Unterschiede

Hier werden die Unterschiede zwischen zwei Versionen angezeigt.

Link zu dieser Vergleichsansicht

Beide Seiten der vorigen Revision Vorhergehende Überarbeitung
Nächste Überarbeitung 		Vorhergehende Überarbeitung
electrical_engineering_1:network_analysis [2023/03/23 14:12]
mexleadmin 		electrical_engineering_1:network_analysis [2023/11/28 00:45] (aktuell)
mexleadmin
Zeile 1: Zeile 1:
-====== 4. Analysis of Networks ======+====== 4 Analysis of Networks ======
  
-<callout> <WRAP> <imgcaption imageNo1 | examples for networks> </imgcaption> {{drawio>Beispiele Netzwerke}} </WRAP>+<callout> <WRAP> <imgcaption imageNo1 | examples for networks> </imgcaption> {{drawio>Beispiele Netzwerke.svg}} </WRAP>
  
 Network analysis plays a central role in electrical engineering.  Network analysis plays a central role in electrical engineering. 
Zeile 26: Zeile 26:
 ==== Preparation of the Circuit ==== ==== Preparation of the Circuit ====
  
-<WRAP> <imgcaption imageNo10 | Preparing the circuit> </imgcaption> {{drawio>VorbereitungDerSchaltung}} </WRAP>+<WRAP> <imgcaption imageNo10 | Preparing the circuit> </imgcaption> {{drawio>VorbereitungDerSchaltung.svg}} </WRAP>
  
 Before the network analysis can be tackled, the circuit must be suitably prepared (cf. <imgref imageNo10 >): Before the network analysis can be tackled, the circuit must be suitably prepared (cf. <imgref imageNo10 >):
Zeile 49: Zeile 49:
 ==== Graphs and Trees ==== ==== Graphs and Trees ====
  
-<WRAP> <imgcaption imageNo11 | Graph of a network> </imgcaption> {{drawio>GraphEinesNetzwerks}} </WRAP>+<WRAP> <imgcaption imageNo11 | Graph of a network> </imgcaption> {{drawio>GraphEinesNetzwerks.svg}} </WRAP>
  
 In the chapter [[:electrical_engineering_1:simple_circuits#nodes_branches_and_loops|2. simple dc_circuits]] the terms nodes, branches, and loops have already been explained. These will now be expanded here to better explain the various network analysis methods in the following. In <imgref imageNo11 > the **graph**  of the example network is drawn. We had already seen this one too, but without knowing that this is called a graph! \\  In the chapter [[:electrical_engineering_1:simple_circuits#nodes_branches_and_loops|2. simple dc_circuits]] the terms nodes, branches, and loops have already been explained. These will now be expanded here to better explain the various network analysis methods in the following. In <imgref imageNo11 > the **graph**  of the example network is drawn. We had already seen this one too, but without knowing that this is called a graph! \\ 
Zeile 80: Zeile 80:
 ===== 4.2 Branch Current Method===== ===== 4.2 Branch Current Method=====
  
-<WRAP> <imgcaption imageNo12 | example circuit> </imgcaption> {{drawio>Beispielschaltung}} </WRAP>+<WRAP> <imgcaption imageNo12 | example circuit> </imgcaption> {{drawio>Beispielschaltung.svg}} </WRAP>
  
 The branch current method (also called branch current method) now "simply times" (almost) all equations of the circuit.  The branch current method (also called branch current method) now "simply times" (almost) all equations of the circuit. 
Zeile 340: Zeile 340:
 <callout title="Example 2 - Spring Force and Displacement"> <callout title="Example 2 - Spring Force and Displacement">
  
-<WRAP> <imgcaption imageNo02 | mechanical spring> </imgcaption> {{drawio>mechanischeFeder}} </WRAP>+<WRAP> <imgcaption imageNo02 | mechanical spring> </imgcaption> {{drawio>mechanischeFeder.svg}} </WRAP>
  
 **Task**:A mechanical, linear spring is displaced due to masses $m_1$ and $m_2$ in the Earth's gravitational field (see <imgref imageNo02 >). What is the magnitude of the deflection if both masses are attached simultaneously? **Task**:A mechanical, linear spring is displaced due to masses $m_1$ and $m_2$ in the Earth's gravitational field (see <imgref imageNo02 >). What is the magnitude of the deflection if both masses are attached simultaneously?
Zeile 389: Zeile 389:
 === Example === === Example ===
  
-<WRAP> <imgcaption imageNo03 | example circuit with superposition> </imgcaption> {{drawio>BeispielschaltungSuperposition}} <WRAP>+<WRAP> <imgcaption imageNo03 | example circuit with superposition> </imgcaption> {{drawio>BeispielschaltungSuperposition.svg}} <WRAP>
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
Zeile 400: Zeile 400:
 ===== Exercises ===== ===== Exercises =====
  
-<panel type="info" title="Exercise 4.5.1 Converting a bipolar signal to a unipolar signal"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+#@TaskTitle_HTML@#4.5.1 Converting a bipolar signal to a unipolar signal <fs medium>(not from written test)</fs>#@TaskText_HTML@# 
 + 
 +Imagine you want to develop a circuit that conditions a sensor signal so that it can be processed by a microcontroller. The sensor signal is in the range $U_{\rm sens} \in [-15...15~\rm V]$, and the microcontroller input can read values in the range $U_{\rm uC} \in [0...3.3~\rm V]$. The sensor can supply a maximum current of $I_{\rm sens, max}=1~\rm mA$. For the internal resistance of the microcontroller, input applies: $R_{\rm uC} \rightarrow \infty$ 
 + 
 +For conditioning, the input signal is to be fed via the series resistor $R_3$ to the center potential of a voltage divider $R_1 - R_2$ with $R_1$ to a supply voltage $U_{\rm s}$. 
 + 
 +The following simulation shows roughly the situation (the resistor values are not correct).
  
 <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCDMCmC0DsIBskB08CcGkA4AsYGADEQKxFIYjnUh6TVxhgBQATiLHjiAEy95kg-oKLgSJdkL4DpkIqOoSiUpMNlrwSMWN4SWAczkK5ekDpYAlaWG03ePHXTGlzboqlIsAbp0K8oE1h-PjwnCDC+Jx1PQz8MAPlBYISoMwsAd3iAgTEUnNkVDnyZZOxStzB9SFxpEQr6iwAjTl5SJHA8ZOZEgnMWAA9W0kSkRFgMCEgMRGFBJoBLAAcAewAbAEM2AB0AZwBlFYBXNgBjaD3t7YA7AApoVANUPY293ehr3ZW2AEpBkFShAYVSohA6cxAln+zF0REQYF4DF4VTooXA0IRfHgwLaWPBaN4-wwwlIoOYfCQAQhUKGwR4enGrmRYghkH+sBq4AwDBSXPxshp4Bwoymtigk1RsiO12W6y2bwW1wA1lc7g8nnsjgBhH5-SDwPLlepcemyCAWWlgHA8K3deAQK2zNGC4KULk8sCkYHlCH7FgrAHiQrODD0sRIVCKAKuFRAA noborder}} </WRAP> <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCDMCmC0DsIBskB08CcGkA4AsYGADEQKxFIYjnUh6TVxhgBQATiLHjiAEy95kg-oKLgSJdkL4DpkIqOoSiUpMNlrwSMWN4SWAczkK5ekDpYAlaWG03ePHXTGlzboqlIsAbp0K8oE1h-PjwnCDC+Jx1PQz8MAPlBYISoMwsAd3iAgTEUnNkVDnyZZOxStzB9SFxpEQr6iwAjTl5SJHA8ZOZEgnMWAA9W0kSkRFgMCEgMRGFBJoBLAAcAewAbAEM2AB0AZwBlFYBXNgBjaD3t7YA7AApoVANUPY293ehr3ZW2AEpBkFShAYVSohA6cxAln+zF0REQYF4DF4VTooXA0IRfHgwLaWPBaN4-wwwlIoOYfCQAQhUKGwR4enGrmRYghkH+sBq4AwDBSXPxshp4Bwoymtigk1RsiO12W6y2bwW1wA1lc7g8nnsjgBhH5-SDwPLlepcemyCAWWlgHA8K3deAQK2zNGC4KULk8sCkYHlCH7FgrAHiQrODD0sRIVCKAKuFRAA noborder}} </WRAP>
  
-Imagine you want to develop a circuit that conditions a sensor signal so that it can be processed by a microcontroller. The sensor signal is in the range $U_{sens} \in [-15...15~\rm V]$, and the microcontroller input can read values in the range $U_{\rm uC} \in [0...3.3~\rm V]$. The sensor can supply a maximum current of $I_{\rm sens, max}=1~\rm mA$. For the internal resistance of the microcontroller, input applies: $R_{\rm uC} \rightarrow \infty$+Questions: 
 + 
 +1. Find the relationship between $R_1$, $R_2$, and $R_3$ using superposition. \\ 
 +  * Determine suitable values for $R_1$, $R_2$, and $R_3$. 
 +  * What values for $R^0_1$, $R^0_2$, and $R^0_3$ from the [[https://de.wikipedia.org/wiki/E-Reihe|E24 series]] can be used to do this? 
 + 
 +#@HiddenBegin_HTML~1,Solution~@# 
 +Using superposition, we create two separate circuits where one source is considered. 
 +For these two circuits, we calculate $U_\rm A^{(1)}$ and $U_\rm A^{(2)}$. \\ 
 +To make the calculation simpler, the resistors $R_3$ and $R_{\rm s}$ will be joined to $R_4 =R_3 +R_{\rm s}$. 
 + 
 +<callout> 
 +=== Circuit 1 : only consider $U_{\rm S}$, ignore $U_{\rm I}$ === 
 +{{drawio>electrical_engineering_1:exc541circ1.svg}} 
 + 
 +\begin{align*} 
 +U_{\rm O}^{(1)}  &= U_{\rm S} \cdot {{R_2||R_4}\over{R_1 + R_2||R_4}}  
 +                  = U_{\rm S} \cdot {{ {{R_2  R_4}\over{R_2 + R_4}} }\over{R_1 + {{R_2  R_4}\over{R_2 + R_4}} }} \\ 
 +                 &= U_{\rm S} \cdot {{ R_2 R_4 }\over{R_1  (R_2 + R_4)+ R_2 R_4 }} \\ 
 +                 &= U_{\rm S} \cdot {{ R_2 R_4 }\over{R_1  R_2 + R_1 R_4 + R_2 R_4 }} \\ 
 +\end{align*} 
 +</callout> 
 + 
 +<callout> 
 +=== Circuit 2 : only consider $U_{\rm I}$, ignore $U_{\rm S}$ === 
 +{{drawio>electrical_engineering_1:exc541circ2.svg}} 
 + 
 +\begin{align*} 
 +U_{\rm O}^{(2)}  &= U_{\rm I} \cdot {{R_1||R_2}\over{R_4 + R_1||R_2}}  
 +                  = U_{\rm I} \cdot {{ {{R_1 R_2}\over{R_1 + R_2}} }\over{R_4 + {{R_1 R_2}\over{R_1 + R_2}} }} \\ 
 +                 &= U_{\rm I} \cdot {{ R_1 R_2 }\over{R_4 (R_1 + R_2)+ R_1 R_2 }} \\ 
 +                 &= U_{\rm I} \cdot {{ R_1 R_2 }\over{R_4 R_1 + R_4 R_2+ R_1 R_2 }} \\ 
 +\end{align*} 
 +</callout> 
 + 
 +<callout> 
 +=== Superposition: Let's sum it up! === 
 +\\ 
 +These two intermediate voltages for the single sources have to be summed up as $U_{\rm O}= U_{\rm O}^{(1)} + U_{\rm O}^{(2)}$. \\ 
 +When deeper investigated, one can see that the denominator for both $U_{\rm O}^{(1)}$ and $U_{\rm O}^{(2)}$ is the same. \\ 
 +We can also simplify further when looking at often-used sub-terms (here: $R_2$) 
 + 
 +\begin{align*} 
 +U_{\rm O}                                            &=  {{ 1 }\over{R_4 R_1 + R_4 R_2+ R_1 R_2 }} \cdot (U_{\rm S} \cdot R_2 R_4  + U_{\rm I} \cdot R_1 R_2 ) \\ 
 +U_{\rm O} \cdot (R_4 R_1 + R_4 R_2+ R_1 R_2 )        &=   U_{\rm S} \cdot R_2 R_4  + U_{\rm I} \cdot R_1 R_2  \\ \\ 
 +U_{\rm O} \cdot ({{R_1 R_4}\over{R_2}} + R_4 + R_1 ) &=   U_{\rm S} \cdot     R_4  + U_{\rm I} \cdot R_1      \tag 1 \\ 
 +\end{align*} 
 + 
 +The formula $(1)$ is the general formula to calculate the output voltage $U_{\rm O}$ for a changing input voltage $U_{\rm I}$, where the supply voltage $U_{\rm S}" is constant. \\ 
 +</callout> 
 + 
 +Now, we can use the requested boundaries: 
 +  - For the minimum input voltage $U_{\rm I}= -15 ~\rm V$, the output voltage shall be $U_{\rm O} =   0 ~\rm V$ 
 +  - For the maximum input voltage $U_{\rm I}= +15 ~\rm V$, the output voltage shall be $U_{\rm O} = 3.3 ~\rm V$ 
 + 
 +This leads to two situations: 
 + 
 +<callout> 
 +=== Situation I : $U_{\rm I,min}= -15 ~\rm V$ shall create $U_{\rm O,min} = 0 ~\rm V$ === 
 +\\ 
 +We put $U_{\rm A} = 0 ~\rm V$ in the formula $(1)$ : 
 +\begin{align*} 
 +0                          &=   U_{\rm S} \cdot     R_4  + U_{\rm I,min} \cdot R_1     \\ 
 +- U_{\rm I,min} \cdot R_1  &=   U_{\rm S} \cdot     R_4       \\ 
 + {{R_1}\over{R_4}}         &=-{{U_{\rm S}}\over {U_{\rm I,min}}} = k_{14} \tag 2 \\ 
 +\end{align*} 
 + 
 +So, with formula $(2)$, we already have a relation between $R_1$ and $R_4$. Yeah 😀 \\ 
 +The next step is situation 2 
 +</callout> 
 + 
 +<callout> 
 +=== Situation II : $U_{\rm I,max}= +15 ~\rm V$ shall create $U_{\rm O,max} = 3.3 ~\rm V$ === 
 +\\ 
 +We use formula $(2)$ to substitute $R_1 = k_{14} \cdot R_4 $ in formula $(1)$, and: 
 +\begin{align*} 
 +U_{\rm O,max} \cdot (k_{14}{{ R_4^2}\over{R_2}} + R_4 + k_{14} R_4 ) &=   U_{\rm S} \cdot     R_4  + U_{\rm I,max} \cdot k_{14} R_4 \\ 
 +U_{\rm O,max} \cdot (k_{14}{{ R_4  }\over{R_2}} + 1   + k_{14}     ) &=   U_{\rm S}                + U_{\rm I,max} \cdot k_{14}  \\ 
 +                     k_{14}{{ R_4  }\over{R_2}}  + 1   + k_{14}      &= {{U_{\rm S}                + U_{\rm I,max} \cdot k_{14} }\over{        U_{\rm O,max} }} \\ 
 +                     k_{14}{{ R_4  }\over{R_2}}                      &= {{U_{\rm S}                + U_{\rm I,max} \cdot k_{14} }\over{        U_{\rm O,max} }}        - (1   + k_{14})\\ 
 +                           {{ R_4  }\over{R_2}}                      &= {{U_{\rm S}                + U_{\rm I,max} \cdot k_{14} }\over{ k_{14} U_{\rm O,max} }} - {{1   + k_{14} }\over{k_{14}}} \tag 3 \\ 
 +\end{align*} 
 + 
 +So, another relation for $R_4$ and $R_2$.  😀 \\ 
 +</callout> 
 + 
 +So, to get values for the relations, we have to put in the values for the input and output voltage conditions. For $k_{14}$ we get by formula $(2)$: 
 +\begin{align*} 
 +k_{14} = {{R_1}\over{R_4}}  =-{{5 ~\rm V}\over {-15 ~\rm V }} = {{1}\over{3}} \\ 
 +\end{align*} 
 + 
 +This value $k_{14}$ we can use for formula $(3)$: 
 +\begin{align*} 
 +{{ R_4  }\over{R_2}} &= {{5 ~\rm V + 15 ~\rm V \cdot {{1}\over{3}} }\over{ 3.3 ~\rm V \cdot {{1}\over{3}} }} - {{1   + {{1}\over{3}} }\over{ {{1}\over{3}} }} \\ 
 +                     &= {{10}\over{1.1}} - 4 \\ 
 +k_{42}               &\approx 5.09 
 +\end{align*} 
 + 
 +We could now - theoretically - arbitrarily choose one of the resistors, e.g., $R_2$, and then calculate the other two. \\ 
 + 
 +But we must consider another boundary, a boundary for $R_{\rm S}$. The maximum voltage and the maximum current are given for the sensor. By this, we can calculate $R_{\rm S}$: 
 +\begin{align*} 
 +R_{\rm S}   &= {{ U_{\rm OC} }\over{ I_{\rm SC} }} = {{ U_{\rm S,max} }\over{ I_{\rm S,max} }} = {{ 15 ~\rm V }\over{ 1 ~\rm mA }} \\ 
 +            &= 15 ~\rm k\Omega 
 +\end{align*} 
 + 
 +Therefore, $R_4 = R_{\rm S} + R_3$ must be larger than this. \\ 
 + 
 +#@HiddenEnd_HTML~1,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result1,Result~@# 
 + 
 +The sensor resistance is 
 +\begin{align*} 
 +R_S &= 15 {~\rm k\Omega}\\ 
 +\end{align*} 
 + 
 +We can choose $R_3$ arbitrarily. Here I choose a nice value to get integer values for $R_3$ and $R_1$: 
 +\begin{align*} 
 +R_3 &= 45 {~\rm k\Omega}\\ 
 +R_1 &= {{1}\over{3}}   (R_3 + 15 {~\rm k\Omega}) = 20   {~\rm k\Omega} \\ 
 +R_2 &= {{1}\over{5.09}}(R_3 + 15 {~\rm k\Omega}) = 11.8 {~\rm k\Omega}  
 +\end{align*} 
 + 
 +Based on the E24 series, the following values are next to the calculated ones: 
 +\begin{align*} 
 +R_3^0 &= 43 {~\rm k\Omega}\\ 
 +R_1^0 &= {{1}\over{3}}   (R_3 + 15 {~\rm k\Omega}) = 20 {~\rm k\Omega} \\ 
 +R_2^0 &= {{1}\over{5.09}}(R_3 + 15 {~\rm k\Omega}) = 12 {~\rm k\Omega}  
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Result1,Result~@# 
 + 
 +2. Find the relationship between $R_1$, $R_2$, and $R_3$ by investigating Kirchhoff's nodal rule for the node where $R_1$, $R_2$, and $R_3$ are interconnected. 
 + 
 +#@HiddenBegin_HTML~Solution2,Solution~@# 
 + 
 +The potential of the node is $U_\rm O$. Therefore the currents are: 
 +  - the current $I_2$ over $R_2$ is flowing to ground: $I_2 = - {{U_\rm O}\over{R_2}} $  
 +  - the current $I_1$ over $R_1$ is coming from the supply voltage $U_{\rm S}$ to the nodal voltage $U_{\rm O}$:  $I_1 = {{U_{\rm S} - U_{\rm O}}\over{R_1}}$ 
 +  - the current $I_4$ over $R_4$ is coming from the input voltage  $U_{\rm I}$ to the nodal voltage $U_{\rm O}$:  $I_4 = {{U_{\rm I} - U_{\rm O}}\over{R_4}}$ 
 + 
 +This led to the formula based on the Kirchhoff's nodal rule:  
 + 
 +\begin{align*} 
 +\Sigma I = 0 &= I_1 + I_2 + I_3 \\ 
 +         0 &= {{U_{\rm S} - U_{\rm O}}\over{R_1}} + {{U_{\rm I} - U_{\rm O}}\over{R_4}} - {{U_\rm O}\over{R_2}}  
 +\end{align*} 
 + 
 +The formula can be rearranged, with all terms containing $ U_{\rm O}$ on the left side:  
 +\begin{align*} 
 +    {{U_{\rm O}}\over{R_1}} + {{U_{\rm O}}\over{R_2}} + {{U_{\rm O}}\over{R_4}}         &=  {{U_{\rm S}}\over{R_1}}  + {{U_{\rm I}}\over{R_4}}  \\ 
 +      U_{\rm O}\cdot \left( {{1}\over{R_1}} + {{1}\over{R_2}} + {{1}\over{R_4}} \right) &=  {{U_{\rm S}}\over{R_1}}  + {{U_{\rm I}}\over{R_4}}  \\         
 +\end{align*} 
 + 
 +Both sides can be multiplied by $\cdot R_1$, $\cdot R_2$, $\cdot R_4$  - in order to get rid of the fractions :  
 +\begin{align*} 
 +      U_{\rm O}\cdot \left( {{R_1 R_2 R_4 }\over{R_1}} + {{R_1 R_2 R_4 }\over{R_2}} + {{R_1 R_2 R_4 }\over{R_4}} \right) &=  R_1 R_2 R_4 \cdot {{U_{\rm S}}\over{R_1}}  + R_1 R_2 R_4 \cdot {{U_{\rm I}}\over{R_4}}  \\         
 +      U_{\rm O}\cdot \left( R_2 R_4 + R_1 R_4 + R_1 R_2 \right) &=  R_2 R_4 \cdot U_{\rm S}  + R_1 R_2 \cdot U_{\rm I} \\         
 +      U_{\rm O} &= {{R_2}\over{R_2 R_4 + R_1 R_4 + R_1 R_2 }}  \left( R_4 \cdot U_{\rm S}  + R_1 \cdot U_{\rm I} \right)\\         
 +\end{align*} 
 + 
 +The last formula was just the result we also got by the superposition but by more thinking. \\ 
 +So, sometimes there is an easier way...  
 +  * Unluckily, there is no simple way to know before, what way is the easiest. 
 +  * Luckily, all ways lead to the correct result. 
 + 
 +#@HiddenEnd_HTML~Solution2,Solution ~@# 
 + 
 +3. What is the input resistance $R_{\rm in}(R_1, R_2, R_3)$ of the circuit (viewed from the sensor)? 
 + 
 +#@HiddenBegin_HTML~Solution3,Solution~@# 
 + 
 +\begin{align*} 
 +R_{\rm in}(R_1, R_2, R_3) &= R_3 + R_1 || R_2 \\ 
 +                          &= R_3 + {{R_1  R_2}\over{R_1 + R_2}} 
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Solution3,Solution~@# 
 + 
 +4. What is the minimum allowed input resistance ($R_{\rm in, min}(R_1, R_2, R_3)$) for the sensor to still deliver current? 
 + 
 +#@HiddenBegin_HTML~Solution4,Solution~@# 
 + 
 +\begin{align*} 
 +R_{\rm in, min} &= {{U_{\rm sense}}\over{I_{\rm sense, max}}} \\ 
 +                &= \rm {{15 V}\over{1 mA}} \\ 
 +                &= 15 k\Omega \\ 
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Solution4,Solution~@# 
 + 
 +#@TaskEnd_HTML@#
  
-For conditioning, the input signal is to be fed via the series resistor $R_3$ to the center potential of a voltage divider $R_1 - R_2$ with $R_1$ against $U_{\rm uC, max}$ (similar circuit see in simulation on the right). 
  
-  - Find the relationship between $R_1$, $R_2$ and $R_3$ using superposition. 
-  - Find the relationship between $R_1$, $R_2$ and $R_3$ using star-delta transformation. 
-  - What is the input resistance $R_{\rm in}(R_1, R_2,R_3)$ of the circuit (viewed from the sensor)? 
-  - What is the maximum allowed input resistance $R_{\rm in}(R_1, R_2,R_3)$ for the sensor to still deliver current? 
-  - Determine suitable values for $R_1$, $R_2$ and $R_3$. 
-  - What values for $R^0_1$, $R^0_2$, and $R^0_3$ from the [[https://de.wikipedia.org/wiki/E-Reihe|E24 series]] can be used to do this? 
  
-</WRAP></WRAP></panel> 
 {{page>aufgabe_4.5.2_mit_rechnung&nofooter}} {{page>aufgabe_4.5.2_mit_rechnung&nofooter}}
 {{page>aufgabe_4.5.3&nofooter}} {{page>aufgabe_4.5.3&nofooter}}