{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture_16-18 - 67 Lecture 16-18 In many practical...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
67 Lecture 16-18 In many practical applications, we are asked to approximate a given signal using a linear combination of a fixed collection of elementary signals. Recall that while studying DT signals, in the first part of the course, we looked at at least 3 ways of representing arbitrary DT signals using elementary signals as building blocks. In those cases we required exact representations. However, if there is some constraint on the available resources, in the sense that the collection of elementary signals is not enough for exact representations, then how to obtain the best representation of a given signal ? This question is answered by projection theorem . Projection Theorem: Let S be a Hilbert space, and let V be a subspace of S . For any vector (arbitrary) x S , 1. There exist a unique solution to the optimization problem arg min v V bardbl x v bardbl where bardbl · bardbl denotes the induced norm. 2. v 0 is the minimizer of the above optimization problem if and only if ( x v 0 ) v, for all v V Intuition from 3-dimensional euclidean space: Let S be a 3D euclidean space. Let V = span { (1,0,0), (0,1,0) } . See Fig. 58. Let ( x, y ) = x 1 y 1 + x 2 y 2 + x 3 y 3 bardbl x y bardbl = radicalbig ( x 1 y 1 ) 2 + ( x 2 y 2 ) 2 + ( x 3 y 3 ) 2 In this case we can look at the optimization problem as: arg min ( v 1 ,v 2 ,v 3 ) V radicalbig ( x 1 v 1 ) 2 + ( x 2 v 2 ) 2 + ( x 3 v 3 ) 2 We can solve this by inspection and obtain the optimizer as V 0 = ( x 1 , x 2 , 0) . Observe that in this case the error vector is given by (0 , 0 , x 3 ) and it is orthogonal to every vector in V. Projection theorem says that this picture can be extended to arbitrary Hilbert space. Proof: We will look at the proof of Statement (b). The proof of Statement (a) is technical in nature and is out of the scope of this course. Necessary Condition: If V 0 is the minimizer then ( x v 0 ) V for all v V Assume: V 0 is the minimizer and there exist a vector v V such that ( x v 0 ) is not V . Let ( ( x v 0 ) , v ) = δ negationslash = 0 . z = v 0 + δ bardbl v bardbl 2 v .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
68 x v e 0 Figure 58: Projection Theorem Since v 0 V and v V , we have z V . ⇒ bardbl x z bardbl 2 = bardbl x v 0 δ bardbl v bardbl 2 v bardbl 2 = bardbl x v 0 bardbl 2 + | δ | 2 bardbl v bardbl 4 bardbl v bardbl 2 + ( x v 0 , δv bardbl v bardbl 2 ) + ( δv bardbl v bardbl 2 , x v 0 ) = bardbl x v 0 bardbl 2 + | δ | 2 bardbl v bardbl 4 + ( δ ) δ bardbl v bardbl 2 + ( δ ) δ bardbl v bardbl 2 = bardbl x v 0 bardbl 2 | δ | 2 bardbl v bardbl 2 ⇒ bardbl x z bardbl 2 < bardbl x v 0 bardbl 2 V 0 cannot be the minimizer. Sufficient Condition: If an arbitrary vector v 0 V satisfies ( x v 0 ) v for all v V , then v 0 is the minimizer. Assume: v 0 V is a vector that satisfies ( x v 0 ) v for all v V . Consider any other vector ˜ v V and ˜ v negationslash = v 0 . Consider: bardbl x ˜ v bardbl 2 = bardbl x v 0 + v 0 ˜ v bardbl 2 = bardbl x v 0 bardbl 2 + bardbl v 0 ˜ v bardbl 2 + ( x v 0 , v 0 ˜ v ) + ( v 0 ˜ v, x v 0 ) = bardbl x v 0 bardbl 2 + bardbl v 0
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern