4)}\\ &= \frac{(2,2kΩ + 220Ω) \cdot (2,2kΩ + 220Ω)}{2,2kΩ + 220Ω + 2,2kΩ + 220Ω}\\ &= 1210Ω\end{split}\end{equation}$
+$\begin{split} R_{ges} &= \frac{(R_1 + R_2) \cdot (R_3 + R_4)}{(R_1 + R_2) + (R_3 + R_4)}\\ &= \frac{(2,2kΩ + 220Ω) \cdot (2,2kΩ + 220Ω)}{2,2kΩ + 220Ω + 2,2kΩ + 220Ω}\\ &= 1210Ω\end{split}$
 </fragment>
 ---
 [question:AD114]
diff --git a/contents/slides/schwingkreis_2.md b/contents/slides/schwingkreis_2.md
index e32acc3..ac9fe96 100644
--- a/contents/slides/schwingkreis_2.md
+++ b/contents/slides/schwingkreis_2.md
@@ -115,7 +115,7 @@ William Thomson, später Lord Kelvin, in 1853
 * gesucht: $f_0$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ &= \frac{1}{2 \cdot \pi \cdot \sqrt{1,2µH \cdot 6,8pF}} = 55,7MHz \end{split}\end{equation}$
+$\begin{split} f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ &= \frac{1}{2 \cdot \pi \cdot \sqrt{1,2µH \cdot 6,8pF}} = 55,7MHz \end{split}$
 </fragment>
 <fragment>
 Widerstand $R$ wird zur Berechnung nicht benötigt.
@@ -129,7 +129,7 @@ Widerstand $R$ wird zur Berechnung nicht benötigt.
 * gesucht: $f_0$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ &= \frac{1}{2 \cdot \pi \cdot \sqrt{10µH \cdot 1nF}} = 1,592MHz \end{split}\end{equation}$
+$\begin{split} f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ &= \frac{1}{2 \cdot \pi \cdot \sqrt{10µH \cdot 1nF}} = 1,592MHz \end{split}$
 </fragment>
 ---
 [question:AD210]
@@ -140,7 +140,7 @@ $\begin{equation}\begin{split}\nonumber f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{
 * gesucht: $f_0$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ = \frac{1}{2 \cdot \pi \cdot \sqrt{100µH \cdot 0,01µF}} = 159kHz \end{split}\end{equation}$
+$\begin{split} f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ &= \frac{1}{2 \cdot \pi \cdot \sqrt{100µH \cdot 0,01µF}} = 159kHz \end{split}$
 </fragment>
 ---
 [question:AD211]
@@ -151,7 +151,7 @@ $\begin{equation}\begin{split}\nonumber f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{
 * gesucht: $f_0$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ &= \frac{1}{2 \cdot \pi \cdot \sqrt{2,2µH \cdot 56pF}} = 14,34MHz$
+$\begin{split} f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ &= \frac{1}{2 \cdot \pi \cdot \sqrt{2,2µH \cdot 56pF}} = 14,34MHz \end{split}$
 </fragment>
 ---
 [question:AD212]
@@ -167,7 +167,7 @@ $\begin{equation}\begin{split}\nonumber f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{
 $C = C_1 + C_2 + C_3 = 0,1nF + 1,5nF + 220pF = 1,82nF$
 </fragment>
 <fragment>
-$\begin{equation}\begin{split}\nonumber f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ &= \frac{1}{2 \cdot \pi \cdot \sqrt{1,2mH \cdot 1,82nF}} = 107,7kHz \end{split}\end{equation}$
+$\begin{split} f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ &= \frac{1}{2 \cdot \pi \cdot \sqrt{1,2mH \cdot 1,82nF}} = 107,7kHz \end{split}$
 </fragment>
 ---
 ### Verändern der Resonanzfrequenz
@@ -271,11 +271,11 @@ $Q = \frac{f_0}{B} = \frac{R_P}{X_L}$
 </right>
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ &= \frac{1}{2 \cdot \pi \cdot \sqrt{100µH \cdot 0,01µF}} = 159,2kHz \end{split}\end{equation}$
+$\begin{split} f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ &= \frac{1}{2 \cdot \pi \cdot \sqrt{100µH \cdot 0,01µF}} = 159,2kHz \end{split}$
 </fragment>
 <fragment>
 $B$ oder $X_L$ ausrechnen
-$\begin{equation}\begin{split}\nonumber X_L = \omega \cdot L = 2 \cdot \pi \cdot f_0 \cdot L\\ &= 2 \cdot \pi \cdot 159,2kHz \cdot 100µH = 100,03Ω \end{split}\end{equation}$ 
+$\begin{split} X_L &= \omega \cdot L = 2 \cdot \pi \cdot f_0 \cdot L\\ &= 2 \cdot \pi \cdot 159,2kHz \cdot 100µH = 100,03Ω \end{split}$
 </fragment>
 <fragment>
 $Q = \frac{X_L}{R_S} = \frac{100,03Ω}{10Ω} \approx 10$
@@ -294,11 +294,11 @@ $Q = \frac{X_L}{R_S} = \frac{100,03Ω}{10Ω} \approx 10$
 </right>
   
 <fragment>
-$\begin{equation}\begin{split}\nonumber f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ &= \frac{1}{2 \cdot \pi \cdot \sqrt{2,2µH \cdot 56pF}} = 14,34MHz \end{split}\end{equation}$
+$\begin{split} f_0 &= \frac{1}{2 \cdot \pi \cdot \sqrt{L \cdot C}}\\ &= \frac{1}{2 \cdot \pi \cdot \sqrt{2,2µH \cdot 56pF}} = 14,34MHz \end{split}$
 </fragment>
 <fragment>
 $B$ oder $X_L$ ausrechnen
-$\begin{equation}\begin{split}\nonumber X_L &= \omega \cdot L = 2 \cdot \pi \cdot f_0 \cdot L\\ &= 2 \cdot \pi \cdot 14,34MHz \cdot 2,2µH = 198,2Ω \end{split}\end{equation}$ 
+$\begin{split} X_L &= \omega \cdot L = 2 \cdot \pi \cdot f_0 \cdot L\\ &= 2 \cdot \pi \cdot 14,34MHz \cdot 2,2µH = 198,2Ω \end{split}$
 </fragment>
 <fragment>
 $Q = \frac{R_P}{X_L} = \frac{1kΩ}{198,2Ω} \approx 5$
@@ -336,7 +336,7 @@ Herleitung nicht gezeigt
 * gesucht: $B$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber B &= \frac{1}{2\cdot \pi \cdot R_P \cdot C}\\ &= \frac{1}{2\cdot \pi \cdot 1kΩ \cdot 56pF} = 2,84MHz \end{split}\end{equation}$
+$\begin{split} B &= \frac{1}{2\cdot \pi \cdot R_P \cdot C}\\ &= \frac{1}{2\cdot \pi \cdot 1kΩ \cdot 56pF} = 2,84MHz \end{split}$
 </fragment>
 
 ---
diff --git a/contents/slides/sender_messungen.md b/contents/slides/sender_messungen.md
index 88a2198..e17b875 100644
--- a/contents/slides/sender_messungen.md
+++ b/contents/slides/sender_messungen.md
@@ -93,7 +93,7 @@ Keine Berechnung notwendig, da es nur eine Antwort mit Dämpfungsglied gibt
 
 
 <fragment>
-$\begin{aligned}R &= (\frac{1}{R_T + R_T} + \frac{1}{R_V} + \frac{1}{R_V})^{-1}\\ &= (\frac{1}{330Ω + 330Ω} + \frac{1}{110Ω} + \frac{1}{110Ω})^{-1}\\ &= 50,77Ω\end{aligned}$
+$\begin{split}R &= (\frac{1}{R_T + R_T} + \frac{1}{R_V} + \frac{1}{R_V})^{-1}\\ &= (\frac{1}{330Ω + 330Ω} + \frac{1}{110Ω} + \frac{1}{110Ω})^{-1}\\ &= 50,77Ω\end{split}$
 </fragment>
 
 --- style="font-size: smaller;"
@@ -103,7 +103,7 @@ $\begin{aligned}R &= (\frac{1}{R_T + R_T} + \frac{1}{R_V} + \frac{1}{R_V})^{-1}\
 * gesucht: $U_A$
 
 <fragment>
-$\begin{aligned}P_E &= \frac{U_{E,eff}^2}{R}\\ \Rightarrow U_{E,eff} &= \sqrt{P_E \cdot R}\\ &= \sqrt{1W \cdot 50,77Ω}\\ &= 7,125V\end{aligned}$
+$\begin{split}P_E &= \frac{U_{E,eff}^2}{R}\\ \Rightarrow U_{E,eff} &= \sqrt{P_E \cdot R}\\ &= \sqrt{1W \cdot 50,77Ω}\\ &= 7,125V\end{split}$
 </fragment>
 
 --- style="font-size: smaller;"
@@ -113,12 +113,12 @@ $\begin{aligned}P_E &= \frac{U_{E,eff}^2}{R}\\ \Rightarrow U_{E,eff} &= \sqrt{P_
 
 <left>
 <fragment>
-$\begin{aligned}U_S &= U_{E,eff} \cdot \sqrt{2}\\ &= 7,071V \cdot 1,414\\ &= 10,07V\end{aligned}$
+$\begin{split}U_S &= U_{E,eff} \cdot \sqrt{2}\\ &= 7,071V \cdot 1,414\\ &= 10,07V\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}U_A &= \frac{U_S}{2}\,-\,U_F\\ &= \frac{10,07V}{2}\,-\,0,23V\\ &= 5,035V\,-\,0,23V\\ &= 4,805V \approx 4,8V\end{aligned}$
+$\begin{split}U_A &= \frac{U_S}{2}\,-\,U_F\\ &= \frac{10,07V}{2}\,-\,0,23V\\ &= 5,035V\,-\,0,23V\\ &= 4,805V \approx 4,8V\end{split}$
 </fragment>
 </right>
 
@@ -146,12 +146,12 @@ $\begin{aligned}U_A &= \frac{U_S}{2}\,-\,U_F\\ &= \frac{10,07V}{2}\,-\,0,23V\\ &
 
 <left>
 <fragment>
-$\begin{aligned}R &= (\frac{1}{R_T + R_T} + \frac{1}{R_1})^{-1}\\ &= (\frac{1}{330Ω + 330Ω} + \frac{1}{54,1Ω})^{-1}\\ &= 50Ω\end{aligned}$
+$\begin{split}R &= (\frac{1}{R_T + R_T} + \frac{1}{R_1})^{-1}\\ &= (\frac{1}{330Ω + 330Ω} + \frac{1}{54,1Ω})^{-1}\\ &= 50Ω\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}U_S &= (U_A + U_F) \cdot 2\\ &= (14,9V + 0,7V) \cdot 2\\ &= 31,2V\end{aligned}$
+$\begin{split}U_S &= (U_A + U_F) \cdot 2\\ &= (14,9V + 0,7V) \cdot 2\\ &= 31,2V\end{split}$
 </fragment>
 </right>
 
@@ -162,12 +162,12 @@ $\begin{aligned}U_S &= (U_A + U_F) \cdot 2\\ &= (14,9V + 0,7V) \cdot 2\\ &= 31,2
 * gesucht: $P_E$
 
 <fragment>
-$\begin{aligned}U_{E,eff}\\ &= \frac{U_S}{\sqrt{2}}\\ &= \frac{31,2V}{1,414}\\ &= 22,06V\end{aligned}$
+$\begin{split}U_{E,eff}\\ &= \frac{U_S}{\sqrt{2}}\\ &= \frac{31,2V}{1,414}\\ &= 22,06V\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}P_E &= \frac{U_{E,eff}^2}{R}\\ &= \frac{(22,06V)^2}{50Ω}\\ &\approx 9,7W\end{aligned}$
+$\begin{split}P_E &= \frac{U_{E,eff}^2}{R}\\ &= \frac{(22,06V)^2}{50Ω}\\ &\approx 9,7W\end{split}$
 </fragment>
 </right>
 
@@ -196,12 +196,12 @@ $\begin{aligned}P_E &= \frac{U_{E,eff}^2}{R}\\ &= \frac{(22,06V)^2}{50Ω}\\ &\ap
 
 <left>
 <fragment>
-$\begin{aligned}R &= (\frac{1}{R_{V1}} + \frac{1}{R_{V2}})^{-1}\\ &= (\frac{1}{R_{56Ω}} + \frac{1}{R_{470Ω}})^{-1}\\ &= 50,04Ω\end{aligned}$
+$\begin{split}R &= (\frac{1}{R_{V1}} + \frac{1}{R_{V2}})^{-1}\\ &= (\frac{1}{R_{56Ω}} + \frac{1}{R_{470Ω}})^{-1}\\ &= 50,04Ω\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}U_S &= \frac{U_A}{2} + U_F\\ &= \frac{15,3V}{2} + 0,23V\\ &= 7,88V\end{aligned}$
+$\begin{split}U_S &= \frac{U_A}{2} + U_F\\ &= \frac{15,3V}{2} + 0,23V\\ &= 7,88V\end{split}$
 </fragment>
 </right>
 
@@ -212,12 +212,12 @@ $\begin{aligned}U_S &= \frac{U_A}{2} + U_F\\ &= \frac{15,3V}{2} + 0,23V\\ &= 7,8
 * gesucht: $P_E$
 
 <fragment>
-$\begin{aligned}U_{E,eff} &= \frac{U_S}{\sqrt{2}}\\ &= \frac{7,88V}{1,414}\\ &= 5,57V\end{aligned}$
+$\begin{split}U_{E,eff} &= \frac{U_S}{\sqrt{2}}\\ &= \frac{7,88V}{1,414}\\ &= 5,57V\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}P_E &= \frac{U_{E,eff}^2}{R}\\ &= \frac{{5,57V}^2}{50,04Ω}\\ &\approx 600mW\end{aligned}$
+$\begin{split}P_E &= \frac{U_{E,eff}^2}{R}\\ &= \frac{{5,57V}^2}{50,04Ω}\\ &\approx 600mW\end{split}$
 </fragment>
 </right>
 
@@ -234,12 +234,12 @@ $\begin{aligned}P_E &= \frac{U_{E,eff}^2}{R}\\ &= \frac{{5,57V}^2}{50,04Ω}\\ &\
 
 <left>
 <fragment>
-$\begin{aligned}U_S &= \frac{U_A}{2} + U_F\\ &= \frac{15,3V}{2} + 0,23V\\ &= 7,88V\end{aligned}$
+$\begin{split}U_S &= \frac{U_A}{2} + U_F\\ &= \frac{15,3V}{2} + 0,23V\\ &= 7,88V\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}U_{E,eff} &= \frac{U_S}{\sqrt{2}}\\ &= \frac{7,88V}{1,414}\\ &= 5,57V\end{aligned}$
+$\begin{split}U_{E,eff} &= \frac{U_S}{\sqrt{2}}\\ &= \frac{7,88V}{1,414}\\ &= 5,57V\end{split}$
 </fragment>
 </right>
   
@@ -249,7 +249,7 @@ $\begin{aligned}U_{E,eff} &= \frac{U_S}{\sqrt{2}}\\ &= \frac{7,88V}{1,414}\\ &=
 * gesucht: $P_E$
 
 <fragment>
-$\begin{aligned}P_E &= \frac{(U_{E,eff} \cdot 10)^2}{R}\\ &= \frac{(5,57V \cdot 10)^2}{50Ω}\\ &\approx 60W\end{aligned}$
+$\begin{split}P_E &= \frac{(U_{E,eff} \cdot 10)^2}{R}\\ &= \frac{(5,57V \cdot 10)^2}{50Ω}\\ &\approx 60W\end{split}$
 </fragment>
 
 --- style="font-size: smaller;"
diff --git a/contents/slides/spannungswandler.md b/contents/slides/spannungswandler.md
index b6de7b1..ac9b12e 100644
--- a/contents/slides/spannungswandler.md
+++ b/contents/slides/spannungswandler.md
@@ -30,7 +30,7 @@ $\eta = \frac{P_{\textrm{AB}}}{P_{\textrm{ZU}}}$
 * gesucht: $\eta$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber \eta &= \frac{P_{\textrm{AB}}}{P_{\textrm{ZU}}} = \frac{U_{\textrm{AB}} \cdot I_{\textrm{AB}}}{U_{\textrm{ZU}} \cdot I_{\textrm{ZU}}}\\ &= \frac{5V \cdot 3A}{12V \cdot 2A} = \frac{15W}{24W} = 0,625 = 62,5\% \end{split}\end{equation}$
+$\begin{split} \eta &= \frac{P_{\textrm{AB}}}{P_{\textrm{ZU}}} = \frac{U_{\textrm{AB}} \cdot I_{\textrm{AB}}}{U_{\textrm{ZU}} \cdot I_{\textrm{ZU}}}\\ &= \frac{5V \cdot 3A}{12V \cdot 2A} = \frac{15W}{24W} = 0,625 = 62,5\% \end{split}$
 </fragment>
 ---
 [question:AB214]
@@ -42,5 +42,5 @@ $\begin{equation}\begin{split}\nonumber \eta &= \frac{P_{\textrm{AB}}}{P_{\textr
 * gesucht: $\eta$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber \eta &= \frac{P_{\textrm{AB}}}{P_{\textrm{ZU}}} = \frac{U_{\textrm{AB}} \cdot I_{\textrm{AB}}}{U_{\textrm{ZU}} \cdot I_{\textrm{ZU}}}\\ &= \frac{12V \cdot 1A}{5V \cdot 3A} = \frac{12W}{15W} = 0,8 = 80\% \end{split}\end{equation}$
+$\begin{split} \eta &= \frac{P_{\textrm{AB}}}{P_{\textrm{ZU}}} = \frac{U_{\textrm{AB}} \cdot I_{\textrm{AB}}}{U_{\textrm{ZU}} \cdot I_{\textrm{ZU}}}\\ &= \frac{12V \cdot 1A}{5V \cdot 3A} = \frac{12W}{15W} = 0,8 = 80\% \end{split}$
 </fragment>
\ No newline at end of file
diff --git a/contents/slides/spiegelfrequenzen.md b/contents/slides/spiegelfrequenzen.md
index 8684299..5715927 100644
--- a/contents/slides/spiegelfrequenzen.md
+++ b/contents/slides/spiegelfrequenzen.md
@@ -58,7 +58,7 @@ Maßnahme für eine möglichst hohe Unterdrückung:
 * gesucht: $f_S$
 
 <fragment>
-$\begin{aligned}f_S &= 2 \cdot f_{OSZ} - f_E\\ &= 2 \cdot 134,9MHz - 145,6MHz\\ &= 124,2MHz\end{aligned}$
+$\begin{split}f_S &= 2 \cdot f_{OSZ} - f_E\\ &= 2 \cdot 134,9MHz - 145,6MHz\\ &= 124,2MHz\end{split}$
 </fragment>
 ---
 [question:AF203]
@@ -69,7 +69,7 @@ $\begin{aligned}f_S &= 2 \cdot f_{OSZ} - f_E\\ &= 2 \cdot 134,9MHz - 145,6MHz\\
 * gesucht: $f_S$
 
 <fragment>
-$\begin{aligned}f_S &= 2 \cdot f_{OSZ} - f_E\\ &= 2 \cdot 39MHz - 28,3MHz\\ &= 49,7MHz\end{aligned}$
+$\begin{split}f_S &= 2 \cdot f_{OSZ} - f_E\\ &= 2 \cdot 39MHz - 28,3MHz\\ &= 49,7MHz\end{split}$
 </fragment>
 ---
 [question:AF204]
@@ -84,7 +84,7 @@ $\begin{aligned}f_S &= 2 \cdot f_{OSZ} - f_E\\ &= 2 \cdot 39MHz - 28,3MHz\\ &= 4
 * gesucht: $f_S$
 
 <fragment>
-$\begin{aligned}f_S &= 2 \cdot f_{OSZ} - f_E\\ &= 2 \cdot 24,94MHz - 14,24MHz\\ &= 35,64MHz\end{aligned}$
+$\begin{split}f_S &= 2 \cdot f_{OSZ} - f_E\\ &= 2 \cdot 24,94MHz - 14,24MHz\\ &= 35,64MHz\end{split}$
 </fragment>
 ---
 [question:AF108]
@@ -96,7 +96,7 @@ $\begin{aligned}f_S &= 2 \cdot f_{OSZ} - f_E\\ &= 2 \cdot 24,94MHz - 14,24MHz\\
 
 <fragment>
 Bei $f_E &lt; f_{OSZ}$:
-$\begin{aligned}f_S &= f_E + 2 \cdot f_{ZF}\\ &= 28,5MHz + 2 \cdot 10,7MHz\\ &= 49,9MHz\end{aligned}$
+$\begin{split}f_S &= f_E + 2 \cdot f_{ZF}\\ &= 28,5MHz + 2 \cdot 10,7MHz\\ &= 49,9MHz\end{split}$
 </fragment>
 ---
 [question:AF109]
diff --git a/contents/slides/spule_2.md b/contents/slides/spule_2.md
index 81643dc..a1e0ef4 100644
--- a/contents/slides/spule_2.md
+++ b/contents/slides/spule_2.md
@@ -57,7 +57,7 @@ $X_{\textrm{L}} = \omega \cdot L = 2\pi \cdot f \cdot L$
 * gesucht: $X_{\textrm{L}}$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber X_{\textrm{L}} &= \omega \cdot L = 2\pi \cdot f \cdot L\\ &= 2\pi \cdot 100MHz \cdot 3\mu H\\ &\approx 1885\Omega \end{split}\end{equation}$
+$\begin{split} X_{\textrm{L}} &= \omega \cdot L = 2\pi \cdot f \cdot L\\ &= 2\pi \cdot 100MHz \cdot 3\mu H\\ &\approx 1885\Omega \end{split}$
 </fragment>
 
 ---
@@ -99,7 +99,7 @@ Deshalb werden (Ring-)Kerne eingesetzt
 * gesucht: $L$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber L &= N^2 \cdot A_{\textrm{L}}\\ &= 14^2 \cdot 1,5nH\\ &= 0,294\mu H \end{split}\end{equation}$
+$\begin{split} L &= N^2 \cdot A_{\textrm{L}}\\ &= 14^2 \cdot 1,5nH\\ &= 0,294\mu H \end{split}$
 </fragment>
 
 ---
@@ -111,7 +111,7 @@ $\begin{equation}\begin{split}\nonumber L &= N^2 \cdot A_{\textrm{L}}\\ &= 14^2
 * gesucht: $L$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber L &= N^2 \cdot A_{\textrm{L}}\\ &= 300^2 \cdot 1250nH\\ &= 112,5mH \end{split}\end{equation}$
+$\begin{split} L &= N^2 \cdot A_{\textrm{L}}\\ &= 300^2 \cdot 1250nH\\ &= 112,5mH \end{split}$
 </fragment>
 
 ---
@@ -123,7 +123,7 @@ $\begin{equation}\begin{split}\nonumber L &= N^2 \cdot A_{\textrm{L}}\\ &= 300^2
 * gesucht: $N$
 
 <fragment>
-$\begin{equation}\begin{align}\nonumber L &= N^2 \cdot A_{\textrm{L}}\\ \nonumber N &= \sqrt{\frac{L}{A_{\textrm{L}}}} = \sqrt{\frac{2mH}{250nH}} \\ \nonumber &= 89\ \textrm{Windungen} \end{align}\end{equation}$
+$\begin{split} L &= N^2 \cdot A_{\textrm{L}}\\ N &= \sqrt{\frac{L}{A_{\textrm{L}}}} = \sqrt{\frac{2mH}{250nH}} \\ &= 89\ \textrm{Windungen} \end{split}$
 </fragment>
 
 ---
@@ -135,7 +135,7 @@ $\begin{equation}\begin{align}\nonumber L &= N^2 \cdot A_{\textrm{L}}\\ \nonumbe
 * gesucht: $N$
 
 <fragment>
-$\begin{equation}\begin{align}\nonumber L &= N^2 \cdot A_{\textrm{L}}\\ \nonumber N &= \sqrt{\frac{L}{A_{\textrm{L}}}} = \sqrt{\frac{12\mu H}{30nH}} \\ \nonumber &= 20\ \textrm{Windungen} \end{align}\end{equation}$
+$\begin{split} L &= N^2 \cdot A_{\textrm{L}}\\ N &= \sqrt{\frac{L}{A_{\textrm{L}}}} = \sqrt{\frac{12\mu H}{30nH}} \\ &= 20\ \textrm{Windungen} \end{split}$
 </fragment>
 
 ---
diff --git a/contents/slides/strom_spannung_messung_3.md b/contents/slides/strom_spannung_messung_3.md
index 5b45d73..8826340 100644
--- a/contents/slides/strom_spannung_messung_3.md
+++ b/contents/slides/strom_spannung_messung_3.md
@@ -28,7 +28,7 @@ Der angezeigte Messwert unterscheidet sich meist vom tatsächlichen Wert
 * gesucht: Abweichung der Leistung $P = U \cdot I$
 
 <fragment>
-$\begin{equation}\begin{split} \nonumber P_{\textrm{Abw}} &= 100\% - (U_{\mathrm{Abw}} \cdot I_{\mathrm{Abw}})\\ &= 100\% - (95\% \cdot 95\%)\\ &= 100\% - 90,25\%\\ &= 9,75\% \end{split}\end{equation}$
+$\begin{split} P_{\textrm{Abw}} &= 100\% - (U_{\mathrm{Abw}} \cdot I_{\mathrm{Abw}})\\ &= 100\% - (95\% \cdot 95\%)\\ &= 100\% - 90,25\%\\ &= 9,75\% \end{split}$
 </fragment>
 
 ---
@@ -47,6 +47,6 @@ $\begin{equation}\begin{split} \nonumber P_{\textrm{Abw}} &= 100\% - (U_{\mathrm
 * gesucht: $I$
 
 <fragment>
-$\begin{equation} \nonumber I = \frac{U}{R} = \frac{0,5V}{10M\Omega} = 50nA \end{equation}$
+$$I = \frac{U}{R} = \frac{0,5V}{10M\Omega} = 50nA$$
 </fragment>
 
diff --git a/contents/slides/transistor_2.md b/contents/slides/transistor_2.md
index e3bdfe3..2e974f9 100644
--- a/contents/slides/transistor_2.md
+++ b/contents/slides/transistor_2.md
@@ -255,7 +255,7 @@ MOSFET: metal oxide semiconductor FET
 </left>
 <right>
 <fragment>
-$\begin{equation}\begin{align}\nonumber \frac{U_{\textrm{R2}}}{U_{\textrm{B}}} &= \frac{R_2}{R_1+R_2}\\ \nonumber \Rightarrow U_{\textrm{R2}} &= \frac{R_2}{R_1+R_2} \cdot U_{\textrm{G}}\\ \nonumber &= \frac{1k\Omega}{10k\Omega+1k\Omega} \cdot 44V\\ \nonumber &= \frac{1}{11} \cdot 44V = 4V \end{align}\end{equation}$
+$\begin{split} \frac{U_{\textrm{R2}}}{U_{\textrm{B}}} &= \frac{R_2}{R_1+R_2}\\ \Rightarrow U_{\textrm{R2}} &= \frac{R_2}{R_1+R_2} \cdot U_{\textrm{G}}\\ &= \frac{1k\Omega}{10k\Omega+1k\Omega} \cdot 44V\\ &= \frac{1}{11} \cdot 44V = 4V \end{split}$
 </fragment>
 </right>
 
@@ -274,7 +274,7 @@ $\begin{equation}\begin{align}\nonumber \frac{U_{\textrm{R2}}}{U_{\textrm{B}}} &
 </left>
 <right>
 <fragment>
-$\begin{equation}\begin{align}\nonumber \frac{U_{\textrm{R1}}}{U_{\textrm{R2}}} &= \frac{R_1}{R_2}\\ \nonumber \Rightarrow R_2 &= R_1 \cdot \frac{U_{\textrm{R2}}}{U_{\textrm{R1}}}\\ \nonumber &= R_1 \cdot \frac{U_{\textrm{R2}}}{U_{\textrm{B}}-U_{\textrm{GS}}}\\ \nonumber &= 10k\Omega \cdot \frac{2,8V}{44V-2,8V}\\ \nonumber &\approx 680\Omega \end{align}\end{equation}$
+$\begin{split} \frac{U_{\textrm{R1}}}{U_{\textrm{R2}}} &= \frac{R_1}{R_2}\\ \Rightarrow R_2 &= R_1 \cdot \frac{U_{\textrm{R2}}}{U_{\textrm{R1}}}\\ &= R_1 \cdot \frac{U_{\textrm{R2}}}{U_{\textrm{B}}-U_{\textrm{GS}}}\\ &= 10k\Omega \cdot \frac{2,8V}{44V-2,8V}\\ &\approx 680\Omega \end{split}$
 </fragment>
 </right>
 ---
diff --git a/contents/slides/uebertrager_2.md b/contents/slides/uebertrager_2.md
index 65b3a67..2cd1482 100644
--- a/contents/slides/uebertrager_2.md
+++ b/contents/slides/uebertrager_2.md
@@ -42,7 +42,7 @@ Die letzte Formel steht so in der Formelsammlung
 * gesucht: $I_P$
 
 <fragment>
-$\begin{equation}\begin{align} \nonumber \frac{U_P}{U_S} &= \frac{I_S}{I_P} \\ \nonumber \Rightarrow I_P &= \frac{I_S \cdot U_S}{U_P} = \frac{1,15A \cdot 6V}{230V} \\ \nonumber &= 30mA \end{align}\end{equation}$
+$\begin{split} \frac{U_P}{U_S} &= \frac{I_S}{I_P} \\ \Rightarrow I_P &= \frac{I_S \cdot U_S}{U_P} = \frac{1,15A \cdot 6V}{230V} \\ &= 30mA \end{split}$
 </fragment>
 ---
 [question:AC303]
@@ -53,7 +53,7 @@ $\begin{equation}\begin{align} \nonumber \frac{U_P}{U_S} &= \frac{I_S}{I_P} \\ \
 * gesucht: $Z_P$
 
 <fragment>
-$\begin{equation}\begin{align} \nonumber ü &= \sqrt{\frac{Z_P}{Z_S}} \\ \nonumber \Rightarrow Z_P &= ü^2 \cdot Z_S = \frac{1^2}{4^2} \cdot 16k\Omega \\ \nonumber &= \frac{16k\Omega}{16} = 1k\Omega \end{align}\end{equation}$
+$\begin{split} ü &= \sqrt{\frac{Z_P}{Z_S}} \\ \Rightarrow Z_P &= ü^2 \cdot Z_S = \frac{1^2}{4^2} \cdot 16k\Omega \\ &= \frac{16k\Omega}{16} = 1k\Omega \end{split}$
 </fragment>
 
 ---
@@ -65,7 +65,7 @@ $\begin{equation}\begin{align} \nonumber ü &= \sqrt{\frac{Z_P}{Z_S}} \\ \nonumb
 * gesucht: $Z_P$
 
 <fragment>
-$\begin{equation}\begin{align} \nonumber ü &= \sqrt{\frac{Z_P}{Z_S}} \\ \nonumber \Rightarrow Z_P &= ü^2 \cdot Z_S = \frac{1^2}{4^2} \cdot 6,4k\Omega \\ \nonumber &= \frac{6,4k\Omega}{16} = 0,4k\Omega \end{align}\end{equation}$
+$\begin{split} ü &= \sqrt{\frac{Z_P}{Z_S}} \\ \Rightarrow Z_P &= ü^2 \cdot Z_S = \frac{1^2}{4^2} \cdot 6,4k\Omega \\ &= \frac{6,4k\Omega}{16} = 0,4k\Omega \end{split}$
 </fragment>
 
 ---
@@ -77,7 +77,7 @@ $\begin{equation}\begin{align} \nonumber ü &= \sqrt{\frac{Z_P}{Z_S}} \\ \nonumb
 * gesucht: $ü$
 
 <fragment>
-$\begin{equation}\begin{split} \nonumber ü &= \sqrt{\frac{Z_P}{Z_S}} = \sqrt{\frac{450\Omega}{50\Omega}} \\ &= \sqrt{\frac{9}{1}} = \frac{3}{1} \end{split}\end{equation}$
+$\begin{split} ü &= \sqrt{\frac{Z_P}{Z_S}} = \sqrt{\frac{450\Omega}{50\Omega}} \\ &= \sqrt{\frac{9}{1}} = \frac{3}{1} \end{split}$
 </fragment>
 
 ---
@@ -89,7 +89,7 @@ $\begin{equation}\begin{split} \nonumber ü &= \sqrt{\frac{Z_P}{Z_S}} = \sqrt{\f
 * gesucht: $ü$
 
 <fragment>
-$\begin{equation}\begin{split} \nonumber ü &= \sqrt{\frac{Z_P}{Z_S}} = \sqrt{\frac{50\Omega}{2,5k\Omega}} \\ &= \sqrt{\frac{1}{50}} \approx \frac{1}{7} \end{split}\end{equation}$
+$\begin{split} ü &= \sqrt{\frac{Z_P}{Z_S}} = \sqrt{\frac{50\Omega}{2,5k\Omega}} \\ &= \sqrt{\frac{1}{50}} \approx \frac{1}{7} \end{split}$
 </fragment>
 
 ---
diff --git a/contents/slides/verkuerzungsfaktor_2.md b/contents/slides/verkuerzungsfaktor_2.md
index 3ba210b..6947884 100644
--- a/contents/slides/verkuerzungsfaktor_2.md
+++ b/contents/slides/verkuerzungsfaktor_2.md
@@ -22,12 +22,12 @@
 
 <left>
 <fragment>
-$\begin{aligned}l_E &= \frac{1}{2} \cdot \frac{\lambda}{2}\\ &= \frac{1}{4} \cdot \frac{c}{f}\\ &= \frac{1}{4} \cdot \frac{3\cdot 10^8\frac{m}{s}}{14,2MHz}\\ &= \frac{1}{4} \cdot 21,13m\\ &= 5,28m\end{aligned}$
+$\begin{split}l_E &= \frac{1}{2} \cdot \frac{\lambda}{2}\\ &= \frac{1}{4} \cdot \frac{c}{f}\\ &= \frac{1}{4} \cdot \frac{3\cdot 10^8\frac{m}{s}}{14,2MHz}\\ &= \frac{1}{4} \cdot 21,13m\\ &= 5,28m\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 0,95 \cdot 5,28m\\ &= 5,02m\end{aligned}$
+$\begin{split}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 0,95 \cdot 5,28m\\ &= 5,02m\end{split}$
 </fragment>
 </right>
 
@@ -49,12 +49,12 @@ $\begin{aligned}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 0
 
 <left>
 <fragment>
-$\begin{aligned}l_E &= \frac{1}{2} \cdot \frac{\lambda}{2}\\ &= \frac{1}{4} \cdot \frac{c}{f}\\ &= \frac{1}{4} \cdot \frac{3\cdot 10^8\frac{m}{s}}{7,1MHz}\\ &= \frac{1}{4} \cdot 42,25m\\ &= 10,56m\end{aligned}$
+$\begin{split}l_E &= \frac{1}{2} \cdot \frac{\lambda}{2}\\ &= \frac{1}{4} \cdot \frac{c}{f}\\ &= \frac{1}{4} \cdot \frac{3\cdot 10^8\frac{m}{s}}{7,1MHz}\\ &= \frac{1}{4} \cdot 42,25m\\ &= 10,56m\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 0,95 \cdot 10,56m\\ &= 10,04m\end{aligned}$
+$\begin{split}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 0,95 \cdot 10,56m\\ &= 10,04m\end{split}$
 </fragment>
 </right>
 
@@ -75,12 +75,12 @@ $\begin{aligned}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 0
 
 <left>
 <fragment>
-$\begin{aligned}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_E &= \frac{l_G}{k_v}\\ &= \frac{20m}{0,95}\\ &= 21,05m\end{aligned}$
+$\begin{split}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_E &= \frac{l_G}{k_v}\\ &= \frac{20m}{0,95}\\ &= 21,05m\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}l_E &= \frac{\lambda}{2}\\ &= \frac{1}{2} \cdot \frac{c}{f}\\ \Rightarrow f &= \frac{1}{2} \cdot \frac{c}{l_E}\\ &= \frac{1}{2} \cdot \frac{3\cdot 10^8\frac{m}{s}}{21,05m}\\&= 7,125MHz\end{aligned}$
+$\begin{split}l_E &= \frac{\lambda}{2}\\ &= \frac{1}{2} \cdot \frac{c}{f}\\ \Rightarrow f &= \frac{1}{2} \cdot \frac{c}{l_E}\\ &= \frac{1}{2} \cdot \frac{3\cdot 10^8\frac{m}{s}}{21,05m}\\&= 7,125MHz\end{split}$
 </fragment>
 </right>
 
@@ -102,12 +102,12 @@ $\begin{aligned}l_E &= \frac{\lambda}{2}\\ &= \frac{1}{2} \cdot \frac{c}{f}\\ \R
 
 <left>
 <fragment>
-$\begin{aligned}l_E &= \frac{\lambda}{4}\\ &= \frac{1}{4} \cdot \frac{c}{f}\\ &= \frac{1}{4} \cdot \frac{3\cdot 10^8\frac{m}{s}}{7,1MHz}\\ &= \frac{1}{4} \cdot 42,25m\\ &= 10,56m\end{aligned}$
+$\begin{split}l_E &= \frac{\lambda}{4}\\ &= \frac{1}{4} \cdot \frac{c}{f}\\ &= \frac{1}{4} \cdot \frac{3\cdot 10^8\frac{m}{s}}{7,1MHz}\\ &= \frac{1}{4} \cdot 42,25m\\ &= 10,56m\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 0,95 \cdot 10,56m\\ &= 10,04m\end{aligned}$
+$\begin{split}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 0,95 \cdot 10,56m\\ &= 10,04m\end{split}$
 </fragment>
 </right>
 
@@ -129,12 +129,12 @@ $\begin{aligned}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 0
 
 <left>
 <fragment>
-$\begin{aligned}l_E &= \frac{5}{8}\lambda\\ &= \frac{5}{8} \cdot \frac{c}{f}\\ &= \frac{5}{8} \cdot \frac{3\cdot 10^8\frac{m}{s}}{14,2MHz}\\ &= \frac{5}{8} \cdot 21,13\\ &= 13,20m\end{aligned}$
+$\begin{split}l_E &= \frac{5}{8}\lambda\\ &= \frac{5}{8} \cdot \frac{c}{f}\\ &= \frac{5}{8} \cdot \frac{3\cdot 10^8\frac{m}{s}}{14,2MHz}\\ &= \frac{5}{8} \cdot 21,13\\ &= 13,20m\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 0,97 \cdot 13,20m\\ &= 12,80m\end{aligned}$
+$\begin{split}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 0,97 \cdot 13,20m\\ &= 12,80m\end{split}$
 </fragment>
 </right>
 
@@ -179,12 +179,12 @@ Ein Verlängerungsfaktor bedeutet <u>nicht</u>, dass sich die Welle mit *Überli
 
 <left>
 <fragment>
-$\begin{aligned}l_E &= \lambda\\ &= \frac{c}{f}\\ &= \frac{3\cdot 10^8\frac{m}{s}}{7,1MHz}\\ &= 42,23m\end{aligned}$
+$\begin{split}l_E &= \lambda\\ &= \frac{c}{f}\\ &= \frac{3\cdot 10^8\frac{m}{s}}{7,1MHz}\\ &= 42,23m\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 1,02 \cdot 42,23m\\ &= 43,10m\end{aligned}$
+$\begin{split}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 1,02 \cdot 42,23m\\ &= 43,10m\end{split}$
 </fragment>
 </right>
 
@@ -230,11 +230,11 @@ $v_\mathrm{k} = \dfrac{1}{\sqrt{\epsilon_\mathrm{r}}}$
 
 <left>
 <fragment>
-$\begin{aligned}l_E &= \lambda\\ &= \frac{c}{f}\\ &= \frac{3\cdot 10^8\frac{m}{s}}{145MHz}\\ &= 2,07m\end{aligned}$
+$\begin{split}l_E &= \lambda\\ &= \frac{c}{f}\\ &= \frac{3\cdot 10^8\frac{m}{s}}{145MHz}\\ &= 2,07m\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 0,66 \cdot 2,07m\\ &= 1,37m\end{aligned}$
+$\begin{split}k_v &= \frac{l_G}{l_E}\\ \Rightarrow l_G &= k_v \cdot l_E\\ &= 0,66 \cdot 2,07m\\ &= 1,37m\end{split}$
 </fragment>
 </right>
\ No newline at end of file
diff --git a/contents/slides/verstaerkungsleistung.md b/contents/slides/verstaerkungsleistung.md
index 180d3ff..1bf2e78 100644
--- a/contents/slides/verstaerkungsleistung.md
+++ b/contents/slides/verstaerkungsleistung.md
@@ -11,7 +11,7 @@
 * gesucht: $g$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber g &= 20\cdot \log_{10}{\left(\frac{U_2}{U_1}\right)}dB\\ &= 20\cdot \log_{10}{\left(\frac{4mV}{1mV}\right)}dB = 12dB \end{split}\end{equation}$
+$\begin{split} g &= 20\cdot \log_{10}{\left(\frac{U_2}{U_1}\right)}dB\\ &= 20\cdot \log_{10}{\left(\frac{4mV}{1mV}\right)}dB = 12dB \end{split}$
 </fragment>
 ---
 [question:AD428]
@@ -22,7 +22,7 @@ $\begin{equation}\begin{split}\nonumber g &= 20\cdot \log_{10}{\left(\frac{U_2}{
 * gesucht: $g$
 
 <fragment>
-$\begin{equation}\begin{split}\nonumber g &= 10\cdot \log_{10}{\left(\frac{P_2}{P_1}\right)}dB\\ &= 10\cdot \log_{10}{\left(\frac{38W}{2,5W}\right)}dB = 11,8dB \end{split}\end{equation}$
+$\begin{split} g &= 10\cdot \log_{10}{\left(\frac{P_2}{P_1}\right)}dB\\ &= 10\cdot \log_{10}{\left(\frac{38W}{2,5W}\right)}dB = 11,8dB \end{split}$
 </fragment>
 ---
 [question:AD426]
diff --git a/contents/slides/vor_rueck_verhaeltnis.md b/contents/slides/vor_rueck_verhaeltnis.md
index 4d06261..7a78e6a 100644
--- a/contents/slides/vor_rueck_verhaeltnis.md
+++ b/contents/slides/vor_rueck_verhaeltnis.md
@@ -41,7 +41,7 @@
 * gesucht: $\frac{Vor}{Rück}$
 
 <fragment>
-$\begin{aligned}\frac{Vor}{Rück} &= 10 \cdot \log_{10}{(\frac{P_V}{P_R})} dB\\ &= 10 \cdot \log_{10}{(\frac{15W}{0,6W})} dB\\ &= 14dB\end{aligned}$
+$\begin{split}\frac{Vor}{Rück} &= 10 \cdot \log_{10}{(\frac{P_V}{P_R})} dB\\ &= 10 \cdot \log_{10}{(\frac{15W}{0,6W})} dB\\ &= 14dB\end{split}$
 </fragment>
 
 ---
@@ -61,12 +61,12 @@ $\begin{aligned}\frac{Vor}{Rück} &= 10 \cdot \log_{10}{(\frac{P_V}{P_R})} dB\\
 
 <left>
 <fragment>
-$\begin{aligned}P_V &= P_{ERP}\\ &= P_S \cdot 10^{\frac{g_d}{10dB}}\\ &= 100W \cdot 10^{\frac{10dB}{10dB}}\\ &= 1000W\end{aligned}$
+$\begin{split}P_V &= P_{ERP}\\ &= P_S \cdot 10^{\frac{g_d}{10dB}}\\ &= 100W \cdot 10^{\frac{10dB}{10dB}}\\ &= 1000W\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}20dB &= 10 \cdot \log_{10}{(\frac{P_V}{P_R})} dB\\ \Rightarrow \frac{P_V}{P_R} &= 10^{\frac{20dB}{10}}\\ &= 100\\ \Rightarrow P_R &= \frac{P_V}{100}\\ &= \frac{1000W}{100}\\ &= 10W\end{aligned}$
+$\begin{split}20dB &= 10 \cdot \log_{10}{(\frac{P_V}{P_R})} dB\\ \Rightarrow \frac{P_V}{P_R} &= 10^{\frac{20dB}{10}}\\ &= 100\\ \Rightarrow P_R &= \frac{P_V}{100}\\ &= \frac{1000W}{100}\\ &= 10W\end{split}$
 </fragment>
 </right>
 ---
@@ -86,12 +86,12 @@ $\begin{aligned}20dB &= 10 \cdot \log_{10}{(\frac{P_V}{P_R})} dB\\ \Rightarrow \
 
 <left>
 <fragment>
-$\begin{aligned}P_V &= P_{ERP}\\ &= P_S \cdot 10^{\frac{g_d}{10dB}}\\ &= 6W \cdot 10^{\frac{15dB}{10dB}}\\ &= 189,7W\end{aligned}$
+$\begin{split}P_V &= P_{ERP}\\ &= P_S \cdot 10^{\frac{g_d}{10dB}}\\ &= 6W \cdot 10^{\frac{15dB}{10dB}}\\ &= 189,7W\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}25dB &= 10 \cdot \log_{10}{(\frac{P_V}{P_R})} dB\\ \Rightarrow \frac{P_V}{P_R} &= 10^{\frac{25dB}{10}}\\ &= 316,2\\ \Rightarrow P_R &= \frac{P_V}{316,2}\\ &= \frac{189,7W}{316,2}\\ &= 0,6W\end{aligned}$
+$\begin{split}25dB &= 10 \cdot \log_{10}{(\frac{P_V}{P_R})} dB\\ \Rightarrow \frac{P_V}{P_R} &= 10^{\frac{25dB}{10}}\\ &= 316,2\\ \Rightarrow P_R &= \frac{P_V}{316,2}\\ &= \frac{189,7W}{316,2}\\ &= 0,6W\end{split}$
 </fragment>
 </right>
 
@@ -112,11 +112,11 @@ $\begin{aligned}25dB &= 10 \cdot \log_{10}{(\frac{P_V}{P_R})} dB\\ \Rightarrow \
 
 <left>
 <fragment>
-$\begin{aligned}g_D &= 20 \cdot \log_{10}{(\frac{U_V}{U_D})} dB\\ &= 20 \cdot \log_{10}{(\frac{300µV/m}{128µV/m})}\\ &= 7,4dB\end{aligned}$
+$\begin{split}g_D &= 20 \cdot \log_{10}{(\frac{U_V}{U_D})} dB\\ &= 20 \cdot \log_{10}{(\frac{300µV/m}{128µV/m})}\\ &= 7,4dB\end{split}$
 </fragment>
 </left>
 <right>
 <fragment>
-$\begin{aligned}\frac{Vor}{Rück} &= 20 \cdot \log_{10}{(\frac{U_V}{U_R})} dB\\ &= 20 \cdot \log_{10}{(\frac{300µV/m}{20µV/m})}\\ &= 23,5dB\end{aligned}$
+$\begin{split}\frac{Vor}{Rück} &= 20 \cdot \log_{10}{(\frac{U_V}{U_R})} dB\\ &= 20 \cdot \log_{10}{(\frac{300µV/m}{20µV/m})}\\ &= 23,5dB\end{split}$
 </fragment>
 </right>
diff --git a/contents/slides/wechselstrom_leistung.md b/contents/slides/wechselstrom_leistung.md
index 125848c..66bef9d 100644
--- a/contents/slides/wechselstrom_leistung.md
+++ b/contents/slides/wechselstrom_leistung.md
@@ -23,7 +23,7 @@ $P = U_{\text{eff}} \cdot I_{\text{eff}}$
 * gesucht: $P$
 
 <fragment>
-$\begin{equation}\begin{split} \nonumber P &=  I^2 \cdot R = \left(\frac{I_{\textrm{max}}}{\sqrt{2}}\right)^2 \cdot R\\ &= \frac{(0,5A)^2}{2} \cdot 20\Omega \\ &= \frac{1}{8}A^2 \cdot 20\Omega = 2,5W \end{split}\end{equation}$
+$\begin{split} P &=  I^2 \cdot R = \left(\frac{I_{\textrm{max}}}{\sqrt{2}}\right)^2 \cdot R\\ &= \frac{(0,5A)^2}{2} \cdot 20\Omega \\ &= \frac{1}{8}A^2 \cdot 20\Omega = 2,5W \end{split}$
 </fragment>
 
 ---
diff --git a/contents/slides/wellenwiderstand.md b/contents/slides/wellenwiderstand.md
index 7161d3c..6789a9a 100644
--- a/contents/slides/wellenwiderstand.md
+++ b/contents/slides/wellenwiderstand.md
@@ -17,7 +17,7 @@ $Z = \dfrac{120Ω}{\sqrt{\epsilon_\mathrm{r}}} \cdot \ln{(\dfrac{2 \cdot a}{d})}
 * gesucht: $Z$
 
 <fragment>
-$\begin{aligned}Z &= \dfrac{120Ω}{\sqrt{\epsilon_\mathrm{r}}} \cdot \ln{(\dfrac{2 \cdot a}{d})}\\ &= \dfrac{120Ω}{\sqrt{1}} \cdot \ln{(\dfrac{2 \cdot 200mm}{2mm})}\\ &\approx 635Ω\end{aligned}$
+$\begin{split}Z &= \dfrac{120Ω}{\sqrt{\epsilon_\mathrm{r}}} \cdot \ln{(\dfrac{2 \cdot a}{d})}\\ &= \dfrac{120Ω}{\sqrt{1}} \cdot \ln{(\dfrac{2 \cdot 200mm}{2mm})}\\ &\approx 635Ω\end{split}$
 </fragment>
 ---
 ## Wellenwiderstand einer Koaxialleitung
@@ -39,7 +39,7 @@ $Z = \dfrac{60Ω}{\sqrt{\epsilon_\mathrm{r}}} \cdot \ln{\dfrac{D}{d}}$
 * gesucht: $Z$
 
 <fragment>
-$\begin{aligned}Z &= \dfrac{60Ω}{\sqrt{\epsilon_\mathrm{r}}} \cdot \ln{(\dfrac{D}{d})}\\ &= \dfrac{60Ω}{\sqrt{1}} \cdot \ln{(\dfrac{5mm}{1mm})}\\ &\approx 97Ω\end{aligned}$
+$\begin{split}Z &= \dfrac{60Ω}{\sqrt{\epsilon_\mathrm{r}}} \cdot \ln{(\dfrac{D}{d})}\\ &= \dfrac{60Ω}{\sqrt{1}} \cdot \ln{(\dfrac{5mm}{1mm})}\\ &\approx 97Ω\end{split}$
 </fragment>
 ---
 [question:AG307]
@@ -51,7 +51,7 @@ $\begin{aligned}Z &= \dfrac{60Ω}{\sqrt{\epsilon_\mathrm{r}}} \cdot \ln{(\dfrac{
 * gesucht: $Z$
 
 <fragment>
-$\begin{aligned}Z &= \dfrac{60Ω}{\sqrt{\epsilon_\mathrm{r}}} \cdot \ln{(\dfrac{D}{d})}\\ &= \dfrac{60Ω}{\sqrt{2,29}} \cdot \ln{(\dfrac{4,4mm}{0,7mm})}\\ &\approx 75Ω\end{aligned}$
+$\begin{split}Z &= \dfrac{60Ω}{\sqrt{\epsilon_\mathrm{r}}} \cdot \ln{(\dfrac{D}{d})}\\ &= \dfrac{60Ω}{\sqrt{2,29}} \cdot \ln{(\dfrac{4,4mm}{0,7mm})}\\ &\approx 75Ω\end{split}$
 </fragment>
 ---
 ### Anpassung von Koaxialleitungen