Re: simpler way to get a particular banded matrix?
- To: mathgroup at smc.vnet.net
- Subject: [mg98103] Re: [mg98079] simpler way to get a particular banded matrix?
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Mon, 30 Mar 2009 04:43:41 -0500 (EST)
- Reply-to: hanlonr at cox.net
H = {h0, h1, h2, h3, h4, h5};
rH = Reverse[H];
z = Table[0, {5}];
Take[#, -6] & /@
NestList[RotateRight[#, 2] &, Join[rH, z], 5]
Table[Take[Join[z, rH, z], {13, 18} - 2 n], {n, 6}]
Bob Hanlon
---- rip pelletier <bitbucket at comcast.net> wrote:
=============
Hi,
The following command constructs a matrix (typically denoted M0, it
arises in filter banks in general, in Burrus, Gopinath, & Guo "intro to
wavelets and wavelet transforms: a primer" in particular).
I'm sure there's a way to do this without typing out all 5 bands. Here's
what i did:
SparseArray[Band[{1,1}]->H,{6,6}]+
SparseArray[Band[{2,3}]->H,{6,6}]+
SparseArray[Band[{3,5}]->H,{6,6}]+
SparseArray[Band[{-2,-3},{1,1},{-1,-1}]->rH,{6,6}]+
SparseArray[Band[{-3,-5},{1,1},{-1,-1}]->rH,{6,6}]
where
H={h0,h1,h2,h3,h4,h5};
rH=Reverse[H];
I get what I want:
h0 0 0 0 0 0
h2 h1 h0 0 0 0
h4 h3 h2 h1 h0 0
0 h5 h4 h3 h2 h1
0 0 0 h5 h4 h3
0 0 0 0 0 h5
Any easier ways to get this?
TIA and vale,
rip
--
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