Confused with Notation
in [DSP]
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From: firepluswater on 17 Apr 2010 03:09 So some of the papers I am looking at represent convolution in 'matrix' form in one of two ways: y[n] = H^H * X where H^H = hermitian (conj transpose) of filter H and y[n] = H^t * X where H^t = transpose of filter H Why is the first formula conjugated? To get a better sense of what I mean..incase my notation above is confusing you. I am trying to understand why on wikipedia the formula for the error of the RLS algorithm (http://en.wikipedia.org/wiki/Recursive_least_squares_filter - located at end under "RLS SUMMARY" section) is e(n) = d(n) - H(n-1)^t * x(n) while here (http://www.researchusa.org/uploads/avi/pVPGuUnUhxGgcpr0UdAn.pdf) on page 4 the formula is e(n) = d(n) - H(n-1)^H * x(n) The equations in these two papers seem to be the exact same..but they differ just on whether H should be conjugated..anyone know why?
From: HardySpicer on 17 Apr 2010 04:13 On Apr 17, 7:09 pm, "firepluswater" <ampierce(a)n_o_s_p_a_m.gmail.com> wrote: > So some of the papers I am looking at represent convolution in 'matrix' > form in one of two ways: > > y[n] = H^H * X where H^H = hermitian (conj transpose) of filter H > and > y[n] = H^t * X where H^t = transpose of filter H > > Why is the first formula conjugated? > > To get a better sense of what I mean..incase my notation above is confusing > you. > > I am trying to understand why on wikipedia the formula for the error of the > RLS algorithm (http://en.wikipedia.org/wiki/Recursive_least_squares_filter > - located at end under "RLS SUMMARY" section) is > e(n) = d(n) - H(n-1)^t * x(n) > > while here > (http://www.researchusa.org/uploads/avi/pVPGuUnUhxGgcpr0UdAn.pdf) on page 4 > the formula is > e(n) = d(n) - H(n-1)^H * x(n) > > The equations in these two papers seem to be the exact same..but they > differ just on whether H should be conjugated..anyone know why? If it is complex then you use conjugate transpose. To be honest, a lot of poser engineer/maths types put Hermitian when it is not actually needed but try and be general. For real data you don't need Hermitian. In some comms problems however you have I and q data so the problem becomes complex. For the rest of us who use real data then it is an unnescessary complication! Hardy
From: Rune Allnor on 17 Apr 2010 06:01 On 17 apr, 09:09, "firepluswater" <ampierce(a)n_o_s_p_a_m.gmail.com> wrote: > So some of the papers I am looking at represent convolution in 'matrix' > form in one of two ways: > > y[n] = H^H * X where H^H = hermitian (conj transpose) of filter H > and > y[n] = H^t * X where H^t = transpose of filter H > > Why is the first formula conjugated? > > To get a better sense of what I mean..incase my notation above is confusing > you. The notation is standard. You need the conjugate if you work with complex-valued data. It's a property of the inner product between complex-valued vectors. For real-valued X the Hermitian X^H and the transpose X^T are equal. Unfortunately, a lot of authors, both on DSP and on linear algebra, only write the real-valued form when they derive results on matrix form. There are cases where one needs to treat the complex-valued case somewhat differently than the real-valued case, that thus are missing from the textbooks. In case you need to 'complexify' a real-valued expression, you might want to be aware that there might be more to it than merely substituting superscript H for superscript T. Rune
From: Tim Wescott on 17 Apr 2010 15:43 HardySpicer wrote: > On Apr 17, 7:09 pm, "firepluswater" <ampierce(a)n_o_s_p_a_m.gmail.com> > wrote: >> So some of the papers I am looking at represent convolution in 'matrix' >> form in one of two ways: >> >> y[n] = H^H * X where H^H = hermitian (conj transpose) of filter H >> and >> y[n] = H^t * X where H^t = transpose of filter H >> >> Why is the first formula conjugated? >> >> To get a better sense of what I mean..incase my notation above is confusing >> you. >> >> I am trying to understand why on wikipedia the formula for the error of the >> RLS algorithm (http://en.wikipedia.org/wiki/Recursive_least_squares_filter >> - located at end under "RLS SUMMARY" section) is >> e(n) = d(n) - H(n-1)^t * x(n) >> >> while here >> (http://www.researchusa.org/uploads/avi/pVPGuUnUhxGgcpr0UdAn.pdf) on page 4 >> the formula is >> e(n) = d(n) - H(n-1)^H * x(n) >> >> The equations in these two papers seem to be the exact same..but they >> differ just on whether H should be conjugated..anyone know why? > > If it is complex then you use conjugate transpose. To be honest, a lot > of poser engineer/maths types put Hermitian > when it is not actually needed but try and be general. For real data > you don't need Hermitian. In some comms problems however > you have I and q data so the problem becomes complex. For the rest of > us who use real data then it is an unnescessary complication! Is it poser or poseur? Would I be a poser to say 'poseur' without knowing French? Or would I be a poseur to say 'poser' without being a rocker? (I have listened to Doug & Dave -- does that mean I can say 'poser'?) -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
From: HardySpicer on 17 Apr 2010 22:56
On Apr 17, 10:01 pm, Rune Allnor <all...(a)tele.ntnu.no> wrote: > On 17 apr, 09:09, "firepluswater" <ampierce(a)n_o_s_p_a_m.gmail.com> > wrote: > > > So some of the papers I am looking at represent convolution in 'matrix' > > form in one of two ways: > > > y[n] = H^H * X where H^H = hermitian (conj transpose) of filter H > > and > > y[n] = H^t * X where H^t = transpose of filter H > > > Why is the first formula conjugated? > > > To get a better sense of what I mean..incase my notation above is confusing > > you. > > The notation is standard. > > You need the conjugate if you work with complex-valued data. > It's a property of the inner product between complex-valued > vectors. For real-valued X the Hermitian X^H and the > transpose X^T are equal. > > Unfortunately, a lot of authors, both on DSP and on linear > algebra, only write the real-valued form when they derive > results on matrix form. There are cases where one needs to > treat the complex-valued case somewhat differently than the > real-valued case, that thus are missing from the textbooks. > > In case you need to 'complexify' a real-valued expression, > you might want to be aware that there might be more to it > than merely substituting superscript H for superscript T. > > Rune All true but for begiiners books it doesn't help one biy. In control theory we used to use '*' to denote Hermition transpose. Looked a little simpler than raising to the power H! Hardy |