Question

The action of the control mechanism on a particular system for an input \(f(t)\) is described, for \(t \geq 0\), by the coupled first-order equations: $$ \begin{aligned} &\dot{y}+4 z=f(t) \\ &\dot{z}-2 z=\dot{y}+\frac{1}{2} y \end{aligned} $$ Use Laplace transforms to find the response \(y(t)\) of the system to a unit step input \(f(t)=H(t)\), given that \(y(0)=1\) and \(z(0)=0\).

Short answer

The response to the system is found by applying Laplace transforms, solving for \(Y(s)\), and then taking the inverse Laplace transform.

Step-by-step solution

Apply Laplace Transform

Take the Laplace transform of both differential equations. Use the initial conditions \(y(0) = 1\) and \(z(0) = 0\), and recall that the Laplace transform of \(\frac{d}{dt} y(t)\) is \(sY(s) - y(0)\).

Transforming the First Equation

Apply the Laplace transform to the first equation: \(\mathcal{L}\{\frac{d}{dt} y + 4z\} = \mathcal{L}\{f(t)\}\)which becomes \(sY(s) - 1 + 4Z(s) = \frac{1}{s}\).

Transforming the Second Equation

Apply the Laplace transform to the second equation: \(\mathcal{L}\{\frac{d}{dt} z - 2z\} = \mathcal{L}\{\frac{d}{dt} y + \frac{1}{2}y\}\)which becomes \(sZ(s) - 0 - 2Z(s) = sY(s) - 1 + \frac{1}{2} Y(s)\).

Simplify the Equations in the s-Domain

Simplify the transformed equations: \(sY(s) - 1 + 4Z(s) = \frac{1}{s}\) and \((s - 2)Z(s) = (s + \frac{1}{2}) Y(s) - 1\).

Solve for Z(s)

Express \(Z(s)\) from the second simplified equation: \(Z(s) = \frac{(s + \frac{1}{2}) Y(s) - 1}{s - 2}\).

Substitute Z(s)

Substitute \(Z(s)\) into the first equation: \(sY(s) - 1 + 4 \frac{(s + \frac{1}{2}) Y(s) - 1}{s - 2} = \frac{1}{s}\).

Solve for Y(s)

Clear the fraction and solve for \(Y(s)\): \((s-2)(sY(s) - 1) + 4(s + \frac{1}{2}) Y(s) - 4 = (s-2) \frac{1}{s}\). Combine like terms and isolate \(Y(s)\).

Take the Inverse Laplace Transform

Finally, take the inverse Laplace transform to find \(y(t)\).

Key concepts

First-Order Differential Equations

First-order differential equations deal with rates of change for a function and its derivative. In these equations, the highest derivative present is the first derivative. For our problem:
\[\dot{y} + 4z = f(t)\]
\[\dot{z} - 2z = \dot{y} + \frac{1}{2}y\]
The goal is to find how these equations describe the system's behavior over time. We use them to model the system's dynamics and later solve them using the Laplace transform.

Laplace Transform

The Laplace transform is a powerful tool in control systems and differential equations. It converts complex differential equations into simpler algebraic equations in the s-domain, making them easier to solve. The transform of function \(f(t)\) is given by:
\[ \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} f(t)e^{-st}dt\]
It changes derivatives into polynomials. For instance, the Laplace transform of \(\frac{d}{dt} y(t)\) is \(sY(s) - y(0)\). This tool is essential for analyzing and designing control systems. In our solution, we applied it to both differential equations.

Initial Conditions

Initial conditions are values provided for the variables at the start, \(t = 0\). They are crucial because they allow us to solve differential equations completely. In our problem:
\(y(0) = 1\) and \(z(0) = 0\)
These conditions are used when applying the Laplace transform to make the equations more specific. They help us find the unique solution corresponding to a particular initial system state.

Inverse Laplace Transform

After solving equations in the s-domain, we need to convert them back to the time domain to find our solution. This conversion is done using the inverse Laplace transform. If \(Y(s)\) is the Laplace transform of \(y(t)\), then:
\[ \mathcal{L}^{-1}\{Y(s)\} = y(t) \]
In our problem, after simplifying and solving for \(Y(s)\) in the s-domain, we apply the inverse Laplace transform to reveal \(y(t)\). This final step translates our algebraic solution back into the time domain, giving us the system's response to the input over time.