The following discrete-time equations result from the numerical integration of the differential equations of an un-damped simple harmonic oscillator with state variables x and y. the integration time step is h.

$\begin{array}{l}\frac{x_{k+1}-x_k}{h}=y_k\\ \frac{y_{k+1}-y_k}{h}=-x_k\end{array}$

For this discrete-time system, which one of the following statements is TRUE?

Discrete-Time Oscillator Stability Analysis

This problem involves analyzing the stability of a discrete-time system that results from numerically integrating the equations of motion for an undamped simple harmonic oscillator. The integration uses a time step denoted by $h$, and the system's state is described by variables $x$ and $y$.

Understanding the Discrete-Time Equations

The differential equations of an undamped simple harmonic oscillator are typically $\dot{x}=y$ and $\dot{y}=-x$. The discrete-time equations provided are derived using a numerical integration method (specifically, the forward Euler method):

- $\frac{x_{k+1}-x_k}{h}=y_k$
- $\frac{y_{k+1}-y_k}{h}=-x_k$

These can be rewritten to express the state at the next time step ($k+1$) in terms of the current state ($k$):

- $x_{k+1}=x_k+h{y}_k$
- $y_{k+1}=y_k-h{x}_k$

Matrix Representation of the System

The system dynamics can be conveniently represented in matrix form. Let the state vector at time step $k$ be $X_k=\left(\begin{array}{c}{x}_k\\ y_k\end{array}\right)$. The system equations become:

$$X_{k+1}=\left(\begin{array}{c}{x}_{k+1}\\ y_{k+1}\end{array}\right)=\left(\begin{array}{cc}1& h\\ -h& 1\end{array}\right)\left(\begin{array}{c}{x}_k\\ y_k\end{array}\right)$$

The matrix $A=\left(\begin{array}{cc}1& h\\ -h& 1\end{array}\right)$ is known as the state transition matrix.

Stability Criterion for Discrete Systems

A fundamental concept in analyzing the stability of discrete-time linear time-invariant (LTI) systems is the eigenvalues of the state transition matrix $A$. For a system described by $X_{k+1}=A{X}_k$, the system is stable if and only if the magnitude of all its eigenvalues ($\lambda$) is strictly less than 1 ($|\lambda |<1$).

Calculating the Eigenvalues

The eigenvalues are found by solving the characteristic equation $\det (A-\lambda I)=0$, where $I$ is the identity matrix.

Calculating the determinant:

$$\det \left(\begin{array}{cc}1-\lambda & h\\ -h& 1-\lambda \end{array}\right)=(1-\lambda )(1-\lambda )-(h)(-h)$$

This simplifies to:

$$(1-\lambda {)}^2+h^2=0$$

Solving for $\lambda$:

$$(1-\lambda {)}^2=-h^2$$

$$1-\lambda =\pm \sqrt{-h^2}=\pm jh(\text{where }j\text{ is the imaginary unit})$$

Thus, the eigenvalues are $\lambda =1\mp jh$. The two eigenvalues are ${\lambda }_1=1-jh$ and ${\lambda }_2=1+jh$.

Analyzing Eigenvalue Magnitudes for Stability

To determine stability, we examine the magnitude of these eigenvalues:

$$|{\lambda }_1|=|1-jh|=\sqrt{1^2+(-h{)}^2}=\sqrt{1+h^2}$$

$$|{\lambda }_2|=|1+jh|=\sqrt{1^2+h^2}=\sqrt{1+h^2}$$

The condition for stability is $|\lambda |<1$. Applying this to our eigenvalues:

$$\sqrt{1+h^2}<1$$

Squaring both sides yields:

$$1+h^2<1^2$$

$$1+h^2<1$$

Subtracting 1 from both sides gives:

$$h^2<0$$

Since $h$ represents a time step, it must be a real number. The square of any non-zero real number ($h^2$) is always positive ($h^2>0$). Therefore, the condition $h^2<0$ cannot be satisfied for any real $h\ne 0$. This implies that the magnitude of the eigenvalues, $\sqrt{1+h^2}$, is always greater than 1 for any $h>0$.

Evaluating the Statements

Our analysis shows that the system is unstable for all positive time steps $h$ because the stability condition $|\lambda |<1$ is never met.

Let's consider the options:

- The system is not stable for $h>0$. (This aligns with our findings.)
- The system is stable for $h>\frac{1}{\pi }$. (This is incorrect.)
- The system is not stable for $0<h>\frac{1}{2\pi }$. (The notation is ambiguous, but even if interpreted as instability in this range, it's incomplete as instability occurs for all $h>0$.)
- The system is not stable for $\frac{1}{2\pi }<h>\frac{1}{\pi }$. (Similar to the previous point, this is incomplete.)

The analysis confirms that the discrete-time system is unstable whenever the time step $h$ is positive.