FIR Filters and Convolution Example
An FIR filter has impulse response
h  n  = – 2  n  + 3  n – 1  +  n – 2 
The input to the filter, x  n  , is
 4 – n, 0  n  3
xn = 
otherwise
 0,
• Find the filter output y  n 
Expected output range
[0+0, 3+2] = [0,5]
sum of products formed between h[k]
and x[n-k] inside red box.
h[k]
n=0
n=1
...
1
y[n]
0
4
3
2
1
0
0
0
...
...
0
0
4
3
2
1
0
0
0
...
...
0
0
4
3
2
1
0
0
0
...
...
0
0
4
3
2
1
0
0
0
...
...
0
0
4
3
2
1
0
0
0
...
...
0
0
4
3
2
1
0
0
0
...
4
...
0
0
4
3
2
1
0
0
0
0
n=3
n=4
n=5
n=6
n=
3
0
n=2
n=7
-2
-2
-1
1
3
15
Flipped and
shifted: x[n-k]
Outputs for n < 0 and
n > 5 are all 0
n=
n=
Start at
n=0
>> filter([-2 3 1],1,[1 2 3 4 0 0])
ans = -2 -1 1 3 15 4
ECE 2610 Example
Page–1
An FIR filter has impulse response
h  n  =  1 1 2 2 
n = 0
The input to the filter is
x  n  =  4 – 2 2 
n = 0
• Find the filter output y  n 
sum of products formed between h[k]
and x[n-k] inside red box.
h[k] 0
Expected output range
[0+0, 3+2] = [0,5]
0
0
0
0
0
1
1
2
2
0
0
0
0
0
0
y[n]
n=0
0
0
0
0
2
-2
4
0
0
0
0
0
0
0
0
0
4
n=1
0
0
0
0
0
2
-2
4
0
0
0
0
0
0
0
0
2
n=2
0
0
0
0
0
0
2
-2
4
0
0
0
0
0
0
0
8
n=3
0
0
0
0
0
0
0
2
-2
4
0
0
0
0
0
0
6
n=4
0
0
0
0
0
0
0
0
2
-2
4
0
0
0
0
0
0
n=5
0
0
0
0
0
0
0
0
0
2
-2
4
0
0
0
0
4
n=6
0
0
0
0
0
0
0
0
0
0
2
-2
4
0
0
0
0
n=7
0
0
0
0
0
0
0
0
0
0
0
2
-2
4
0
0
0
n=8
0
0
0
0
0
0
0
0
0
0
0
0
2
-2
4
0
0
n=9
n =10
Flipped and
shifted: x[n-k]
Outputs for n < 0 and
n > 5 are all 0
n = 11
n =12
n =13
MATLAB
Check
ECE 2610 Example
>> filter([1 1 2 2],1,[4 -2 2 0 0 0 0])
ans = 4 2 8 6 0 4 0
n=0
n=5
Page–2