Question

Show that if\(L\{ g\} (s) = {\left[ {(s + \alpha )\left( {1 - {e^{ - Ts}}} \right)} \right]^{ - 1}}\), where\(T > 0\)is fixed, then

\(\begin{array}{c}g(t) = {e^{ - \alpha t}} + {e^{ - \alpha (t - T)}}u(t - T)\\ + {e^{ - \alpha (t - 2T)}}u(t - 2T)\\ + {e^{ - \alpha (t - 3T)}}u(t - 3T) + L\end{array}\)

[Hint: Use the fact that\(\left. {1 + x + {x^2} + L = 1/(1 - x).} \right]\)

Short answer

By using the geometric series in the given function, it is showed that if\(\mathcal{L}\{ g\} (s) = {\left[ {(s + \alpha )\left( {1 - {e^{ - {\gamma _x}}}} \right)} \right]^{ - 1}}\) , where\(T > 0\)is fixed, then

\(\begin{array}{c}g(t) = {e^{ - \alpha t}} + {e^{ - \alpha (t - T)}}u(t - T)\\ + {e^{ - \alpha (t - 2T)}}u(t - 2T)\\ + {e^{ - \alpha (t - 3T)}}u(t - 3T) + \cdots \end{array}\)

Step-by-step solution

Define geometric series

The geometric series can be in the form of,\(1 + x + {x^2} + \cdots = 1/(1 - x)\).

Prove the given function by using the geometric series

Thus, when \(0 < {e^{ - Ts}}\left\langle {1,s,T} \right\rangle 0\) , the geometric series can be written as,

\(\frac{1}{{1 - {e^{ - Ts}}}} = 1 + {e^{ - Ts}} + {e^{ - 2Ts}} + \cdots \)

Hence, we have

\(\begin{array}{c}{\left[ {(s + \alpha )\left( {1 - {e^{ - Ts}}} \right)} \right]^{ - 1}} = \frac{1}{{1 + \alpha }}\frac{1}{{1 - {e^{ - Ts}}}}\\ = \frac{1}{{a + \alpha }}\left( {1 + {e^{ - Ts}} + {e^{ - 2Ts}} + \cdots } \right)\\ = \frac{1}{{s + \alpha }} + \frac{{{e^{ - Ts}}}}{{s + \alpha }} + \frac{{{e^{ - 2Ts}}}}{{s + \alpha }} + \cdots \end{array}\)

Use inverse Laplace transform and linearity to obtain solution,

\(\begin{array}{c}{\mathcal{L}^{ - 1}}\left\{ {{{\left[ {(s + \alpha )\left( {1 - {e^{ - Ts}}} \right)} \right]}^{ - 1}}} \right\} = {\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{s + \alpha }} + \frac{{{e^{ - Ts}}}}{{s + \alpha }} + \frac{{{e^{ - 2Ts}}}}{{s + \alpha }} + \cdots } \right\}\\ = {\mathcal{L}^{ - 1}}\left\{ {\frac{1}{{s + \alpha }}} \right\} + {\mathcal{L}^{ - 1}}\left\{ {\frac{{{e^{ - Ts}}}}{{s + \alpha }}} \right\} + {\mathcal{L}^{ - 1}}\left\{ {\frac{{{e^{ - 2Ts}}}}{{s + \alpha }}} \right\} + \cdots \\ = {e^{ - \alpha t}} + {e^{ - \alpha (t - T)}}u(t - T) + {e^{ - \alpha (t - 2T)}}u(t - 2T) + \cdots \end{array}\)

Therefore,\(\mathcal{L}\left\{ {{e^{ - \alpha t}} + {e^{ - \alpha (t - T)}}u(t - T) + {e^{ - \alpha (t - 2T)}}u(t - 2T) + \cdots } \right\} = {\left[ {(s + \alpha )\left( {1 - {e^{ - Ts}}} \right)} \right]^{ - 1}}\)

Hence, by using geometric series, the given function is proved.