Consider a discrete-time linear time-invariant (LTI) system, $\boldsymbol{S}$, where
$$ y[n]=S\{x(\mathrm{n})\} $$
$$Let\,\,\,\, S\{\delta[n]\}=\left\{\begin{array}{lc} 1, & n \in\{0,1,2\} \\ 0, & \text { otherwise } \end{array}\right. $$
where $\delta[n]$ is the discrete-time unit impulse function. For an input signal $x[n]$, the output $y[n]$ is
Consider a continuous-time signal
$$ x(t)=-t^2\{u(t+4)-u(t-4)\} $$
where $u(t)$ is the continuous-time unit step function. Let $\delta(t)$ be the continuous-time unit impulse function. The value of
$$ \int_{-\infty}^{\infty} x(t) \delta(t+3) d t $$
is
A continuous time periodic signal $x(t)$ is
$$ x(t)=1+2 \cos 2 \pi t+2 \cos 4 \pi t+2 \cos 6 \pi t $$
If $T$ is the period of $x(t)$, then $\frac{1}{T} \int_0^T|x(t)|^2 d t=$________(round off to the nearest integer).
Let continuous-time signals $x_1(t)$ and $x_2(t)$ be
$x_1(t)=\left\{\begin{array}{cc}1, & t \in[0,1] \\ 2-t, & t \in[1,2] \\ 0, & \text { otherwise }\end{array}\right.$
and $x_2(t)=\left\{\begin{array}{cc}t, & t \in[0,1] \\ 2-t, & t \in[1,2] \\ 0, & \text { otherwise }\end{array}\right.$.
$y(t)=x_1(t) * x_2(t)$. Then $\int\limits_{-\infty}^{\infty} y(t) d t$ is :