I'm currently studying the Heaviside step function in an engineering mathematics course and am having some trouble understanding some basic concepts. Perhaps I'll go right into the example that caused this trouble:
Write the following function using unit step functions and find its transform.
$$f(t) = \begin{cases}\begin{align}&2 & 0 \lt t \lt 1 \\ &\frac{1}{2}t^2 & 1 \lt t \lt \frac{\pi}{2} \\ &\cos{(t)} & t \gt \frac{\pi}{2}\end{align}\end{cases}$$
The textbook gives the solution in the following steps:
Step 1
Write out the function in terms of unit step functions:
$$f(t) = 2(1 - u(t - 1)) + \frac{1}{2}t^2(u(t - 1) - u(t - \frac{\pi}{2}) + (\cos{(t)}) u(t - \frac{\pi}{2})$$
I found the intuition or idea behind how to elegantly write this out at first, but I suppose that comes with practice.
Step 2
Next, the textbook states that "we must write each term in $f(t)$ in the form $f(t - a)u(t - a)$. Thus, $2(1 - u(t - 1))$ remains as it is and gives the transform $2(1 - e^{-s})/s$. Then:
$$ \begin{align} \mathscr{L} \left( \frac{1}{2}t^2 u(t - 1) \right) & = \mathscr{L} \left( \left( \frac{1}{2}(t - 1)^2 + (t - 1) + \frac{1}{2} \right) u(t - 1) \right) \\ & = \left( \frac{1}{s^3} + \frac{1}{s^2} + \frac{1}{2s} \right) e^{-s} \end{align} $$
$$ \begin{align} \mathscr{L} \left( \frac{1}{2}t^2 u\left( t - \frac{\pi}{2} \right) \right) & = \mathscr{L} \left( \left( \frac{1}{2} \left( t - \frac{\pi}{2} \right)^2 + \frac{\pi}{2} \left( t - \frac{\pi}{2} \right) + \frac{\pi^2}{8} \right) u\left( t - \frac{\pi}{2} \right)\right) \\ & = \left( \frac{1}{s^3} + \frac{\pi}{2s^2} + \frac{\pi^2}{8s} \right) e^{-\pi s / 2} \end{align} $$
I'll omit the rest of the solution because this is where I'm mainly getting confused. More specifically, what are the steps or intuition behind going from $f(t)$ to $f(t - a)$?
If you look at the two Laplace transform equations above, the two equations convert:
$$ \begin{align} \frac{1}{2}t^2 & = \frac{1}{2}(t - 1)^2 + (t - 1) + \frac{1}{2} \\ \frac{1}{2}t^2 & = \frac{1}{2}\left(t - \frac{\pi}{2}\right)^2 + \frac{\pi}{2}\left(t - \frac{\pi}{2}\right) + \frac{\pi^2}{8} \end{align} $$
When I first read the statement "change $f(t)$ to $f(t - a)$, my approach was the following:
$$ \begin{align} \frac{1}{2}t^2 & = \frac{1}{2}t^2 -t + \frac{1}{2} + t - \frac{1}{2} \\ & = \left( \frac{1}{2}t^2 - t + \frac{1}{2} \right) + t - \frac{1}{2} \\ & = \frac{1}{2}(t - 1)^2 + t - \frac{1}{2} \\ \frac{1}{2}t^2 & = \frac{1}{2}t^2 - \frac{\pi}{2}t + \frac{\pi^2}{8} + \frac{\pi}{2}t - \frac{\pi^2}{8} \\ & = \left( \frac{1}{2}t^2 -\frac{\pi}{2}t + \frac{\pi^2}{8} \right) + \frac{\pi}{2}t - \frac{\pi^2}{8} \\ & = \frac{1}{2}\left(t - \frac{\pi}{2} \right)^2 + \frac{\pi}{2} \left( t - \frac{\pi}{4} \right) \end{align} $$
and as you can clearly see, my approach and results are wrong. What exactly are we supposed to be doing here? The textbook's explanation of "changing $f(t)$ to $f(t - a)$" is insufficient.
Any tips or advice are greatly appreciated. Thank you.