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Radio Receiver. A simplified version of the front end of an AM radio receiver is shown in Figure 2.13. This receiver consists of an \(R L C\) tuned circuit containing a resistor, capacitor, and an inductor connected in series. The RLC circuit is connected to an external antenna and ground as shown in the figure. The tuned circuit allows the radio to select a specific station out of all the stations transmitting on the AM band. At the resonant frequency of the circuit, essentially all of the signal \(V_{0}\) appearing at the antenna appears across the resistor, which represents the rest of the radio. In other words, the radio receives its strongest signal at the resonant frequency. The resonant frequency of the LC circuit is given by the equation $$ f_{0}=\frac{1}{2 \pi \sqrt{L C}} $$ where \(L\) is inductance in henrys (H) and \(C\) is capacitance in farads (F). Write a program that calculates the resonant frequency of this radio set given specific values of \(L\) and \(C\). Test your program by calculating the frequency of the radio when \(L=0.1 \mathrm{mH}\) and \(C=0.25 \mathrm{nF}\).

Short Answer

Expert verified
The resonant frequency of the AM radio receiver with \(L = 0.1\,\text{mH}\) and \(C = 0.25\,\text{nF}\) is approximately 1,006,230 Hz or 1.0062 MHz.

Step by step solution

01

Understand the given formula

In this problem, we are given a formula to calculate the resonant frequency of an RLC circuit: $$f_{0} = \frac{1}{2\pi\sqrt{LC}}$$ f₀ represents the resonant frequency we are looking to determine. L represents inductance in henrys (H) and C represents capacitance in farads (F).
02

Converting given values

We are given the values of L and C: \(L = 0.1\,\text{mH}\) and \(C = 0.25\,\text{nF}\). We need to convert these values to henrys (H) and farads (F) respectively: 1 milli-Henry (mH) = 0.001 Henry (H), so \(L = 0.1\,\text{mH} = 0.0001\, \text{H}\). 1 nano-Farad (nF) = 0.000000001 Farad (F), so \(C = 0.25\,\text{nF} = 0.00000000025\,\text{F}\).
03

Calculate the resonant frequency

Now, we have the values of L and C in henrys and farads, respectively. We can use the given formula to calculate the resonant frequency: $$f_{0} = \frac{1}{2\pi\sqrt{LC}}$$ Substitute the values of L and C: $$f_{0} = \frac{1}{2\pi\sqrt{(0.0001)(0.00000000025)}}$$
04

Solve and find the resonant frequency

Now, we can solve the equation for f₀: $$f_{0} = \frac{1}{2\pi\sqrt{0.000000000025}} \approx 1,006,230.086\,\text{Hz}$$ So, the resonant frequency of the receiver is approximately 1,006,230 Hz or 1.0062 MHz.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

RLC Circuit
An RLC circuit is an electrical circuit consisting of a resistor (R), an inductor (L), and a capacitor (C) connected in series or parallel. In this context, the RLC circuit is used to tune into specific radio frequencies in a receiver.
The resonant frequency is the frequency at which the circuit naturally oscillates. This is important in radio receivers as it allows you to isolate a particular frequency while ignoring others. The resonant frequency in an LC circuit is calculated using the formula:
  • \[ f_{0} = \frac{1}{2\pi\sqrt{LC}} \]
At this frequency, the circuit minimizes impedance, allowing maximum signal transfer. Inductance (L) and capacitance (C) are critical here as they determine the frequency range the circuit can tune into.
It's essential to convert units properly:
  • 1 milli-Henry (mH) = 0.001 Henry (H)
  • 1 nano-Farad (nF) = 0.000000001 Farad (F)
These conversions ensure accurate calculations, as seen in converting 0.1 mH to 0.0001 H and 0.25 nF to 0.00000000025 F.
Electrical Engineering Education
Understanding RLC circuits and resonant frequency is fundamental in electrical engineering. This knowledge is key when designing circuits for specific tasks like filtering and signal selection.
In educational settings, learning about RLC circuits provides students with a foundation for understanding AC circuit behavior. It bridges theoretical concepts and practical applications, giving insights into how devices like radios and televisions operate.
Students often start with learning how components like resistors, inductors, and capacitors work individually before combining them in circuits. Hands-on labs and simulations help solidify these concepts, allowing students to see the impact of different configurations and parameters on circuit behavior.
Mastering these fundamentals helps electrical engineering students apply these principles to more complex systems, paving the way for innovation in communication technologies.
MATLAB Programming
MATLAB is a powerful tool for solving mathematical problems, making it ideal for calculating the resonant frequency of an RLC circuit. Writing a MATLAB program requires you to define variables, perform calculations, and display results.
Here's a simple outline for a MATLAB script to calculate resonant frequency:
  • Define the inductance \( L \) and capacitance \( C \) in henrys and farads.
  • Use the formula \( f_{0} = \frac{1}{2\pi\sqrt{LC}} \) to calculate \( f_{0} \).
  • Display the result using MATLAB's built-in functions.
For example:```matlabL = 0.0001; % in HenriesC = 0.00000000025; % in Faradsf0 = 1 / (2 * pi * sqrt(L * C));disp(['Resonant frequency: ', num2str(f0), ' Hz']);```This script performs the calculations efficiently and shows the output in a user-friendly format, demonstrating MATLAB's advantages in handling numerical analyses and complex calculations in electrical engineering contexts.

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Most popular questions from this chapter

Assume that array arrayl is defined as shown, and determine the contents of the following sub-arrays. \(\operatorname{array} 1=\left[\begin{array}{rrrrr}1.1 & 0.0 & 2.1 & -3.5 & 6.0 \\\ 0.0 & 1.1 & -6.6 & 2.8 & 3.4 \\ 2.1 & 0.1 & 0.3 & -0.4 & 1.3 \\ -1.4 & 5.2 & 0.0 & 1.1 & 0.0\end{array}\right]\) a. array \(1(3, t)\) b. array \(1(:, 3)\) c. array \(1\left(1: 2: 3,\left[\begin{array}{lll}3 & 3 & 4\end{array}\right]\right)\) d. array \(\left.1\left(\begin{array}{ll}1 & 1\end{array}\right], 4\right)\)

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?

Hyperbolic cosine. The hyperbolic cosine function is defined by the equation $$ \cosh x=\frac{e^{x}+e^{-x}}{2} $$ Write a program to calculate the hyperbolic cosine of a user-supplied value \(x\). Use the program to calculate the hyperbolic cosine of \(3.0\). Compare the answer that your program produces to the answer produced by the MATLAB intrinsic function cosh \((x)\). Also, use MATLAB to plot the function \(\cosh (x)\). What is the smallest value that this function can have? At what value of \(x\) does it occur?

Solve the following system of simultaneous equations for \(x\). \(\begin{aligned}-2.0 x_{1}+5.0 x_{2}+1.0 x_{3}+3.0 x_{4}+4.0 x_{5}-1.0 x_{6}=0.0 \\ 2.0 x_{1}-1.0 x_{2}-5.0 x_{3}-2.0 x_{4}+6.0 x_{5}+4.0 x_{6}=& 1.0 \\\\-1.0 x_{1}+6.0 x_{2}-4.0 x_{3}-5.0 x_{4}+3.0 x_{5}-1.0 x_{6}=&-6.0 \\\ 4.0 x_{1}+3.0 x_{2}-6.0 x_{3}-5.0 x_{4}-2.0 x_{5}-2.0 x_{6}=10.0 \\\\-3.0 x_{1}+6.0 x_{2}+4.0 x_{3}+2.0 x_{4}-6.0 x_{5}+4.0 x_{6}=&-6.0 \\ 2.0 x_{1}+4.0 x_{2}+4.0 x_{3}+4.0 x_{4}+5.0 x_{5}-4.0 x_{6}=&-2.0 \end{aligned}\)

Assume that \(a, b, c\), and \(a\) are defined as follows, and calculate the results of the following operations if they are legal. If an operation is itlegal, explain why it is illegal. $$ \begin{array}{ll} a=\left[\begin{array}{rr} 2 & -2 \\ -1 & 2 \end{array}\right] & b=\left[\begin{array}{rr} 1 & -1 \\ 0 & 2 \end{array}\right] \\ c=\left[\begin{array}{r} 1 \\ -2 \end{array}\right] & d=\text { eye }(2) \end{array} $$ a. result \(=a+b\); b. result \(=a * d t\) c. result \(=a \cdot d\) i d. result \(=a \cdot c\) z e. reault \(=a \cdot * c\) ? f. result \(=a \backslash b\); g. result \(=a .1 \mathrm{~b}\) : h. result \(=a \cdot A \mathrm{~b}\);

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