eleg--305: digital signal processing - university of...

ELEG–305: Digital Signal ProcessingLecture 23: Design of Digital IIR Filters (Part II)

Kenneth E. Barner

Department of Electrical and Computer EngineeringUniversity of Delaware

Fall 2008

K. E. Barner (Univ. of Delaware) ELEG–305: Digital Signal Processing Fall 2008 1 / 25

Outline

1 Review of Previous Lecture

2 Lecture Objectives

3 Design of IIR Filters from Analog FiltersCharacteristics of Commonly Used Analog FiltersButterworth FiltersAnalog Filters Comparison

4 Frequency TranslationFrequency Transformations in the Analog DomainFilter Design Steps & Examples

5 IIR Filter Design Using Matlab


Review of Previous Lecture

Review of Previous LectureIIR Filter Design – Map well understood analog filters to thediscrete domain (s to z plane mapping)Approximation of Derivatives Method for IIR Filter design –

z =1

1− sTs =

1− z−1

TMaps s LHP to right half of z unit circle; No highpass filtersImpulse Invariance Method for IIR Filter design – z = esT ; Maps sLHP to inside of z unit circle; T small to avoid aliasing; poles &zeroes have different mappingsBilinear Transformation Method for IIR Filter design –

s =2T

(1− z−1

1 + z−1

)z =

1 + T2 s

1− T2 s

Ω =2T

tan (ω/2) ω =2 arctan (ΩT/2)

Maps s LHP to inside of z unit circle; No aliasingK. E. Barner (Univ. of Delaware) ELEG–305: Digital Signal Processing Fall 2008 3 / 25

Lecture Objectives

Lecture Objectives

Objective Analyze analog filter families that can be mapped todiscrete IIR filters; Develop analog filter design methods;Present frequency translation methods for transformingfilter types; Present Matlab IIR filter design routines

Reading Chapters 10 (10.3–10.4); Next lecture, multirate signalprocessing (Chapter 11)


Design of IIR Filters from Analog Filters Characteristics of Commonly Used Analog Filters

Objectives: Design an analog filter and perform a H(s) to H(z)conversion that preserves filter characteristics

Only lowpass analog filters considered – frequency translationconverts lowpass prototypes to high, band, stop/pass filtersConsider several lowpass analog filters with different pass/stopband characteristics

Aside: Ωc is defined as the -3 dB, or cutoff, frequency of H(s), i.e.,

|H(Ωc)|dB = 20 log |H(Ωc)| = 10 log |H(Ωc)|2 = 10 log(0.5) = −3dB

where it is assumed that |H(Ω)| = 1 in the passband

Recall also the passband andstopband frequencies andripple definitions

(shown in the discrete domain)

IH(w)1

o

01 - Passband ripple02 - Stopband ripplew - Passband edge frequencypW s - Stopband edge frequency

: Stopband, f----------.:------ _:-4

Transition I... ..

band


Design of IIR Filters from Analog Filters Butterworth Filters

Case 1: Butterworth Filters are all-pole, with the lowpass filter given by

|H(Ω)|2 =1

1 + (Ω/Ωc)2N =1

1 + ε2(Ω/Ωp)2N (∗)

⇒ Ωc is the cutoff frequency⇒ Ωp is the passband frequency⇒ 1− δ1 = 1/

√1 + ε2 is determined by the passband ripple

Recall that|H(Ω)|2 = H(s)H(−s)|s=jΩ

Using Ω = s/j substitution in (∗)

H(s)H(−s) =1

1 + ((s/j)/Ωc)2N

=1

1 + ((s/j)2/Ωc2)N

=1

1 + (−s2/Ωc2)N



H(s)H(−s) = 11+(−s2/Ωc

2)N has poles defined by

−s2

Ω2c

= (−1)1/N = ej(2k+1)π/N , k = 0,1, . . . ,N − 1

⇒ sk = Ωcejπ/2ej(2k+1)π/2N , k = 0,1, . . . ,N − 1

Observations: The poles of H(s)H(−s) are equally spaced around acircle of radius Ωc

Note: Poles of H(s) are inthe LHP and incomplex–conjugate pairs

(stable system with realvalued coefficients)

Poles ofH(s)

Poles ofH(s)

N=4

(a)

\

\\

\

\\

\

\\

\\

\\

N= 5

Poles ofH(-s)

Poles ofH(-s)

Butterworth pole positions (N = 5)



Consider the Butterworth filterresponse |H(Ω)|2

Observation: |H(Ω)| is monotonicin the pass and stop bands

Note: The required filter order Nmust still be determined

Set N to achieve the desiredstopband attenuation

|H(Ωs)|2 =1

1 + ε2(Ωs/Ωp)2N = δ22

IH(Q)12

1.1

1 1.01 + E2

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

o

Frequency response of Butterworth filtersRearranging gives

N =log[(1/δ2)− 1]

2 log(Ωs/Ωp)=

log(δ/ε)

log(Ωs/Ωp)

where we define δ =√

1/δ22 − 1



Butterworth Filter Summary

Note: A Butterworth filter is completely characterized by N, δ2, ε,Ωcand the ratio Ωs/Ωp

Set the filter order

N =log(δ/ε)

log(Ωs/Ωp), where δ =

√1/δ2

2 − 1

Determine the poles

sk = Ωcejπ/2ej(2k+1)π/2N , k = 0,1, . . . ,N − 1

Set the filter

H(s) =N−1∏k=0

1s − sk

Where H(s) is then mapped to H(z), e.g., through the bilineartransformation



ExampleDesign a digital Butterworth filter to achieve the following:

Passband Corner Frequency Fp = 15 kHz with ε = .2Cutoff Frequency Fc = 17 kHz

Stopband Corner Frequency Fs = 22 kHz with 40 dB of attenuationSampling Rate SR = 44kHz

Utilize the bilinear transformation s → z domain mapping

Fist normalize (convert) to digital frequencies

ωp = 2πFp/SR = 2.14ωc = 2πFc/SR = 2.43 and ωs = 2πFs/SR = 3.14

Next, warp the frequencies using Ω = 2T tan (ω/2); Final result is

indifferent to the T value; Set T = 1/SR

Ωp = 2SR tan (ωp/2) = 1644Ωc = 2SR tan (ωc/2) = 1866 and Ωs = 2SR tan (ωs/2) = 2413



Set the stopband ripple terms

10 log(δ22) = −40 dB

⇒ δ22 = 10−4, ⇒ δ =

√1δ2

2− 1 ≈ 100

The necessary filter order is

N =log(δ/ε)

log(Ωs/Ωp)=

log 100/.2log 2413/1644

=2.70

0.1667= 16.2⇒ N = 17

Determine the poles, map resulting transform to the digital domain

sk = Ωcejπ/2ej(2k+1)π/2N , k = 0,1, . . . ,16

⇒ H(s) =16∏

k=0

1(s − sk )

⇒ H(z) = H(s)|s=2SR

(1−z−1

1+z−1

)K. E. Barner (Univ. of Delaware) ELEG–305: Digital Signal Processing Fall 2008 11 / 25

Design of IIR Filters from Analog Filters Analog Filters Comparison

Analog Filter Characteristic Comparison

Note: Full specs for common analog filters are given in Chapter 10.3

Butterworth FiltersAll poleMonotonic frequency response in the pass and stop bands

Type I Chebyshev FiltersAll poleEquiripple passband frequency responseMonotonic stopband frequency response

Type II Chebyshev FiltersPole/ZeroMonotonic passband frequency responseEquiripple stopband frequency response

Elliptic FiltersPole/ZeroEquiripple pass/stop band frequency response


Design of IIR Filters from Analog Filters Analog Filters Comparison

Chebyshev (Type I and II) & Elliptic Filter Comparison

IH(D.)j2

11 +€2

Type I Chebyshev Filter Type II Chebyshev Filter Elliptic Filter

Observations:Note the different pass/stop band characteristics (monotonic vs.equiripple)The elliptic filter has the sharpest transition region


Frequency Translation Frequency Transformations in the Analog Domain

Observation: Polynomial expressions are given for lowpass filters only

Approach: Use frequency translation to obtain other filters

Example: The 2nd order Butterworth LP filter is

H(s) =1

s2 +√

2s + 1Note that for Ω = 1 or s = j (for frequency evaluation)

|H(s)|2 =1

| − 1 + j√

2 + 1|2=

12

⇒ Ωp = 1 (define passband edge frequency as the -3 dB frequency)

Question: How do we define a new LP filter with different cutoff freq.?

Solution: Perform the mapping

s →Ωp

Ω′ps

where Ωp and Ω′p are the old and new cutoff frequenciesK. E. Barner (Univ. of Delaware) ELEG–305: Digital Signal Processing Fall 2008 14 / 25

Frequency Translation Frequency Transformations in the Analog Domain

Thus utilizing s → ΩpΩ′p

s in the previous case (noting Ωp = 1)

H ′(s) =1(

ΩpΩ′p

s)2

+√

2(

ΩpΩ′p

s)

+ 1=

1(1

Ω′p

)2s2 +

√2

Ω′ps + 1

Analog Frequency Translation SummaryBand edge

Type of frequencies oftransformation Transformation new filter

Lowpass s → ΩpΩ′p

s Ω′p

Highpass s → ΩpΩ′ps Ω′p

Bandpass s → Ωps2+Ωl Ωus(Ωu−Ωl )

Ωl ,Ωu

Bandstop s → Ωps(Ωu−Ωl )s2+Ωl Ωu

Ωl ,Ωu


Frequency Translation Filter Design Steps & Examples

Summary of Filter Design Steps

Approach: Given a set of design criteria (pass/stop band frequencies,attenuation and ripple parameters, etc.), perform the following steps

1 Determine order (N) from specs (ripple, attenuation, etc.)2 Look up the (lowpass) polynomial H(s) for that order3 Transform to desired filter type (high pass, band pass/stop, etc.)4 Map to the digital domain (H(s)→ H(z) by bilinear transformation

Note: Filter type transformation can also be performed in the discretedomain (see Table 10.8)

In that case, skid step #3 and perform the type transformation in thedigital domain after step #4



ExampleDetermine a digital filter that meets or exceeds the following specs

Passband Frequency Fp = 4 kHz with passband ripple ≤ 1 dBStopband Frequency Fs = 6 kHz with ≥ 40 dB of attenuation

Sampling Rate SR = 24 kHz

Normalizing the frequency specifications to digital frequencies yields

ωp = 2πFp/SR = π/3 ωs = 2πFs/SR = π/2

Next, warp the frequencies using T = 2 for simplicity

Ωp = 2/T tan(ωp/2) = tan(π/6) = 0.509Ωs = tan(ωs/2) = tan(π/4) = 1

Next, utilize the given passband ripple (δ1) to determine ε

−(passband ripple) dB = 10 log(

11 + ε2

)K. E. Barner (Univ. of Delaware) ELEG–305: Digital Signal Processing Fall 2008 17 / 25


Thus

log(1 + ε2) =passband ripple

10= 0.1

⇒ ε =√

100.1 − 1 = 0.509

Now determine the other ripple parameters, δ2 and δ,

10 log(δ22) = −40 ⇒ δ2

2 = 10−4

⇒ δ =

√1δ2

2− 1 =

√104 − 1 = 99.995

Finally, determine the order necessary for different filter types

Case 1: Butterworth Filter order is given by

N ≥ log(δ/ε)

log(Ωs/Ωp)

=log(99.995/0.509)

log(1/0.509)

= 8.39⇒ N = 9K. E. Barner (Univ. of Delaware) ELEG–305: Digital Signal Processing Fall 2008 18 / 25


Case 2: Chebyshev Filter order is given by

N ≥ cosh−1(δ/ε)

cosh−1(Ωs/Ωp)

= 4.9⇒ N = 5

Case 3: Elliptic Filter requires complex integration, but yields N = 4

Observations:Required order decreases for Butterworth→ Chebyshev→ EllipticFilters that require smaller order allow passband/stopband ripple(or both); elliptic filters have the least linear phase response

Once order is found, there are two paths that can be taken:1 Use the exact polynomial and map to the discrete domain by the

bilinear transformation2 Look up low pass polynomial of correct order

- This is usually low pass with Ωc = 1- Use low→ low/band/high pass translation- Map to the discrete domain by the bilinear transformation


IIR Filter Design Using Matlab


Command: butter - design Butterworth IIR Filter

[B,A] = butter(N,Wn)

Gives N th order lowpass IIR Butterworth filter0 < Wn < 1 is the cutoff freq. (1 is half the sampling rate, or π)If Wn=[W1,W2], the order 2N bandpass filter is returned

[B,A] = butter(N,Wn,’high’)

Gives N th order highpass IIR Butterworth filter[B,A] = butter(N,Wn,’stop’)

Gives the order 2N band stop IIR Butterworth filter (Wn=[W1,W2])[z,p,k] = butter(N,Wn)

Gives the poles, zeros, and gain term, i.e.,

H(z) = k∏M

k=1(1− p(k)z−1)∏Nk=0(1− z(k)z−1)



Command: buttord - determines the necessary Butterworth filter order

[N,Wn] = buttord(Wp,Ws,Rp,Rs)

Inputs:Wp – passband frequencyWs – stopband frequencyRp – maximum passband loss (ripple)Rs – minimum stop band attenuationNote: Rp and Rs are in dB

Outputs:N – filter orderWn – filter cutoff (3 dB) frequency

Note: All frequencies are normalized (0<W<1). Also, order value isreturned for band pass/stop filters when vector frequencies are used(e.g., Wp=[W1,W2], etc.)

Usage: Determine (1) filter order (N) and cutoff frequency (Wn), then(2) filter parameters ([B,A]), i.e., apply buttord followed by butter



Chebyshev Type I Filters

Command: cheby1 - determines the Chebyshev Type I filter

[B, A] = cheby1(N, R, Wn)N – filter orderR – decibels of ripple in the passbandWn – cut-off frequency (3dB point)

Command: cheby1ord - determines the Chebyshev Type I filter order

[N, Wn] = cheb1ord(Wp, Ws, Rp, Rs)Wp – passband edge frequencyWs – stopband edge frequencyRp – decibels of loses allowed in the passbandRs – decibels of attenuation in the stopbandN – filter orderWn – cut off frequency

Note: Highpass and band pass/stop filters can be returned (same as inprevious case)



Chebyshev Type II Filters

Command: cheby2 - determines the Chebyshev Type II filter

[B, A] = cheby2(N, R, Wn)N – filter orderR – decibels of ripple in the stopbandWn – cutoff frequency (3dB point)

Command: cheby2ord - determines the Chebyshev Type II filter order

[N, Wn] = cheb2ord(Wp, Ws, Rp, Rs)Wp – passband edge frequencyWs – stopband edge frequencyRp – allowable passband ripple (dBs)Rs – minimum stopband attenuation (dBs)N – filter orderWn – cut off frequency

Note: Highpass and band pass/stop filters can be returned (same as inprevious case)



Elliptical Filters

Command: ellip - determines the elliptical filter

[B, A] = ellip(N, Rp, Rs, Wn)N – filter orderRp – passband ripple (dBs)Rs – stopband attenuation (dbs)Wn – cutoff frequency (3dB point)

Command: ellipord - determines the elliptical filter order

[N, Wn] = ellipord(Wp, Ws, Rp, Rs)Wp – passband edge frequencyWs – stopband edge frequencyRp – allowable passband ripple (dBs)Rs – minimum stopband attenuation (dBs)N – filter orderWn – cut off frequency


Lecture Summary

Lecture SummaryAnalog Filter Families –Butterworth Filters: All pole; Monotonic frequency response in thepass and stop bandsChebyshev Type I Filters: All pole; Equiripple passband frequencyresponse; Monotonic stopband frequency responseChebyshev Type II Filters: Pole/Zero; Monotonic passbandfrequency response; Equiripple stopband frequency responseElliptic Filters: Pole/Zero; Equiripple pass/stop band responseFrequency Translation – Only lowpass filter expressions given;Use frequency translation to obtain highpass, band stop/passfilters; Performed in analog or digital domains (Tables 10.7 & 10.8)Filter Design – (1) Determine order (N); (2) Look up the (lowpass)polynomial H(s); (3) Transform to desired filter type; (4) Map tothe digital domain by bilinear transformationMatlab IIR Filter Design – butter, buttord; cheby1,cheb1ord; cheby2, cheb2ord; ellip, ellipord


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