Digital Signal Processing Reference
In-Depth Information
y[n] =
adx[n]
+ bdx[n1]
+ cdx[n2]
+ ddx[n3]
+ acx[n1]
+ bcx[n2]
+ ccx[n3]
+ cdx[n4]
+ abx[n2]
+ bbx[n3]
+ bcx[n4]
+ bdx[n5]
+ aax[n3]
+ abx[n4]
+ acx[n5]
+ adx[n6]
adx[n]
+ acx[n1]abx[n2]
+ aax[n3]
+ bdx[n1]bcx[n2]
+ bbx[n3]abx[n4]
cdx[n2]
+ ccx[n3]bcx[n4]
+ acx[n5]
+ ddx[n3]cdx[n4]
+ bdx[n5]adx[n6]:
Repeating the above equation, only eliminating the parts that cancel:
y[n] = bdx[n1]
+ ddx[n3]
+ acx[n1]
+ ccx[n3]
+ bbx[n3]
+ bdx[n5]
+ aax[n3]
+ acx[n5]
+ acx[n1]
+ aax[n3]
+ bdx[n1]
+ bbx[n3]
+ ccx[n3]
+ acx[n5]
+ ddx[n3]
+ bdx[n5]:
The terms x[n1] and x[n5] are pesky, since we want each output of y[n]
to depend upon only one input of x[n]. But if ac happens to equalbd, then these
terms will cancel each other out. Therefore, we will make this a requirement. If
ac+bd = 0, then the x[n1] and x[n5] terms are eliminated, and we are left with:
y[n] = 2(aa + bb + cc + dd)x[n3]
or that y[n] is a delayed version of x[n], multiplied by a constant. Of course, we
could have a second requirement that this constant be 1. The Daubechies coe-
cients satisfy both of these criteria: (These are the lowpass coecients from db2,
obtained from MATLAB)
a =0:1294
b = 0:2241
c = 0:8365
d = 0:4830
ac =0:1083
