Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine , where the periodic function is described by its graph.

Step-by-Step Solution

Answer

The value of\(\mathcal{L}\{ f\} \), where the periodic function is described by its graph is\(F(s) = \frac{1}{{1 - {e^{ - as}}}} \cdot \left[ {\frac{{ - as{e^{ - as}} - {e^{ - as}} + 1}}{{{s^2}}}} \right]\).

Step 1: Define the transform of periodic function theorem

If \(f\)has period\(T\)and is piecewise continuous on\([0,T]\) then the Laplace transforms,

are related by,

\({F_T}(s) = F(s)\left[ {1 - {e^{ - sT}}} \right]{\rm{\;or\;}}F(s) = \frac{{{F_T}(s)}}{{1 - {e^{ - sT}}}}\).

 

Step 2: Determine the solution by using the theorem and the given graph

Consider the graph,

Thus, from the graph, the function has a period \(T = a\) ,

\(f(t) = t,\quad 0 < t < a\)

Hence, we know

 

Simplify by using parts rule,

\(\begin{array}{c}{f_T}(s) = \left[ {\frac{{ - s{e^{ - st}}t - {e^{ - st}}}}{{{s^2}}}} \right]_0^a\\ = \left[ {\frac{{ - as{e^{ - as}} - {e^{ - as}} + 1}}{{{s^2}}}} \right]\end{array}\)

Apply the theorem,

\(\begin{array}{c}F(s) = \frac{{{f_T}(s)}}{{1 - {e^{ - sT}}}}\\ = \frac{{\left[ {\frac{{ - as{e^{ - as}} - {e^{ - as}} + 1}}{{{s^2}}}} \right]}}{{1 - {e^{ - as}}}}\\ = \frac{1}{{1 - {e^{ - as}}}} \cdot \left[ {\frac{{ - as{e^{ - as}} - {e^{ - as}} + 1}}{{{s^2}}}} \right]\end{array}\)

Therefore, the value of\(\mathcal{L}\{ f\} \), where the periodic function is described by the given graph is \(F(s) = \frac{1}{{1 - {e^{ - as}}}} \cdot \left[ {\frac{{ - as{e^{ - as}} - {e^{ - as}} + 1}}{{{s^2}}}} \right]\).