Double Tuned Filters

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﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿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 $R_{L}$.
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
$$f_{s}=\frac{1}{2 \pi \sqrt{L C}}$$
and
$$L_{s}=\frac{1}{4 \pi^{2} f_{s}^{2} C}=\frac{1}{4 \pi^{2}(600 \mathrm{kHz})^{2}(500 \mathrm{pF})}=140.7 \mu \mathrm{H}$$
At $200 \mathrm{kHz}$,
$$X_{L_{s}}=\omega L=2 \pi f_{s} L_{s}=(2 \pi)(200 \mathrm{kHz})(140.7 \mu \mathrm{H})=176.8 \Omega$$
and
$$X_{C}=\frac{1}{\omega C}=\frac{1}{(2 \pi)(200 \mathrm{kHz})(500 \mathrm{pF})}=1591.5 \Omega$$
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),
$$X_{L_{p}}=X_{C}^{\prime}$$
and
$$L_{p}=\frac{X_{L_{P}}}{\omega}=\frac{1414.7 \Omega}{(2 \pi)(200 \mathrm{kHz})}=1.13 \mathrm{mH}$$

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