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Write f(x) in terms of the unit step function u(x)

Alexmahone

Active member
Jan 26, 2012
268
Write f(x) in terms of the unit step function u(x).

$u(x)=\left\{ \begin{array}{rcl} 1\ &\text{if}& \ x\geq 0 \\ 0\ &\text{if}& \ x<0\end{array} \right.$

$f(x)=\left\{ \begin{array}{rcl} 1\ &\text{if}& \ 2n\le x\le 2n+1 \\ 0\ &\text{elsewhere}\end{array} \right.$
 

chisigma

Well-known member
Feb 13, 2012
1,704
A more useful definition of 'Haeviside Step Function' or 'Unit Step Function' is...

$\mathcal{u}(x)=\begin{cases}1 &\text{if}\ x>0\\ \frac{1}{2} &\text{if}\ x=0\\ 0 &\text{if}\ x<0\end{cases}$ (1)

... and the function...

$f(x)=\begin{cases} 1 &\text{if}\ 2n<x<2n+1\\ \frac{1}{2} &\text{if}\ x=n\\ 0 &\text{elsewhere}\end{cases}$ (2)

... can be written as...

$\displaystyle f(x)= \sum_{n=0}^{\infty} (-1)^{n} \mathcal{u}(x-n)$ (3)

Kind regards

$\chi$ $\sigma$
 

Alexmahone

Active member
Jan 26, 2012
268
A more useful definition of 'Haeviside Step Function' or 'Unit Step Function' is...
But I need to use the definition in post #1.

I get $\displaystyle f(x)=\sum_{-\infty}^\infty u(x-2n)u(2n+1-x)$
 

chisigma

Well-known member
Feb 13, 2012
1,704
A more useful definition of 'Haeviside Step Function' or 'Unit Step Function' is...

$\mathcal{u}(x)=\begin{cases}1 &\text{if}\ x>0\\ \frac{1}{2} &\text{if}\ x=0\\ 0 &\text{if}\ x<0\end{cases}$ (1)

... and the function...

$f(x)=\begin{cases} 1 &\text{if}\ 2n<x<2n+1\\ \frac{1}{2} &\text{if}\ x=n\\ 0 &\text{elsewhere}\end{cases}$ (2)

... can be written as...

$\displaystyle f(x)= \sum_{n=0}^{\infty} (-1)^{n} \mathcal{u}(x-n)$ (3)

Kind regards

$\chi$ $\sigma$
It is curious the fact that the function...

$\displaystyle f_{a}(x) =\begin{cases} 1 &\text{if}\ 2n \le x \le 2n+1\\ 0 &\text{elsewhere}\end{cases}$ (1)

... has Laplace Transform...

$\displaystyle \mathcal{L}\{f_{a}(x)\}= F(s)= \frac{1}{s\ (1+e^-s)}$ (2)

... and the (2) has Inverse Laplace Transform...

$\displaystyle \mathcal{L}^{-1}\{F(s)\}= f_{b}(x)=\sum_{n=0}^{\infty} (-1)^{n}\ \mathcal{u}(x-n)$ (3)

... so that for the unicity of the inverse L-transform $f_{a}(x)$ and $f_{b}(x)$ are 'pratically' the same function... where 'pratically' means that the difference between then is a 'null function'...

Kind regards

$\chi$ $\sigma$