Question

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

Answer 1

$y={\int }_{-\infty }^{t}x\left(\tau \right)cos\left(3\tau \right)d\tau$
$y(t-t_0)={\int }_{-\infty }^{t-t_0}x\left(\tau \right)cos\left(3\tau \right)d\tau$

$\text{y'(t) for input x(t- t}{}_0)\text{is}$

$y^{\prime }\left(t\right)={\int }_{-\infty }^{t}x(\tau -t_o)cos3\tau d\tau$

$y^{\prime }\left(t\right)={\int }_{-\infty }^{(t-t_o)}x\left(\tau \right)cos3(\tau +t_o)d\tau$

$y^{\prime }\left(t\right)\ne y(t-t_o)$ so system is not time invariant for input $x\left(\tau \right)=cos\left(3\tau \right)$ (bounded input)

$y\left(t\right)={\int }_{-\infty }^{t}co{s}^2\left(3\tau \right)d\tau \to \infty$ as $t\to \infty$

So, for bounded input, output is not bounded. Therefore system is not stable.