Laplace Transform Practice Questions

The unilateral Laplace transform of $f(t)$ is $\frac{1}{s^{2}+s+1}$ . The unilateral Laplace transform of $t f(t)$ is

An input x(t)=exp(-2t)u(t)+ $\delta$ (t-6) is applied to an LTI system with impulse response h(t)=u(t). The output is

If the unit step response of a network is $(1-e^{-at})$ , then its unit impulse response is

If $F(s)=L[f(t)]=\frac{2(s+1)}{s^{2}+4s+7}$ then the initial and final values of $f(t)$ are respectively

Given f(t)=l^{-1}\[\frac{3s+1}{s^{3}+4s^{2}+(K-3)s}]$.$If$\lim_{t\rightarrow \infty }f(t)=1$ , then the value of K is

A continuous time LTI system is described by $\frac{d^{2}y(t)}{dt^{2}}+4\frac{dy(t)}{dt}+3y(t)=2\frac{dx(t)}{dt}+4x(t)$ 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-side Laplace transform of $f(t)$ , the Laplace transform of $\int_{0}^{t}f(\tau )d\tau$ is

If the Laplace transform of a signal y(t) is $Y(s)=\frac{1}{s(s-1)}$ , then its final value is

Consider the function f(t) having Laplace transform $F(s)=\frac{\omega _{0}}{s^{2}+\omega ^{2}_{0}}\; \; \; \; Re[s] \gt 0$ The final value of f(t) would be

In what range should Re(s) remain so that the Laplace transform of the function $e^{(a+2)t+5}$ exits.

The Laplace transform of i(t) is given by $I(s)=\frac{2}{s\left ( 1+s \right )}$ At $t\rightarrow \infty$ The value of i(t) tends to

The Laplace transform of a continuous-time signal x(t) is $X(s)=\frac{5-s}{s^{2}-s-2}$ . If the Fourier transform of this signal exists, then x(t) is

The transfer function of a system is given by $H(s)=\frac{1}{s^2(s-2)}$ . The impulse response of the system is: (* denotes convolution, and U(t) is unit step function)