Scaled unit impulse/step sequences

[freezer]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

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]<br /> \delta[n] = \left\{\begin{matrix}<br /> 0 &amp; n&lt;0\\ <br /> 8 &amp; n = 0\\ <br /> 10 &amp; n = 1 \\ <br /> 12 &amp; n = 2\\ <br /> 20 &amp; n \geq 3 <br /> \end{matrix}\right.<br />

Is this how you would answer this question?

 

[rude man]

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


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

 

[freezer]

The general difference equation for a causal FIR is:

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

and then

<br /> \delta[n] = \left\{\begin{matrix}<br /> 0 &amp; n&lt;0\\ <br /> 8 &amp; n = 0\\ <br /> 10 &amp; n = 1 \\ <br /> 12 &amp; n = 2\\ <br /> 20 &amp; n \geq 3 <br /> \end{matrix}\right.<br />

 

[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.