(Solved): All the same question, just expanding upon how to do it Problem 3) RLC band pass filter The circuit ...

Problem 3) RLC band pass filter The circuit in Figure 3 is an RLC band pass filter. It passes frequencies in a narrow band around a center frequency \( f_{0} \) and attenuates signals at other frequencies. The center frequency is the resonance frequency of the RLC circuit. The frequency response of this circuit depends on the Figure 3. RLC band pass filter with a model values of RLC, but also on the series resistance \( R_{5 .} \) All for a practical inductor. practical inductors have a small series resistance due to the resistance in the wire winding. In the Figure 3 , the circuit in the dashed box is a model for a practical inductor. \( \mathrm{R}_{\mathrm{s}} \) was neglected in problem 2, but it is important in this problem. Unfortunately, deriving the transfer function for figure 3 is messy; it requires two node equations and a lot of algebra. The circuit in Figure 4 is an approximation for Figure Figure 4. Circuit from Figure 3 with the shunt 3 where the small series resistance is replaced by a large parallel resistance, \( \mathrm{R}_{\mathrm{p}} \). This makes the transfer function much \( \mathrm{R} / \mathrm{L} \) model for the practical inductor. easier to derive; only one node equation is necessary. To get \( R_{p} \) from \( R_{s} \), use the following, where \( Q_{L} \). is called the quality factor of the inductor. It varies with frequency. For the \( 1.0 \mathrm{mH} \) inductor you are using in this lab, \( Q_{L} \) is about 20 at \( 50 \mathrm{KHz}, 40 \) at \( 100 \mathrm{KHz} \) and 60 at \( 150 \mathrm{KHz}, \mathrm{R}_{\text {s }} \), is about 2 ohms. Approximating the effect of \( R_{s} \) using \( R_{p} \) works very well for (1) frequencies in the vicinity of \( f_{o} \) and \( (2) \) when \( Q_{L} \) is much greater than 1 . See the appendix for more details. For this problem, do the following, a) Derive the transfer function, \( \mathrm{G}(\omega)=\mathrm{V}_{\alpha} / \mathrm{V}_{\mathrm{s}} \) for the circuit in Figure 4. (This will be an approximation of the circuit in Figure 3.) The transfer function will be an equation b) Find the expression for \( G\left(\omega_{0}\right) \) where \( \omega_{0} \) is the resonance frequency of the L,C combination. \( \omega_{0}=2 \pi \mathrm{f}_{0}=(\mathrm{LC})^{-1 / 2} \). Your answer will be in terms of \( \mathrm{R} \) and \( \mathrm{R}_{\mathrm{p}} \). Replace \( \mathrm{R}_{\mathrm{p}} \) with \( \mathrm{R}_{\mathrm{s}} \mathrm{QL}^{2} \). c) Make \( \mathrm{L}=1.0 \mathrm{mH} \) and choose a capacitor from the available parts in the lab to get an \( \mathrm{f}_{0} \) within \( 10 \% \) of your characteristic frequency, \( f_{x} \). d) Simulate the circuit in Figure 3 with PSpice or CircuitLab. Use \( \mathrm{R}_{\mathrm{s}}=2 \mathrm{ohms} \) and \( \mathrm{R}= \) \( 4.7 \) kilohm. Plot \( |G(\omega)| \) and \( <\mathrm{G}(\omega) \) versus frequency for \( 10 \mathrm{~Hz}<\mathrm{f}<10^{6} \mathrm{~Hz} \). Use a log scale for frequency and note the frequency where \( |\mathrm{G}(\omega)| \) is maximum. The frequency of the maximum should be the same as the \( f_{o} \) you had in part \( c \). If not, then you need to find the problem. Evaluate the \( \mathrm{G}\left(\omega_{0}\right) \) expression you got in part \( \mathrm{b} \) and compare it to the maximum \( G(\omega) \) you got in your simulation. You will need all the results and plots from problems 1, 2,3 for your lab Appendix I: Inductor \( Q \) and the high \( Q \) approximation. Practical inductors are coils of wire that have a small resistance. They are modeled as an ideal inductor in series with an ideal resistor as in Figure A1 (a). A good inductor has the series resistance, \( \mathrm{R}_{\mathrm{s}} \), much smaller than the inductor reactance, \( \omega \mathrm{L} \). A typical measure of this is the inductor quality factor, \( \mathrm{Q}_{\mathrm{L}}=\omega \mathrm{L} / \mathrm{R}_{\mathrm{s}} \), which depends on frequency. (The inductor quality factor is not necessarily the same as the \( \mathrm{Q} \) of a resonant circuit that uses the inductor since there may be other sources of loss in the circuit.) \( \begin{array}{ll}\text { (a) } & \text { (b) } \\ \text { Figure } \mathrm{Al} \text { (a) model of practical }\end{array} \) inductor, (b) approximation of model. When doing a circuit calculation, the model in \( \mathrm{Al}(\mathrm{a}) \) is sometimes awkward to use. When \( \mathrm{Q}_{\mathrm{L}} \) is large, the circuit \( \mathrm{Al} \) (b) can approximate \( \mathrm{Al} \) (a). The admittance of the two circuits is, \[ Y_{a}=\frac{1}{R_{s}+j \omega L} \quad Y_{b}=\frac{1}{j \omega L}+\frac{1}{R_{b}} \] The expression for \( Y_{a} \) can be made to look like \( Y_{b} \) when \( Q_{L} \) is large \( { }^{1} \). \[ Y_{\mathrm{a}}=\frac{1}{R_{s}+j \omega L}=\frac{1}{j \omega L} \frac{1}{\frac{1}{j Q_{L}}+1}=\frac{1}{j \omega L}\left(1-\frac{1}{j Q_{L}}\right)=\frac{1}{j \omega L}+\frac{1}{\omega L Q_{L}}=\frac{1}{j \omega L}+\frac{1}{R_{s} Q_{L}^{2}} \] By identifying the last term with \( \mathrm{Y}_{\mathrm{b}} \), you get \( \mathrm{R}_{\mathrm{p}} \) is approximately \( \mathrm{R}_{\mathrm{s}} \mathrm{Q}_{\mathrm{L}}{ }^{2} \).