What is the solution for a periodic signal x(t) as shown in the figure?

Solution:

Where,  \(\rightleftharpoons\) F₁(ω) = -2T₁ sin (ωT₁)

= \(\rm -2T_1\frac{\sin(ω T_1)}{ω T_1}\)

\(\rm -\frac{2}{ω}\sin(ω T_1)\)

By applying FT on (1),

jωX(ω) = F₁(ω) + 2T₁

⇒ \(\rm X(\omega)=\frac{-\frac{2}{\omega}\sin(\omega T_1)+2T_1}{j\omega}\)

\(=\frac{-j}{\omega}\left[\frac{-2}{\omega}\sin(\omega T_1)+2T_1\right]\)

\(=j\left[\frac{2\sin(\omega T_1)}{\omega^2}-\frac{2T_1}{\omega}\right]\)