Question 5.7: The parameters of an RLC parallel circuit excited by a curre......
The parameters of an RLC parallel circuit excited by a current source are R = 40 Ω, L = 2 mH and C = 3 µF. Determine the resonant frequency, quality factor, bandwidth and cut-off frequencies.
Given that, R = 40 Ω, L = 2 mH and C = 3 µF
Angular resonant frequency, \omega_{ r }=\frac{1}{\sqrt{ LC }}
=\frac{1}{\sqrt{2 \times 10^{-3} \times 3 \times 10^{-6}}}=12909.9445 \approx 12910~ rad / s
Resonant frequency , f_{ r }=\frac{\omega_{ r }}{2 \pi}=\frac{12910}{2 \pi}=2054.7 ~Hz =\frac{2054.7}{1000} kHz =2.0547~ kHz
Quality factor at resonance , Q _{ r }=\frac{ R }{\omega_{ r } L }=\frac{40}{12910 \times 2 \times 10^{-3}}=1.5492
Bandwidth, \beta=\frac{1}{ RC }=\frac{1}{40 \times 3 \times 10^{-6}}=8333.3333 ~rad / s \approx 8333~ rad / s
Bandwidth in Hz =\frac{\beta}{2 \pi}=\frac{8333.3333}{2 \pi}=1326.3 ~Hz =\frac{1326.3}{1000}~ kHz =1.3263~ kHz
Higher cut – off frequecy, f _{ h }= f _{ r }\left[\frac{1}{2 Q _{ r }}+\sqrt{1+\frac{1}{4 Q _{ r }^2}}\right]
\begin{aligned} & =2.0547 \times\left[\frac{1}{2 \times 1.5492}+\sqrt{1+\frac{1}{4 \times 1.5492^2}}\right] \\ & =2.8222~ kHz \end{aligned}
\begin{aligned}\text{Lower cut – off frequecy,} f _l & = f _{ r }\left[-\frac{1}{2 Q _{ r }}+\sqrt{1+\frac{1}{4 Q_{ r }^2}}\right] \\ & =2.0547 \times\left[-\frac{1}{2 \times 1.5492}+\sqrt{1+\frac{1}{4 \times 1.5492^2}}\right] \\ & =1.4959 ~kHz \end{aligned}