How to Write a/an Heaviside/Unit Step Function as a Piece wise Function?

Question

My professor makes us convert our final answers to piecewise functions. Since the textbook we use leaves the final answers as is, I can't check my work. Can someone tell me what this answer would look like as a piecewise function?

Answer 1

Well, that third term only applies for $t>\pi$.

So then, you have:

$t<\pi: y=\frac{1}{4}\sin(4t)+\frac{1}{8}t\sin(4t)$

$t\geq \pi:y=\frac{1}{4}\sin(4t)+\frac{1}{8}t\sin(4t)-\frac{1}{8}(t-\pi)\sin(4(t-\pi))$.

Simplifying the third term, we have $-\frac{1}{8}(t-\pi)\sin(4(t-\pi))=-\frac{1}{8}(t-\pi)\sin(4t)=-\frac{1}{8}t\sin(4t)+\frac{\pi}{8}\sin(4t)$.

So finally, the piecewise form is

$t<\pi: y=\frac{1}{4}\sin(4t)+\frac{1}{8}t\sin(4t)$

$t\geq \pi:y=\frac{2+\pi}{8}\sin(4t)$.

Answer 2

The unit step function is defined by

$$u(t-\pi)=\begin{cases}0 ,& t < \pi \\1,& \ t \ge \pi\end{cases}$$

therefore $-\dfrac{1}{8}(t-\pi)\sin 4(t-\pi)u(t-\pi) = 0$ when $t<\pi$ and you can rewrite the answer as

$$y(t)=\begin{cases}\dfrac{1}{4}\sin{4t}+\dfrac{1}{8}t\sin{4t} ,& t < \pi \\\dfrac{1}{4}\sin{4t}+\dfrac{1}{8}t\sin{4t}-\dfrac{1}{8}(t-\pi)\sin 4(t-\pi),& \ t \ge \pi\end{cases}$$

where $$\sin 4(t-\pi)=\sin(4t-4\pi)=\sin{4t}\cos{4\pi}-\cos{4t}\sin{4\pi}=\sin{4t}$$ thus $$-\frac{1}{8}(t-\pi)\sin 4(t-\pi)=\left(-\frac{t}{8}+\frac{\pi}{8}\right)\sin 4t=-\frac{1}{8}t\sin 4t +\frac{\pi}{8}\sin 4t$$ hence the piecewise defined function may be written as $$y(t)=\begin{cases}\dfrac{1}{4}\sin{4t}+\dfrac{1}{8}t\sin{4t} ,& t < \pi \\\dfrac{1}{4}\sin 4t+\dfrac{\pi}{8}\sin 4t,& \ t \ge \pi\end{cases}$$