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Exercise E11 Signal Analysis
(written test, approx. 6 % of a 120-minute written test, SS2021)

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 $U$, $I$

Path

  • 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 ${{1}\over{2}}\sqrt{2}$

Result

Amplitude values:

  • $\hat{U} = 50{~\rm V}$
  • $\hat{I} = 30{~\rm A}$

RMS values:

  • $U = 35.4{~\rm V}$
  • $I = 21.2{~\rm A}$

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

Path

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*}

Result

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

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

Result

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