Digital Signal Processing Reference
In-Depth Information
only with dierent coecients.
W[2; 0] =
34
2
= 17
W[2; 1] =
174
2
= 87
W[2; 2] =
80
2
= 40
W[2; 3] = 0
Looking at octave 3:
W
h
[3;n] = W[2; 0]g[2n0] + W[2; 1]g[2n1] + W[2; 2]g[2n2] + W[2; 3]g[2n3]
+W[2; 4]g[2n4] + W[2; 5]g[2n5] + W[2; 6]g[2n6] + W[2; 7]g[2n7]
W
h
[3;n] = 17g[2n0] + 87g[2n1] + 40g[2n2]
17
p
2
W
h
[3; 0] = 17g[0] + 87g[1] + 40g[2] =
4087
p
47
p
W
h
[3; 1] = 17g[2] + 87g[1] + 40g[0] =
=
2
2
W
h
[3; 2] = 17g[4] + 87g[3] + 40g[2] = 0:
Finding the lowpass outputs for this octave:
17
p
2
W[3; 0] = 17h[0] + 87h[1] + 40h[2] =
4087
p
47
p
W[3; 1] = 17h[2] + 87h[1] + 40h[0] =
=
2
2
W[3; 2] = 17h[4] + 87h[3] + 40h[2] = 0:
We stop here, because we get no more useful information after this point. Why?
Because our original signal has only 8 samples, and each octave uses only half the
data of the previous octave. In other words, since we have 2
3
values, we should not
perform more than 3 octaves of resolution on the signal.
We end this chapter with a program to show the frequency magnitude response
