Problem 48 - Advanced Engineering Mathematics

Question

Solve equation (10) subject to \(i(0)=0\) with \(L, R, C\), and \(E(t)\) as given. Use a graphing utility to graph the solution for \(0 \leq t \leq 3\). $$ \begin{aligned} &L=0.005 \mathrm{~h}, R=1 \Omega, C=0.02 \mathrm{f} \\ &E(t)=100[t-(t-1) ?(t-1)] \end{aligned} $$

Step-by-Step Solution

Answer

Solve the differential equation using Laplace Transforms to find \(i(t)\), then graph for \(0 \leq t \leq 3\).

Step 1: Understand the given values and equation

The equation given is part of an RLC circuit problem. The specified values are \(L = 0.005\,\mathrm{h}\), \(R = 1\,\Omega\), \(C = 0.02\,\mathrm{f}\), and \(E(t) = 100[t-(t-1)\cdot ?(t-1)]\). The "?" in \(?(t-1)\) seems to be intended as a unit step function, often represented as \(u(t-1)\). So, \(E(t)\) is 100 for \(t\geq 1\) and 0 otherwise.

Step 2: Set up the differential equation

The equation describing the current \(i(t)\) in an RLC circuit is given by:\[ L \frac{di}{dt} + Ri + \frac{1}{C}\int{i(t) dt} = E(t) \]With initial condition \(i(0) = 0\). Substituting the provided values gives:\[ 0.005 \frac{di}{dt} + 1i +\frac{1}{0.02}\int{i(t) dt} = 100u(t-1) \]

Step 3: Simplify and solve the differential equation

Rearrange and differentiate the equation to convert it from an integral format:\[ 0.005\frac{d^2i}{dt^2} + \frac{di}{dt} = 100u(t-1) - i \]This is a second-order non-homogeneous linear differential equation with a unit step function forcing term. Use techniques like Laplace Transform to solve for \(i(t)\) subject to the initial condition.

Step 4: Apply Laplace Transform

Take the Laplace Transform of both sides:\[ 0.005s^2I(s) + (sI(s) - i(0)) = \frac{100}{s} e^{-s} - I(s) \]With \(i(0)=0\), rearrange terms to solve for \(I(s)\), the Laplace Transform of \(i(t)\). Simplify equation and apply inverse Laplace Transform to determine \(i(t)\).

Step 5: Obtain and interpret the solution

After simplifying, the solution will be in a form that allows you to use a known table of inverse transforms or partial fraction decomposition. The solution \(i(t)\) will provide the current over time, considering the step function from \(E(t)\).

Step 6: Graph the Solution

Use a graphing calculator or software (such as Desmos or GeoGebra) to graph \(i(t)\) from the solution obtained in previous steps for the range \(0 \leq t \leq 3\). Pay attention to \(t=1\), where \(E(t)\) steps up.

Key Concepts

Laplace Transform

The Laplace Transform is an essential tool in analyzing circuits, especially when dealing with differential equations like those found in RLC circuits. It helps transform complex time-domain functions into simpler, algebraic expressions in the frequency domain.

• Transforms differential equations into algebraic equations, making them easier to handle.
• Particularly useful for evaluating systems with initial conditions, as with this RLC circuit where the initial current is zero.

Let's break down a simple process for using the Laplace Transform: 1. **Transform the time domain equation**: You start by taking the Laplace Transform of both sides of the given differential equation. For example, when transforming the second-order differential equation obtained from the RLC circuit, each derivative `d/dt` becomes multiplication by `s` in the Laplace domain. 2. **Solve for the variable in the Laplace domain**: Once transformed, you solve the equation for the variable of interest, usually represented as `I(s)` for the current. 3. **Apply inverse Laplace Transform**: Finally, return to the time domain by applying the inverse Laplace Transform, obtaining the solution `i(t)`. The Laplace Transform thus provides a bridge between the time and frequency domains, simplifying the analysis and solutions for electrical circuits.

Second-Order Differential Equation

Second-order differential equations often come into play when analyzing RLC circuits due to the presence of both a capacitor and an inductor. These types of equations involve second derivatives, reflecting the dynamic behavior and energy storage in capacitors and inductors.

• Models the behavior of electrical circuits, considering both resistance and reactance.
• Requires initial conditions for a complete solution, such as `i(0) = 0` in our exercise.

The process includes: - **Formulating the equation**: Begin by setting up the differential equation based on Kirchhoff's laws and the components in the circuit. - **Simplifying the equation**: In the case of our problem, simplifying leads to a form that can be solved using standard methods, like solving the homogeneous and particular solution separately. - **Solving the equation**: By applying techniques such as the Laplace Transform, you can handle terms like the unit step function, incorporating them directly into the equation. Overall, understanding second-order differential equations opens up the dynamics of how RLC circuits respond over time and under specific driving conditions.

Unit Step Function

The unit step function, often denoted as `u(t)`, is a critical mathematical tool in circuit analysis, particularly for describing switching actions and non-continuous inputs.

• Represents signals that abruptly change from 0 to 1 at a specified time, such as turning a power source on at `t=1`.
• Simplifies modeling of piecewise continuous inputs, allowing us to use standard analytical techniques.

In this exercise, the unit step function is used in `E(t)`, ensuring the step occurs at `t=1`, effectively switching the input to 100 volts. - **Inclusion in equations**: When it appears in differential equations, the step function's sudden change can introduce non-homogeneous terms that need special treatment, as they generate discontinuities. - **Handling with transforms**: The Laplace Transform is well-suited for such functions, seamlessly integrating them into the transformed domain and enabling easier solutions. Overall, understanding how to incorporate the unit step function in circuit equations is crucial for modeling real-world systems where inputs change over time.