In the circuit shown in the figure, the switch is closed at time $t=0$, while the capacitor is initially charges to -5 V (i.e., $\left(\theta_C(0)=-5 \mathrm{~V}\right)$.

The time after which the voltage across the capacitor becomes zero (rounded off to three decimal places) is $\_\_\_\_$ ms .

Consider a real-valued base-band signal $x(t)$. band limited to 10 kHz . The Nyquist rate for the signal $y(t)=x(t) \times \left(1+\frac{t}{2}\right)$ is

Consider the signals $x[n]=2^{n-1} u[-n+2]$ and $y[n]=2^{-n+2} u[n+1]$, where $u[n]$ is the unit step sequence. Let $X\left(e^{j \omega}\right)$ and $Y\left(e^{j \omega}\right)$ be the discrete-time Fourier transform of $x[n]$ and $y[n]$, respectively. The value of the integral

$$ \frac{1}{2 \pi} \int_o^{2 \pi} X\left(e^{j \omega}\right) Y\left(e^{-j \omega}\right) d \omega $$

(round off to one decimal place) is $\_\_\_\_$