Correct Answer - Option 2 :
\(\frac{\omega}{(s-\theta)^2+ω^2 }\)
Concept:
Bilateral Laplace transform:
\(L\left[ {x\left( t \right)} \right] = x\left( s \right) = \;\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - st}}dt\)
Unilateral Laplace transform:
\(L\left[ {x\left( t \right)} \right] = x\left( s \right) = \;\mathop \smallint \limits_0^\infty x\left( t \right){e^{ - st}}dt\)
Frequency shifting property:
If X(s) is the Laplace transform of x(t), then
\({e^{at}}X\left( s \right) \leftrightarrow X\left( {s - a} \right)\)
Calculation:
f(t) = eθt sin (ωt)
\(L[sin \ ω_0t]= \frac{\omega}{{{s^2} + \omega^2}}\)
By using frequency shifting property,
The Laplace transform of f(t) is
\(F\left( s \right) = \frac{\omega}{(s-\theta)^2+ω^2 }\)
Some important Laplace transforms:
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f(t)
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f(s)
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ROC
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1.
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δ(t)
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1
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Entire s-plane
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2.
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e-at u(t)
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\(\frac{1}{{s + a}}\)
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s > - a
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3.
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e-at u(-t)
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\(\frac{1}{{s + a}}\)
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s < - a
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4.
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cos ω0 t u(t)
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\(\frac{s}{{{s^2} + \omega _0^2}}\)
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s > 0
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5.
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te-at u(t)
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\(\frac{1}{{{{\left( {s + a} \right)}^2}}}\)
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s > - a
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6.
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sin ω0t u(t)
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\(\frac{{{\omega _0}}}{{{s^2} + \omega _0^2}}\)
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s > 0
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7.
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u(t)
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1/s
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s > 0
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