The Laplace transform of the causal periodic square wave of period T shown in the figure below is
A signal $2\cos\left(\frac{2\mathrm\pi}3t\right)-\;\cos\;\left(\mathrm{πt}\right)$ is the input to an LTI system with the transfer function
$H\left(s\right)=e^s+e^{-s}$
The bilateral Laplace transform of a function $f\left(t\right)=\left\{\begin{array}{ll}1& ifa\le t\le b\\ 0& \mathrm{Other}\mathrm{wise}\end{array}\right.$ is
Let the signal f(t) = 0 outside the interval [T_{1}, T_{2}], where T_{1} and T_{2} are finite. Furthermore, $\left|f\left(t\right)\right|<\infty $ . The region of convergence (RoC) of the signal’s bilateral Laplace transform F(s) is
Consider the differential equation $\frac{dx}{dt}$ = 10-0.2x with initial condition x(0) = 1. The response x(t) for t>0 is
Input x(t) and output y(t) of an LTI system are related by the differential equation y''(t) - y-(t) - 6y(t) = x(t). If the system is neither causal nor stable, the impulse response h(t) of the system is
Let x(t) = $\alpha $ s(t) +s(–t) with s(t) = $\beta $e^{-4t} u(t) , where u(t) is unit step function . If the bilateral Laplace transform of x(t) is
$X\left(s\right)=\frac{16}{{s}^{2}-16}-4Re\left\{s\right\}4;$
Then the value of $\beta $ is______.
A system is described by the following differential equation, where u(t) is the input to the system and y(t) is the output of the system
$\stackrel{.}{y}\left(t\right)+5y\left(t\right)=u\left(t\right)$
When y(0) =1 and u(t) is a unit step function, y(t) is
The impulse response of a continuous time system is given by h(t) = $\delta $(t -1) + $\delta $(t - 3) . The value of the step response at t = 2 is
A system is described by the differential equation $\frac{{d}^{2}y}{d{t}^{2}}+5\frac{dy}{dt}+6y(t)=x(t)$. Let x(t) be a rectangular pulse given by $x(t)=\{\begin{array}{ll}1& 0<t<2\\ 0& otherwise\end{array}$
Assuming that y(0) = 0 and $\frac{dy}{dt}=0$ at t = 0, the Laplace transform of y(t) is
The unilateral Laplace transform of f (t) is $\frac{1}{{s}^{2}+s+1}$. The unilateral Laplace transform of t f (t) is
If the unit step response of a network is ${\left(1-{e}^{-\alpha t}\right)}^{}$ then its unit impulse response is
An input x (t) = exp (−2t)u(t) + δ (t − 6) is applied to an LTI system with impulse response h(t) = u(t) . The output is is
If $F\left(S\right)=L\left[f\left(t\right)\right]=\frac{2\left(s+1\right)}{{s}^{2}+4s+7}$ then the initial and final values of f(t) are respectively
A continuous time LTI system is described by
$\frac{{d}^{2}y\left(t\right)}{d{t}^{2}}+4\frac{dy\left(t\right)}{dt}+3y\left(t\right)=2\frac{dx\left(t\right)}{dt}+4x\left(t\right)$
Assuming zero initial conditions, the response y(t) of the above system for the input x(t)=e^{-2t} u(t) is given by
Given that F(s) is the one-sided Laplace transform of f(t), the Laplace transform of $\style{font-size:14px}{\int\limits_0^tf\left(\tau\right)\operatorname{d}\tau}$ is
If the Laplace transform of a signal y(t) is $Y\left(s\right)=\frac{1}{s\left(s-1\right)}$, then its final value is