Digital Signal Processing Reference
In-Depth Information
bd = 0:1083:
Errors due to precision come about when using wavelets. The coecients above are
actually an approximation of the following Daubechies coecients (obtained from
[3]):
a =
1
p
2
; b =
3
p
p
p
3
2
; c =
3 +
3
2
; d =
1 +
3
3
2
:
Note that the reference gives these as the highpass lter coecients, which means
that the b and d values above were negated. So,
p
p
p
p
4
4
4
4
p
p
2
3
; bd =
2
3
32
:
ac =
32
Therefore:
ac =bd:
Using these coecients, we see that:
p
p
p
p
a
2
=
42
3
; b
2
=
126
3
; c
2
=
12 + 6
3
; d
2
=
4 + 2
3
:
32
32
32
32
Finding the sum aa + bb + cc + dd leads to:
p
p
p
p
a
2
+ b
2
+ c
2
+ d
2
=
42
3 + 126
3 + 12 + 6
3 + 4 + 2
3
32
a
2
+ b
2
+ c
2
+ d
2
=
4 + 12 + 12 + 4
32
a
2
+ b
2
+ c
2
+ d
2
= 1:
This is what we should expect|that the values for a, b, c, and d are carefully chosen
so that the output y[n] is just a delayed version of the input x[n]. But above we
found that y[n] = 2(aa + bb + cc + dd)x[n3], so these coecients mean that
y[n] = 2x[n3]. Why is the 2 there? The answer for this lies in the down-sampling
operation. We have not yet looked at the eect of the down-sampling operation,
but it gives us the same outputs, only at 1/2 the scale. That is, if we use the above
Daubechies coecients, but with down-samplers and up-samplers in place, then we
get y[n] = x[n3]. This is why aa + bb + cc + dd = 1, instead of 1=2. This is shown
in section 9.5.
