Output of FIR Filter for Different Inputs

[freezer]

Homework Statement

Consider the FIR filter with {bk} = {3, 4, -4, -3}. These are the filter coefficients for k = 0, 1, 2, and 3, respectively. Sketch the output y[n] where the input is:

x[n] = δ[n]
x[n] = (u[n] - u[n-2])
x[n] - u[n]

Homework Equations

\sum^{M}_{k=0}b_k x[n-k]

The Attempt at a Solution

I am not sure how to process this form.
I would build a table


for the first one x[0] = 3, x[2] = 4, etc..
for second x[0] = 3,..., x[3]=(3-4) = -1

I my notes don't have anything with u[n]

 

[collinsmark]

Homework Statement

Consider the FIR filter with {bk} = {3, 4, -4, -3}. These are the filter coefficients for k = 0, 1, 2, and 3, respectively. Sketch the output y[n] where the input is:

x[n] = δ[n]
x[n] = (u[n] - u[n-2])
x[n] - u[n]

Homework Equations

\sum^{M}_{k=0}b_k x[n-k]

The Attempt at a Solution

I am not sure how to process this form.
I would build a tablefor the first one x[0] = 3, x[2] = 4, etc..

That's the right idea, with the exception of: don't you mean " y[0] = 3, y[1] = 4. etc.."? (As opposed to x[0] = 3, x[2] = 4?)

for second x[0] = 3,..., x[3]=(3-4) = -1

I my notes don't have anything with u[n]

I don't think that's quite right.

u[n] is the unit step function.
<br /> u[n] =<br /> \begin{cases}<br /> 1 &amp; \text{if } n \geq 0 \\<br /> 0 &amp; \text{if } n &lt; 0<br /> \end{cases}<br />

I suggest making a table giving u[n] as a function of n, for n = -1 to around 7 or so. Then do the same thing for -u[n-2]. Then again for u[n] - u[n-2]. And finally make another table for y[n] with that input.