Question 6.6: Frequency Response of Band-Pass Filter Compute the frequency...
Frequency Response of Band-Pass Filter
Compute the frequency response of the band-pass filter of Figure 6.19 for two sets of component values.
Known Quantities:
(a) R=1 \mathrm{k} \Omega ; C=10 \mu \mathrm{F} ; L=5 \mathrm{mH}.
(b) R=10 \Omega; C=10 \mu \mathrm{F} ; L=5 \mathrm{mH}.
Find: The frequency response, H_{V}(j \omega).
Assumptions: None.
Analysis: We write the frequency response of the band-pass filter as in equation 6.29:
\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}(j \omega)=\frac{j \omega C R}{1+j \omega C R+(j \omega)^{2} L C} (6.29)
\begin{aligned}H_{V}(j \omega) &=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}(j \omega)=\frac{j \omega C R}{1+j \omega C R+(j \omega)^{2} L C} \\&=\frac{\omega C R}{\sqrt{\left(1-\omega^{2} L C\right)^{2}+(\omega C R)^{2}}} \angle\left[\frac{\pi}{2}-\arctan \left(\frac{\omega C R}{1-\omega^{2} L C}\right)\right]\end{aligned}
We can now evaluate the response for two different values of the series resistance. The frequency response plots for case a (large series resistance) are shown in Figure 6.21. Those for case b (small series resistance) are shown in Figure 6.22. Let us calculate some quantities for each case. Since L and C are the same in both cases, the resonant frequency of the two circuits will be the same:
\omega_{0}=\frac{1}{\sqrt{L C}}=4.47 \times 10^{3} \mathrm{rad} / \mathrm{s}
On the other hand, the quality factor, Q, will be substantially different:
\begin{array}{ll}Q_{a}=\omega_{0} C R \approx 0.45 & \text { case } a \\Q_{b}=\omega_{0} C R \approx 45 & \text { case } b\end{array}
From these values of Q we can calculate the approximate bandwidth of the two filters:
\begin{array}{ll}B_{a} & =\frac{\omega_{0}}{Q_{a}} \approx 10,000 \mathrm{rad} / \mathrm{s} \quad \text { case } a \\B_{b} & =\frac{\omega_{0}}{Q_{b}} \approx 100 \mathrm{rad} / \mathrm{s} \quad \text { case } b\end{array}
The frequency response plots in Figures 6.21 and 6.22 confirm these observations.
Comments: It should be apparent that, while at the higher and lower frequencies most of the amplitude of the input signal is filtered from the output, at the mid-band frequency (4,500 \mathrm{rad} / \mathrm{s}) \mathrm{most} of the input signal amplitude passes through the filter. The first band-pass filter analyzed in this example would “pass” the mid-band range of the audio spectrum, while the second would pass only a very narrow band of frequencies around the center frequency of 4,500 \mathrm{rad} / \mathrm{s}. Such narrow-band filters find application in tuning circuits, such as those employed in conventional AM radios (although at frequencies much higher than that of the present example). In a tuning circuit, a narrow-band filter is used to tune in a frequency associated with the carrier of a radio station (for example, for a station found at a setting of “AM 820,” the carrier wave transmitted by the radio station is at a frequency of 820 \mathrm{kHz} ). By using a variable capacitor, it is possible to tune in a range of carrier frequencies and therefore select the preferred station. Other circuits are then used to decode the actual speech or music signal modulated on the carrier wave; some of these will be discussed in Chapter 8 .
