# Scaled unit impulse/step sequences

1. Nov 10, 2013

### freezer

1. The problem statement, all variables and given/known data

$b_{k} = {4, 1, 1, 4}$

x[n] = 2u[n]

Write your answer using scaled unit impulse sequences and scaled unit step sequences. Write explicitly.

2. Relevant equations

3. The attempt at a solution

Code (Text):

4114
2222222....
------------
8228
8228
8228
8228
8228
.....
------------
8, 10, 12, 20, 20, 20, ...

$x[n] = 8\delta[n] + 10\delta[n-1] + 12\delta[n-2] + \sum^{\infty}_{k=3} 20 \delta[n-k]$

$\delta[n] = \left\{\begin{matrix} 0 & n<0\\ 8 & n = 0\\ 10 & n = 1 \\ 12 & n = 2\\ 20 & n \geq 3 \end{matrix}\right.$

Is this how you would answer this question?

Last edited: Nov 10, 2013
2. Nov 10, 2013

### rude man

You haven't stated the problem fully. What is the difference equation, exactly?

In any case, u[n] is always 1.

3. Nov 10, 2013

### freezer

The general difference equation for a causal FIR is:

$y[n] = \sum^{\infty}_{k=0} b_{k} x[n-k]$

and then

$\delta[n] = \left\{\begin{matrix} 0 & n<0\\ 8 & n = 0\\ 10 & n = 1 \\ 12 & n = 2\\ 20 & n \geq 3 \end{matrix}\right.$

Last edited: Nov 10, 2013
4. Nov 10, 2013

### rude man

Your expression for x[n] is correct. But I don't quite understand your table, probably because I can't make out the column after the 1st equal sign.

But δ[n] = 1, n = 0
= 0, n > 0 always.

BTW your equation is for a non-recursive filter, which is not necessarily an FIR filter.
FIR filters can also be recursive, and IIR filters can be non-recursive. However, saying FIR → non-recursive and IIR → recursive is almost universal, if misleading.