Given the parallel RLC circuit in Fig. 12.34,

a. Derive the expression for the resonant frequency, the half-power frequencies, the bandwidth, and the quality factor for the transfer characteristic $V _{\text {out }} / I _{\text {in }}$ in terms of the circuit parameters R, L, and C.
b. Compute the quantities in part (a) if R = 1 kΩ, L = 10 mH, and C = 100 μF.

Question

Given the parallel RLC circuit in Fig. 12.34,

a. Derive the expression for the resonant frequency, the half-power frequencies, the bandwidth, and the quality factor for the transfer characteristic [latex]V _{ ext {out }} / I _{ ext {in }}[/latex] in terms of the circuit parameters R, L, and C.
b. Compute the quantities in part (a) if R = 1 kΩ, L = 10 mH, and C = 100 μF.

Accepted Answer

a. The output voltage can be written as
[latex]V _{ ext {out }}=\frac{ I _{ ext {in }}}{ Y _{T}}[/latex]
and, therefore, the magnitude of the transfer characteristic can be expressed as
[latex]\left|\frac{ V _{ ext {out }}}{ I _{ ext {in }}} ight|=\frac{1}{\sqrt{\left(1 / R^{2} ight)+(\omega C-1 / \omega L)^{2}}}[/latex]
The transfer characteristic is a maximum at the resonant frequency
[latex]\omega_{0}=\frac{1}{\sqrt{L C}}[/latex] 12.41
and at this frequency
[latex]\left|\frac{ V _{ out }}{ I _{ in }} ight|_{\max }=R[/latex] 12.42
As demonstrated earlier, at the half-power frequencies the magnitude is equal to [latex] 1 / \sqrt{2}[/latex] of its maximum value, and hence the half-power frequencies can be obtained from the expression
[latex]\frac{1}{\sqrt{\left(1 / R^{2} ight)+(\omega C-1 / \omega L)^{2}}}=\frac{R}{\sqrt{2}}[/latex]
Solving this equation and taking only the positive values of ω yields
[latex]\omega_{ LO }=-\frac{1}{2 R C}+\sqrt{\frac{1}{(2 R C)^{2}}+\frac{1}{L C}}[/latex] 12.43
and
[latex]\omega_{ HI }=\frac{1}{2 R C}+\sqrt{\frac{1}{(2 R C)^{2}}+\frac{1}{L C}}[/latex] 12.44
Subtracting these two half-power frequencies yields the bandwidth
[latex]BW =\omega_{ HI }-\omega_{ LO }[/latex] 12.45
[latex]=\frac{1}{R C}[/latex]
Therefore, the quality factor is
[latex]Q=\frac{\omega_{0}}{ BW }[/latex]
[latex]=\frac{R C}{\sqrt{L C}}[/latex] 12.46
[latex]=R \sqrt{\frac{C}{L}}[/latex]
Using Eqs. (12.41), (12.45), and (12.46), we can write Eqs. (12.43) and (12.44) as
[latex]\omega_{ LO }=\omega_{0}\left[\frac{-1}{2 Q}+\sqrt{\frac{1}{(2 Q)^{2}}+1} ight][/latex] 12.47
[latex]\omega_{ HI }=\omega_{0}\left[\frac{1}{2 Q}+\sqrt{\frac{1}{(2 Q)^{2}}+1} ight][/latex] 12.48
b. Using the values given for the circuit components, we find that
[latex]\omega_{0}=\frac{1}{\sqrt{\left(10^{-2} ight)\left(10^{-4} ight)}}=10^{3} rad / s[/latex]
The half-power frequencies are
[latex]\omega_{ LO }=\frac{-1}{(2)\left(10^{3} ight)\left(10^{-4} ight)}+\sqrt{\frac{1}{\left[(2)\left(10^{-1} ight) ight]^{2}}+10^{6}}[/latex]
= 995 rad/s
and
[latex]\omega_{ HI }=1005 rad / s[/latex]
Therefore, the bandwidth is
[latex]BW =\omega_{ HI }=\omega_{ LO }=10 rad / s[/latex]
and
[latex]Q=10^{3} \sqrt{\frac{10^{-4}}{10^{-2}}}[/latex]
= 100

Question 12.14: Given the parallel RLC circuit in Fig. 12.34, a. Derive the ...

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