Analog LTI system Digital LTI system

Size: px

Start display at page:

Download "Analog LTI system Digital LTI system"

Patricia West
5 years ago
Views:

1 Sampling Decimation Seismometer Amplifier AAA filter DAA filter Analog LTI system Digital LTI system

2 Filtering (Digital Systems) input output filter xn [ ] X ~ [ k] Convolution of Sequences hn [ ] yn [ ] = h[n] *x[n] Discrete Fourier Transform (DFT) T ~ [ k] Y ~ [ k] = T ~ [ k]x ~ [ k] Xz [ ] z-transform T[ z] Y[ z] = T[ z]xz [ ]

3 The z-transform Z{ x[ n] } = xn [ ]z n = n = Xz ( ) Properties x 1 [ n]*x 2 [ n] = x 1 [ m]x 2 [ n m] X 1 ( z) X 2 ( z) x[ n n 0 ] z n m = (convolution theorem) 0 Xz ( ) (shifting theorem) x[ n] X( 1 z) Tz ( ) Transfer Function Z{ y[ n] } = = Z{ x[ n] } Y( z) Xz ( )

4 Rational Transfer Function Linear Difference Equation Tz ( ) Yz ( ) = = X( z) M l = 0 N b l z l a k z k k = 0 N a k yn k k = 0 M [ ] = b l xn [ l] l = 0

5 Recursive/Non-Recursive Filters N M a yn [ k] = k b l x[ n l] k = 0 l = 0 N M a 0 y[ n] + a k y[ n k] = b l xn [ l] k = 1 l = 0 y[ n] = N a M k b l a y[ n k] x[ n l] a 0 k = 1 l = 0 recursive non-recursive ( IIR Filters ) FIR Filters ( ) a 0 = 1 and a k = 0 for k 1 M y[ n] = b x[ n l] l l = 0

6 Rational Transfer Function Tz ( ) Y( z) = = X( z) M l = 0 N b l z l a k z k k = 0 Special Case: FIR filter ( a 0 = 1and a k = 0 for k 1) M Tz ( ) = b l z l = b 0 ( 1 c l z 1 ) l = 0 M l = 1 = b 0 ( z c l ) z M M l = 1 Linear Difference Equation M yn [ ] b l xn [ l] = = l = 0 x[ n]*b[l]

7 FIR filters : + Always stable. - Steep filters need many coefficients. + Both causal and noncausal filters can be implemented. + Filters with given specifications are easy to implement! IIR filters : - Potentially unstable and subject to quantization errors. + Steep filters can easily be implemented with a few coefficients. Speed. - Filters with given specifications are in general, difficult, if not impossible, to implement exactly(!).

9 -t1 -t2

10 How can we remove these effects? Recursive/Non-Recursive Filters N M a yn [ k] = k b l x[ n l] k = 0 l = 0 N M a 0 y[ n] + a k y[ n k] = b l xn [ l] k = 1 l = 0 Back to Transfer function: y[ n] = N a M k b l a y[ n k] x[ n l] a 0 k = 1 l = 0 recursive non-recursive ( IIR Filters ) FIR Filters ( ) a 0 = 1 and a k = 0 for k 1 M y[ n] = b x[ n l] l l = 0

12 Frequency

13 flip time axis Waveform Properties and Root Positions minimum delay/phase maximum delay/phase mixed delay/phase (zero phase) mixed delay/phase mixed delay/phase

14 Zero Phase FIR Filter Problem: Two-Sided IR Cure: Change IR into Minimum Phase Methods: 1) Add phase of Minimum Phase Filter to trace spectrum 2) Recursive Filtering of time inverted trace

15 Removing the acausal response of a Zero Phase FIR Filter Linear Phase and Zero Phase Filter: 0 F( z) time delay lp X( z) Y( z) Linear Phase FIR filter lp 0 X( z) F( z) Linear Phase FIR filter z lp linear phase shift correction Yz ( ) Linear-phase FIR Filter: Zero-phase FIR Filter: Fz ( ) Fz ( ) = Fz ( ) z lp z lp = Time delay correction by lp samples

16 General rule: Any filter can be expressed by convolution of its minimum phase and maximum phase component. roots within UC roots outside UC => Linear phase FIR Filter: F z ( ) = F max ( z) F min ( z) maximum phase component -> left-sided (acausal) component minimum phase component -> right-sided (causal) component

17 Removal of acausal response: Replace maximum phase component F max (z) by its minimum phase equivalent MinPhase{F max (z)}. MinPhase { F max ( z) } = minimum phase maximum phase Answer: flip maximum phase component in time. In terms of z-transform: => MinPhase F max ( z) { } = F max ( 1 z)

18 z-transform representation of digital seismogram LP FIR LP correction Y ~ ( z) = F( z) z lp X ~ ( z) digital seismogram after FIR filtering but before decimation 0P FIR digital seismogram before FIR filtering x ~ [ n] = unfiltered digital seismogram ~ X( z) = z-transform of the input signal x ~ [ n] F( z) = z-transform of the linear phase FIR filter y ~ [ n] = filtered digital seismic trace before decimation 1 ( ) = z-transform of y ~ n Y ~ z [ ] 1 This is a fictitious signal since decimation is commonly done while filtering.

19 F( z) z lp corresponds to a zero phase filter in which the linear phase component of F(z) is corrected. In practice: Treat time shift z lp separately from F(z)

20 Removing the maximum phase component of a FIR filter Principle: F max ( z) -> MinPhase{ F max ( z) } Corrected seismogram Y(z): Yz ( ) = F F max ( z) max ( 1 z) Y ~ ( z) Problem: Since F max (z) has only zeros outside the unit circle, 1/ F max (z) will have poles outside the unit circle.

21 Solution: Flip time axis. Y( 1 z) = F max ( 1 z) F ( z ) Y ~ ( 1 z) max Impulse response corresponding to 1/ F max (z) becomes a stable causal sequence in nominal time and the deconvolution of the maximum phase component F max (z) poses no stability problems.

22 The difference equation F ( max 1 z) Y ( 1 z) = F ( max z) Y ~ ( 1 z ) Rewrite to A' ( z) Y' ( z) = B' ( z) X' ( z) A' ( z) <=> F max ( 1 z) Y' ( z) <=> Y( 1 z) B' ( z) <=> F max ( z) X' ( z) Y ~ ( 1 z) <=> Written as convolution sum: k = a' [ k] y' [ i k] = l = b' [] l x' [ i l]

23 Assumption: F(z) contains mx zeros outside the unit circle => wavelets a [k] and b [k] will be of length mx + 1. mx k = 0 Rearrange to a' [ k] y' [ i k] = mx l = 0 b' [] l x' [ i l] y' [] i a' [ 0] mx + a' [ k] y' [ i k] = b' [] l x' [ i l] k = 1 mx l = 0 which is equivalent to y' [] i = mx k = 1 a' [ k] y' [ i k] a' [ 0] + mx l = 0 b' [] l x' [ i l] a' [ 0]

24 This is ak [ ] a' [ k] = = a' [ 0] f max [ mx k] f max [ mx] for k = 1 to mx and bl [] b' [] l = = a' [ 0] f max [ l] f max [ mx] for l = 0 to mx The convolution sum mx y' [] i = ak [ ] y' [ i k] k = 1 + mx l = 0 b[] l x' [ i l] y' []= i the time reversed corrected sequence. To obtain y[i] flip y [i] back in time.

25 Corrected seismogram: Y( z) = F ~ ( z) Y ~ ( z) 0 lp 0 lp Y ~ F ~ ( z) ( z) Yz ( ) Correction filter F ~ ( z) = changes the linear phase filter F(z) into a minimum phase filter => onset of output signal is advanced by lp samples.

26 To account for this time shift in the corrected seismogram: a) change time tag of corrected trace or b) delay corrected trace by lp samples.

27 What is needed for correction? mx + 1 coefficients of the maximum phase portion of the a linear phase FIR filter How to get maximum phase component? ( ) = b l z l = b 0 ( 1 c l z l ) F z m l = 0 m l = 1 mn zeros inside UC (c i min ) <=> F min (z) mx zeros outside UC (c i max ) <=> F max (z) F max z mx max ( ) f i z i max = = b 0 ( 1 c i z i ) i = 0 mx i = 1 From zeros ouside UC -> polynomial F max (z). f max [l]for l = 0 to mx: coefficients of polynomial F max (z).

30 .... initial sampling frequency F s Correction Procedure Decimation stages intermediate data streams final sampling frequency f n yn [ ] digital signal Stage 1 F s :f 1 f 1 f Stage 2 2 f 1 :f 2... decimation while filtering Stage n f n 1 :f n xn [ ] decimated digital signal Full Correction (not always necessary): interpolation -> f n 1 -> correction for FIR filter stage n interpolation -> f n 2 -> correction for FIR filter stage n-1 interpolation -> f 1 -> correction for FIR filter stage 2 interpolation -> F s -> correction for FIR filter stage 1 finally: decimation back to f n

38 a) b)

39 a) b)

41 Conclusions FIR filter generated precursory artefacts: can become impossible to be identified visually can have similar scaling properties as nucleation phases Zero - phase FIR filters in general affect the determination of all onset properties (onset times, onset polar ities) For the interpretation of onset properties (onset times, onset po larities, nucleation phases, etc.) the acausal response of the zero-phase FIR filter has to be removed but not for waveform analysis. Consequence

Responses of Digital Filters Chapter Intended Learning Outcomes:

Responses of Digital Filters Chapter Intended Learning Outcomes: (i) Understanding the relationships between impulse response, frequency response, difference equation and transfer function in characterizing