a) Show that the RLC circuit in Fig. 14.22 is also a bandpass filter by deriving an expression for the transfer function H\left(s\right).
b) Compute the center frequency, \omega _{o}.
c) Calculate the cutoff frequencies, \omega _{c1} and \omega _{c2}, the bandwidth, \beta, and the quality factor, Q.
d) Compute values for R and L to yield a bandpass filter with a center frequency of 5 kHz and a bandwidth of 200 Hz,using a 5 \mu F capacitor.
a) Begin by drawing the s-domain equivalent of the circuit in Fig. 14.22, as shown in Fig. 14.23. Using voltage division, we can compute the transfer function for the equivalent circuit if we first compute the equivalent impedance of the parallel combination of L and C, identified as Z_{eq}\left(s\right) in Fig.14.23:
Z_{eq}\left(s\right) =\frac{\frac{L}{C} }{sL+\frac{1}{sC} }.
Now,
H\left(s\right)=\frac{\frac{s}{RC} }{s^{2}+\frac{s}{RC}+\frac{1}{LC}}.
b) To find the center frequency, , we need to calculate where the transfer function magnitude is maximum. Substituting s=j\omega in H\left(s\right),
\mid H\left(j\omega\right)\mid =\frac{\frac{\omega }{RC} }{\sqrt{\left(\frac{1}{LC}-\omega ^{2} \right) ^{2} +\left(\frac{\omega }{RC} \right) ^{2}} }
=\frac{1}{\sqrt{1+\left(\omega RC-\frac{1 }{\omega \frac{L }{R}} \right) ^{2} }} .
The magnitude of this transfer function is maximum when the term
\left(\frac{1}{LC}-\omega ^{2} \right) ^{2}is zero. Thus,
\omega _{o} =\sqrt{\frac{1}{LC} }and
H_{max} =\mid H\left(j\omega o\right)\mid=1.
c) At the cutoff frequencies, the magnitude of the transfer function is \left(1/ \sqrt{2} \right) H_{max}= \left(1/ \sqrt{2} \right). Substituting this constant on the left-hand side of the magnitude equation and then simplifying, we get
\left[\omega _{c} RC-\frac{1 }{\omega _{c} \frac{L }{R}}\right]=\pm 1.
Squaring the left-hand side of this equation once again produces two quadratic equations for the cutoff frequencies, with four solutions. Only two of them are positive and therefore have physical significance:
\omega _{c1}=-\frac{1}{2RC} +\sqrt{\left(\frac{1}{2RC}\right) ^{2} +\frac{1}{LC}},
\omega _{c2}=\frac{1}{2RC} +\sqrt{\left(\frac{1}{2RC}\right) ^{2} +\frac{1}{LC}}.
\blacktriangle Cutoff frequencies for parallel RLC filters
We compute the bandwidth from the cutoff frequencies:
\beta =\omega _{c2}-\omega _{c1}
=\frac{1}{RC}.
Finally, use the definition of quality factor to calculate Q:
Q=\omega _{o}/ \beta
=\sqrt{\frac{R^{2} C}{L} }.
Notice that once again we can specify the cutoff frequencies for this bandpass filter in terms of its center frequency and bandwidth:
\omega _{c1}=-\frac{\beta }{2} +\sqrt{\left(\frac{\beta }{2} \right) ^{2}+\omega _{o}^{2} },
\omega _{c2}=\frac{\beta }{2} +\sqrt{\left(\frac{\beta }{2} \right) ^{2}+\omega _{o}^{2} }.
d) Use the equation for bandwidth in (c) to compute a value for R, given a capacitance of 5 \mu F. Remember to convert the bandwidth to the appropriate units:
R=\frac{1}{\beta C}=\frac{1}{\left(2\pi \right)\left(200\right) \left(5\times 10^{-6} \right) }
=159.15 \Omega.
Using the value of capacitance and the equation for center frequency in (c), compute the inductor value:
L=\frac{1}{\omega ^{2}_{o}C }=\frac{1}{\left[2\pi \left(5000\right) \right] ^{2}\left(5\times 10^{-6} \right)}
=202.64 \mu H.
