Representing a piece-wise function in terms of the unit step

Question

$f(t) = \begin{cases} 4, & \text{if 0 $\le$ t $\le$ 2} \\ t, & \text{if $t \ge 2$} \end{cases}$

I need to express $f(t)$ in terms of the unit step function, $u(t)$.
By purely visualizing, my attempt is as follows:
$f(t) = 4[u(t-0)-u(t-2)] + t[u(t-2)-u(t-\infty)]$
$f(t) = 4[(u(t)-u(t-2)] +t[u(t-2)]$

The answer to this from the book is:
$f(t) = 4 + u(t-2)[t-4] = 4+(t-2)u(t-2)-2u(t-2)$

I am completely unable to follow this solution. Could someone please help explain this as simply as possible? Thanks!

Answer 1

By looking at $$f(t) = \begin{cases} 4, & \text{if 0 $\le$ t $\le$ 2} \\ t, & \text{if $t \ge 2$} \end{cases}$$ we can deduce that function $f(t)$ must be of form $$f(t)=(t+a)\cdot u(t-2)+b$$ where $u(t-2)$ is $$u(t-2) = \begin{cases} 0, & \text{if }\ t < 2 \\ 1, & \text{if }\ t > 2 \end{cases}$$

Clearly from $f(t)=4, \text{if 0 $\le$ t $\le$ 2}$ we get $b=4$ and by the second case of $f(t)$ we get a = $-4$.

Hence $$f(t)=(t-4)\cdot u(t-2)+4$$

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Edit:

Assume $$f(t) = \begin{cases} l, & \text{if 0 $\le$ t $\le$ 2} \\ mt+n, & \text{if $t \ge 2$} \end{cases}$$ where $l,m,n$ are constants.

Since maximum degree of $f(t)$ is $1$, we take $$f(t)=(c\cdot u(t-2)+d)\cdot t+a\cdot u(t-2)+b$$ where we have to find $a,b,c,d$.

From first case we get $b=l$ and $d=0$.

From second case we get $c=m$ and $a=-b=-l$.

So we get $f(t)$ of form $(ct+a)\cdot u(t-2)+b$. In the question's case $c=m=1$.