Recent questions tagged laplace-transform

1 1 vote
1 1 answer
387
387 views
The Laplace Transform of the signal $x(t)=u(t-2) *(t u(t))$ is given by which of the following expressions?["*" represents convolution operator]$\dfrac{e^{-2 s}}{s^{2}(s-...
2 2 votes
0 0 answers
2.8k
2.8k views
​​​In the feedback control system shown in the figure below $G(s)=\frac{6}{s(s+1)(s+2)}$.$\text{R(s), Y(s)}$, and $\text{E(s)}$ are the Laplace transforms of $r(t), y(t)$...
1 1 vote
0 0 answers
449
449 views
What is the Laplace transform $F(s)$ of the signal $f(t), t \geq 0$ defined below? In $t \in[0,1),$\[f(t)=\left\{\begin{array}{ll}1, & t \in\left[0, \frac{1}{2}\right) \\...
0 0 votes
0 0 answers
349
349 views
The Laplace Transform of $e \alpha t \cos (\alpha \; t)$ is equal to$\frac{(s-\alpha)}{(s-\alpha)^2+\alpha^2}$$\frac{(s+\alpha)}{(s-\alpha)^2+\alpha^2}$$\frac{1}{(s-\alph...
0 0 votes
0 0 answers
600
600 views
Given that$\mathcal{L}[f(t)]=\frac{s+2}{s^2+1}, \mathcal{L}[f(t)]=\frac{s^2+1}{(s+3)(s+2)}, $ $h(t)=\int_0^1 f(\tau) g(t-\tau) d \tau, \mathcal{L}[h(t)]$ is$\frac{s^2+1}{...
0 0 votes
0 0 answers
347
347 views
If $[f(t)]=\mathrm{F}(s)$, then $[f(t-\mathrm{T})]$ is equal to$e^{s \mathrm{T}} \mathrm{~F}(s)$$e^{-s \mathrm{T}} \mathrm{~F}(s)$$\frac{\mathrm{F}(s)}{1-e^{s T}}$$\frac{...
0 0 votes
0 0 answers
342
342 views
The circuit shown in the figure, operating in steady-state with switch $S_{1}$ closed. The switch $S_{1}$ is opened at $t=0$Find $i_{\mathrm{L}}\left(0^{+}\right)$.Find $...
0 0 votes
0 0 answers
532
532 views
The Laplace transform of a continuous-time signal $x(t)$ is $X(s)=\frac{5-s}{s^{2}-s-2}$. If the Fourier transform of this signal exists, then $x(t)$ is$e^{2 t} u(t)-2 e^...
0 0 votes
0 0 answers
326
326 views
The Laplace transform of $i(t)$ is given by $$ I(s)=\frac{2}{s(1+s)} $$ As $t \rightarrow \infty$, the value of $i(t)$ lends to$0$$1$$2$$\infty$
0 0 votes
0 0 answers
340
340 views
The circuit for $\text{Q. 33-34}$ are given in the figure. For both are the questions, assume that the switch $S$ is in position $1$ for a long time and thrown to positio...
0 0 votes
0 0 answers
304
304 views
The circuit shown in the figure has initial current $i_{t}\left(0^{-}\right)=1 \mathrm{~A}$ through the inductor and an initial voltage $v_{c}\left(0^{-}\right)=-1 \mathr...
0 0 votes
0 0 answers
338
338 views
In what range should $\operatorname{Re}(s)$ remain so that the Laplace transform of the function $e^{(n+2)t+5}$ exits?$\operatorname{Re}(s)>a+2$$\operatorname{Re}(\mathrm...
0 0 votes
0 0 answers
328
328 views
If $L[f(t)]=\frac{2(s+1)}{s^{2}+2 s+5}$ then $f(0+)$ and $f(\infty)$ are given by$0, 2$ respectively$2, 0$ respectively$0,1$ respectively$2/5, 0$ respectively[Note : 'L' ...
0 0 votes
0 0 answers
379
379 views
The inverse Laplace transform of the function$\frac{s+5}{(s+1)(s+3)}$ is$2 e^{-t}-e^{-3 t}$$2 e^{-t}+e^{-3 t}$$e^{-t}-2 e^{-3 t}$$e^{-t}+e^{-3 t}$
1 1 vote
0 0 answers
610
610 views
Consider the function $f(t)$ having Laplace transform $$ \text{F}(s)=\frac{\omega_0}{s^2+\omega_0^2} \operatorname{Re}[s]>0 $$ The final value of $f(t)$ would be$0$$1$$-1...
1 1 vote
0 0 answers
705
705 views
A $2 \; \mathrm{mH}$ inductor with some initial current can be represented as shown below, where $s$ is the Laplace Transform variable. The value of initial current is$0...
0 0 votes
0 0 answers
385
385 views
The Laplace transform of a unit ramp function starting at $t=a$, is$\frac{1}{(s+a)^{2}}$$\frac{e^{-as}}{(s+a)^{2}}$$\frac{e^{-as}}{s^{2}}$$\frac{a}{s^{2}}$
1 1 vote
1 1 answer
526
526 views
If the Laplace transform of a signal $y(t)$ is $Y(s)=\dfrac{1}{s(s-1)},$ then its final value is$-1$$0$$1$Unbounded
0 0 votes
0 0 answers
297
297 views
The Laplace transform of the periodic function $f(t)$ described by the curve below, i.e.$f(t)=\left\{\begin{array}{c}\sin t \text { if }(2 n-1) \pi \leq t \leq 2 n \pi(n=...
0 0 votes
0 0 answers
268
268 views
If $\tau \mathrm{F}(\mathrm{s})=[f(t)]=\frac{\mathrm{K}}{(\mathrm{s}+1)\left(\mathrm{s}^{2}+4\right)}$ then $\lim _{t \rightarrow \infty} f(t)$ is given by$\mathrm{K} / 4...
1 1 vote
0 0 answers
359
359 views
Given that $F(s)$ is the one-sided Laplace transform of $f(t),$ the Laplace transform of $\displaystyle{}\int_{0}^{t} f(\tau) d \tau$ is$s F(s)-f(0)$$\dfrac{1}{s} F(s)$$\...
1 1 vote
0 0 answers
270
270 views
Given $f(t)=\mathscr{L}^{-1}\left[\dfrac{3 s+1}{s^{3}+4 s^{2}+(K-3) s}\right]$. If $\displaystyle{}\lim _{t \rightarrow \infty} f(t)=1$, then the value of $K$ is$1$$2$$3$...
0 0 votes
0 0 answers
512
512 views
The voltage across an impedance in a network is $V(s)=z(s) I(s)$, where $V(s)$, $Z(s)$ are the Laplace transforms of the corresponding time function $v(t), z(t)$ and $i(t...
0 0 votes
0 0 answers
460
460 views
(a) Find the Laplace transform of the waveform $x(t)$ shown in figure.(b) The network shown in figure is initially under steady state condition with the switch in positio...
1 1 vote
0 0 answers
262
262 views
If $F(s)=L[f(t)]=\dfrac{2(s+1)}{s^2+4 s+7}$ then the initial and final values of $f(t)$ are respectively$0,2$$2,0$$0, 2 / 7$$2 / 7,0$
1 1 vote
0 0 answers
424
424 views
The block diagram of a closed-loop control system is shown in the figure. $R(s), Y(s),$ and $D(s)$ are the Laplace transforms of the time-domain signals $r(t), y(t),$ and...
0 0 votes
0 0 answers
466
466 views
Which one of the following is a property of the solutions to the Laplace equation: $\nabla^2f = 0$?The solutions have neither maxima nor minima anywhere except at the bo...
0 0 votes
0 0 answers
948
948 views
The Laplace transform of the casual periodic square wave of period $T$ shown in the figure below is$F(S) = \frac{1}{1+e^{-sT/2}} \\$$F(S) =\frac{1}{s(1+e^{-sT/2})} \\$$F(...
0 0 votes
0 0 answers
605
605 views
The bilateral Laplace transform of a function $f(t) = \begin{cases} 1 & \text{if } a \leq t \leq b \\ 0 & \text{otherwise} \end{cases}$ is$\dfrac{a-b}{s} \\$$\dfrac{e^{s...
0 0 votes
0 0 answers
526
526 views
Let the signal ݂$f(t) = 0$ outside the interval $[T_{1},T_{2}]$, where ܶ$T_{1}$ and ܶ$T_{2}$ are finite. Furthermore, $\mid f(t) \mid < \infty$. The region of convergence...