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The input $x(t)$ and output $y(t)$ of a system are related as $y(t)=\int_{-\infty}^{t} x(\tau)\cos(3\tau)\,d\tau$. The system is

1. time-invariant and stable
2. stable and not time-invariant
3. time-invariant and not stable
4. not time-invariant and not stable

Correct answer: not time-invariant and not stable

Solution

The kernel contains an explicit factor $\cos(3\tau)$, so shifting the input does not simply shift the output; hence the system is not time-invariant. For bounded input like $x(t)=1$, the output becomes $\int_{-\infty}^{t}\cos(3\tau)d\tau$, which does not remain bounded in the usual sense, so the system is not stable.

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