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So, I might be completely over thinking, or over looking this, but I am having a hard time understanding the linear combinations of step functions.

Example: $g(t) = u_{1}(t) + 2u_{3}(t) - 6u_{4}(t)$

I understand the concept of a single unit step function, for example, $ 6u_{4}(t) =\begin{cases}0, & t < 4\\6, & t \geq 4 \end{cases} $

So, my question is, how do I combine these types of functions?

EDIT Going off of Alan's answer, and my example above:

$g(t) = \begin{cases}0, & t < 1\\1, & 1 \leq t < 3\\ 3, & 3 \leq t < 4\\ -3, & t\geq4\end{cases} $ .. is this correct?

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    Ah! That helped me from a graphing standpoint. Thank you very much.2012-10-21

1 Answers 1

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The sum of functions is defined pointwise. For example, say you have two step functions

$u_1(t) = \begin{cases} 0, & t < 1 \\ 1, & t \ge 1 \end{cases}$

and

$u_2(t) = \begin{cases} 0, & t < 2 \\ 1, & t \ge 2 \end{cases}$.

Then a linear combination of them, say $au_1(t) + bu_2(t)$ would be

$au_1(t) + bu_2(t) = \begin{cases} 0, & t < 1 \\ a, & 1 \le t < 2 \\ a+b, & t \ge 2 \end{cases}$.

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    Alan, thanks for your answer. I've updated my question. I believe I understand it now, but please correct me if I'm wrong.2012-10-21