The antenna of an FM radio picks up stations across the entire FM frequency range-approximately 87.5 MHz to 108 MHz. The radio's circuitry must have the capability to first reject all of the stations except the one that the listener wants to hear and then to boost the minute antenna signal.

Question

The antenna of an FM radio picks up stations across the entire FM frequency range-approximately 87.5 MHz to 108 MHz. The radio's circuitry must have the capability to first reject all of the stations except the one that the listener wants to hear and then to boost the minute antenna signal. A tuned amplifier incorporating parallel resonance can perform both tasks simultaneously.

The network in Fig. 12.75a is a circuit model for a single-stage tuned transistor amplifier where the resistor, capacitor, and inductor are discrete elements. Let us find the transfer function [latex]\mathbf{V}_{o}(s) / \mathbf{V}_{A}(s)[/latex], where [latex]\mathbf{V}_{A}(s)[/latex] is the antenna voltage and the value of C for maximum gain at 91.1 MHz. Finally, we will simulate the results.

Accepted Answer

Since [latex]\mathbf{V}(s)=\mathbf{V}_{A}(s)[/latex], the transfer function is

[latex]\begin{aligned}&\frac{\mathbf{V}_{o}(s)}{\mathbf{V}_{A}(s)}=-\frac{4}{1000}\left[R / / s L / / \frac{1}{s C} ight] \\&\frac{\mathbf{V}_{o}(s)}{\mathbf{V}_{A}(s)}=-\frac{4}{1000}\left[\frac{s / C}{s^{2}+\frac{s}{R C}+\frac{1}{L C}} ight]\end{aligned}[/latex]

The parallel resonant network is actually a band-pass filter. Maximum gain occurs at the center frequency, [latex]f_{0}[/latex]. This condition corresponds to a minimum value in the denominator. Isolating the denominator polynomial, D(s), and letting s = jω, we have

[latex]\mathbf{D}(j \omega)=\frac{R}{L C}-\omega^{2}+\frac{j \omega}{C}[/latex]

which has a minimum value when the real part goes to zero, or

[latex]\frac{1}{L C}-\omega_{0}^{2}=0[/latex]

yielding a center frequency of

[latex]\omega_{0}=\frac{1}{\sqrt{L C}}[/latex]

Thus, for a center frequency of 91.1 MHz, we have

[latex]2 \pi\left(91.1 imes 10^{6} ight)=\frac{1}{\sqrt{L C}}[/latex]

and the required capacitor value is

[latex]C=3.05 \mathrm{pF}[/latex]

The Bode plot for the tuned amplifier, as shown in Fig. 12.75b, confirms the design, since the center frequency is 91.1 MHz, as specified.

Question 12.36: The antenna of an FM radio picks up stations across the enti...

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