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f ( t ) , f ˜ k ( t ) = δ ( - k )

Because this type of “orthogonality" requires two sets of vectors, the expansion set and the dual set, the system is called biorthogonal . Using [link] with the expansion in [link] gives

g ( t ) = k g ( t ) , f ˜ k ( t ) f k ( t )

Although a biorthogonal system is more complicated in that it requires, not only the original expansion set, but the finding, calculating, andstorage of a dual set of vectors, it is very general and allows a larger class of expansions. There may, however, be greater numerical problemswith a biorthogonal system if some of the basis vectors are strongly correlated.

The calculation of the expansion coefficients using an inner product in [link] is called the analysis part of the complete process, and the calculation of the signal from the coefficients and expansion vectorsin [link] is called the synthesis part.

In finite dimensions, analysis and synthesis operations are simply matrix–vector multiplications. If the expansion vectors in [link] are a basis, the synthesis matrix has these basis vectors as columns and the matrix is square and non singular. If the matrix is orthogonal, itsrows and columns are orthogonal, its inverse is its transpose, and the identity operator is simply the matrix multiplied by its transpose. If itis not orthogonal, then the identity is the matrix multiplied by its inverse and the dual basis consists of the rows of the inverse. If thematrix is singular, then its columns are not independent and, therefore, do not form a basis.

Matrix examples

Using a four dimensional space with matrices to illustrate the ideas of this chapter, the synthesis formula g ( t ) = k a k f k ( t ) becomes

g ( 0 ) g ( 1 ) g ( 2 ) g ( 3 ) = a 0 f 0 ( 0 ) f 0 ( 1 ) f 0 ( 2 ) f 0 ( 3 ) + a 1 f 1 ( 0 ) f 1 ( 1 ) f 1 ( 2 ) f 1 ( 3 ) + a 2 f 2 ( 0 ) f 2 ( 1 ) f 2 ( 2 ) f 2 ( 3 ) + a 3 f 3 ( 0 ) f 3 ( 1 ) f 3 ( 2 ) f 3 ( 3 )

which can be compactly written in matrix form as

g ( 0 ) g ( 1 ) g ( 2 ) g ( 3 ) = f 0 ( 0 ) f 1 ( 0 ) f 2 ( 0 ) f 3 ( 0 ) f 0 ( 1 ) f 1 ( 1 ) f 2 ( 1 ) f 3 ( 1 ) f 0 ( 2 ) f 1 ( 2 ) f 2 ( 2 ) f 3 ( 2 ) f 0 ( 3 ) f 1 ( 3 ) f 2 ( 3 ) f 3 ( 3 ) a 0 a 1 a 2 a 3

The synthesis or expansion [link] or [link] becomes

g = F a ,

with the left-hand column vector g being the signal vector, the matrix F formed with the basis vectors f k as columns, and the right-hand vector a containing the four expansion coefficients, a k .

The equation for calculating the k t h expansion coefficient in [link] is

a k = g ( t ) , f ˜ k ( t ) = f ˜ k T g

which can be written in vector form as

a 0 a 1 a 2 a 3 = f ˜ 0 ( 0 ) f ˜ 0 ( 1 ) f ˜ 0 ( 2 ) f ˜ 0 ( 3 ) f ˜ 1 ( 0 ) f ˜ 1 ( 1 ) f ˜ 1 ( 2 ) f ˜ 1 ( 3 ) f ˜ 2 ( 0 ) f ˜ 2 ( 1 ) f ˜ 2 ( 2 ) f ˜ 2 ( 3 ) f ˜ 3 ( 0 ) f ˜ 3 ( 1 ) f ˜ 3 ( 2 ) f ˜ 3 ( 3 ) g ( 0 ) g ( 1 ) g ( 2 ) g ( 3 )

where each a k is an inner product of the k t h row of F ˜ T with g and analysis or coefficient [link] or [link] becomes

a = F ˜ T g

which together are [link] or

g = F F ˜ T g .

Therefore,

F ˜ T = F - 1

is how the dual basis in [link] is found.

If the columns of F are orthogonal and normalized, then

F F T = I .

This means the basis and dual basis are the same, and [link] and [link] become

g = F F T g

and

F ˜ T = F T

which are both simpler and more numerically stable than [link] .

The discrete Fourier transform (DFT) is an interesting example of a finite dimensional Fourier transform with orthogonal basis vectors where matrixand vector techniques can be informative as to the DFT's characteristics and properties. That can be found developed in several signal processingbooks.

Fourier series example

The Fourier Series is an excellent example of an infinite dimensional composition (synthesis) and decomposition (analysis). The expansionformula for an even function g ( t ) over 0 < x < 2 π is

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Wavelets and wavelet transforms. OpenStax CNX. Aug 06, 2015 Download for free at https://legacy.cnx.org/content/col11454/1.6
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