The signal x(t), whose Laplace transform is defined as \(X\left(
A. <span class="math-tex">\({e^{ - 2t}}u\left( { - t} \right) + 2{e^{ - 3t}}u\left( t \right)\)</span>
B. <span class="math-tex">\(- {e^{ - 2t}}u\left( { - t} \right) + 2{e^{ - 3t}}u\left( t \right)\)</span>
C. <span class="math-tex">\(- {e^{ - 2t}}u\left( { - t} \right) - 2{e^{ - 3t}}u\left( t \right)\)</span>
D. <span class="math-tex">\({e^{ - 2t}}u\left( { - t} \right) - 2{e^{ - 3t}}u\left( t \right)\)</span>
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Right Answer is: A
SOLUTION
Concept:
Laplace Transform representation of a signal includes the expression and the Region of Convergence (ROC).
\({e^{ - at}}u\left( t \right)\mathop \leftrightarrow \limits^{L.T.} \frac{1}{{s + a}}\;with\;ROC > - a\)
\(and - {e^{ - at}}u\left( t \right)\mathop \leftrightarrow \limits^{L.T.} \frac{1}{{s + a}}\;with\;ROC < - a\)
Calculation:
For the given Laplace transform X(s), the expression is simplified using Partial Fraction as shown:
\(X\left( s \right) = \frac{{s + 1}}{{s + 5s + 6}} = \frac{{s + 1}}{{\left( {s + 2} \right)\left( {s + 3} \right)}}\)
\(\Rightarrow \frac{{s + 1}}{{\left( {s + 2} \right)\left( {s + 3} \right)}} = \frac{A}{{s + 2}} + \frac{B}{{s + 3}}\)
⇒ s + 1 = A(s + 3) + B(s + 2)
⇒ Putting s = -2,
⇒ -1 = A
Putting s = -3
-2 = -B
B = 2
\(X\left( s \right) = \frac{{s + 1}}{{\left( {s + 2} \right)\left( {s + 3} \right)}} = - \frac{1}{{s + 2}} + \frac{2}{{s + 3}}\)
The inverse Laplace Transform of the above for the region of convergence given as -3 < ROC < -2 will be:
\(= - \left[ { - {e^{ - 2t}}u\left( { - t} \right)} \right] + 2{e^{ - 3t}}u\left( t \right)\)
= e-2tu(-t) + 2e-3t u(t)