Q9E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine \(L\{ f\} \), 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\(F(s) = \frac{1}{{1 - {e^{ - 2\pi s}}}} \cdot \left[ {\frac{{{e^{ - s\pi }} + 1}}{{{s^2} + 1}}} \right]\).

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: Obtain the solution by using the given graph and by applying the theorem

Consider the given graph,

Thus, the function having the period\(T = 2\pi \)is,

\(\begin{array}{c}f(t) = \sin t,\quad 0 < t < \pi \\ = 0\quad \pi   < t < 2\pi \end{array}\)

Since, we know

 

Simplify using the parts rule,

\(\begin{array}{c}{f_T}(s) = \left[ {\frac{{ - {e^{ - st}}\cos t - s{e^{ - st}}\sin t}}{{{s^2} + 1}}} \right]_0^\pi \\ = \left[ {\frac{{ - {e^{ - s\pi }}\cos \pi  - s{e^{ - s\pi }}\sin \pi  + {e^{ - 0s}}\cos 0 + s{e^{ - 0s\pi }}\sin 0}}{{{s^2} + 1}}} \right]\\ = \left[ {\frac{{{e^{ - s\pi }} + 1}}{{{s^2} + 1}}} \right]\end{array}\)

 

Thus, apply the theorem

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

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