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#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Analyzing a Scope Plot        \\ <fs medium>(written test, approx. 12 % of a 60-minute written test, SS2023)</fs>
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On an oscilloscope is the following plot visible. \\
The measured current curve shall be visible as a dashed line. \\
**The continuous line shows the voltage.**

{{drawio>electrical_engineering_1:exbtaqnp18vn0griCircuit.svg}}

Use the correct symbols and units in your answers!

1. Calculate the frequency $f$ of the periodic signals.

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Frequency $f$ is given by the period $T$. The period can be measured in the imagine of the scope. 
  - The sine waves repeat after $6 ~\rm divisions$ (e.g. from falling turning point to falling turning point of one curve) 
  - The scale is $0.1 ~\rm ms/Div$

\begin{align*}
f   &= {{1} \over {T}} \\
T   &= 6 ~\rm Div \cdot 0.1 ~ms/Div \\
\rightarrow f &= {{1} \over {6 ~\rm Div \cdot 0.1 ~ms/Div}}
\end{align*}
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\begin{align*}
f   = 1.66... ~{\rm kHz} \rightarrow f   &= 1.67 ~\rm kHz
\end{align*}

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2. Write down the amplitude values for the voltage and current.

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Similar to 1. the amplitude can be derived. The amplitude is given by the voltage difference between maximum and turning point. 
  - The amplitude of the voltage is $3 ~\rm divisions$
  - The scale is $5 ~\rm V/Div$

\begin{align*}
\hat{U} &= 3 ~\rm Div \cdot 5 ~\rm V/Div
\end{align*}

Similar for the current. 
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\begin{align*}
\hat{U} &= 15 ~\rm V \\
\hat{I} &= 0.5 ~\rm A 
\end{align*}
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3. What are the RMS values of voltage and current?

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The RMS values for sine waves are given as \begin{align*}
\square_{\rm RMS} ={{1} \over {2} } \sqrt{2} \cdot \hat{\square} 
\end{align*}

\begin{align*}
U_{\rm RMS} & ={{1} \over {2} } \sqrt{2} \cdot \hat{U} ={{1} \over {2} } \sqrt{2} \cdot 15 ~\rm V  \\
I_{\rm RMS} & ={{1} \over {2} } \sqrt{2} \cdot \hat{I} ={{1} \over {2} } \sqrt{2} \cdot 0.5 ~\rm A  
\end{align*}

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\begin{align*}
U_{\rm RMS} & = 10.606... ~\rm V \rightarrow U_{\rm RMS} = 10.6 ~\rm V  \\
I_{\rm RMS} &= 0.3535... ~\rm A \rightarrow I_{\rm RMS} = 0.35 ~\rm A   
\end{align*}

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4. What is the phase shift on the system under test (in radiant and degree)?


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The phase shift $\varphi$ is given as the angle between the current and voltage phasor. The phase shift is **negative** when the **voltage lags the current**. \\
A full period has an angle of $360°$ or $2\pi$ in $\rm radian$. \\
One has to find out the phase shift as a fraction of a period to get the value of the phase shift:

\begin{align*}
\varphi &= \rm -{{0.5 ~Div} \over { 6 ~Div}} \cdot 2\pi 
\end{align*}

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\begin{align*}
\varphi &= -{{1} \over { 6}} \pi \\
\varphi &= - 30°
\end{align*}

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