Topic: Signals and Systems
The input $x(t)$ and output $y(t)$ of a system are related as $y(t)=\int_{-\infty}^{t}x(t)\cos(4t)dt$. The system is
- A. time-invariant and stable
- B. stable and not time-invariant
- C. time-invariant and not stable
- D. not time-invariant and not stable
Answer: D. not time-invariant and not stable
Explanation: $y(t-t_0) = \int_{-\infty}^{t-t_0}x(t)\cos(4t)dt$. However, $y^{\prime}(t) = \int_{-\infty}^{t}x(t-t_0)\cos(4t)dt$. Since $y(t-t_0) \ne y^{\prime}(t)$, the system is not time-invariant. Furthermore, if $x(t)$ is bounded, $\cos(4t)$ is also bounded, but integrating a bounded non-zero signal from $-\infty$ to $t$ can result in an unbounded signal, making it not stable.