Signal Analysis. Filter Banks and. One application for filter banks is to decompose the input signal into different bands or channels

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1 Filter banks Multi dimensional Signal Analysis A common type of processing unit for discrete signals is a filter bank, where some input signal is filtered by n filters, producing n channels Channel 1 Lecture 2G Filter Banks and Multi resolution Analysis (I) Input signal Channel 2 This an example of how to create n channels Signals and filters are discrete! Channel n 2 Filter banks In order to have the same amount of data (on average per time unit) before and after the filtering: down sample the filter outputs by a factor n Keep every n th sample Filter banks, applications One application for filter banks is to decompose the input signal into different bands or channels Here band is rather imprecise and can mean frequency band but also more general subspaces 3 4

2 Filter banks, applications Filter banks, applications Each hband dis processed independently d relative lti to the others Filtering, e.g., LP filter to reducenoise Data compression, to reduce data Numerical resolution, to reducedatadata Even: removal, to reduce data Each band is then stored or transmitted From each band we want to be able to reconstruct the input signal again Possibly only to some degree of approximation Or perfect reconstruction: output = input (+ shift!) In general, something like this: Input signal Possibly less data than the input signal Output tsignal Input signal 5 6 The 2 channel filter bank The simplest type of filter bank is the 2 channel filter bank Down sample = Remove every second sample Analysing part Same amount of samples as the input signal s Reconstructing part Only specific choices of filters can make s = s Up sample = Insert 0 between every sample 7 The 2 channel filter bank Let the DFTs of h 0, h 1, g 0, g 1 be denoted as H 0, H 1, G 0, G 1 (they are 2π periodic!) We can show that necessary and sufficient conditions for s = s are given by H 1 (u)h 0 (u)+g 1 (u)g 0 (u) =2 H 1 (u)h 0 (u + π)+g 1 (u)g 0 (u + π) =0 FB1: No distortion FB2: No aliasing from the sub sampling (why?) (FB1) (FB2) Ex

3 A practical note Assuming that all four filters are causal: The filters introduce delays in the output signal s This means that, at best, we can accomplish s [k] = s[k d] for some positive integer d We will ignore this practical matter for now and consider the requirement s [k] = s[k] As a consequence, some filters may become non causal Can be fixed by time shift if they also are FIR Output will then be delayed accordingly We want the four filters to be FIR Finding thefour filters Bad idea: choose two of the filters arbitrarily and solve the other two from FB1 and FB2 (why?) Better idea: Apply some design principles for the four filters that produce useful filters Many possibilities exists! 9 10 A particular choice of filters FB2 is automatically satisfied if H 0 (u) = H 1 (u) G iu 1 (u) = e H 1 (u + π) G 0 (u) = e iu H 1 (u + π) = G 1 (u) FB1 then becomes H 1 (u) 2 + H 1 (u + π) 2 =2 (O) Called: Conjugate Mirror filters Ex Alternating flip The conjugate mirror filter condition can also be formulated in the time domain as h 0 [k] = h 1 [ k] g ( 1) 1 k 1 [k] = h 1 [1 k] Called: Alternating flip g = ( 1) k+1 0 [k] h 1 [k + 1] = g 1 [ k] Ex

4 An orthogonal set We can show that the condition (O) implies that hh 1 [ ] h 1 [ 2 m]i = δ[m] h 1 is always orthogonal to itself when shifted by multiples l of 2 (unless m = 0) Ex Convolution and linear combinations We have already seen that the convolution operation can be interpreted in terms of producing linear combinations Here: u 2 [k] = X a[n] h 1 [k 2 n] n= The signal u 2 is a linear combination of shifted versions of h 1 Shift by multiples of 2 We just saw that they form an ON set! 14 More orthogonality In the same way we can show that the conjugate mirror filter bank leads to hg 1 [ ] g 1 [ 2 m]i = δ[m] hg 1 [ ] h 1 1[ [ 2 m]i =0 Ex (why?) 15 More orthogonality This means that also v 2 [k] is formed as a linear combination of orthonormal discrete functions g 1 [k 2 m] These are functions of k, one for each m Finally, it also means that the two orthogonal sets h 1 [k 2 m] and g 1 [k 2 m] are mutually orthogonal These are functions of k, one for each m 16

5 Reconstruction The output signal from the filter bank, s = s, is formed as the linear combination of two orthogonal sets of functions (an ON basis) Means: thecoefficients in this linear combination are the coordinates relative the reconstructing basis functions. These are a[m] for the basis functions h 1 [k 2m] 1 d[m] for the basis functions g 1 [k 2m] Analysis In the filter bank, a[m] is formed as a[m] =(s h 0 )[2 m] = X = k= s[k] h 0 [2 m k] X X s[k] [ ] h1 [k 2 m] ] = hs[ [ ] h 1 [ 2 m]i k= Analysis Similarly: d[m] = h s[ ] g 1 [ 2 m] i This result is expected! We are analysing s[k] [ ] using an ON basis The corresponding coordinates are then given as the scalar products between s[k] [ ] and the basis functions They are their own duals! 19 Orthogonal filter bank In summary: we choose H 1 such that 2 2 H 1 (u) 2 + H 1 (u + π) 2 =2 and then the other three filters such that H 0 (u) = H 1 (u) G 1 (u) =e iu H 1 (u + π) G 0 (u) =e iu H 1 (u + π) =G 1 (u) This produces an orthogonal filter bank 20

6 Orthogonal filter bank If it is possible to choose h 1 as a FIR filter, then h 0, g 1, g 0 also become FIR 0 g 1 g 0 Is this possible? Yes: h 1 = 2 1/2 (1, 1) is the simplest possible choice g 1 = 2 1/2 ( 1, 1) 1/2 Called: h 0 = 2 (1, 1) Haar filter bank also exist! g 0 = 2 1/2 (1, 1) Other choices Splitting ofthesignal space The signal u 2 is constructed as a linear combination of 2 shifts of h 1 Span some subspace V 0 of the signal space V The signal v 2 is constructed as a linear combination of 2 shifts of g 1 Span some subspace W 0 of the signal space V The output signal s V is u 2 + v 2 V = V 0 W 0, V 0 W Filter banks The conjugate mirror filter bank is only one special solution to FB1 and FB2! There are other ways of obtaining s = s Still leads to perfect reconstruction ti Some approaches do not lead to an orthogonal filter bank Can be useful anyway! Scaling function Consider a function φ(t) ) that satisfies ifi 1. φ(t k), k Z, is an orthogonal basis of some function space V 0 2. φ(t) can be written as a linear combination of the functions φ(2t k), k Z Such a function φ is called a scaling function These relatively simple and innocent assumptions lead to a framework for defining a discrete wavelet transform in terms of orthogonal filter banks 23 24

7 Example The simplest example of a scaling function is φ(t) ) = ( 1 0 t<1 0 otherwise Example In this case: V 0 is the space of piece wise constant functions on integer intervals The functions φ(2t k) look like k = 0 k = 1 φ(t) = φ(2 t) + φ(2 t 1) k = Consequences The fact that φ(t k) is an orthogonal basis for k Z means Z hφ(t) φ(t k)i = φ(t) φ(t k) dt = δ[k], k Z From this follows that also 2 1/2 φ(2t k), k Z, form an orthogonal basis Ex 22.1 Spans a function space V 1 V 0 Ex Consequences (II) Property 2. of the scaling functions leads to φ(t) = X hk 2 1/2 φ(2 t k) k= for some sequence of scalars h k Since we have an ON basis: Coordinates of φ relative to the ON basis 2 1/2 φ(2x k), k Z Z h k = hφ(t) 2 1/2 φ(2 t k)i =2 1/2 φ(t) ) φ(2 t k) ) dt 28

8 Consequences (III) In fact, any φ(x ( l), ) l Z, can be written as a linear combination of the basis 2 1/2 φ(2 t k): φ(t l) = X m= h m 2l 2 1/2 φ(2 t m) Ex Consequences (IV) Notice that: t The functions φ(t k) are shifted by multiples of 1 relative to t Spans a space denoted V 0 The functions 2 1/2 φ(2t k) are shifted by multiples of 1/2 relative to t Functions in V Spans a space denoted V 1 can 1 contain finer details than V 0 Both thbases are orthonormal within each space V 0 V 1 Since V 1 has a basis that spans every basis vector in V 0 30 Consequences (V) Returning to property 1. we have seen that hφ(t) φ(t k)i = δ[k] This can also be formulated as X Φ(v +2πk) 2 =1 Ex 22.4 k where Φ is the Fourier transform of φ 31 The sequence h The sequence of scalars h[k], k Z, depends d only on the choice of scaling function φ Remember that φ itself has particular properties, p described by 1. and 2. Also h[k] has particular properties! h[k] can be interpreted as the coefficients of a discrete filter with Fourier transform X X H(u) = hk e iuk This is a 2π periodic k= function 32

9 The function H From 1. and 2. follows that H must be related to Φ as ³ u ³ u Φ(u) = 2 1/2 H Φ 2 2 Ex 22.5 The function H The fact that φ has special properties and that H is related to φ leads to 2 2 H(u) 2 + H(u + π) 2 =2 Ex 22.6 or H(u) =2 1/2 Φ(u) This must be a 2π periodic Φ(2 u) function! We recognise this as condition (O) on the filter h 1 in a conjugate mirror filter bank! The function ψ(x) Define a new function ψ(t), with a Fourier transform Ψ given by Where or 1 ³ u ³ u Ψ(u) = G Φ G(u) =e iu H(u + π) g[n] ] = ( 1) 1 n h[1 [ n] ] This is the alternating fp flip condition in a conjugate mirror filter bank The sequence g The sequence g is derived from h But, h depends only on φ: g depends only on φ Also ψ depends only on the choice of φ Finally, we can derive 1/2 hψ(t) 2 1/2 φ(2t n)i = g[n] Ex

10 The sequences h and g We summarise: h[k] = hφ(t) 2 1/2 φ(2 t k)ii Coordinates of φ relative basis in V 1 g[k] = hψ(t) 2 1/2 φ(2 t k)i Coordinates of ψ or, more generally: hφ(t l) 2 1/2 φ(2 t k)i = h[k 2 l] hψ(t l) 2 1/2 φ(2 t k)i = g[k 2 l] relative basis in V 1 At least if we can prove that ψ V 1 We can show that: ψ(t l) = X k Properties of ψ, (I) ( 1) k h k 2 1/2 φ(2t k +2l +1) which means that all ψ(t l) V 1, l Z Ex 22.8 g n are the coordinates of ψ in the ON basis of V Example For the simple example of a scaling function ( 1 0 t < 1 φ(t) = 0 otherwise we get g[k] = {2 1/2, 2 1/2 } and the corresponding function ψ(t): Called Haar wavelet Properties of ψ, (II) Using the previous relations, we can also prove that Ex 22.9 hψ(t m) φ(t l)i =0 forl,, m Z which means that all ψ(t l) ) are orthogonal to the space V 0 since φ(t m) is a basis for this space ψ(t m) V 1 and ψ(t m) V

11 Properties of ψ, (III) We can also show hat hψ(t n) ψ(t)i = δ[n] which means that the set ψ(t n) is orthonormal for n Z In summary: ll ψ( ) All ψ(t n) are to V 0 They form an orthonormal set Ex (why?) Properties of V 1 Let f be an arbitrary function in V 1. It can then be written as f(t) = X s[n] 2 1/2 φ(2t n) n for some sequence s[n] since 2 1/2 φ(2t n) is an ON basis for V Properties of V 1 Using the previous results, we can show that it is also possible to write this f V 1 as p f 1 f(t) = X a[n] φ(t n) + X d[n] ψ(t n) n n Ex for some sequences a[k] and d[k] This means that φ(t n) and ψ(t n) together span V 1 The difference space W 0 V 0 is a subspace of V 1 Any f V 1 can be written as a unique linear combination of a component in V 0 and a component in V 1, and this component is orthogonal to V 0 There is a difference space W 0 characterised by o W 0 V 1 V 0 o W 0 V 0 o V 1 = V 0 W 0 V 1 o For any f V 1 : f = f 0 + w 0, f 0 V 0, w 0 W 0 W

12 The difference space W 0 V 1 has an ON basis 2 1/2 φ(2t k), k Z V 0 has an ON basis φ(t k), k Z W 0 has an ON basis ψ(t k), k Z W 0 V 0 V 1 = V 0 W 0 Coordinates of f V 1 are given by s[k] [ ] Coordinates of f V 0 W 0 given by a[k] and d[k] Sequences a, b and c Since f V 1 is completely represented either by sequence s[k], [ ] or by sequences a[k] [ ] and d[k] [ ] : There must be a (linear) mapping s (a, d) There must be a (linear) mapping (a, d) s A linear mapping a b & c A linear mapping b & c a From previous results follows: From previous results also follows: Ex Ex a[k] = (s[ ] h[ ])[2 k] d[k] =(s[ ] g[ ])[2 k] Convolve and skip every second sample (the odd ones) s[k] = X X n= (a[n] h[k 2 n] + d[n] g[k 2 n]) Insert zeros between every sample and convolve 47 48

13 A 2 channel filter bank We can summarise these results as Coordinates of f relative to basis in V 1 Coordinates of f relative to basis in V 0 Coordinates of f relative to basis in W 0 An orthogonal 2 channel filter bank with h 0 = h g 0 = g (as alternating flip of h 0 ) What you should know The anatomy of the 2 channel lfilter bank The 2 constraints on the 4 filters to produce s =s The conjugate mirror filter (CMF) bank Orthogonal 2 channel filter bank How a CMF decomposes the signal space into V 0 and W 0, and how they are related Definition of a scaling function φ Leads to a unique function ψ How φ and ψ decompose the signal space into V 0 and W 0, and how they are related 49 50