The unilateral laplace transform of -t-t-t0- e-d- is 1s2-Ps-13-1s-12- Then the value of P is-

E^ tu(t)L.T.⟷1s+1 t.e^ tu(t)L.T.⟷ dds(1s+1) t.e^ tu(t)L.T.⟷1(s+1)^2 ∫^t_0β.e^ βu(β).dβ nbsp;L.T.⟷1s.1(s+1)^2 ∫^t_0β.e^ βdβ nbsp;L.T.⟷1s(s+1)^2 ∫^t_0β.e ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Geometric Interpretation of Def.Int as Limit of Sum

The unilatera...

Question

The unilateral laplace transform of $ϕ (t) = t \int_{0}^{t} β e^{- β} d β i s \frac{1}{s^{2}} + \frac{P}{(s + 1)^{3}} - \frac{1}{(s + 1)^{2}} .$ Then the value of P is________.
-2

Open in App

Solution

The correct option is A -2
$e^{- t} u (t) L . T . ⟷ \frac{1}{s + 1}$

$t . e^{- t} u (t) L . T . ⟷ - \frac{d}{d s} (\frac{1}{s + 1})$

$t . e^{- t} u (t) L . T . ⟷ \frac{1}{(s + 1)^{2}}$

$\int_{0}^{t} β . e^{- β} u (β) . d β L . T . ⟷ \frac{1}{s} . \frac{1}{(s + 1)^{2}}$

$\int_{0}^{t} β . e^{- β} d β L . T . ⟷ \frac{1}{s (s + 1)^{2}}$

$\int_{0}^{t} β . e^{- β} d β L . T . ⟷ \frac{1}{s} - \frac{1}{(s + 1)} - \frac{1}{(s + 1)^{2}}$

$t \int_{0}^{t} β . e^{- β} d β L . T . ⟷ - \frac{d}{d s} [\frac{1}{s} - \frac{1}{(s + 1)} - \frac{1}{(s + 1)^{2}}]$

$t \int_{0}^{t} β . e^{- β} d β L . T . ⟷ \frac{1}{s^{2}} - \frac{1}{(s + 1)^{2}} - \frac{2}{(s + 1)^{3}}$

Suggest Corrections

Similar questions

Q. A signal x(t) having unilateral Laplace transform as,

X (s) = \frac{1}{e^{s + 3}} . \frac{s^{2}}{(s + 1) (s + 2)}

A function G(s) having relation with X(s) is defined as X(s) =

\frac{1}{e^{s + 3}} G (s)

.

Select the correct statement(s).

Q. The inverse Laplace transform of

F (s) = \frac{s + 3}{s^{2} + 2 s + 1}

for

t \geq 0

Q. The inverse Laplace transform of

\frac{s + 9}{s^{2} + 6 s + 13}

Q. If the Laplace transform of the voltage across a capacitor of value of

\frac{1}{2} F

V_{C} (s) = \frac{s + 1}{s^{3} + s^{2} + s + 1}

,
the value of the current through the capacitor at

t = 0^{+}

Q. Laplace transform of x(t) is

X (s) = \frac{s - 2}{s (s + 4)}

, then the final value of x(t) is _____.

Join BYJU'S Learning Program

The unilateral laplace transform of ϕ(t)=t∫t0βe−βdβ is 1s2+P(s+1)3−1(s+1)2. Then the value of P is________.-2

The unilateral laplace transform of $ϕ (t) = t \int_{0}^{t} β e^{- β} d β i s \frac{1}{s^{2}} + \frac{P}{(s + 1)^{3}} - \frac{1}{(s + 1)^{2}} .$ Then the value of P is________.
-2