Decimation Interpolation M4 Bandwidth 4 LSB 4 Figure

Decimation & Interpolation (M=4) Bandwidth - /4 LSB /4 Figure 12 USB 0 /4 /2 /4 M=4 M M

Decimation & Interpolation Still OK. Lets Try a sequence with wider Bandwidth

Decimation & Interpolation (M=4) Bandwidth > /4 Figure 13 LSB USB /4 0 /4 /2 /4 M M

Decimation & Interpolation Distortion (commonly known as “aliasing error”) is present if the bandwidth of the input sequence exceeds + /M When aliasing error is present, the signal cannot be recovered with interpolation Even if aliasing error is absent, there are multiple images in the interpolated spectrum

Decimation Filter To avoid aliasing error, a low pass filter is used to restricted the signal bandwidth to within + /M x(n) h(n) v(n) M y(n) (14 a) (14 b)

Action of Anti-aliasing Filter (M=4) Low Pass anti-aliasing Filter /4 0 /4 /2 /4 Figure 14 /2 /4 M

Interpolation Filter (M=4) x(n) M M v(n) g(n) y(n) -2 -1 0 1 2 Figure 15 a -4 -3 -2 -1 0 1 2 3 4

Interpolation Filter (M=4) x(n) M M v(n) g(n) -2 -1 0 1 2 y(n) The I-filter fills the missing gaps in the output sequence Figure 15 b -4 -3 -2 -1 0 1 2 3 4

Interpolation Band Pass Filter (M=4) Figure 16 /2 /4 0 /4 /2 /4 Interpolation Filter G(z) M

Action of Bandpass Filter (M=4) Band Pass Filter Figure 17 /2 /4 0 /4 /2 /4

Action of Bandpass Filter (M=4) Band Pass Filter /2 /4 0 /4 /2 /4 Figure 18 /4 /2 /4

Action of Bandpass Filter (M=4) Band Pass Filter /2 /4 0 /4 /2 /4 Figure 19 /4 /2 /4 M

Interpolation Band Pass Filter (M=4) /2 /4 0 /4 /2 /4 M Figure 20

What’s the use of learning all this?

The power of Multi-rate Signal Processing Real world signal usually only occupies part of the frequency spectrum Sometimes only certain part(s) of the spectrum is/are useful Decimation, interpolation and filtering forms the backbone for efficient extraction of useful signal spectrum

Information extraction with Milti-band filtering Band 0 yo(n) Band 1 y 1(n) Band 2 y 2(n) x(n) Band M-1 y. M-1(n) Figure 21

Applications of Bandpass filtering Subband Coding - Audio and Image data usually concentrates in narrow bands in the spectrum Number of data samples x(n)= [x(0), x(1), . . . , x(N-1)] N Decompose x(n) into M bands MN Downsample each band by M M(N/M) = N Select only the essential bands where information is concentrated K(N/M) < N (e. g. select K bands where K<M)

Applications of Bandpass filtering Subband Coding - A more practical scheme Number of data samples x(n)= [x(0), x(1), . . . , x(N-1)] N Decompose x(n) into M bands MN Downsample each band by M M(N/M) = N A coder to allocate more bits for (finer quantization) for more important bands and vice versa

Applications of Bandpass filtering Speech Recognition with Multi-band filtering Band 0 zero crossing detector Z 0 Band 1 zero crossing detector Z 1 Band 2 zero crossing detector Z 2 Band M-1 zero crossing detector ZM-1 x(n) Figure 22

Applications of Bandpass filtering Speech Recognition with Multi-band filtering Decompose x(n) into M bands M<<N Compute the number of zerocrossings in each band Each word is represented by the sequence of zero crossing Z = [z 1, z 2, . . . , z. M]

Applications of Bandpass filtering Speech Recognition with Multi-band filtering Word 1 - Z 1 Unknown Word x. U(n) Word 2 - Z 2 Word 3 - Z 3 ZU =[z 1, . . , z. M] BEST MATCH Word P MATCH Word K - ZK Figure 23

Subband Coding o(n) H 0(z) vo(n) M Coder 1(n) H 1(z) x(n) HP(z) P(n) f 1(n) y 1(n) G 1(z) M v. P(n) f. P(n) Coder M v. M-1(n) f. M-1(n) y. M-1(n) Coder GM-1(z) M M GP(z) y. P(n) M M-1(n) Analysis filter Coder yo(n) G 0(z) M v 1(n) M HM-1(z) fo(n) Synthesis filter + y(n) Figure 24

Subband Coding Forget the coder for the time being Can the input signal x(n) be fully recovered at the output?

Suband Decomposition and Reconstruction o(n) H 0(z) vo(n) H 1(z) x(n) HP(z) v 1(n) P(n) v. P(n) Analysis filter v. M-1(n) M y 1(n) G 1(z) f. P(n) M M M-1(n) HM-1(z) f 1(n) M M yo(n) G 0(z) M M 1(n) fo(n) GP(z) y. P(n) f. M-1(n) y. M-1(n) GM-1(z) M Synthesis filter + y(n) Figure 25

Subband Coding Forget the coder for the time being Can the input signal x(n) be fully recovered at the output? Answer: No WHY?

Non-perfect Reconstruction in Subband Coding Filters are never perfect, causing aliasing error Ideal Low Pass anti-aliasing Filter Figure 26 Non-ideal response /4 M 0 /4 /2 /4

Perfect Reconstruction (PR) in Subband Coding M = 2 o(n) H 0(z) vo(n) fo(n) G 0(z) 2 2 yo(n) y(n) x(n) 1(n) H 1(z) v 1(n) 2 f 1(n) 2 y 1(n) G 1(z) + Figure 27

o(n) H 0(z) vo(n) fo(n) G 0(z) 2 2 yo(n) y(n) x(n) 1(n) H 1(z) v 1(n) f 1(n) 2 2 y 1(n) G 1(z) + Figure 27 (15) Y(z) will not be equal to X(z) unless 1. T(z) = cz -K (c and K are constant, and 2. S(z) = 0 for all z

(15) (16) Eqn. (16) can be satisfied simply by selecting G 0(z) = -H 1(-z) and G 1(z) = H 0(-z) (17)

(15) G 0(z) = -H 1(-z) and G 1(z) = H 0(-z) (17) Eqn. (15) and (17) gives (18)

(15) G 0(z) = -H 1(-z) and G 1(z) = H 0(-z) (17) Eqn. (15) and (17) gives (18) Let H 1(z) = z-(N-1)H 0(-z-1) and N is even, (19)

(19) For perfect reconstruction, Q(z) = constant The term H 0(z)H 0(z-1) is the autocorrelation function, i. e. , (20) It can be easily proved that (21) * Prove (20) and (21) as an exercise

For Perfect Reconstruction, R(z)+R(-z) = constant (22) if R(z)+R(-z) = 1 the PR filter is known as a ‘paraunitary’ solution, i. e. (23)

Paraunitary Perfect Reconstruction Filter for M = 2 o(n) H 0(z) vo(n) fo(n) G 0(z) 2 2 yo(n) y(n) x(n) 1(n) H 1(z) 1. G 0(z) = -H 1(-z) v 1(n) 2 and f 1(n) 2 y 1(n) + G 1(z) = H 0(-z) 2. H 1(z) = z-(N-1)H 0(-z-1) 3. R(z) = H 0(z) H 0(z-1) 4. R(z) + R(-z) = 1 (24)

Multiband Paraunitary Perfect Reconstruction Filter The 2 band PR structure can be extended to multiband H 0(z) H 1(z) x[n] H 0(z) H 1(z) G 0(z) G 1(z) y[n] G 0(z) G 1(z)

Paraunitary PR Filter - a physical view 1. G 0(z) = -H 1(-z) and G 1(z) = H 0(-z) 2. H 1(z) = z-(N-1)H 0(-z-1) 3. R(z) = H 0(z) H 0(z-1) 4. R(z) + R(-z) = 1 Let H(z) = Ho(z)

Paraunitary PR Filter - a physical view Note: all odd terms are zero except for n=0, hence By definition,

Paraunitary PR Filter - a physical view 1. G 0(z) = -H 1(-z) and G 1(z) = H 0(-z) 2. H 1(z) = z-(N-1)H 0(-z-1) 3. R(z) = H 0(z) H 0(z-1) 4. R(z) + R(-z) = 1 Let H(z) = Ho(z) (25) Filters satisfying these requirement: Power Complementary filters

Simple Power Complementary Filter 1. G 0(z) = -H 1(-z) and G 1(z) = H 0(-z) 2. H 1(z) = z-(N-1)H 0(-z-1) 3. R(z) = H 0(z) H 0(z-1) 4. R(z) + R(-z) = 1 Let H(z) = Ho(z) (27)

Simple Power Complementary Filter (28) However, H(z) = Ho(z) and H 1(z) = z-(N-1)Ho(-z-1) (29)

A physical interpretation 1. G 0(z) = -H 1(-z) and G 1(z) = H 0(-z) 2. H 1(z) = z-(N-1)H 0(-z-1) 3. R(z) = H 0(z) H 0(z-1) 4. R(z) + R(-z) = 1 Ho(ej ) - Low pass filter H 1(ej ) - High pass filter H 1(ej ) Go(ej ) - LPF which passes Ho(ej ) and reject H 1(ej ) /2 G 1(ej ) - Opposite of G 0(ej )

Equivalent Structure o(n) H 0(z) vo(n) fo(n) G 0(z) 2 2 yo(n) y(n) x(n) 1(n) H 1(z) v 1(n) f 1(n) 2 2 y 1(n) + G 1(z) 2 channel PR-QMF subband filter c(n) b(n) a(n) M H (z) d(n) a(n) H (z. M) c(n) M

Equivalent Structure c(n) b(n) a(n) M d(n) a(n) H (z. M) c(n) M and (30 a)

Equivalent Structure c(n) b(n) a(n) M ? d(n) a(n) H (z. M) c(n) M All pass filter Decimation discards samples from a(n) b(n) different from a(n) and d(n) If b(n) different from d(n), how can they give the same result c(n)? Answer: a matter of perspective Lets study from the other point of view

Equivalent Structure d(n) a(n) H (z. M) c(n) M c(n) b(n) a(n) M H (z) All pass filter d(n) = a(n) c(n) is the decimation of a(n) and d(n) We look at the alternative representation c(n) = b(n) Hence c(n) is the decimation of a(n) Same

Equivalent Structure d(n) a(n) H (z) c(n) M c(n) b(n) a(n) M H (z. M) (30 b)

Polyphase Decomposition (30 c)

Polyphase Decomposition (30 c) Decimating H(z) by M times, associated with different delay terms.

Polyphase Decomposition b(n) a(n) H (z) c(n) b(n) a(n) H 0 (z. M) M z-1 H 1 (z. M) z-1 H 2 (z. M) z-1 Decimator HM-1(z. M) c(n) M

Polyphase Decomposition b(n) a(n) H 0 (z. M) c(n) a(n) M z-1 M H 0 (z) M H 1 (z) M H 2 (z) M HM-1(z) z-1 H 1 (z. M) z-1 H 2 (z. M) Decimator z-1 HM-1 (z. M) z-1 According to equation (30 a) c(n)

Polyphase Decomposition - a simplification in filter design H(ej ) /M 0 , where /M

Polyphase Decomposition - a simplification in filter design H(ej ) /M 0 /M

Polyphase Decomposition H(ej ) /M 0 /M which is just a series of all-pass linear phase filter

All Pass Filter An approximation of linear phase APF

Uniform Bank Filter Applying Polyphase Decomposition

Uniform Bank Filter

Uniform Bank Filter DFT Matrix Polyphase decomposition of H 0(z)

Polyphase Decomposition b(n) a(n) M c(n) a(n) G(z) b(n) M c(n) G 0 (z. M) z-1 G 1 (z. M) z-1 G 2 (z. M) z-1 Interpolator GM-1(z. M)

Polyphase Decomposition b(n) a(n) M c(n) a(n) G(z) b(n) M c(n) G 0 (z. M) z-1 G 1 (z. M) z-1 G 2 (z. M) GM-1 Interpolator (z. M) z-1

Polyphase Decomposition a(n) b(n) M c(n) a(n) G 0 (z. M) G 0 (z) M z-1 G 1 (z. M) z-1 G 1 (z) M z-1 G 2 (z. M) GM-1 (z. M) z-1 G 2 (z) z-1 GM-1(z) Interpolator According to equation (30 b) M M z-1

Polyphase Decomposition of 2 channels QMF banks o(n) H 0(z) vo(n) fo(n) G 0(z) 2 2 yo(n) y(n) x(n) 1(n) H 1(z) v 1(n) 2 f 1(n) 2 y 1(n) G 1(z) +

Polyphase Decomposition of 2 channels QMF banks For PR condition, This can be , satisfied by choosing so that

Polyphase Decomposition of 2 channels QMF banks

Polyphase Decomposition of 2 channels QMF banks o(n) H 0(z) vo(n) 2 fo(n) 2 yo(n) H 0(z) x(n) 1(n) H 0(-z) 2 v 1(n) f 1(n) y 1(n) -H 0(-z) 2 y(n) +

Polyphase Decomposition of 2 channels QMF banks o(n) H 0(z) vo(n) 2 fo(n) 2 yo(n) H 0(z) x(n) 1(n) H 0(-z) 2 v 1(n) Polyphase decomposition: f 1(n) y 1(n) -H 0(-z) 2 y(n) +

Polyphase Decomposition of 2 channels QMF banks o(n) H 0(z) 2 vo(n) fo(n) yo(n) G 0(z) 2 y(n) x(n) 1(n) H 1(z) Consider H 0(z) 2 v 1(n) f 1(n) 2 y 1(n) G 1(z) +

Polyphase Decomposition of 2 channels QMF banks x(n) Equivalent to x(n) H 0(z) E 0(z 2) 2 2 vo(n) z-1 E 1(z 2) Equivalent to x(n) 2 E 0(z) 2 E 1(z) z-1 vo(n)

Polyphase Decomposition of 2 channels QMF banks x(n) Equivalent to x(n) H 1(z) 2 E 0(z 2) z-1 E 1(z 2) Equivalent to v 1(n) x(n) 2 -1 2 E 0(z) z-1 2 v 1(n) E 1(z) v 1(n) -1