Exercise E2 Analyzing a Scope Plot
(written test, approx. 12 % of a 60-minute written test, SS2023)

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.

electrical_engineering_1:exbtaqnp18vn0gricircuit.svg

Use the correct symbols and units in your answers!

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

Solution

Frequency $f$ is given by the period $T$. The period can be measured in the imagine of the scope.

  1. The sine waves repeat after $6 ~\rm divisions$ (e.g. from falling turning point to falling turning point of one curve)
  2. 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*}

Result

\begin{align*} f = 1.66... ~{\rm kHz} \rightarrow f &= 1.67 ~\rm kHz \end{align*}

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

Solution

Similar to 1. the amplitude can be derived. The amplitude is given by the voltage difference between maximum and turning point.

  1. The amplitude of the voltage is $3 ~\rm divisions$
  2. 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.

Result

\begin{align*} \hat{U} &= 15 ~\rm V \\ \hat{I} &= 0.5 ~\rm A \end{align*}

3. What are the RMS values of voltage and current?

Solution

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*}

Result

\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*}

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

Solution

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*}

Result

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