The bilateral Laplace transform of $${e^t}\cos 2tu\left( { - t} \right) + {e^{ - t}}u\left( t \right) + {e^{\frac{t}{2}}}u\left( t \right)$$ is
A. $$\frac{{1 - s}}{{{{\left( {s - 1} \right)}^2} + 4}} + \frac{1}{{s + 1}} + \frac{1}{{s - 0.5}},\,0.5 < \operatorname{Re} \left( s \right) < 1$$
B. $$\frac{{1 - s}}{{{{\left( {s - 1} \right)}^2} + 4}} + \frac{1}{{s + 1}} + \frac{1}{{s - 0.5}},\, - 1 < \operatorname{Re} \left( s \right) < 1$$
C. $$\frac{{s - 1}}{{{{\left( {s - 1} \right)}^2} + 4}} + \frac{1}{{s + 1}} + \frac{1}{{s - 0.5}},\,0.5 < \operatorname{Re} \left( s \right) < 1$$
D. $$\frac{{s - 1}}{{{{\left( {s - 1} \right)}^2} + 4}} + \frac{1}{{s + 1}} + \frac{1}{{s - 0.5}},\, - 1 < \operatorname{Re} \left( s \right) < 1$$
Answer: Option A
This Question Belongs to Engineering Maths >> Transform Theory
A. $$\frac{1}{{{\text{s}} + {\text{a}}}}$$
B. $$\frac{1}{{{\text{s}} - {\text{a}}}}$$
C. $$\frac{1}{{{\text{a}} - {\text{s}}}}$$
D. $$\infty $$
Evaluate $$\int\limits_0^\infty {\frac{{\sin {\text{t}}}}{{\text{t}}}{\text{dt}}} $$
A. $$\pi $$
B. $$\frac{\pi }{2}$$
C. $$\frac{\pi }{4}$$
D. $$\frac{\pi }{8}$$
A. $$\frac{{1 + {{\text{s}}^2}}}{{{{\left( {{{\text{s}}^2} - 1} \right)}^2}}}$$
B. $$\frac{{{\text{st}}}}{{\left( {{{\text{s}}^2} - 1} \right)}}$$
C. $$\frac{{1 - {{\text{s}}^2}}}{{{{\left( {{{\text{s}}^2} - 1} \right)}^2}}}$$
D. $$\frac{{1 + {{\text{s}}^2}}}{{1 - {{\text{s}}^2}}}$$
A. $$\frac{2}{{{\text{s}} + 1}}$$
B. $$\frac{4}{{{\text{s}} + 1}}$$
C. $$\frac{4}{{{{\text{s}}^2} + 1}}$$
D. $$\frac{2}{{{{\text{s}}^2} + 1}}$$
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