Question

Given the differential equations $$ \begin{aligned} &\dot{x}_{1}(t)=x_{2}(t) \\ &\dot{x}_{2}(t)=-x_{1}(t)-x_{2}^{2}(t)+u(t) \end{aligned} $$ and the output function \(y(t)=x_{1}(t)\). Show that for \(u(t)=\cos ^{2}(t)\) a solution of the differential equations is \(x_{1}=\sin t, x_{2}=\cos t\). Linearize the state equations and the output function around this solution and write the result in matrix form. Is the linearized system time-invariant?

Short answer

The linearized system is not time-invariant due to time-dependent \(\cos t\) in the matrix.

Step-by-step solution

Verify Solution

To verify that \(x_1 = \sin t\) and \(x_2 = \cos t\) is a solution given \(u(t) = \cos^2(t)\), substitute into the equations:1. Calculate \(\dot{x}_1(t)\): \[ \dot{x}_1(t) = \frac{d}{dt}(\sin t) = \cos t = x_2(t). \] This satisfies the first equation.2. Calculate \(\dot{x}_2(t)\): \[ \dot{x}_2(t) = \frac{d}{dt}(\cos t) = -\sin t = -x_1(t). \] Substitute into the second equation: \[ \begin{align*} -x_1(t) - x_2^2(t) + u(t) &= -\sin t - (\cos t)^2 + \cos^2(t) \ &= -\sin t. \end{align*} \] This satisfies the second equation. Hence, the given functions are indeed solutions.

Linearization around Solution

To linearize, express the system as small deviations from the solution: \(x_1 = \sin t + \delta x_1\), \(x_2 = \cos t + \delta x_2\), and \(u = \cos^2(t) + \delta u\).1. Linearize \(x_1(t)\): \[ \dot{\delta x_1} = \delta x_2. \]2. Linearize \(x_2(t)\): \[ \dot{\delta x_2} = -\delta x_1 - 2\cos t \cdot \delta x_2 + \delta u. \]The linearized system is thus:\[\begin{align*}\dot{\delta x_1} &= \delta x_2, \\dot{\delta x_2} &= -\delta x_1 - 2\cos t \cdot \delta x_2 + \delta u.\end{align*}\]

Write in Matrix Form

Express the linearized system in matrix form:\[\begin{bmatrix}\dot{\delta x_1} \\dot{\delta x_2}\end{bmatrix} =\begin{bmatrix}0 & 1 \-1 & -2\cos t\end{bmatrix}\begin{bmatrix}\delta x_1 \\delta x_2\end{bmatrix} +\begin{bmatrix}0 \1\end{bmatrix}\delta u.\]The state matrix is\[A(t) = \begin{bmatrix}0 & 1 \-1 & -2\cos t\end{bmatrix}.\]

Check for Time-Invariance

Observe that the matrix \(A(t)\) contains \(\cos t\), which explicitly depends on time. Therefore, the linearized system is not time-invariant because the dynamics change with time.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. These are used to describe various phenomena, such as motion, heat, and sound. In this problem, we have a system of differential equations that defines the change over time for two functions, \(x_1(t)\) and \(x_2(t)\). More specifically, these are ordinary differential equations because they involve derivatives with respect to a single variable, time \(t\). This system is designed to model a dynamic or time-evolving process.

To solve such differential equations, we often need specific conditions, similar to our given function \(u(t)\). In our instance, verifying the solution involves substituting potential solutions (like \(x_1 = \sin t\) and \(x_2 = \cos t\)) back into the original equations to check if they satisfy these equations.

• First, we derive each of the potential solutions with respect to time, and check whether these derivatives fulfill the equations provided.
• The verification confirms that the solutions indeed satisfy our equations, given the input \(u(t) = \cos^2(t)\).

Understanding differential equations is crucial as they form the basis of modeling real-world dynamic systems.

Time-Invariant Systems

Time-invariance in systems refers to a property where the system's behavior and rules do not change over time. For a system to be time-invariant, its governing equations should not explicitly depend on time. This means if we delay the input, the output is simply delayed by the same amount without altering its form.

In the linearization process of our example, we explore whether the linearized system exhibits time-invariance. When we expressed the linearized equations, the presence of \(\cos t\) in the coefficients indicates that the properties of the system vary with time.

Consequently, our system is deemed time-varying, not time-invariant, because the coefficients of the system depend explicitly on time \(t\). These changes mean the response of the system varies at different times, affecting predictions and simulations.

Grasping the concept of time-invariance is integral when designing systems intended for consistent behavior over time, such as tools and machinery that need predictable outputs.

Matrix Form of System Equations

Converting a system of equations into matrix form provides a neat and efficient way to analyze and solve multidimensional linear systems. This abstraction is especially useful in fields like control systems and state-space analysis.

For our linearized system, which involves the deviations \(\delta x_1\) and \(\delta x_2\), we write these equations using matrices. The approach entails forming a matrix equation of the type:

• State vector \(\begin{bmatrix} \delta x_1 \ \delta x_2 \end{bmatrix}\) represents deviations from our original solution.
• The rate of change or the derivative vector \(\begin{bmatrix} \dot{\delta x_1} \ \dot{\delta x_2} \end{bmatrix}\) is a product of the state matrix \(A(t)\) with this state vector, plus any input influence.

The matrix \(A(t) = \begin{bmatrix} 0 & 1 \ -1 & -2\cos t \end{bmatrix}\) encapsulates the system dynamics and identifies how each state variable interacts with another or itself over time.

By utilizing this matrix form, it becomes easier to analyze, simulate, and visualize complex state interactions. Such clarification is instrumental when designing control strategies and finding system stability.