Question

Radio Receiver. The voltage across the resistive load in Figure \(2.13\) varies as a function of frequency according to Equation (2.18). $$ V_{R}=\frac{R}{\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}} V_{\theta} $$ where \(\omega=2 \pi f\) and \(f\) is the frequency in hertz. Assume that \(L=0.1 \mathrm{mH}\), \(C=0.25 \mathrm{nF}, R=50 \Omega\), and \(V_{O}=10 \mathrm{mV}\). a. Plot the voltage on the resistive load as a function of frequency. At what frequency does the voltage on the resistive load peak? What is the voltage on the load at this frequency? This frequency is called the resonant frequency \(f_{0}\) of the circuit. b. If the frequency is changed to \(10 \%\) greater than the resonant frequency, what is the voltage on the load? How selective is this radio receiver? c. At what frequencies will the voltage on the load drop to half of the voltage at the resonant frequency?

Short answer

a. The resonant frequency \(f_0\) is determined by observing the voltage-frequency plot, where the voltage on the resistive load peaks. At this frequency, read the maximum voltage value on the load from the plot. b. To calculate the voltage on the load when the frequency is \(10 \%\) greater than the resonant frequency, substitute this new frequency value into the voltage equation and solve for the voltage on the load. Then, calculate the selectivity of the radio receiver as follows: \(\text{Selectivity} = \frac{V_R(f_0)}{V_R(1.1f_0)}\). c. To find the frequencies (\(f_1\) and \(f_2\)) at which the voltage on the load is half of the voltage at the resonant frequency (\(V_R(f_0)\)), solve the voltage equation for \(f_1\) and \(f_2\) and substitute the value of half the voltage at the resonant frequency into the equation: $$\frac{1}{2} V_R(f_0) = \frac{R}{\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}}V_{O}$$

Step-by-step solution

a. Plotting voltage as a function of frequency

Using the given formula for the voltage across the resistive load, we can calculate the voltage at different frequencies. Since we want to plot the voltage as a function of frequency, let's first create a table with frequency values and the corresponding voltage values.

i. Calculating voltage values

For each frequency value, from \(10^3\) Hz to \(10^7\) Hz, substitute the given values of the circuit components \((L = 0.1\,\text{mH}, C = 0.25\,\text{nF}, R = 50\,\Omega, V_{O} = 10\,\text{mV})\) and the frequency, \(f\), into the equation to calculate the voltage across the resistive load: $$V_{R}=\frac{R}{\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}}V_{O}$$

ii. Plotting the voltage-frequency curve

Plot the calculated voltage values against their corresponding frequency values on a graph. This will show us the behavior of the voltage on the resistive load as a function of frequency. Observe the curve and find the frequency at which the voltage across the resistive load peaks. This frequency is the resonant frequency, \(f_{0}\).

a. Determining resonant frequency and voltage on the load

At the resonant frequency \(f_{0}\), the voltage on the load is the maximum. Read the corresponding voltage value from the plot at this frequency.

b. Changing frequency to \(10 \%\) greater than the resonant frequency

Increase the frequency to \(10 \%\) greater than the resonant frequency, and substitute this new frequency value into the equation for the voltage across the resistive load. Calculate the voltage on the load at this frequency: $$V_{R}=\frac{R}{\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}}V_{O}$$

b. Calculating the selectivity of the radio receiver

The selectivity of the radio receiver is given by the ratio of the voltage on the load at the resonant frequency to the voltage on the load at a frequency \(10 \%\) greater than the resonant frequency. Calculate the selectivity as follows: $$\text{Selectivity} = \frac{V_R(f_0)}{V_R(1.1f_0)}$$

c. Calculating frequencies with half voltage at the resonant frequency

To calculate the frequencies at which the voltage on the load is half of the voltage at the resonant frequency, we have to find the frequencies (\(f_1\) and \(f_2\)) for which: $$\frac{1}{2} V_R(f_0) = V_{R}(f_1), \;V_R(f_2)$$ You can do this by solving the voltage equation for \(f_1\) and \(f_2\) and substituting the value of half the voltage at the resonant frequency into the equation: $$\frac{1}{2} V_R(f_0) = \frac{R}{\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}}V_{O}$$

Key concepts

Radio Receiver Circuit Analysis

When engineers analyze radio receiver circuits, they examine the behavior of the circuit in response to different radio frequencies. By studying the equation for voltage across the resistive load, as shown in the given problem, one can determine how voltage changes with frequency.

Radio receivers must be able to discern between closely spaced signals in order to function properly, which is an aspect of their 'selectivity'. The given formula for voltage across a resistive load encapsulates the interaction of resistance (R), inductance (L), and capacitance (C) within an RLC circuit - a common type of radio receiver circuit - and how these parameters influence the voltage output in response to varying frequencies.

Engineers would plot voltage as a function of frequency to determine important performance metrics like the resonant frequency at which the receiver is most responsive, and the selectivity, which shows how well the circuit can differentiate between frequencies that are close together.

Frequency Response of RLC Circuits

An RLC circuit, which is essential to radio receivers, includes a resistor (R), an inductor (L), and a capacitor (C). The frequency response of such a circuit describes how the output signal's amplitude, in this case, voltage, varies with frequency.

The formula given in the exercise embodies the frequency response of an RLC circuit, capturing the interaction between the impedance contributions of L and C and the resistance R. Notably, at a specific frequency, the impedance caused by the inductor and the capacitor will cancel each other out, leading to a peak in the circuit’s response. This peak is indicative of 'resonance'.

By calculating and plotting these values for a range of frequencies, we can visualize the behavior of the RLC circuit, showing peaks and troughs that signify how effective the circuit is at different frequencies. This visualization is key for engineers when tuning radio receivers to achieve optimal performance.

Resonant Frequency Calculation

The resonant frequency, often denoted as \( f_0 \), is a crucial parameter in radio receiver circuit analysis. At this frequency, the voltage across the resistive load of an RLC circuit reaches its maximum value because the reactive effects of the inductor and capacitor offset each other, resulting in a purely resistive circuit impedance.

The circuit given in the exercise responds maximally at the resonant frequency, which can be derived mathematically from the formula provided. Since the impedance contributions from L and C cancel at \( f_0 \), the formula simplifies and allows us to solve for that frequency. Knowing \( f_0 \) is important in applications such as tuning the radio to the desired station and ensuring that it can receive signals at that specific frequency with maximum efficiency.

Voltage across Resistive Load

The voltage across a resistive load, \( V_R \), in an RLC circuit is important because it indicates how much of the input signal is being effectively transferred to the resistance without being dropped across the inductor or the capacitor. In the case of the RLC radio receiver circuit, the voltage across the resistor is what would be 'heard' or 'seen' as the output.

The formula \( V_R = \frac{R}{\sqrt{R^{2} + (\omega L - \frac{1}{\omega C})^{2}}} V_{\theta} \) allows us to determine this voltage across the resistive load at any given frequency. Here, \( \omega \) is the angular frequency, which is directly related to the standard frequency (\( f \)) by the relationship \( \omega = 2\pi f \).

This relationship is crucial because it means that the voltage will be at its highest when the receiver is at resonance and will vary as the frequency deviates from the resonant frequency, which informs us about the circuit's ability to handle signals at different frequencies.

Signal Selectivity in Radio Receivers

Signal selectivity is the capability of a radio receiver to differentiate between multiple signals of different frequencies. It is a vital feature that allows a radio to tune in to a specific station without interference from the signals of others.

The problem given provides insight into how selectivity is quantified - by comparing the voltage across the resistive load at the resonant frequency to that at a slightly different frequency (in this case, 10% higher). The less the voltage drops when the frequency changes, the more selective the receiver is, meaning it can more precisely filter out unwanted signals.

For example, if a circuit has a high selectivity, it will be more capable of isolating a signal at a desired frequency while suppressing others. The calculation provided shows the method to evaluate this characteristic of radio receivers, thus ensuring clear reception of the chosen signal amidst the myriad of radio waves traveling through the air.