1 RC low-pass filter, 1st order

The output voltage of an RC low-pass filter should only be 10What should the cut-off frequency be?

1.1 Lösung:

\begin {align*} \underline {F}(\mathrm {j}\omega )&=\frac {\underline {U}_a}{\underline {U}_e} = \frac {\frac {1}{j\omega C}}{\frac {1}{j\omega C} + R} = \frac {1}{1 + j\omega RC}\\ A(\omega )&=\frac {U_a}{U_e} = \frac {1}{\sqrt {1+(\omega RC)^2}}\\ A(\omega _g)&=\frac {1}{\sqrt {1+(\omega _g RC)^2}} \overset {!}{=} \frac {1}{\sqrt {2}} \Rightarrow \omega _g = \frac {1}{RC} \\ A(\omega )&\overset {!}{=} 0,1 \Rightarrow \sqrt {1+(\omega RC)^2} \overset {!}{=} 10\\ \Leftrightarrow 1 + (\omega RC)^2 &= 100 \Leftrightarrow (\omega RC)^2 = 99 \Rightarrow \omega _g = \frac {\omega }{\sqrt {99}}\\ \Rightarrow f_g &= \frac {f}{\sqrt {99}} = \frac {1\,\mathrm {kHz}}{\sqrt {99}} = 100,5\,\mathrm {Hz} \end {align*}

2 1st order RC high pass filter

What is the phase shift \(\varphi \) in a first-order RC high-pass filter if the stopband attenuation (ratio of output voltage to input voltage) is \(-6dB\)?

2.1 Lösung:

\begin {align*} -6\,\mathrm{d}B &= 20\cdot \lg \left (\frac {U_a}{U_e}\right )\,\mathrm{d}B \\ \Rightarrow -0,3 &= \lg \left (\frac {U_a}{U_e}\right )\\ \Rightarrow \frac {U_a}{U_e} &= 10^{-0,3} = 0,501\\ \frac {\underline {U}_a}{\underline {U}_e} &= \frac {R}{R+\frac {1}{j\omega C}} = \frac {1}{1-j\frac {1}{\omega CR}}\\ \frac {U_a}{U_e} &= \frac {1}{\sqrt {1+\frac {1}{(\omega CR)^2}}} \overset {!}{=} 0,501\\ \Leftrightarrow \sqrt {1 + \frac {1}{(\omega CR)^2}} &= \frac {1}{0,501}\\ \Leftrightarrow \frac {1}{(\omega CR)^2} &= 2,981 \Leftrightarrow \frac {1}{\omega CR} = \sqrt {2,981} = 1,727\\ \frac {\underline {U}_a}{\underline {U}_e} &= \frac {1+j\frac {1}{\omega CR}}{1+\frac {1}{(\omega CR)^2}}\\[0.75em] \varphi &= \arctan \left (\frac {1}{\omega CR}\right ) = \arctan \left (\sqrt {2,981}\right ) = 59,92^\circ \end {align*}

3 Frequency response of an RLC series resonant circuit

Given is the circuit shown on the right with \(L = 20\,\mathrm {mH}\), a resistor \(R = 150\,\Omega \) and a capacitor with \(C = 1 \mu \mathrm {F}\).

• Calculate and sketch the value of the frequency response \(\frac {\underline {U}_2}{\underline {U}_1}\) in semi-logarithmic representation. Give the magnitude for \(\omega = 0, 100\,\mathrm {s}^{-1}\), \(2000\,\mathrm {s}^{-1}\), \(3500\,\mathrm {s}^{-1}\), \(\omega _0\) , \(10000\,\mathrm {s}^{-1}\) and \(\infty \).

• Does an inductance of 10 mH lead to voltage overvoltage in the event of resonance?

3.1 Lösung:

• The resonance frequency \(\omega _0\) is given by the resonance condition \(\Im \{\underline {Z}=0\}\). \begin {align*} \underline {Z} &= R + \mathrm {j}\omega L + \frac {1}{\mathrm {j} \omega C} = R + \mathrm {j}\cdot \left (\omega L + \frac {1}{\omega C}\right ) &&\Rightarrow \omega _0 L - \frac {1}{\omega _0 C} \overset {!}= 0 \\ \Rightarrow \omega _0 &= \frac {1}{\sqrt {LC}} = \frac {1}{\sqrt {20\,\mathrm {mH} \cdot 1\,\mu \mathrm {F}}} = 7071,07\,\mathrm {s}^{-1} &&\Rightarrow f_0 = \frac {\omega _0}{2\pi } = 1125,4\,\mathrm {Hz} \end {align*}

  Since the circuit is a series connection of a resistor, capacitor and coil, it is a voltage resonance.

• The voltage divider rule provides the frequency response: \begin {align*} \frac {\underline {U}_2}{\underline {U}_1} = \underline {F}(\mathrm {j}\omega ) &= \frac {\underline {Z}_{\mathrm {C}}}{\underline {Z}_{\mathrm {R}} + \underline {Z}_{\mathrm {L}} + \underline {Z}_{\mathrm {C}}} \\ &= \frac {\frac {1}{\mathrm {j}\omega C}}{R + \mathrm {j}\omega L + \frac {1}{\mathrm {j}\omega C}} &&\bigg | \cdot \frac {\mathrm {j}\omega C}{\mathrm {j}\omega C} \\ &= \frac {1}{\mathrm {j}\omega CR + \mathrm {j}^2 \omega ^2 LC + 1} \\ &= \frac {1}{1 - \omega ^2 LC + \mathrm {j} \omega R C} \end {align*}

• The amount of the frequency response (amplitude response \(A(\omega )\)) is calculated as follows: \begin {align*} \left | \frac {\underline {U}_2}{\underline {U}_1} \right | = A(\omega ) &= \frac {1}{\sqrt {(1-\omega ^2 LC)^2 + (\omega R C)^2 }} \end {align*}

  The amount of frequency response is shown in the following figure.
  

  The corresponding values are:
  

  \(\omega \)	\(0\)	\(100\,\mathrm {s}^{-1}\)	\(2000\,\mathrm {s}^{-1}\)	\(3500\,\mathrm {s}^{-1}\)	\(\omega _0\)	\(10000\,\mathrm {s}^{-1}\)	\(\omega \to \infty \)
  \(A(\omega )\)	\(1,000\)	\(1,009\)	\(1,033\)	\(1,087\)	\(0,943\)	\(0,555\)	\(0\)

• The quality of a series resonant circuit indicates the voltage increase across \(L\) or \(C\) in the case of resonance. \begin {align*} \frac {U_2(\omega _0)}{U_1(\omega _0)} = Q_S &= \frac {X_k}{R} = \frac {\omega _0 L}{R} = \frac {1}{R}\cdot \sqrt {\frac {L}{C}}\\ &= \frac {1}{150\,\Omega }\cdot \sqrt {\frac {10\,\mathrm {mH}}{1\,\mu \mathrm {F}}}= 0,\overline {66} \end {align*}

  This \(L=10\,\mathrm {mH}\) ist \(Q_S < 1\), means that there is no voltage increase in the event of resonance.

4 Frequency response of a four-pole circuit consisting of R, L and C

The circuit shown on the right is to be examined. The general values \(\mathrm {R}_1\), \(\mathrm {R}_2\), L, C are to be used for the components.

4.1 Lösung:

• The voltage divider ratio is calculated as follows: \begin {align*} \underline {F}(\mathrm {j}\omega ) & = \frac {\underline {U}_2}{\underline {U}_1} \\ &= \frac {R_2}{R_1 + \underline {Z}_L || \underline {Z}_C + R_2} &&\text {mit} \underline {Z}_L || \underline {Z}_C = \frac {\mathrm {j}\omega L \cdot \frac {1}{\mathrm {j}\omega C}}{\mathrm {j}\omega L - \mathrm {j}\frac {1}{\omega C}} = -\mathrm {j}\cdot \left (\frac {1}{\omega C - \frac {1}{\omega L}}\right )\\ &= \frac {R_2}{R_1 + R_2 - \mathrm {j}\left (\frac {1}{\omega C - \frac {1}{\omega L}}\right )} \end {align*}

• The voltage divider ratio is generally calculated as follows: \begin {equation*} \left | \underline {F}(\mathrm {j}\omega ) \right | = A(\omega ) = \frac {\mathrm {R}_2}{\sqrt {(\mathrm {R}_1 + \mathrm {R}_2)^2 + \left (\frac {1}{\omega \mathrm {C} - \frac {1}{\omega \mathrm {L}}}\right )^2}} \end {equation*} The parallel connection of \(L\) and \(C\) behaves like a short circuit for \(\omega \to 0\) and for \(\omega \to \infty \), since one of the two components has an impedance of zero in both cases. This results in the following for \(\omega \to 0\) and \(\omega \to \infty \): \begin {equation*} A(\omega \to 0) = A(\omega \to \infty ) = \frac {R_2}{R_1+R_2} \end {equation*} In the case of resonance (\(\omega = \omega _0 = \sqrt {\frac {1}{LC}}\)), the admittances of \(L\) and \(C\) (parallel resonant circuit) cancel each other out. With \(Y_C + Y_L \to 0\), it follows that \(Z_L||Z_C \to \infty \) for \(\omega \to \omega _0\). \begin {equation*} A(\omega _0) = \frac {R_2}{\sqrt {(\mathrm {R}_1 + \mathrm {R}_2)^2 + \infty ^2}} = \frac {1}{\infty } = 0 \end {equation*}

  The phase results in: \begin {align*} \varphi (\omega )&= \underbrace { \arctan \left (\frac {0}{R_2}\right ) }_{\varphi _{\text {Zähler}}} - \underbrace { \arctan \left (\frac {-\left (\frac {1}{\omega C - \frac {1}{\omega L}}\right )}{R_1 + R_2}\right ) }_{\varphi _{\text {Nenner}}} = \arctan \left (\frac {\frac {1}{\omega C - \frac {1}{\omega L}}}{R_1+R_2}\right ) \end {align*}

  With: \begin {align*} \varphi (\omega \to 0) &= \arctan \left (0\right ) = 0\\ \varphi (\omega \to \infty ) &= \arctan \left (0\right ) = 0\\ \varphi (\omega _0) &= \arctan \left (\infty \right ) = \pm \frac {\pi }{2} \\ \varphi (\omega < \omega _0) &< 0\\ \varphi (\omega > \omega _0) &> 0 \end {align*}

  The amplitude response and phase response are shown in the following Bode diagram.

• The output voltage is calculated as follows: \begin {align*} \underline {U}_2 &= \underline {F}(\mathrm {j}\omega ) \cdot \underline {U}_1 \\ u_2(t) &= A(\omega ) \cdot 5\,\mathrm {V} \cdot \cos \left (\omega t + \varphi (\omega )\right ) &&\text {mit} \omega = 2\pi \cdot 10\,\mathrm {kHz}\cdot \end {align*}

5 RLC parallel resonant circuit

• Calculate the resonance frequency \(f_0\) as well as the frequencies \(f_{\mathrm {u}}\) and \(f_{\mathrm {o}}\) at which the magnitude of \(\underline {Z}(\omega )\) has dropped to \(\frac {Z_0}{\sqrt {2}}\), where \(Z_0\) is the magnitude of the total impedance at resonance. Also determine the difference between the two frequencies. This difference represents the bandwidth \(B\). For this part of the problem, assume \(R = 300\,\Omega \), \(L = 200\,\mathrm {mH}\), and \(C = 1\,\mu \mathrm {F}\).
• The bandwidth is also determined using the formula \(B = \frac {f_0}{Q}\). Compare the result with that from part c) of the exercise.
• Are the upper and lower cut-off frequencies symmetrical around \(f_0\)?

5.1 Lösung:

• The total impedance of the resonant circuit is calculated as follows: \begin {align*} \underline {Y} &= \frac {1}{R} + \mathrm {j}\omega C + \frac {1}{\mathrm {j}\omega L} \\[2pt] \underline {Z} &= \frac {1}{\underline {Y}} = \frac {1}{\frac {1}{R} + \mathrm {j}\omega C + \frac {1}{\mathrm {j}\omega L}} \\[2pt] &= \frac {\frac {1}{R}-\mathrm {j}\cdot \left (\omega C - \frac {1}{\omega L}\right )}{\frac {1}{R^2} + \left (\omega C - \frac {1}{\omega L}\right )^2} \end {align*}

• Amount of total impedance in semi-logarithmic representation:

  The total impedance is at its maximum at the resonance frequency with \(\underline {Z}(\omega _0) = R\).

• Resonance: \begin {align*} \Im \{\underline {Y}\} = \omega _0 C - \frac {1}{\omega _0 L} \overset {!}{=} 0 \Leftrightarrow \omega _0 &= \frac {1}{\sqrt {LC}} \Rightarrow f_0 = \frac {\omega _0}{2\pi }\\ |\underline {Z}(\omega _0)| = \frac {1}{\sqrt {\frac {1}{R^2} + \Big (\smash {\underbrace { \omega _0 C - \frac {1}{\omega _0 L} }_{={}{0}}}\Big )^2 } } &= \frac {1}{\sqrt {\frac {1}{R^2}}} = R \end {align*}

  Cutoff frequencies, bandwidth: \begin {align*} {|\underline {Z}(\omega )| \overset {!}{=} \frac {R}{\sqrt {2}} \Leftrightarrow } \frac {1}{\sqrt {\frac {1}{R^2} + \left (\omega C - \frac {1}{\omega L}\right )^2}} &= \frac {R}{\sqrt {2}} \\ \frac {R}{\sqrt {1 + R^2\cdot \left (\omega C - \frac {1}{\omega L}\right )^2}} &= \frac {R}{\sqrt {2}} \\ 1 + R^2\cdot \left (\omega C - \frac {1}{\omega L}\right )^2 &= 2 \\ \left (\omega C - \frac {1}{\omega L}\right )^2 &= \frac {1}{R^2} \\ \omega C - \frac {1}{\omega L} &= \pm \frac {1}{R} \\ \omega ^2 LC \pm \omega \cdot \frac {L}{R} - 1 &= 0 \\ \omega ^2 \pm \omega \cdot \frac {1}{RC} - \frac {1}{LC} &= 0 \\ x^2 + p \cdot x + q &= 0 \vphantom {\frac {1}{LC}} \end {align*}

  \begin {align*} x_{1,2} &= -\frac {p}{2} \pm \sqrt {\left (\frac {p}{2}\right )^2 - q} \\ \Rightarrow \omega &= \pm \frac {1}{2 \cdot RC} \pm \sqrt {\frac {1}{(2 \cdot RC)^2} + \frac {1}{LC}}\\ \frac {1}{2 \cdot RC} &= 1,6667 \cdot 10^3\ \mathrm {s}^{-1} \qquad \frac {1}{LC} = 5 \cdot 10^6\ \mathrm {s}^{-2} \\[4pt] \omega &= \pm 1,6667 \cdot 10^3\ \mathrm {s}^{-1} \pm 2,7889 \cdot 10^3\ \mathrm {s}^{-1} \\[4pt] \omega {g,o} &= 4,4556 \cdot 10^3 \mathrm {s}^{-1} \qquad f_{g,o} = 709,12\ \mathrm {Hz}\\ \omega {g,u} &= 1,1222 \cdot 10^3 \mathrm {s}^{-1} \qquad f_{g,u} = 178,60\ \mathrm {Hz}\\[4pt] B &= f_{g,o} - f_{g,u} = 530,5\,\mathrm {Hz} \end {align*}

• Determination of bandwidth via resonance frequency and quality factor: \begin {align*} B &= \frac {f_0}{Q} & \text {mit} Q &= \sqrt {\frac {C}{L}}\cdot R = \sqrt {\frac {1\ \mu \mathrm {F}}{200\ \mathrm {mH}}} \cdot 300\ \Omega = 0,6708 \\ B &= \frac {355,88\ \mathrm {Hz}}{0,6708} = 530,5\ \mathrm {Hz} & \text {mit} f_0 &= \frac {1}{2\pi \sqrt {LC}} = \frac {1}{2\pi \sqrt {200\ \mathrm {mH} \cdot 1\ \mu \mathrm {F}}} = 355,88\ \mathrm {Hz} \\ \end {align*}
• With a logarithmic frequency axis, the curve of the impedance magnitude \(|\underline {Z}(\omega )|\) is symmetrical around the resonance frequency \(\omega _0\), i.e. \(f_{g,o}\) and \(f_{g,u}\) are symmetrical around \(f_0\). \begin {equation*} \left (\omega C - \frac {1}{\omega L}\right )^2 = \bigg (\underbrace { \frac {\omega }{\omega _0} - \frac {\omega _0}{\omega } }_{=\vartheta (\omega ){,\ \vartheta ^2\ \text {symmetrisch um }\omega _0\text { bei faktorieller Änderung}}} \bigg )^2\cdot \frac {C}{L} \qquad \Longrightarrow \qquad \vartheta ^2(n\cdot \omega _0) = \vartheta ^2\left (\frac {1}{n}\cdot \omega _0\right ) \end {equation*}

6 Second-order RC filter

The following applies to the RC elements connected in series below \(R_1 = R_2 = R = 1000\,\Omega \) and \(C_1 = C_2 = C = 1\,\mu \mathrm F\).

• Determine the frequency response of the circuit. \(\underline {F}(\mathrm {j}\omega ) = \frac {\underline {U}_2}{\underline {U}_1}\).
• Plot the amplitude response and phase response in logarithmic representation. Compare the result with the simple RC circuit with the same components.

6.1 Lösung:

• Frequency response: \begin {align*} \underline {F}(\mathrm {j}\omega ) &= \frac {\underline {U}_2}{\underline {U}_1} = \frac {\underline {U}_2}{\underline {U}_m} \cdot \frac {\underline {U}_m}{\underline {U}_1}\\ \frac {\underline {U}_m}{\underline {U}_1} &= \frac {\underline {Z}_{\mathrm {ers}}}{R + \underline {Z}_{\mathrm {ers}}} = \frac {1}{\frac {R}{\underline {Z}_{\mathrm {ers}}} + 1}\\ \underline {Z}_{\mathrm {ers}} &= \left ( R + \frac {1}{\mathrm {j}\omega C} \right ) \bigg |\bigg |\, \frac {1}{\mathrm {j}\omega C} = \left ( \frac {1}{R + \frac {1}{\mathrm {j}\omega C}} + \mathrm {j}\omega C\right )^{-1}\\ \frac {\underline {U}_m}{\underline {U}_1} &= \frac {1}{R \cdot \left (\frac {1}{R + \frac {1}{\mathrm {j}\omega C}} + \mathrm {j}\omega C\right ) + 1} \\ &= \frac {1}{\frac {R\cdot \mathrm {j}\omega C}{R\cdot \mathrm {j}\omega C + 1} + R\cdot \mathrm {j}\omega C + 1} \\ &= \frac {R\cdot \mathrm {j}\omega C + 1}{R\cdot \mathrm {j}\omega C + \left ( R\cdot \mathrm {j}\omega C + 1 \right )^2} \\ &= \frac {R\cdot \mathrm {j}\omega C + 1}{R\cdot \mathrm {j}\omega C - R^2\omega ^2 C^2 + 2\cdot \mathrm {j}\omega C R + 1 } \\ &= \frac {1}{1 - \omega ^2 C^2 R^2 + \mathrm {j}3\omega C R} \\ &= \frac {1 + \mathrm {j}\Omega }{1 - \Omega ^2 + \mathrm {j}3\Omega } \qquad \text {mit} \qquad \Omega = \omega C R\\[4pt] \frac {\underline {U}_2}{\underline {U}_m} &= \frac {\frac {1}{\mathrm {j}\omega C}}{R + \frac {1}{\mathrm {j}\omega C}} = \frac {1}{1 + \mathrm {j}\omega C R} = \frac {1}{1 + \mathrm {j}\Omega }\\[4pt] \frac {\underline {U}_2}{\underline {U}_1} &= \frac {\underline {U}_2}{\underline {U}_m} \cdot \frac {\underline {U}_m}{\underline {U}_1} = \frac {1}{1 + \mathrm {j}\omega C R} = \frac {1}{1 + \mathrm {j}\Omega } \cdot \frac {1 + \mathrm {j}\Omega }{1 - \Omega ^2 + \mathrm {j}3\Omega } \\ &= \frac {1}{1 - \Omega ^2 + \mathrm {j}3\Omega } = \frac {1}{1 - (\omega C R)^2 + \mathrm {j}3\omega C R} \end {align*}

  Comparison of simple low-pass filter: \begin {align*} \underline {F}(\mathrm {j}\omega ) = \frac {\frac {1}{\mathrm {j}\omega C}}{R + \frac {1}{\mathrm {j}\omega C}} = \frac {1}{1 + \mathrm {j}\omega C R} = \frac {1}{1 + \mathrm {j}\Omega } \end {align*}

  Due to coupling effects, the frequency response of the simple low-pass filter cannot simply be multiplied in series connection.

• Bode diagram (amplitude and phase response) for 1st and 2nd order RC low-pass filters:

  The attenuation in the blocking range is twice as high (\(40\,\mathrm{d}B /\mathrm {dec}\)) for the second order (blue) compared to the first order (red) (\(20\,\mathrm{d}B /\mathrm {dec}\)). The phase shift ranges from \(0^\circ \) to \(-180^\circ \) for the second order, in contrast to the first order, which only ranges from \(0^\circ \) to \(-90^\circ \). The cut-off frequency shifts to the left (smaller for the second order) due to the higher attenuation.

7 Impedance and admittance locus curve of an RL element

Given is the series connection of an inductance \(L = 10\,\mathrm {mH}\) and a resistance \(R = 5\,\mathrm {k}\Omega \) at a variable frequency with \(f_\mathrm {min} = 5\,\mathrm {kHz}\) and \(f_\mathrm {max} = 50\,\mathrm {kHz}\).

• Determine the impedance location curve and represent it quantitatively. Mark the parameter at least three meaningful points.
• Draw the inverted locus curve (admittance locus curve). First, consider the basic curve and then use some numerical values from the previous task a) to determine the locus curve points.

7.1 Lösung:

8 Admittance locus curve of mixed series/parallel connection

• Determine the admittance for \(p=0\), \(p=1\), \(p=2\) and \(p\to \infty \).
• Sketch the course of the admittance curve and plot the values. (Recommended display range: \(0 + \mathrm {j}0\) bis \(0,25\,\mathrm {S}+ \mathrm {j}0,25\,\mathrm {S}\))
• For what value of \(p\) does the total current \(\underline {I}\) of an applied alternating voltage \(\underline {U}\) lead by \(45^\circ \)? Determine the value for \(\underline {Y}\) by drawing a diagram.
• Check the result from c) mathematically.

8.1 Lösung:

• Admittance \(\underline {Y}\) für \(p=0\), \(p=1\), \(p=2\) und \(p\to \infty \):

9 AC measuring bridge

• Is the measured voltage \(\underline {U}_B\) from task b) frequency-dependent? Justify your answer.
• How does the measurement voltage behave? \(\underline {U}_B\) bei \(X_1 = X_2\)?
• How does the measurement voltage behave? \(\underline {U}_B\) bei \(X_1 = -X_2\)?

9.1 Lösung:

• Measured voltage from potential difference of voltage dividers: \begin {align*} \underline {U}_B &= U_0 \cdot \left ( \frac {\underline {Z}_2}{\underline {Z}_1 + \underline {Z}_2} - \frac {\underline {Z}_4}{\underline {Z}_3 + \underline {Z}_4} \right )\\ &= U_0 \cdot \left ( \frac {\underline {Z}_2\cdot \underline {Z}_3+\cancel {\underline {Z}_2\cdot \underline {Z}_4}-\underline {Z}_1\cdot \underline {Z}_4-\cancel {\underline {Z}_2\cdot \underline {Z}_4}}{\left (\underline {Z}_1+\underline {Z}_2\right )\cdot \left (\underline {Z}_3+\underline {Z}_4\right )} \right ) \\ &= U_0 \cdot \left ( \frac {\underline {Z}_2\cdot \underline {Z}_3\ -\ \underline {Z}_1\cdot \underline {Z}_4}{\left (\underline {Z}_1+\underline {Z}_2\right )\cdot \left (\underline {Z}_3+\underline {Z}_4\right )} \right ) \end {align*}
• As in a) only with \(\underline {Z}_1 = \mathrm {j}X_1\), \(\underline {Z}_2 = \mathrm {j}X_2\), \(\underline {Z}_3 = \underline {Z}_4 = R\). \begin {align*} \underline {U}_B &= U_0 \cdot \left ( \frac {X_2 - X_1}{\left (X_1+X_2\right )\cdot \left (1+1\right )} \right ) \\ \underline {U}_B &= U_0 \cdot \left ( \cancel {\frac {\mathrm {j}R}{\mathrm {j}R}} \cdot \frac {\mathrm {j}X_2\cdot R - \mathrm {j}X_1\cdot R}{\left (\mathrm {j}X_1+\mathrm {j}X_2\right )\cdot \left (R+R\right )} \right ) \\ &= \frac {U_0}{2}\cdot \frac {X_2-X_1}{X_2+X_1} \end {align*}
• The measured voltage \(\underline {U}_B\) is independent of the frequency, provided that there are two capacitive (\(X_C \sim \frac {1}{\omega }\)) or two inductive reactances (\(X_L \sim \omega \)). Then the frequency-dependent terms of the reactances \(X_1\) and \(X_2\) cancel each other out.
• For \(X_1=X_2\), \(\underline {U}_B=0\). Both voltage dividers (for \(\underline {U}_B^+\) and \(\underline {U}_B^-\)) yield \(U_0/2\).
• For \(X_1=-X_2\), the resonance condition is satisfied (of the ideal LC series resonant circuit consisting of \(X_1\) and \(X_2\)). The measured voltage \(\underline {U}_B\) approaches infinity (since it is undamped).