Digital Signal Processing Reference
In-Depth Information
n is even-even:
0
w d [n]
=
adw d [n]
+ bdw d [n1]
+ cdw d [n2]
+ ddw d [n3]
+ abw d [n2]
+ bbw d [n3]
+ bcw d [n4]
+ bdw d [n5]
adw d [n]
+ acw d [n1]abw d [n2]
+ aaw d [n3]
cdw d [n2]
+ ccw d [n3]bcw d [n4]
+ acw d [n5]:
n is odd-even:
0
=
acw d [n1]
w d [n]
+ bcw d [n2]
+ ccw d [n3]
+ cdw d [n4]
+ aaw d [n3]
+ abw d [n4]
+ acw d [n5]
+ adw d [n6]
+ bdw d [n1]bcw d [n2]
+ bbw d [n3]abw d [n4]
+ ddw d [n3]cdw d [n4]
+ bdw d [n5]
adw d [n6]:
Canceling out terms gives us:
n is even-even:
0
w d [n]
= bdw d [n1]
+ ddw d [n3]
+ bbw d [n3]
+ bdw d [n5]
+ acw d [n1]
+ aaw d [n3]
+ ccw d [n3]
+ acw d [n5]:
n is odd-even:
0
w d [n]
= acw d [n1]
+ ccw d [n3]
+ aaw d [n3]
+ acw d [n5]
+ bdw d [n1]
+ bbw d [n3]
+ ddw d [n3]
+ bdw d [n5]:
Examining the above equations, we see that we have the same expression for
when n is even-odd as when it is even-even. Therefore, we can rewrite the above
expression, with the note that n must be even. We know from the previous section
that ac =bd, so the terms with w d [n1] and w d [n5] will cancel out. This leaves
us with w d [n]
0
= (aa+bb+cc+dd)w d [n3]. As previously noted, (aa+bb+cc+dd) =
1, so we get the nal expression for w d [n]
0
0
is
the same as a delayed version of the original w d [n]. This is no surprise, since we have
= w d [n3]. The reconstructed w d [n]
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