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rewrite: $(A\sin \omega*t$)u(t) - $(A\sin \omega*t$)u(t-T/2)
u(t) is the unit step function
L(f1(t)) -> $(\frac{A*\omega}{s^2 + \omega^2}$) - $(\frac{A*\omega}{s^2 + \omega^2})(\frac{e^(-s*\tau)}{s}$)
tau = T/2
All I can think to do is take the Laplace transform. What steps would I take to solve the problem? I don't understand what the problem is asking me to do. Where does the + sign come from in the second expression? Thanks
You do not need to make it that complicated.
$$f_1(t)=A \sin(\omega t)[u(t)-u(t-\frac{T}2)]$$
$$f_2(t)=A \sin(\omega t) u(t) + A \sin(\omega (t-\frac{T}2))u(t-\frac{T}2)$$
The problem is to show how $f_1=f_2$.
Just simplify the second part of $f_2$:
$$\sin(\omega (t-\frac{T}2))=\sin(\omega t-\frac{\omega T}2)=\sin(\omega t -\frac{2\pi}2)=\sin(\omega t -\pi)=-\sin(\omega t)$$
$$f_2(t)=A \sin(\omega t) u(t) + A [-\sin(\omega t)]u(t-\frac{T}2)$$ $$=A \sin(\omega t) u(t) - A \sin(\omega t) u(t-\frac{T}2)$$$$=A \sin(\omega t) [u(t) - u(t-\frac{T}2)]=f_1(t)$$