Double-Tuned Filters

Some network configurations display both a pass-band and a stop-band characteristic, such as shown in Fig. 1. Such networks are called double-tuned filters. For the network of Fig. 1(a), the parallel resonant circuit will establish a stop-band for the range of frequencies not permitted to establish a significant $ V_{L} $. The greater part of the applied voltage will appear across the parallel resonant circuit for this frequency range due to its very high impedance compared with $ R_{L} $.
Double-tuned networks.Double-tuned networks.
Fig. 1: Double-tuned networks.
For the pass-band, the parallel resonant circuit is designed to be capacitive (inductive if $ L_{s} $ is replaced by $ C_{s} $ ). The inductance $ L_{s} $ is chosen to cancel the effects of the resulting net capacitive reactance at the resonant circuit. The applied voltage will then appear across $ R_{L} $ at this frequency.
For the network of Fig. 1(b), the series resonant circuit will still determine the pass-band, acting as a very low impedance across the parallel inductor at resonance. At the desired stop-band resonant frequency, the series resonant circuit is capacitive. The inductance $ L_{p} $ is chosen to establish parallel resonance at the resonant stop-band frequency.
The high impedance of the parallel resonant circuit will result in a very low load voltage $ V_{L} $. For rejected frequencies below the pass-band, the networks should appear as shown in Fig. 1. For the reverse situation, $ L_{s} $ in Fig. $ 1 (a) $ and $ L_{p} $ in Fig. 1(b) are replaced by capacitors.
Example 1: determine $ L_{s} $ and $ L_{p} $ for a capacitance $ C $ of $ 500 \mathrm{pF} $ if a frequency of $ 200 \mathrm{kHz} $ is to be rejected and a frequency of $ 600 \mathrm{kHz} $ accepted. Solution: For series resonance, we have
$$ f_{s}=\frac{1}{2 \pi \sqrt{L C}} $$
$$ L_{s}=\frac{1}{4 \pi^{2} f_{s}^{2} C}=\frac{1}{4 \pi^{2}(600 \mathrm{kHz})^{2}(500 \mathrm{pF})}=140.7 \mu \mathrm{H} $$
At $200 \mathrm{kHz} $,
$$ X_{L_{s}}=\omega L=2 \pi f_{s} L_{s}=(2 \pi)(200 \mathrm{kHz})(140.7 \mu \mathrm{H})=176.8 \Omega $$
$$ X_{C}=\frac{1}{\omega C}=\frac{1}{(2 \pi)(200 \mathrm{kHz})(500 \mathrm{pF})}=1591.5 \Omega $$
For the series elements,
$$j\left(X_{L_{s}}-X_{C}\right)=j(176.8 \Omega-1591.5 \Omega)=-j 1414.7 \Omega=-j X_{C}^{\prime}$$
At parallel resonance ( $ Q_{I} \geq 10 $ assumed),
$$L_{p}=\frac{X_{L_{P}}}{\omega}=\frac{1414.7 \Omega}{(2 \pi)(200 \mathrm{kHz})}=1.13 \mathrm{mH}$$