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Digital Filters IIR & FIR

© 2019 School of Information Technology and Electrical Engineering at The University of Queensland

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Lecture Schedule: Week Date Lecture Title 27-Feb Introduction 1 1-Mar Systems Overview 6-Mar Systems as Maps & Signals as Vectors 2 8-Mar Systems: Linear Differential Systems 13-Mar Sampling Theory & Data Acquisition 3 15-Mar Aliasing & Antialiasing 20-Mar Discrete Time Analysis & Z-Transform 4 22-Mar Second Order LTID (& Convolution Review) 27-Mar Frequency Response 5 29-Mar Filter Analysis 6 3-Apr Digital Filters (IIR) & Filter Analysis 5-Apr PS 1: Q & A 10-Apr (FIR) & Digital Windows 7 12-Apr FFT 8 17-Apr Active Filters & Estimation & Holiday 19-Apr 24-Apr Holiday 26-Apr 1-May Introduction to Feedback Control 9 3-May Servoregulation/PID 8-May PID & State-Space 10 10-May State-Space Control 15-May Digital Control Design 11 17-May Stability 22-May State Space Control System Design 12 24-May Shaping the Dynamic Response 29-May System Identification & Information Theory 13 31-May Summary and Course Review

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1 Follow Along Reading:

Today B. P. Lathi • Chapter 10 and linear systems (Discrete-Time System Analysis 1998 Using the z-Transform) TK5102.9.L38 1998 – § 10.3 Properties of DTFT – § 10.5 Discrete-Time Linear System analysis by DTFT – § 10.7 Generalization of DTFT to the 풵 –Transform • Chapter 12 (Frequency Response and Digital Filters) • § 12.1 Frequency Response of Discrete-Time Systems • § 12.3 Digital Filters • § 12.4 Criteria • § 12.7 Nonrecursive Filters

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ELEC3004 is 흅풊 −풆

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2 Periodic Signals: Writing them in the Fourier Domain & z-Domain • Synthesis:

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ퟏ Euler’s Identity: 풔풊풏 흎풕 = 풆풊⋅ 흎풕 − 풆−풊⋅ 흎풕 ퟐ풊 • The Discrete-Time Fourier Transform of a sinusoid

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3

Recap: IIR Filters = “Analog Filters” in Digital Form 

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Filter Specification in the Where:

1 = (dB) |H(ω)| 2 = stopband attenuation (dB) ωp = passband edge (Hz) ωst = stopband edge (Hz) 1 ωc = cutoff frequency (@ 3dB)

N: filter type/order to meet

specification

2

ω ωp ωc ωst

Passband Transition Stopband

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4 Butterworth Filters

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Butterworth Filters • Butterworth: Smooth in the pass-band • The amplitude response |H(jω)| of an nth order Butterworth low pass filter is given by:

• The normalized case (ωc=1)

Recall that:

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5 Analog Filter Summary 

Passband Stopband Transition MATLAB Design Filter Type Ripple Ripple Band Command Butterworth No No Loose butter Chebyshev Yes No Tight cheby Chebyshev Type II No Yes Tight cheby2 (Inverse Chebyshev) Eliptic Yes Yes Tightest ellip

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IIR Filter Design Methods

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6 IIR Filter Design Methods • Normally based on analogue prototypes – Butterworth, Chebyshev, Chebyshev 2, Elliptic, etc.

• Then transform 퐻 푠 → 퐻(푧)

• Three popular methods: 1. Impulse invariant – 퐻(푧): whose impulse response is a sampled version of ℎ(푡) (also step invariant) 2. Matched z–Transform – poles/zeros 퐻(푠) directly mapped to poles/zeros 퐻(푧) 3. Bilinear z – transform – left hand s – plane mapped to unit circle in z – plane

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Impulse Invariant Simplest approach, proceeds as follows:

1. Select prototype analogue filter 2. Determine H(s) for desired 휔 푐 and 휔 푠 푐푢푡표푓푓 푓푟푒푞. 푠푡표푝 푓푟푒푞. 3. Inverse Laplace • Calculate impulse response, ℎ(푡)

4. Sample impulse response ℎ 푡 푡=푛Δ푡푑 • ℎ[푛] = Δ푡푑 ℎ(푛Δ푡푑) 5. Take z-Transform of ℎ 푛 ⇒ 퐻(푧) – Poles: 푝푖 maps to exp (푝푖Δ푡푑) – Zeros: have no simple mapping 

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7 Impulse Invariant [2] Useful approach when: • Impulse (or step) invariance is required – e.g., control applications • Designing Lowpass or Bandpass filters

Has problems when: • 퐻 휔 does not  0 as 휔 → ∞ • Ex: highpass or bandstop filters • If H(w) is not bandlimited, aliasing occurs!

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Matched z - transform • Maps poles/zeros in s – plane directly –to poles/zeros in z – plane

• No great virtues/problems

• Fairly old method –not commonly used –so we won’t consider it further 

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8 Bilinear z - transform • Maps complete imaginary s –plane () – to unit circle in z -plane

• That is: map analogue frequency 휔푎 to discrete frequency 휔푑

• Uses continuous transform:

2 d t a  tan  t 2

흎 This compresses (warps) 흎 to have finite extent  풔 풂 ퟐ  this removes possibility of any aliasing 

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tan transform maps ωa to ωd ω t/2 Analogue ωa d Filter

|H(ωa)| 0 - 0  2 3 4 ωdt/2

Spectral compression |H(ωd)| due to the bilinear z -transform

Digital Filter

-ωs/2 0 ωs/2 ωs 3ωs/2 2ωs ωd Note,ELEC H(3004:ω Systemsd) periodic, due to sampling 3 April 2019 18

9

The bilinear transform

Transforming to s-domain

Remember: s = ja and tan = sin/cos

Where  = dt/2

Using Euler’s relation This becomes… (note: j terms cancel)

Multiply by exp(-j)/exp(-j)

As z = exp(sdt) = exp(jdt)

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Bilinear Transform • Convert H(s)  H(z) by • However, this substituting, transformation compresses the analogue frequency response, which means 21 z 1  – digital cut off frequency s  will be lower than the t1 z 1  analogue prototype

Note: this comes directly from tan transform • Therefore, analogue filter must be “pre-warped” prior to transforming H(s)  H(z)

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10 Bilinear Pre-warping

2 d t a  tan  t 2

a = d

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Bilinear Transform: Example • Design digital Butterworth lowpass filter – order, n = 2, cut off frequency 휔푑 = 628 rad/s – sampling frequency 휔푠 = 5024 rad/s (800Hz)

• Pre-warp to find 휔푎 that gives desired 휔푑  2 628 Note: ωd < ωa a tan 663 rad/s 1 2 800 due to compression 800

• Butterworth prototype (unity cut off) is, 1 H(s)  s2  2s 1

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11 Bilinear Transform: Example [2] 푠 • De-normalised analogue prototype (푠’ = ) 휔푐 – 휔푐 = 663 rad/s (required 휔푎 to give desired 휔푑) 1 H(sd )  2  s  2s    1  663 663 1 – Convert H(s)  H(z) by substituting 21 z  s  1 t1 z  1 H(z)  2  2800(1 z 1)   2800(1 z 1)     2  1  1   1   663(1 z )   663(1 z )  0.098z 2  0.195z  0.098 Note: H(z) has both H(z)  poles and zeros z 2  0.942z  0.333 H(s) was all-pole

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Bilinear Transform: Example [3]

Y (z) 0.098z 2  0.195z  0.098 H (z)   X (z) z 2  0.942z  0.333

• Multiply out and make causal: Y (z)(z 2  0.942z  0.333)  X (z)(0.098z 2  0.195z  0.098)

Y (z)(1 0.942z 1  0.333z 2 )  X (z)(0.098  0.195z 1  0.098z 2 )

• Finally, apply inverse z-transform to yield the difference equation: y[ n ] 0.098 x [ n ]  0.195 x [ n  1]  0.098 x [ n  2] 0.942y [ n  1]  0.333 y [ n  2]

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12 Bilinear Transform: Example [4] Magnitude response

c

Digital compared to Analog: 1. Increased roll off and attenuation in stopband 휔 2. Nearly  attenuation at 푠 2

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Bilinear Transform: Example [5]

Pole/Zero Plot

1

0.8

0.6

0.4

0.2

0

Imaginary Part Imaginary -0.2

-0.4

-0.6

-0.8

-1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Real Part

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13 Bilinear Transform: Example [6]

Phase response Increased phase delay

Bilinear transform has effectively increased digital filter order (by adding zeros)

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Bilinear Transform: Example [7]

Impulse Response 0.35

0.3

0.25

0.2

0.15 Amplitude

0.1

0.05

0

0 2 4 6 8 10 12 14 16 Samples

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14 Bilinear Transform: Example [8]

x[n]  Canonical Implementation

0.942 -0.333 y[n]  0.098y'[n] 0.195y'[n 1] 0.098y'[n  2] y'[n]  x[n] 0.942y'[n 1] 0.333y'[n  2]

z -1 z -1 y[n]’ of the difference equation 0.098 0.195 0.098 y[n]  0.098x[n] 0.195x[n 1] 0.098x[n  2]  0.942y[n 1] 0.333y[n  2] y[n] 

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Bilinear Design Summary • Calculate pre-warping analogue cutoff frequency

• De-normalise filter using pre-warping cut-off

• Apply bilinear transform and simplify

• Use inverse z-transform to obtain difference equation

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15

BREAK

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Tutorial 3 (Week 6 & 7!) • Week 6: PS 1 Review • Week 7: FIR Filters • DCT, FFT & More!

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16 Fun Demo! Resize a “Screen Shot” • 10:1 Reduction of Picture of an LCD Screen…

Without First

WITH Gaussian Filter (σ=2)

• Moire ∵ of Aliasing of a Non Band-limited Signal

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Direct Synthesis (in the Z-Domain)

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17 Direct Synthesis • Not based on analogue prototype – But direct placement of poles/zeros

• Useful for – First order lowpass or highpass • simple smoothers

– and equalisers • Single frequency amplification/removal

– Comb and notch filters • Multiple frequency amplification/removal

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First Order Filter: Example • General first order transfer function – Gain, G, zero at –b, pole at a (a, b both < 1)

Remember: H(ω) = H(z)|z = exp(jwt)

G1 bz 1  H(z)  1 az 1 

with a +ve & b –ve z = exp(jω) this is a lowpass filter i.e., G1 b o x H (0)  exp(j) = -1 1= exp(j0) 1 a -b a G1 b H ( )  s/2 1 a

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18 First Order Filter: Example • Possible design criteria – cut-off frequency, 휔푐 • 3푑퐵 = 20 log ( 퐻 (휔푐) )  • e.g., at 휔 = , 푐 2 1+푏 • = 2 1+푎

– stopband attenuation • assume 휔푠푡표푝 =  (Nyquist frequency) 퐻() 1 • e.g.,  = = i.e., two unknowns (a,b) 2 퐻(0) 21 two (simultaneous) design equations. H( ) (1b)(1 a) 1   H(0) (1 b)(1 a) 21

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Digital Design Prototype

• Second order ‘resonator’ – single narrow peak frequency response – i.e., peak at resonant frequency, 휔0

|H(ω)|2 1

x R 1/2 ω = 3dB width w 0

x

0 ω0 /2  ω

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19 Quality factor (Q-factor) • Dimensionless parameter that compares – Time constant for oscillator decay/bandwidth () to – Oscillation (resonant) period/frequency (0) • High Q = less energy dissipated per cycle  f Q  0  0  f • Alternative to factor () as

1  2  2 Q  0 0 H(s)  2 2  2 s  2 s  2  2 0 0 s  0 s  Q 0 • Note: Q < ½ overdamped (not an oscillator)

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Digital Resonator Design • To make a peak at 휔0 place pole – Inside unit circle (for stability) – At angle 휔0 distance R from origin • i.e., at location 푝 = 푅 exp (푗휔0) – R controls ω » Closer to unit circle  sharper peak • plus complex conj pole at p* = R exp(-jw0)

1 Hz()  11 1R  exp( j00 ) z 1  R  exp(  j ) z  1  1 2 2 1R exp j00  exp  j z  R z G  12 1a12 z a z Where (via Euler’s relation) 2 a1 2 R cos( 0 ) and a 2  R

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20 Digital Resonator Design

• Frequency response H(w) = H(z)|z = exp(jw) G H ()  1R  exp( j00 )exp(  j  ) 1  R  exp(  j  )exp(  j  ) G  1a12 exp(  j )  a exp(  2 j )

Note: we know the form of the 2nd equation from the previous slide

And a1 and a2 remain the same

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Digital Resonator Design

• Fixing unity gain at w0 (pole frequency)

– i.e., |H(w0)| = 1

G Hw(0 ) 1 1R  exp( j0 )exp(  j  0 ) 1  R  exp(  j  0 )exp(  j  0 ) solving for G ,

2 GRRR(1  ) 1  2 cos(20 )  Design relationship between gain G and pole radius R

(at resonant frequency w0)

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21 Digital Resonator Design

• Magnitude squared response is given by

2 2 G H ()  22 1 2RRRR cos(  00 )  1  2 cos(    )   • The 3dB bandwidth, w, occurs when 2 – |H(ω)| = ½ (remember G selected for |H(ω0)| = 1)

– two points ω1 and ω2 (on either side of ω0)

– ω = ω1 - ω2 • when pole, p, is close to the unit circle (R  1)  2(1  R ) i.e., closer pole is to unit circle, the sharper the peak Note: (1 - R) is pole distance to unit circle

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Resonator Design “Formula”

1. For specified resonant frequency ω0 – and 3 dB bandwidth ω

2. Calculate pole angle  = 2ω0/ωs   – i.e.,  0 s 2 휔 3. Calculate pole radius 푅 = 1 − 2

4. Calculate 2 G  (1 R) 1 2Rcos(2w0 )  R

5. Calculate filter coefficients (푎1, 푎2) 2 a1 2 R cos( 0 ) and a 2  R

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22 Digital Resonator: Example • Design a 2-pole resonator with • peak, 푓0 = 500퐻푧 • 3dB width, 푓 = 32퐻푧 • sampling frequency 푓푠 = 10푘퐻푧

• Normalise specification 푓0 • 휔0 = 2 = 0.1 푓푠 푓 • 휔 = 2 = 0.02 푓푠

• Calculate R (from 휔  2(1 − 푅)) • R = 0.99

• Then calculate G and 푎1 and 푎2 • 푮 = 0.0062, 푎1 = −1.8831 and 푎2 = 0.9801

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Discrete Filter Transformations • By convention, design Lowpass filters – transform to HighPass, BandPass, BandStop, etc.

• Simplest transformation – Lowpass H(z’ )  highpass H(z) – 퐻퐻푃(푧) = 퐻퐿푃 푧 푧’  −푧 휔 • reflection about imaginary axis ( 푠) 4 • changing signs of poles and zeros

• LP cutoff frequency, 휔퐶퐿푃 becomes • HP cut-in frequency: 휔 = 1 − 휔 퐶퐻푃 2 퐶퐿푃

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23 Lowpass  highpass (z’ = -z) z - plane Lowpass Highpass prototype transform

wCLP wCHP o x x o

pL = ¼, zL = -1 pH = -¼, zH = 1

Poles/zeros reflected in imaginary axis: wCHP = ½ - wCLP Same gain @ ws/4 (i.e., /4) |H(wHP)| = |H(/2 - wLP)|

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Discrete Filter Transformations • Lowpass H(z’ )  highpass • Lowpass H(z’ )  Bandpass H(z) H(z) – Cut-off (3dB) frequency = – Centre frequency = w0 & wc (remains same) 3dB bandwidth = wc

cosw t z z' c 1 coswctz z  z 2 cos(w t) z'   0 z 1 cos(wct)

Note: these are not the only possible BP and BS transformations!

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24 Discrete Filter Transformations • Lowpass H(z’ )  Bandstop H(z) – Centre frequency = 휔0 3dB bandwidth = 휔푐

z2 2/( k  1) z  (1  k )/(1  k ) z '  1 2/( k  1) z  (1  k )/(1  k ) z2

cos(0t ) 2 kt tan (c  ) cos(ct )

Note: order doubles for bandpass/bandstop transformations

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z - plane Lowpass Highpass prototype transform

o x x o

Bandpass Bandstop transform transform x o

o x x

x o

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25 Notch and Comb Filters • By positioning multiple pole/zero pairs – equally spaced around the unit circle

• Design a filter that removes/amplifies – frequencies at 푛 ⋅ 휔0 – i.e., frequency harmonics

• Can also remove/amplify multiple arbitrary frequencies

• Notch filter – removes multiple/single frequencies

• Comb filter – amplifies multiple/single frequencies

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Comb and Notch Filters

Comb Filter |H(ω)|

x o x o o x o x 0 /2  ω |H(ω)| Notch Filter o x o x x o x o 0 /2  ω

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26 Summary • Digital Filter Structures – Direct form (simplest) – Canonical form (minimum memory) • IIR filters – Feedback and/or feedforward sections • FIR filters – Feedforward only • Filter design – Bilinear transform (LP, HP, BP, BS filters) – Direct form (resonators and notch filters) – Filter transformations (LP  HP, BP, or BS) • Stability & Precision improved – Using cascade of 1st/2nd order sections

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Next Time…  • Digital Windows 

& FIR!

• Review: – Chapter 10 of Lathi

• A signal has many signals  [Unless it’s bandlimited. Then there is the one]

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