IPhO 1984 · Theory problem 3

Electronic frequency filter

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An electronic frequency filter consists of four components coupled as in the upper figure. The impedance of the source can be neglected and the impedance of the load can be taken as infinite. The filter should be such that the voltage ratio \(U_{out}/U_{in}\) has a frequency dependence shown in the lower where \(U_{in}\) is the input voltage and \(U_{out}\) is the output voltage. At frequency \(f_0\) the phase lag between the two voltages is zero.

Figure 1

In order to build the filter you can choose from the following components:

2 resistors, 10 kΩ
2 capacitors, 10 nF
2 solenoids, 160 mH (iron-free and with neglible resistance)

Figure 2

Construct, by combining four of these components, a filter that fulfils the stated conditions. Determine the frequency \(f_0\) and the ratio \(U_{out}/U_{in}\) at this frequency for as many component combinations as possible.

Official English text of the problem.

Answers

  • Problem 3. There are 12 valid component placements, interpreting the plotted dip as the magnitude of \(U_{\rm out}/U_{\rm in}\), as specified in the prose.

Define the positions, in order, as \[ (Z_1,Z_2,Z_3,Z_4) = (\text{left horizontal},\text{right horizontal}, \text{upper bridge},\text{vertical shunt}). \] Each \(R\) denotes a \(10\,\mathrm{k\Omega}\) resistor, each \(C\) a \(10\,\mathrm{nF}\) capacitor, and each \(L\) a \(160\,\mathrm{mH}\) solenoid.

Component placement \((Z_1,Z_2,Z_3,Z_4)\) \(f_0\) \(U_{\rm out}/U_{\rm in}\) at \(f_0\)
\((R,R,C,C)\) or \((C,C,R,R)\) \(1.59\,\mathrm{kHz}\) \(2/3 \approx 0.667\)
\((R,R,L,L)\) or \((L,L,R,R)\) \(9.95\,\mathrm{kHz}\) \(2/3 \approx 0.667\)
\((R,L,R,C)\) \(5.63\,\mathrm{kHz}\) \(4/29 \approx 0.138\)
\((R,C,R,L)\) \(2.81\,\mathrm{kHz}\) \(4/29 \approx 0.138\)
\((C,R,L,R)\) \(5.63\,\mathrm{kHz}\) \(25/29 \approx 0.862\)
\((L,R,C,R)\) \(2.81\,\mathrm{kHz}\) \(25/29 \approx 0.862\)
\((L,C,R,L)\) or \((L,C,C,R)\) \(5.63\,\mathrm{kHz}\) \(1/2\)
\((C,L,R,C)\) or \((C,L,L,R)\) \(2.81\,\mathrm{kHz}\) \(1/2\)

For every placement, the output and input are in phase at \(f_0\).

Solution

The problem has no numbered subparts. We first derive the response of the given circuit, then identify the component placements and calculate their minima.

Use ideal components with \[ R=1.00\times10^4\,\Omega,\qquad C=1.00\times10^{-8}\,\mathrm F,\qquad L=0.160\,\mathrm H. \] In sinusoidal steady state, using phasors with time dependence \(e^{i\omega t}\), \[ Z_R=R,\qquad Z_L=i\omega L,\qquad Z_C=\frac{1}{i\omega C},\qquad \omega=2\pi f. \] The negligible source impedance fixes the input voltage. The infinite load impedance means that no external current leaves the output terminal.

Let \(V_m\) be the voltage at the junction of \(Z_1,Z_2,Z_4\), measured relative to the common lower conductor. Kirchhoff’s current law gives \[ \frac{V_m-V_{\rm in}}{Z_1} +\frac{V_m-V_{\rm out}}{Z_2} +\frac{V_m}{Z_4}=0, \] \[ \frac{V_{\rm out}-V_m}{Z_2} +\frac{V_{\rm out}-V_{\rm in}}{Z_3}=0. \] Eliminating \(V_m\), the complex voltage ratio is \[ H(\omega)\equiv\frac{V_{\rm out}}{V_{\rm in}} = \frac{Z_1Z_2+Z_4(Z_1+Z_2+Z_3)} {Z_1Z_2+Z_4(Z_1+Z_2+Z_3)+Z_1Z_3}. \] It is convenient to write this as \[ \boxed{H=\frac{K}{K+1}}, \] where \[ \boxed{ K=\frac{Z_2}{Z_3} +\frac{Z_4}{Z_3} +\frac{Z_2Z_4}{Z_1Z_3} +\frac{Z_4}{Z_1}. } \]

Component placements. Define the dimensionless constant \[ \kappa=\frac{L}{R^2C}. \] For each of the following placements, substitution into \(K\) gives \[ K=\alpha+i\left(\beta\omega-\frac{\gamma}{\omega}\right), \qquad \alpha,\beta,\gamma>0. \] The component codes in the table are written in the order \(Z_1Z_2Z_3Z_4\).

Placement \(\alpha\) \(\beta\) \(\gamma\)
\(RRCC,\ CCRR\) \(2\) \(RC\) \(1/(RC)\)
\(RRLL,\ LLRR\) \(2\) \(L/R\) \(R/L\)
\(RLRC\) \(\kappa\) \(L/R\) \(2/(RC)\)
\(RCRL\) \(\kappa\) \(2L/R\) \(1/(RC)\)
\(CRLR\) \(1/\kappa\) \(RC\) \(2R/L\)
\(LRCR\) \(1/\kappa\) \(2RC\) \(R/L\)
\(LCRL\) \(1\) \(L/R\) \(2/(RC)\)
\(LCCR\) \(1\) \(RC\) \(2R/L\)
\(CLRC\) \(1\) \(2L/R\) \(1/(RC)\)
\(CLLR\) \(1\) \(2RC\) \(R/L\)

For example, the placement \(RLRC\) means \[ Z_1=R,\quad Z_2=i\omega L,\quad Z_3=R,\quad Z_4=\frac1{i\omega C}. \] It gives \[ K=\frac{i\omega L}{R} +\frac{1}{i\omega CR} +\frac{L}{CR^2} +\frac{1}{i\omega CR} = \kappa+i\left(\frac{\omega L}{R}-\frac{2}{\omega CR}\right), \] as listed.

Frequency dependence and phase. Put \[ u=\beta\omega-\frac{\gamma}{\omega}. \] Then \[ H=\frac{\alpha+iu}{\alpha+1+iu}, \qquad |H|^2=\frac{\alpha^2+u^2}{(\alpha+1)^2+u^2}. \] Since \[ \frac{d|H|^2}{d(u^2)} = \frac{2\alpha+1}{\bigl[(\alpha+1)^2+u^2\bigr]^2}>0, \] the magnitude is smallest when \(u=0\). Moreover, \[ \frac{du}{d\omega}=\beta+\frac{\gamma}{\omega^2}>0, \] so there is exactly one minimum at positive frequency: \[ \boxed{\omega_0=\sqrt{\frac{\gamma}{\beta}}}, \qquad \boxed{f_0=\frac{1}{2\pi}\sqrt{\frac{\gamma}{\beta}}}. \] At this frequency, \[ \boxed{H(\omega_0)=\frac{\alpha}{\alpha+1}>0}. \] Thus the voltage ratio is real and positive: the phase difference is zero, as required.

At both frequency limits, \[ \omega\to0:\quad u\to-\infty, \qquad \omega\to\infty:\quad u\to+\infty, \] and consequently \[ \lim_{\omega\to0}H=\lim_{\omega\to\infty}H=1. \] Every listed placement therefore has the required curve: its magnitude starts at unity, decreases smoothly to a single positive minimum at \(f_0\), and then rises towards unity.

Numerical frequencies and voltage ratios. The table produces four distinct frequencies: \[ f_{RC}=\frac{1}{2\pi RC} =1.5915\times10^3\,\mathrm{Hz} \approx\boxed{1.59\,\mathrm{kHz}}, \] \[ f_{RL}=\frac{R}{2\pi L} =9.9472\times10^3\,\mathrm{Hz} \approx\boxed{9.95\,\mathrm{kHz}}, \] \[ f_+=\frac{1}{2\pi}\sqrt{\frac{2}{LC}} =5.6270\times10^3\,\mathrm{Hz} \approx\boxed{5.63\,\mathrm{kHz}}, \] \[ f_-=\frac{1}{2\pi\sqrt{2LC}} =2.8135\times10^3\,\mathrm{Hz} \approx\boxed{2.81\,\mathrm{kHz}}. \]

Also, \[ \kappa =\frac{0.160}{(10^4)^2(10^{-8})} =0.160=\frac4{25}. \] The possible minimum ratios are therefore \[ \alpha=2:\qquad H_0=\frac23, \] \[ \alpha=\kappa:\qquad H_0=\frac{\kappa}{1+\kappa} =\frac{L}{R^2C+L} =\frac4{29}, \] \[ \alpha=\frac1\kappa:\qquad H_0=\frac{1}{1+\kappa} =\frac{R^2C}{R^2C+L} =\frac{25}{29}, \] \[ \alpha=1:\qquad H_0=\frac12. \] Combining these expressions with the placement table gives all the answers above.

Completeness of the placements. There are 54 assignments of four component types to the four positions, allowing at most two of each type: 18 use two components of each of two types, and 36 use all three types.

The unity limits at both ends require \(K\) to diverge as both \(\omega\to0\) and \(\omega\to\infty\). This gives a simple way to check all assignments. Write each impedance as a positive constant times \(s^p\), with \[ s=i\omega,\qquad p_R=0,\quad p_L=1,\quad p_C=-1. \] The four terms of \(K\) have powers \[ p_2-p_3,\quad p_4-p_3,\quad p_2+p_4-p_1-p_3,\quad p_4-p_1. \] The endpoint condition holds precisely when these powers include both a positive and a negative value. It leaves 22 placements: the 12 already obtained and the following ten.

Remaining placements Reason they fail
\(RLCC,\ RCLL,\ LRCC,\ CRLL\) Their only finite positive frequency with real \(H\) is \(\omega=1/\sqrt{LC}\), where \(H=0\). The output phase is then undefined, and the minimum is zero.
\(LLRC,\ LLCR,\ CCRL,\ CCLR\) Their only finite positive frequency with real \(H\) gives \(H=-1\): the phase difference is \(180^\circ\).
\(LLCC,\ CCLL\) They have undamped resonant poles and do not give the illustrated smooth dip.

For the last two placements, writing \(x=\omega\sqrt{LC}\) explicitly gives \[ H=\frac{(1-x^2)^2}{1-3x^2+x^4}, \] whose denominator vanishes at \[ x^2=\frac{3\pm\sqrt5}{2}. \] Thus the 12 listed placements exhaust the possibilities satisfying the interpreted graph and the zero-phase condition.

The solution was checked against the official solution.