Question

A simple \(R C\)-circuit with emf \(V(t)=3 \cos (\omega t)\) is modeled by the initial value problem $$ R C V_{C}^{\prime}+V_{C}=3 \cos (\omega t), \quad V_{C}(0)=0, $$ where \(R\) is the resistance, \(C\) the capacitance, \(\omega\) is the driving frequency, and \(V_{C}\) is the voltage response across the capacitor. Show that the solution is $$ V_{C}(t)=\frac{R C \omega \sin \omega t+\cos \omega t-e^{-t /(R C)}}{1+R^{2} C^{2} \omega^{2}} $$ Create a function M-file for this solution as follows. function \(V=f(t, R, C\), w \()\) \(V=(R * C * w * \sin (w * t)+\cos (w * t)-\exp (-t /(R * C))) /\left(1+R^{\wedge} 2 * C^{\wedge} 2 * w^{\wedge} 2\right) ;\) Now, call this function with the following script. \(\mathrm{R}=1.2 ; \mathrm{C}=1 ; \mathrm{w}=1 ;\) \(t=\) linspace \((0,2 *\) pi, 1000\() ;\) \(\mathrm{R}=1.2 ; \mathrm{C}=1 ; \mathrm{w}=1 ;\) \(\mathrm{t}=1\) inspace \((0,2 * \mathrm{pi}, 1000) ;\) \(\mathrm{V}=\mathrm{f}(\mathrm{t}, \mathrm{R}, \mathrm{C}, \mathrm{w}) ;\) \(\mathrm{plot}(\mathrm{t}, \mathrm{V})\) \(\mathrm{V}=\mathrm{f}(\mathrm{t}, \mathrm{R}, \mathrm{C}, \mathrm{w})\); plot (t,V) Run this script for \(\omega=1,2,4\), and 8 , keeping \(R=1.2 \mathrm{ohms}\) and \(C=1\) farad constant. What happens to the amplitude of the voltage response across the capacitor as the frequency of the emf is increased? Why do you think this circuit is called a low pass filter?

Short answer

As frequency \( \omega \) increases, the voltage response amplitude decreases. The circuit attenuates higher frequencies, thus it is called a low pass filter.

Step-by-step solution

Understand the Initial Value Problem

We are given the initial value problem for the RC circuit: \( R C V_{C}^{\prime}+V_{C}=3 \cos (\omega t), \quad V_{C}(0)=0 \). This is a first-order linear differential equation representing how the voltage response \( V_C \) changes over time.

General Solution of The Differential Equation

To solve the differential equation \( R C V_{C}^{\prime} + V_{C} = 3 \cos(\omega t) \), we notice it's a linear first-order differential equation. We use the integrating factor method where the integrating factor is \( \mu(t) = e^{t/(RC)} \). Multiply both sides of the differential equation by this factor.

Solve for Voltage Response

Upon multiplying by the integrating factor \( e^{t/(RC)} \) and integrating both sides, the equation transforms to a form involving the derivative of a product. After integrating, we solve for \( V_{C}(t) \) and apply the initial condition \( V_{C}(0) = 0 \) to find the constant of integration.

Derive the Given Solution

Using the steps above, verify that the particular solution provided: \( V_{C}(t)=\frac{R C \omega \sin (\omega t)+\cos (\omega t)-e^{-t /(R C)}}{1+R^{2}C^{2} \omega^{2}} \) satisfies the original differential equation and the initial conditions.

Implement the Function M-file

The function M-file is specified by the problem as: ```matlabfunction V=f(t, R, C, w)V=(R * C * w * sin(w * t)+cos(w * t)-exp(-t/(R * C))) / (1+R^2 * C^2 * w^2);```This script calculates \( V_C(t) \) for given parameters.

Script a MATLAB Simulation

The script to run the function is ```matlab R=1.2; C=1; w=1; t=linspace(0,2*pi,1000); V=f(t,R,C,w); plot(t,V); ``` which plots the voltage response across a time interval for the given initial conditions and frequencies.

Analyze Amplitude Changes

Run the script for different values of \( \omega = 1, 2, 4, 8 \), while keeping \( R \) and \( C \) constant. Observe the effects: as \( \omega \) increases, the amplitude of the voltage response across the capacitor decreases.

Explain the Low Pass Filter Concept

This circuit is called a low pass filter because it allows signals with a frequency lower than a certain cutoff frequency to pass through while attenuating (reducing) the amplitude of signals with frequencies higher than this cutoff.

Key concepts

First-Order Differential Equations

In RC circuit analysis, first-order differential equations often model the behavior of the voltage or current over time. An RC circuit, consisting of a resistor and capacitor in series, can be represented by such an equation because it describes the rate of change of voltage across the capacitor. The form of this first-order differential equation arises from Kirchhoff's voltage law, which states that the sum of the electrical voltages around any closed network is zero. In this context, the first-order differential equation is expressed as \( R C V_{C}^{\prime} + V_{C} = 3 \cos(\omega t), \ V_{C}(0)=0 \), where \( V_{C} \) is the voltage response across the capacitor.
Understanding this equation is crucial because it predicts how the voltage develops over time, taking into account the initial condition \( V_{C}(0)=0 \). The solution of this equation shows us how the input voltage affects the output, allowing us to further analyze and design circuits with specific frequency responses.

Integrating Factor Method

The integrating factor method is a technique used to solve first-order linear differential equations. In the given RC circuit equation, we utilize this method by first identifying the integrating factor \( \mu(t) = e^{t/(RC)} \). This factor transforms the original differential equation into a simpler form that can be easily integrated.
By multiplying the entire differential equation by the integrating factor, the left side becomes the derivative of the product of the integrating factor and the voltage response \( V_{C} \).

• The new form: \( \frac{d}{dt} \left[ e^{t/(RC)} V_{C} \right] = 3 e^{t/(RC)} \cos(\omega t) \).

Integrating both sides with respect to \( t \), and then solving for \( V_{C} \), we apply the initial condition to find any constants of integration. This method is powerful because it turns the problem into a straightforward integration task, making it easier to handle complex equations.

MATLAB Programming

MATLAB programming simplifies the numerical analysis and visualization of RC circuits for different parameter values. With MATLAB, we define a function file using the provided relation for \( V_{C}(t) \). This function ```function V=f(t, R, C, w)V=(R * C * w * sin(w * t)+cos(w * t)-exp(-t/(R * C))) / (1+R^2 * C^2 * w^2);```computes the voltage across the capacitor for each time instance.
We use scripts to automate parameter changes and analyze the response at various frequencies. For instance, with the command `t=linspace(0, 2*pi, 1000)`, we create a fine time vector from 0 to \( 2\pi \) to simulate the response over one period. By plotting the results, we visually assess the impact of changes in frequency \( \omega \).
This approach provides quick and accurate results, aiding in understanding and validating theoretical predictions and allowing experimentation with different circuit conditions.

Low Pass Filter

A low pass filter is an electronic circuit that passes low-frequency signals with little attenuation while attenuating high-frequency signals. In our RC circuit, the inherent nature of capacitive reactance changes with frequency, which dictates its behavior as a low pass filter. As the frequency increases, the impedance of the capacitor decreases, thus reducing the amplitude of the output voltage \( V_{C} \).
This property of the circuit becomes evident when we observe the voltage response across the capacitor; the amplitude of \( V_{C} \) decreases as the frequency \( \omega \) increases. This behavior is why such circuits are called low pass filters, as they effectively 'filter out' or reduce high-frequency components of signals passing through them.
Understanding this concept is crucial in applications such as audio electronics, where it is desirable to allow low-frequency signals (like bass sounds) to pass through, while blocking unwanted high-frequency noise.