# Connexions

You are here: Home » Content » Orthonormal Basis Expansions

### Recently Viewed

This feature requires Javascript to be enabled.

### Tags

(What is a tag?)

These tags come from the endorsement, affiliation, and other lenses that include this content.

# Orthonormal Basis Expansions

Module by: Michael Haag, Justin Romberg. E-mail the authors

Summary: The module looks at decomposing signals through orthonormal basis expansion to provide an alternative representation. The module presents many examples of solving these problems and looks at them in several spaces and dimensions.

Note: You are viewing an old version of this document. The latest version is available here.

## Main Idea

When working with signals many times it is helpful to break up a signal into smaller, more manageable parts. Hopefully by now you have been exposed to the concept of eigenvectors and there use in decomposing a signal into one of its possible basis. By doing this we are able to simplify our calculations of signals and systems through eigenfunctions of LTI systems.

Now we would like to look at an alternative way to represent signals, through the use of orthonormal basis. We can think of orthonormal basis as a set of building blocks we use to construct functions. We will build up the signal/vector as a weighted sum of basis elements.

### Example 1

The complex sinusoids 1Tei ω 0 nt 1 T ω 0 n t for all <n< n form an orthonormal basis for L 2 0 T L 2 0 T .

In our Fourier series equation, ft= n = c n ei ω 0 nt f t n c n ω 0 n t , the c n c n are just another representation of ft f t .

### note:

For signals/vectors in a Hilbert Space, the expansion coefficients are easy to find.

## Alternate Representation

Recall our definition of a basis: A set of vectors b i b i in a vector space S S is a basis if

1. The b i b i are linearly independent.
2. The b i b i span S S. That is, we can find α i α i , where α i C α i (scalars) such that
x ,xS:x=i α i b i x x S x i α i b i
(1)
where xx is a vector in S S, αα is a scalar in C, and b b is a vector in S S.

Condition 2 in the above definition says we can decompose any vector in terms of the b i b i . Condition 1 ensures that the decomposition is unique (think about this at home).

### note:

The α i α i provide an alternate representation of xx.

### Example 2

Let us look at simple example in 2 2 , where we have the following vector: x=12 x 1 2 Standard Basis: e 0 e 1 =10T01T e 0 e 1 1 0 0 1 x= e 0 +2 e 1 x e 0 2 e 1 Alternate Basis: h 0 h 1 =11T1-1T h 0 h 1 1 1 1 -1 x=32 h 0 +-12 h 1 x 3 2 h 0 -1 2 h 1

In general, given a basis b 0 b 1 b 0 b 1 and a vector x 2 x 2 , how do we find the α 0 α 0 and α 1 α 1 such that

x= α 0 b 0 + α 1 b 1 x α 0 b 0 α 1 b 1
(2)

## Finding the Alphas

Now let us address the question posed above about finding α i α i 's in general for 2 2 . We start by rewriting Equation 2 so that we can stack our b i b i 's as columns in a 2×2 matrix.

( x )= α 0 ( b 0 )+ α 1 ( b 1 ) x α 0 b 0 α 1 b 1
(3)
( x )=( b 0 b 1 )( α 0 α 1 ) x b 0 b 1 α 0 α 1
(4)

### Example 3

Here is a simple example, which shows a little more detail about the above equations.

( x0 x1 )= α 0 ( b 0 0 b 0 1 )+ α 1 ( b 1 0 b 1 1 )=( α 0 b 0 0+ α 1 b 1 0 α 0 b 0 1+ α 1 b 1 1 ) x 0 x 1 α 0 b 0 0 b 0 1 α 1 b 1 0 b 1 1 α 0 b 0 0 α 1 b 1 0 α 0 b 0 1 α 1 b 1 1
(5)
( x0 x1 )=( b 0 0 b 1 0 b 0 1 b 1 1 )( α 0 α 1 ) x 0 x 1 b 0 0 b 1 0 b 0 1 b 1 1 α 0 α 1
(6)

### Simplifying our Equation

To make notation simpler, we define the following two items from the above equations:

• Basis Matrix: B=( b 0 b 1 ) B b 0 b 1
• Coefficient Vector: α=( α 0 α 1 ) α α 0 α 1
This gives us the following, concise equation:
x=Bα x B α
(7)
which is equivalent to x= i =01 α i b i x i 1 0 α i b i .

#### Example 4

Given a standard basis, ( 1 0 )( 0 1 ) 1 0 0 1 , then we have the following basis matrix: B=( 01 10 ) B 0 1 1 0

To get the α i α i 's, we solve for the coefficient vector in Equation 7

α=B-1x α B x
(8)
Where B-1 B is the inverse matrix of BB.

### Examples

#### Example 5

Let us look at the standard basis first and try to calculate α α from it. B=( 10 01 )=I B 1 0 0 1 I Where II is the identity matrix. In order to solve for α α let us find the inverse of BB first (which is obviously very trivial in this case): B-1=( 10 01 ) B 1 0 0 1 Therefore we get, α=B-1x=x α B x x

#### Example 6

Let us look at a ever-so-slightly more complicated basis of ( 1 1 )( 1 -1 )= h 0 h 1 1 1 1 -1 h 0 h 1 Then our basis matrix and inverse basis matrix becomes: B=( 11 1-1 ) B 1 1 1 -1 B-1=( 1212 12-12 ) B 1 2 1 2 1 2 -1 2 and for this example it is given that x=( 3 2 ) x 3 2 Now we solve for α α α=B-1x=( 1212 12-12 )( 3 2 )=( 2.5 0.5 ) α B x 1 2 1 2 1 2 -1 2 3 2 2.5 0.5 and we get x=2.5 h 0 +0.5 h 1 x 2.5 h 0 0.5 h 1

#### Exercise 1

Now we are given the following basis matrix and xx: b 0 b 1 =( 1 2 )( 3 0 ) b 0 b 1 1 2 3 0 x=( 3 2 ) x 3 2 For this problem, make a sketch of the bases and then represent xx in terms of b 0 b 0 and b 1 b 1 .

##### Solution

In order to represent xx in terms of b 0 b 0 and b 1 b 1 we will follow the same steps we used in the above example. B=( 12 30 ) B 1 2 3 0 B-1=( 012 13-16 ) B 0 1 2 1 3 -1 6 α=B-1x=( 1 23 ) α B x 1 2 3 And now we can write xx in terms of b 0 b 0 and b 1 b 1 . x= b 0 +23 b 1 x b 0 2 3 b 1 And we can easily substitute in our known values of b 0 b 0 and b 1 b 1 to verify our results.

#### note:

A change of basis simply looks at xx from a "different perspective." B-1 B transforms xx from the standard basis to our new basis, b 0 b 1 b 0 b 1 . Notice that this is a totally mechanical procedure.

## Extending the Dimension and Space

We can also extend all these ideas past just 2 2 and look at them in n n and n n . This procedure extends naturally to higher (> 2) dimensions. Given a basis b 0 b 1 b n 1 b 0 b 1 b n 1 for n n , we want to find α 0 α 1 α n 1 α 0 α 1 α n 1 such that

x= α 0 b 0 + α 1 b 1 ++ α n 1 b n 1 x α 0 b 0 α 1 b 1 α n 1 b n 1
(9)
Again, we will set up a basis matrix B=( b 0 b 1 b 2 b n 1 ) B b 0 b 1 b 2 b n 1 where the columns equal the basis vectors and it will always be an n×n matrix (although the above matrix does not appear to be square since we left terms in vector notation). We can then proceed to rewrite Equation 7 x=( b 0 b 1 b n 1 )( α 0 α n 1 )=Bα x b 0 b 1 b n 1 α 0 α n 1 B α and α=B-1x α B x

## Content actions

### Give feedback:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

#### Definition of a lens

##### Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

##### What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

##### Who can create a lens?

Any individual member, a community, or a respected organization.

##### What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks