Question

An electronic system has two inputs, to each of which a constant unit signal is applied, but starting at different times. The equations governing the system thus take the form $$ \begin{aligned} &\dot{x}+2 y=H(t) \\ &\dot{y}-2 x=H(t-3). \end{aligned} $$ Initially (at \(t=0\) ), \(x=1\) and \(y=0\); find \(x(t)\) at later times.

Short answer

The solution for \(x(t)\) is \( x(t) = 1 + 2 e^{-4(t-3)} H(t-3)\).

Step-by-step solution

- Understand the Differential Equations

The system of differential equations is given by: \[ \dot{x} + 2y = H(t) \] \[ \dot{y} - 2x = H(t-3). \] where \(H(t)\) is the Heaviside step function which equals 0 for \(t < 0\) and 1 for \(t \geq 0\). The initial conditions are \(x(0) = 1\) and \(y(0) = 0\).

- Solve the First Equation for x(t)

Considering \(H(t) = 1\) for \(t \geq 0\), the first equation becomes \( \dot{x} + 2y = 1 \). Taking the Laplace transform of both sides yields: \[ sX(s) - x(0) + 2Y(s) = \frac{1}{s} \] Substituting the initial conditions \(x(0) = 1\): \[ sX(s) - 1 + 2Y(s) = \frac{1}{s}. \]

- Solve the Second Equation for y(t)

Considering \(H(t-3) = 1\) for \(t \geq 3\), the second equation becomes \( \dot{y} - 2x = 1 \). Taking the Laplace transform of both sides yields: \[ sY(s) - y(0) - 2X(s) = e^{-3s} \cdot \frac{1}{s} \] Substituting the initial condition \(y(0) = 0\): \[ sY(s) - 2X(s) = \frac{e^{-3s}}{s}. \]

- Solve the System of Equations

We now have the transformed equations: \[ sX(s) - 1 + 2Y(s) = \frac{1}{s} \] \[ sY(s) - 2X(s) = \frac{e^{-3s}}{s}. \] Rearranging the first equation: \[ X(s) = \frac{1}{s} + \frac{1}{s} - 2Y(s)\rightarrow X(s) = \frac{2}{s} - 2Y(s). \] Substituting into the second equation: \[ sY(s) - 2 \left( \frac{2}{s} - 2Y(s) \right) = \frac{e^{-3s}}{s} \] Expanding and simplifying terms: \[ sY(s) - 4 \frac{1}{s} + 4Y(s) = \frac{e^{-3s}}{s} \rightarrow (s + 4)Y(s) = \frac{4}{s} + \frac{e^{-3s}}{s}. \]

- Inverse Laplace Transform to Find y(t)

Solving for \(Y(s)\): \[ Y(s) = \frac{4}{s(s+4)} + \frac{e^{-3s}}{s(s+4)} \] Performing partial fraction decomposition and taking the inverse Laplace transform, it results in: \[ y(t) = 1 - e^{-4t}[1 - H(t-3)(1 - e^{4(t-3)})]. \]

- Inverse Laplace Transform to Find x(t)

Using the previous transformation for \(X(s) = \frac{2}{s} - 2Y(s)\), substitute \(Y(s)\) and find the inverse Laplace transform of the remaining terms: \[ x(t) = 2 - 2 \left( 1-e^{-4t} \right) + 2 e^{-4(t-3)}H(t-3). \]

Final Answer

Simplifying we get the final expression for \(x(t)\): \[ x(t) = 1 + 2 e^{-4(t-3)} H(t-3). \]

Key concepts

Laplace transform

The Laplace transform is a powerful integral transform that converts a function of time, typically denoted as \( f(t) \), into a function of a complex variable, \( s \), typically denoted as \( F(s) \). This transformation is useful for solving differential equations because it turns differentiation into algebraic operations. The Laplace transform of a function \( f(t) \) is given by: \[ F(s) = \frac{1}{s} \text{lap}(f(t),t,s_|) } \] When solving a system of differential equations, take the Laplace transform of each equation. Remember to incorporate initial conditions like \( x(0) \) and \( y(0) \) when transforming. After transforming the entire system, solve for the Laplace transforms of the unknown functions \( X(s) \) and \( Y(s) \). Finally, convert your transformed equations back into time-domain functions \( x(t) \) and \( y(t) \) using the inverse Laplace transform, which often involves partial fraction decomposition.

Initial conditions

Initial conditions are the values of the dependent variables (like \( x \) or \( y \)) at the start of your observation, usually at \( t = 0 \). They provide necessary information to obtain a unique solution to differential equations. For instance, in our given problem, we have initial conditions \( x(0) = 1 \) and \( y(0) = 0 \). When you take the Laplace transform of the system of differential equations, the initial conditions help simplify and solve the transformed algebraic equations. So, make sure to substitute these conditions right after transforming the differential equations to address the constants added by Laplace.

Heaviside step function

The Heaviside step function, denoted as \( H(t) \), is a piecewise function used to represent signals that turn on at a specific moment. It is defined as: \[ H(t) = \begin{cases} 0, & t < 0\ 1, & t \text{ like } \rightarrow t \rightarrow \text{ like } > 0\ end{cases} \] In differential equations, the Heaviside function models sudden changes in input or forcing functions, like in traffic lights or electronic systems starting at time \( t = 0 \ \). For our problem, \( H(t) \) and \( H(t-3) \) signal two different moments when the inputs become active. Incorporate Heaviside functions properly in your differential equations before applying the Laplace transform.

Partial fraction decomposition

Partial fraction decomposition helps break down complex rational functions into simpler fractions, making them easier to work with, especially when finding inverse Laplace transforms. Take a fraction of the form: \[ \frac{4}{s(s+4)} \] Decomposition splits it into simpler parts, such as: \[ \frac{A}{s} + \frac{B}{s+4} \] Solve for constants \( A \) and \( B \) and transform each term individually. In our differential equations problem, partial fraction decomposition is used to find the inverses of \( Y(s) \) and \( X(s) \), turning complex expressions back into functions of time, \( y(t) \) and \( x(t) \), respectively. This method simplifies the final steps of solving differential equations.