Output of FIR Filter for Different Inputs

  • #1
freezer
76
0

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



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


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]
 
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  • #2
freezer said:

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



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

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.
[tex]
u[n] =
\begin{cases}
1 & \text{if } n \geq 0 \\
0 & \text{if } n < 0
\end{cases}
[/tex]

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.
 
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