{{tag>signal_analysis RMS exam_ee2_SS2021}}{{include_n>1260}}
 

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~~#@ee1_taskctr~~ Signal Analysis \\
<fs medium>(written test, approx. 6 % of a 120-minute written test, SS2021)</fs> 
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At an AC consumer, the following functions for voltage and current were measured:  
  * $u(t) = 50{~\rm V} \cdot \cos (6000 {{1}\over{\rm s}} \cdot t + 4) $
  * $i(t) = 30{~\rm A} \cdot \sin (6000 {{1}\over{\rm s}} \cdot t + 5) $

a) Determine the amplitude values $\hat{U}$, $\hat{I}$ and the RMS values $U$, $I$ 


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  * The amplitude values $\hat{U}$, $\hat{I}$ are given directly by the coefficient of the cosine and sine functions
  * For the RMS values of sinusoidal functions the amplitudes have to be multiplied with ${{1}\over{2}}\sqrt{2}$

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Amplitude values:
  * $\hat{U} = 50{~\rm V}$ 
  * $\hat{I} = 30{~\rm A}$

RMS values:
  * $U = 35.4{~\rm V}$ 
  * $I = 21.2{~\rm A}$


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b) Determine the frequency $f$ and the phase angle $\varphi$ in degrees (°). (Independent)  


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The frequency can be derived by the term in the sine function:
\begin{align*}
\omega       &= 6000 {{1}\over{\rm s}} \\
2\pi \cdot f &= 6000 {{1}\over{\rm s}} \\
           f &= {{6000}\over{2\pi}} {{1}\over{\rm s}} \\
           f &= 954.93... ~\rm Hz \\
\end{align*}

For the phase $\varphi$, we have to subtract $\varphi_i $ from $\varphi_u$. \\
But to get these values, both the $u(t)$ and $i(t)$ need to have the same sinusoidal function!
Therefore: 
  * $\varphi_i = 5$
  * $\varphi_u = 4 + {{\pi}\over{2}}$

By this we get for $\varphi$
\begin{align*}
\varphi &= \varphi_u - \varphi_i \\
        &=  4 + {{\pi}\over{2}} - 5  \\
        &= 2.14159...  \\
\end{align*}

Converted in degree: 
\begin{align*}
\varphi &=  2.14159... \cdot {{360°}\over{2\pi}} \\
        &= 32.7042...°  \\
\end{align*}

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  * $f = 955 ~\rm Hz$
  * $\varphi = +32.7°$

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c) Is the measured element resistive-capacitive or resistive-inductive? \\ The quantities are available in the consumer arrow system. (hard)  


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The phase shift is positive - therefore, the element is resistive-inductive.

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