Given that h(t) = 10e^-10u(t), and e(t) = sin10t.u(t), the Laplace transform of the signal $$f\left( t \right) = \int\limits_{\tau = 0}^t {h\left( {t - \tau } \right)e\left( \tau \right)d\tau } $$ is given by

A) $$\frac{{10}}{{\left( {s + 10} \right)\left( {{s^2} + 100} \right)}}$$ B) $$\frac{{10\left( {s + 10} \right)}}{{\left( {{s^2} + 100} \right)}}$$ C) $$\frac{{100}}{{\left( {s + 10} \right)\left( {{s^2} + 100} \right)}}$$ D) $$\frac{1}{{\left( {s + 10} \right)\left( {{s^2} + 100} \right)}}$$

Answer & Solution

Correct Answer: Option C

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Given that h(t) = 10e-10u(t) and e(t) = sin10t u(t) the...

Given that h(t) = 10e-10u(t), and e(t) = sin10t.u(t), the Laplace transform of the signal $$f\left( t \right) = \int\limits_{\tau = 0}^t {h\left( {t - \tau } \right)e\left( \tau \right)d\tau } $$ is given by

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Given that h(t) = 10e^-10u(t), and e(t) = sin10t.u(t), the Laplace transform of the signal $$f\left( t \right) = \int\limits_{\tau = 0}^t {h\left( {t - \tau } \right)e\left( \tau \right)d\tau } $$ is given by