I'm studying for a final in a wireless communications class, and one of the topics we're expected to be able to work is the calculation of the minimum mean squared error (MMSE) for an adaptive equalizer, given a certain input vector and equalizer filter structure. In the class examples we've done so far, the equalizer is always a two-tap finite impulse response filter, which gives rise to an input correlation matrix, R, that is 2x2. The formula we are supposed to use in calculating the MMSE thus involves inverting the 2x2 matrix R, and manipulating it along with other system parameters to arrive at a final result. To make a long story short, the example we were provided with results in an input correlation matrix | 1/2 cos(2pi/N) | R =| | | cos(2pi/N) 1/2 | I need to invert this, and am having the damndest time. I undertand how to invert a matrix in general (although I admit I had to go back and look at some old textbooks). I just concatenate an identity matrix of the same size as the matrix to be inverted with the original matrix; reduce it using elementary row operations so that the identity matrix stands on the left; and now the inverted matrix appears in the right half of the concatenated matrix. But for some reason I can't seem to actually get the identity matrix to show up on th left hand side of the concatenated matrix for R. I tested out my method using simple numbers and got the right answers, but for the life of me I can't seem to make it work for this slightly more complicated scenario, even though since N is fixed (N>2), the value of cos(2pi/N) is also just a number. I'd be grateful is someone could point me in the right direction. BTW, to allay anyone's fears about giving away an answer to a final - this is just a preparation question, and the final undoubtedly will have different form (although probably similar). Thanks in advance, J

# symmetric matrix inversion problems

Started by ●April 25, 2004

Reply by ●April 25, 20042004-04-25

Never mind - I've solved the inversion problem for a 2x2 matrix in general and have reproduced the correct answer for my problem as given in the solutions. In case anyone is interested, the inverse of a matrix of form 1 N N 1 is: -1/(N^2-1) N/(N^2-1) N/(N^2-1) -1/(N^2-1) This is what I used to solve my specific problem. Amazing what one can do when one actually spells out the problem in a public forum! J "James Calivar" <amheiserbush@yahoo.com.au> wrote in message news:c6hiem$emm$1@home.itg.ti.com...> I'm studying for a final in a wireless communications class, and one ofthe> topics we're expected to be able to work is the calculation of the minimum > mean squared error (MMSE) for an adaptive equalizer, given a certain input > vector and equalizer filter structure. In the class examples we've doneso> far, the equalizer is always a two-tap finite impulse response filter,which> gives rise to an input correlation matrix, R, that is 2x2. The formulawe> are supposed to use in calculating the MMSE thus involves inverting the2x2> matrix R, and manipulating it along with other system parameters to arrive > at a final result. > > To make a long story short, the example we were provided with results inan> input correlation matrix > > | 1/2 cos(2pi/N) | > R =| | > | cos(2pi/N) 1/2 | > > I need to invert this, and am having the damndest time. > > I undertand how to invert a matrix in general (although I admit I had togo> back and look at some old textbooks). I just concatenate an identitymatrix> of the same size as the matrix to be inverted with the original matrix; > reduce it using elementary row operations so that the identity matrixstands> on the left; and now the inverted matrix appears in the right half of the > concatenated matrix. But for some reason I can't seem to actually get the > identity matrix to show up on th left hand side of the concatenated matrix > for R. I tested out my method using simple numbers and got the right > answers, but for the life of me I can't seem to make it work for this > slightly more complicated scenario, even though since N is fixed (N>2),the> value of cos(2pi/N) is also just a number. > > I'd be grateful is someone could point me in the right direction. > > BTW, to allay anyone's fears about giving away an answer to a final - this > is just a preparation question, and the final undoubtedly will have > different form (although probably similar). Thanks in advance, > > J > >

Reply by ●April 25, 20042004-04-25

Hello James, I'm glad you solved your problem. In a more general case, the matrix from an autocorellation is Teoplitz and a very efficient solution is had by the Durbin algo. Since this is a recursive method, this is sometimes called Durbin's recursion. Clay -- Clay S. Turner, V.P. Wireless Systems Engineering, Inc. Satellite Beach, Florida 32937 (321) 777-7889 www.wse.biz csturner@wse.biz "James Calivar" <amheiserbush@yahoo.com.au> wrote in message news:c6hlco$igq$1@home.itg.ti.com...> Never mind - I've solved the inversion problem for a 2x2 matrix in general > and have reproduced the correct answer for my problem as given in the > solutions. > > In case anyone is interested, the inverse of a matrix of form > > 1 N > N 1 > > is: > > -1/(N^2-1) N/(N^2-1) > N/(N^2-1) -1/(N^2-1) > > This is what I used to solve my specific problem. > > Amazing what one can do when one actually spells out the problem in apublic> forum! > > J > > "James Calivar" <amheiserbush@yahoo.com.au> wrote in message > news:c6hiem$emm$1@home.itg.ti.com... > > I'm studying for a final in a wireless communications class, and one of > the > > topics we're expected to be able to work is the calculation of theminimum> > mean squared error (MMSE) for an adaptive equalizer, given a certaininput> > vector and equalizer filter structure. In the class examples we've done > so > > far, the equalizer is always a two-tap finite impulse response filter, > which > > gives rise to an input correlation matrix, R, that is 2x2. The formula > we > > are supposed to use in calculating the MMSE thus involves inverting the > 2x2 > > matrix R, and manipulating it along with other system parameters toarrive> > at a final result. > > > > To make a long story short, the example we were provided with results in > an > > input correlation matrix > > > > | 1/2 cos(2pi/N) | > > R =| | > > | cos(2pi/N) 1/2 | > > > > I need to invert this, and am having the damndest time. > > > > I undertand how to invert a matrix in general (although I admit I had to > go > > back and look at some old textbooks). I just concatenate an identity > matrix > > of the same size as the matrix to be inverted with the original matrix; > > reduce it using elementary row operations so that the identity matrix > stands > > on the left; and now the inverted matrix appears in the right half ofthe> > concatenated matrix. But for some reason I can't seem to actually getthe> > identity matrix to show up on th left hand side of the concatenatedmatrix> > for R. I tested out my method using simple numbers and got the right > > answers, but for the life of me I can't seem to make it work for this > > slightly more complicated scenario, even though since N is fixed (N>2), > the > > value of cos(2pi/N) is also just a number. > > > > I'd be grateful is someone could point me in the right direction. > > > > BTW, to allay anyone's fears about giving away an answer to a final -this> > is just a preparation question, and the final undoubtedly will have > > different form (although probably similar). Thanks in advance, > > > > J > > > > > >

Reply by ●April 26, 20042004-04-26

James Calivar wrote:> Never mind - I've solved the inversion problem for a 2x2 matrix in general > and have reproduced the correct answer for my problem as given in the > solutions.> In case anyone is interested, the inverse of a matrix of form> 1 N > N 1> is:> -1/(N^2-1) N/(N^2-1) > N/(N^2-1) -1/(N^2-1)For a 2x2 matrix a b c d the inverse is d -b -c a divided by the determinant of either matrix, ad-bc. A real symmetric matrix has a real symmetric inverse. -- glen

Reply by ●April 26, 20042004-04-26

"Clay S. Turner" <CSTurner@WSE.Biz> wrote in message news:p3_ic.79805$UC4.28786@bignews2.bellsouth.net...> Hello James, > > I'm glad you solved your problem. In a more general case, the matrix froman> autocorellation is Teoplitz and a very efficient solution is had by the > Durbin algo. Since this is a recursive method, this is sometimes called > Durbin's recursion. > > Clay > >And comments like that are the reason I will remain at the protocol level when it comes to wireless communications systems! This stuff gets too abstract and complex for my relatively tiny brain to comprehend. Thanks to this class, I've got a pretty good feel for how difficult it is to ensure reasonable performance in a radio environment - and I appreciate all the work that is done (and has been done to date) in order to realize these performance levels. But I also am a pragmatist, and I know that I will *never* get to a level of some of these papers and articles I had to read. One man's 5-minute article scan is another man's PhD. thesis... James

Reply by ●April 26, 20042004-04-26

James Calivar wrote: ...> And comments like that are the reason I will remain at the protocol level > when it comes to wireless communications systems! This stuff gets too > abstract and complex for my relatively tiny brain to comprehend. > > Thanks to this class, I've got a pretty good feel for how difficult it is to > ensure reasonable performance in a radio environment - and I appreciate all > the work that is done (and has been done to date) in order to realize these > performance levels. But I also am a pragmatist, and I know that I will > *never* get to a level of some of these papers and articles I had to read. > One man's 5-minute article scan is another man's PhD. thesis...Have hope. When my son began to play the guitar -- he eventually got very good -- he despaired of ever learning to tune it himself. That all seems silly now. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������