dsp & digital filters

DSP and Digital Filters (2015-6724) Introduction: 1 1 / 16

DSP & Digital Filters

Mike Brookes

1: Introduction

1: IntroductionOrganization

Signals

Processing

Syllabus

Sequences

Time Scaling

z-TransformRegion ofConvergence

z-Transform examples

Rational z-Transforms

Rational example

Inverse z-Transform

MATLAB routines

Summary


Organization

1: Introduction

OrganizationSignals

Processing

Syllabus

Sequences

Time Scaling




Rational example

Inverse z-Transform

MATLAB routines

Summary


18 lectures: feel free to ask questions

Textbooks: (a) Mitra Digital Signal Processing ISBN:0071289461 41 covers

most of the course except for some of the multirate stuff (b) Harris Multirate Signal Processing ISBN:0137009054 49

covers multirate material in more detail but less rigour thanMitra

Lecture slides available via Blackboard or on my website:http://www.ee.ic.ac.uk/hp/staff/dmb/courses/dspdf/dspdf.htm quite dense - ensure you understand each line email me if you dont understand or dont agree with anything

Prerequisites: 3rd year DSP - attend lectures if dubious

Exam + Formula Sheet (past exam papers + solutions on website)

Problems: Mitra textbook contains many problems at the end of eachchapter and also MATLAB exercises

Signals

1: Introduction

Organization

SignalsProcessing

Syllabus

Sequences

Time Scaling




Rational example

Inverse z-Transform

MATLAB routines

Summary


A signal is a numerical quantity that is a function of one or moreindependent variables such as time or position.

Real-world signals are analog and vary continuously and takecontinuous values.

Digital signals are sampled at discrete times and are quantized to afinite number of discrete values

We will mostly consider one-dimensionsal real-valued signals withregular sample instants; except in a few places, we will ignore thequantization. Extension to multiple dimensions and complex-valued signals

is straighforward in many cases.

Examples:

Processing

1: Introduction

Organization

Signals

ProcessingSyllabus

Sequences

Time Scaling




Rational example

Inverse z-Transform

MATLAB routines

Summary


Aims to improve a signal in some way or extract some informationfrom it

Examples:

Modulation/demodulation

Coding and decoding

Interference rejection and noise suppression

Signal detection, feature extraction

We are concerned with linear, time-invariant processing

Syllabus

1: Introduction

Organization

Signals

Processing

SyllabusSequences

Time Scaling




Rational example

Inverse z-Transform

MATLAB routines

Summary


Main topics:

Introduction/Revision

Transforms

Discrete Time Systems

Filter Design

FIR Filter Design

IIR Filter Design

Multirate systems

Multirate Fundamentals

Multirate Filters

Subband processing

Sequences

1: Introduction

Organization

Signals

Processing

Syllabus

SequencesTime Scaling




Rational example

Inverse z-Transform

MATLAB routines

Summary


We denote the nth sample of a signal as x[n] where < n < + andthe entire sequence as {x[n]} although we will often omit the braces.

Special sequences:

Unit step: u[n] =

{1 n 0

0 otherwise

Unit impulse: [n] =

{1 n = 0

0 otherwise

Condition: condition[n] =

{1 condition is true

0 otherwise(e.g. u[n] = n0)

Right-sided: x[n] = 0 for n < Nmin Left-sided: x[n] = 0 for n > Nmax Finite length: x[n] = 0 for n / [Nmin, Nmax] Causal: x[n] = 0 for n < 0, Anticausal: x[n] = 0 for n > 0

Finite Energy:

n= |x[n]|2

Time Scaling

1: Introduction

Organization

Signals

Processing

Syllabus

Sequences

Time Scalingz-TransformRegion ofConvergence



Rational example

Inverse z-Transform

MATLAB routines

Summary


For sampled signals, the nth sample is at time t = nT = nfs

where fs =1T

is the sample frequency.

Often easiest to scale time so that fs = 1Hz. E.g. to design a 1 kHzlow-pass filter for fs = 44.1 kHz we can design a 0.0227Hz filter forfs = 1Hz.

To scale back to real-world values: Every quantity of dimension (Time)n ismultiplied by Tn (or equivalently by fns ). Thus all times are multiplied byT and all frequencies and angular frequencies by T1.

We use for real angular frequencies and for normalized angularfrequency. The units of are radians per sample.

Power is measured in units of energy per sample: to convert back toenergy per second you must multiply by T1 = fs.

Warning: Several MATLAB routines scale time so that fs = 2 Hz. Weird,non-standard and irritating.

z-Transform

1: Introduction

Organization

Signals

Processing

Syllabus

Sequences

Time Scaling




Rational example

Inverse z-Transform

MATLAB routines

Summary


The z-transform converts a sequence, {x[n]}, into a function, X(z), of anarbitrary complex-valued variable z.

Why do it?

Complex functions are easier to manipulate than sequences

Useful operations on sequences correspond to simple operations onthe z-transform:

addition, multiplication, scalar multiplication, time-shift,convolution

Definition: X(z) =+

n= x[n]zn

Region of Convergence

1: Introduction

Organization

Signals

Processing

Syllabus

Sequences

Time Scaling

z-Transform

Region ofConvergence



Rational example

Inverse z-Transform

MATLAB routines

Summary


The set of z for which X(z) converges is its Region of Convergence (ROC).

Complex analysis : the ROC of a power series (if it exists at all) is alwaysan annular region of the form 0 Rmin < |z| < Rmax .

X(z) will always converge absolutely inside the ROC and may converge onsome, all, or none of the boundary.

converge absolutely +

n= |x[n]zn|



The sample at n = 0 is indicated by an open circle.

u[n] 11z1 1 < |z|

x[n] 2z2 + 2 + z1 0 < |z|


1: Introduction

Organization

Signals

Processing

Syllabus

Sequences

Time Scaling



Rationalz-Transforms

Rational example

Inverse z-Transform

MATLAB routines

Summary


Most z-transforms that we will meet are rational polynomials with realcoefficients, usually one polynomial in z1 divided by another.

G(z) = g

Mm=1(1zmz

1)

Kk=1(1pkz

1)= gzKM

Mm=1(zzm)Kk1(zpk)

Completely defined by the poles, zeros and gain.

The absolute values of the poles define the ROCs:R + 1 different ROCs

where R is the number of distinct pole magnitudes.

Note: There are K M zeros or M K poles at z = 0 (easy to overlook)

Rational example

1: Introduction

Organization

Signals

Processing

Syllabus

Sequences

Time Scaling




Rational exampleInverse z-Transform

MATLAB routines

Summary


G(z) = 82z1

44z13z2

Poles/Zeros: G(z) = 2z(z0.25))(z+0.5)(z1.5) Poles at z = {0.5,+1.5)},

Zeros at z = {0,+0.25}

Partial Fractions: G(z) = 0.751+0.5z1 +1.25

11.5z1

ROC ROC 0.751+0.5z11.25

11.5z1 G(z)

a 0 |z| < 0.5

b 0.5 < |z| < 1.5

c 1.5 < |z|

Inverse z-Transform

1: Introduction

Organization

Signals

Processing

Syllabus

Sequences

Time Scaling




Rational example

Inversez-Transform

MATLAB routines

Summary


g[n] = 12pijG(z)zn1dz where the integral is anti-clockwise around a

circle within the ROC, z = Rej.

Proof:1

2pij

G(z)zn1dz= 12pij

(m= g[m]z

m)zn1dz

(i)=

m= g[m]1

2pij

znm1dz

(ii)=

m= g[m][nm]= g[n]

(i) depends on the circle with radius R lying within the ROC

(ii) Cauchys theorem: 12pijzk1dz = [k] for z = Rej anti-clockwise.

dzd

= jRej 12pijzk1dz = 12pij

2pi=0

Rk1ej(k1) jRejd

= Rk

2pi

2pi=0

ejkd

= Rk(k)= (k) [R0 = 1]

In practice use a combination of partial fractions and table of z-transforms.

MATLAB routines

1: Introduction

Organization

Signals

Processing

Syllabus

Sequences

Time Scaling




Rational example

Inverse z-Transform

MATLAB routinesSummary


tf2zp,zp2tf b(z1)

a(z1) {zm, pk, g}

residuez b(z1)

a(z1)

krk

1pkz1

tf2sos,sos2tf b(z1)

a(z1)

l

b0,l+b1,lz1+b2,lz

2

1+a1,lz1+a2,lz2

zp2sos,sos2zp {zm, pk, g}

l

b0,l+b1,lz1+b2,lz

2

1+a1,lz1+a2,lz2

zp2ss,ss2zp {zm, pk, g}

{x = Ax+Bu

y = Cx+Du

tf2ss,ss2tf b(z1)

a(z1)

{x = Ax+Bu

y = Cx+Du

Summary

1: Introduction

Organization

Signals

Processing

Syllabus

Sequences

Time Scaling




Rational example

Inverse z-Transform

MATLAB routines

Summary


Time scaling: assume fs = 1 Hz so <

z-transform: X(z) =+

n= x[n]n

ROC: 0 Rmin < |z| < Rmax Causal: ROC Absolutely summable: |z| = 1 ROC

Inverse z-transform: g[n] = 12pijG(z)zn1dz

Not unique unless ROC is specified Use partial fractions and/or a table

For further details see Mitra:1 & 6.

2: Three Different Fourier Transforms

2: Three DifferentFourier Transforms

Fourier TransformsConvergence ofDTFT

DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines

DSP and Digital Filters (2015-6724) Fourier Transforms: 2 1 / 13

Fourier Transforms



DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


Three different Fourier Transforms:

Continuous-Time Fourier Transform (CTFT): x(t) X(j) Discrete-Time Fourier Transform (DTFT): x[n] X(ej) Discrete Fourier Transform (DFT): x[n] X[k]

Forward Transform Inverse Transform

CTFT X(j) = x(t)e

jtdt x(t) = 12piX(j)e

jtd

DTFT X(ej) = x[n]e

jn x[n] = 12pi pipiX(e

j)ejnd

DFT X[k] =N1

0 x[n]ej2pi kn

N x[n] = 1N

N10 X[k]e

j2pi knN

We use for real and = T for normalized angular frequency.Nyquist frequency is = 2 fs2 =

piTand = .

For power signals (energy time interval), CTFT & DTFT areunbounded.Fix this by normalizing:

X(j) = limA12A

AA x(t)e

jtdt

X(ej) = limA12A

AA x[n]e

jn

Convergence of DTFT


Fourier Transforms

Convergence ofDTFT

DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


DTFT: X(ej) = x[n]e

jn does not converge for all x[n].

Consider the finite sum: XK(ej) =

KK x[n]e

jn

Strong Convergence:x[n] absolutely summable X(ej) converges uniformly

|x[n]|

[DTFT Convergence Proofs]

DSP and Digital Filters (2015-6724) Fourier Transforms: 2 note 1 of slide 3

(1) Strong Convergence: [these proofs are not examinable]

We are given that |x[n]| 0, N such that

|n|>N |x[n]| <

For K N , supX(ej)XK(ej)= sup |n|>K x[n]ejn

sup

(|n|>K

x[n]ejn)=|n|>K |x[n]| < (2) Weak Convergence:

We are given that |x[n]|

2 0, N such that|n|>N |x[n]|

2 <

Define y[K][n] =

{0 |n| K

x[n] |n| > Kso that its DTFT is, Y [K](ej) =

y

[K][n]ejn

We see that X(ej)XK(ej) =

x[n]e

jn KK x[n]e

jn

=|n|>K x[n]e

jn = y

[K][n]ejn = Y [K](ej)

From Parsevals theorem,

y[K][n]2 = 12pi

pipi

Y [K](ej)2 d= 1

2pi

pipi

X(ej)XK(ej)2 dHence for K N , 1

2pi

pipi

X(ej)XK(ej)2 d = y[K][n]2 =|n|>N |x[n]|2 <

DTFT Properties



DTFT PropertiesDFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


DTFT: X(ej) = x[n]e

jn

DTFT is periodic in : X(ej(+2mpi) = X(ej) for integer m.

DTFT is the z-Transform evaluated at the point ej:X(z) =

x[n]z

n

DTFT converges iff the ROC includes |z| = 1.

DTFT is the same as the CTFT of a signal comprising impulses atthe sample times (Dirac functions) of appropriate heights:

x(t) =

x[n](t nT )= x(t) (t nT )

Equivalent to multiplying a continuous x(t) by an impulse train.

Proof: X(ej) = x[n]e

jn

n= x[n] (t nT )e

j tT dt

(i)=

n= x[n](t nT )e

j tT dt

(ii)=

x(t)e

jtdt

(i) OK if |x[n]|

DFT Properties



DTFT Properties

DFT PropertiesSymmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


DFT: X[k] =N1

0 x[n]ej2pi kn

N

DTFT: X(ej) = x[n]e

jn

Case 1: x[n] = 0 for n / [0, N 1]

DFT is the same as DTFT at k =2piNk.

The {k} are uniformly spaced from = 0 to = 2N1N

.DFT is the z-Transform evaluated at N equally spaced pointsaround the unit circle beginning at z = 1.

Case 2: x[n] is periodic with period N

DFT equals the normalized DTFT

X[k] = limKN

2K+1 XK(ejk)

where XK(ej) =

KK x[n]e

jn

Symmetries



DTFT Properties

DFT Properties

SymmetriesParsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


If x[n] has a special property then X(ej)and X[k] will have correspondingproperties as shown in the table (and vice versa):

One domain Other domain

Discrete PeriodicSymmetric Symmetric

Antisymmetric AntisymmetricReal Conjugate Symmetric

Imaginary Conjugate AntisymmetricReal + Symmetric Real + Symmetric

Real + Antisymmetric Imaginary + Antisymmetric

Symmetric: x[n] = x[n]X(ej) = X(ej)X[k] = X[(k)mod N ] = X[N k] for k > 0

Conjugate Symmetric: x[n] = x[n]Conjugate Antisymmetric: x[n] = x[n]

Parsevals Theorem



DTFT Properties

DFT Properties

Symmetries

ParsevalsTheorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


Fourier transforms preserve energy

CTFT|x(t)|2 dt = 12pi

|X(j)|2 d

DTFT |x[n]|

2 = 12pi pipi

X(ej)

2 d

DFTN1

0 |x[n]|2 = 1

N

N10 |X[k]|

2

More generally, they actually preserve complex inner products:

N10 x[n]y

[n] = 1N

N10 X[k]Y

[k]

Unitary matrix viewpoint for DFT:

If we regard x and X as vectors, then X = Fx where F isa symmetric matrix defined by fk+1,n+1 = e

j2pi knN .

The inverse DFT matrix is F1 = 1NFH

equivalently, G = 1NF is a unitary matrix with GHG = I.

Convolution



DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

ConvolutionSampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


DTFT: Convolution Productx[n] = g[n] h[n]=

k= g[k]h[n k]

X(ej) = G(ej)H(ej)

DFT: Circular convolution Productx[n] = g[n]N h[n]=

N1k=0 g[k]h[(n k)modN ]

X[k] = G[k]H[k]

DTFT: Product Circular Convolution 2y[n] = g[n]h[n] Y (ej) = 12piG(e

j)H(ej) = 12pi pipi G(e

j)H(ej())d

DFT: Product Circular Convolution Ny[n] = g[n]h[n]

X[k] = 1NG[k]N H[k]

g[n] : h[n] : g[n] h[n] : g[n]3 h[n]

Sampling Process


Time Time Frequency

AnalogCTFT

Low PassFilter

* CTFT

Sample DTFT

Window DTFT

DFTDFT

Zero-Padding



DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-PaddingPhase Unwrapping

Summary

MATLAB routines


Zero padding means added extra zeros onto the end of x[n] beforeperforming the DFT.

Time x[n] Frequency |X[k]|

WindowedSignal

With zero-padding

Zero-padding causes the DFT to evaluate the DTFT at more valuesof k. Denser frequency samples.

Width of the peaks remains constant: determined by the length andshape of the window.

Smoother graph but increased frequency resolution is an illusion.

Phase Unwrapping



DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase UnwrappingSummary

MATLAB routines


Phase of a DTFT is only defined to within an integer multiple of 2.

-2 0 20

2

4

6

(rad/sample)

x[n] |X[k]|

-2 0 2

-2

0

2

(rad/sample)

-2 0 2-40

-20

0

(rad/sample)

X[k] X[k] unwrapped

Phase unwrapping adds multiples of 2 onto each X[k] to make thephase as continuous as possible.

Summary



DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

SummaryMATLAB routines


Three types: CTFT, DTFT, DFT

DTFT = CTFT of continuous signal impulse train

DFT = DTFT of periodic or finite support signal

DFT is a scaled unitary transform

DTFT: Convolution Product; Product Circular Convolution

DFT: Product Circular Convolution

DFT: Zero Padding Denser freq sampling but same resolution

Phase is only defined to within a multiple of 2.

Whenever you integrate frequency components you get a scale factor

12pi for CTFT and DTFT or

1N

for DFT

Inverse transform, Parseval, frequency domain convolution

For further details see Mitra: 3 & 5.

MATLAB routines



DTFT Properties

DFT Properties

Symmetries

Parsevals Theorem

Convolution

Sampling Process

Zero-Padding

Phase Unwrapping

Summary

MATLAB routines


fft, ifft DFT with optional zero-paddingfftshift swap the two halves of a vectorconv convolution or polynomial multiplication (not

circular)x[n]y[n] real(ifft(fft(x).*fft(y)))unwrap remove 2 jumps from phase spectrum

3: Discrete Cosine Transform

3: Discrete CosineTransform

DFT Problems

DCT

DCT of sine waveDCT/DFTEquivalence

DCT Properties

IDCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform

MDCT (ModifiedDCT)

IMDCT & TDAC

Summary

MATLAB routines

DSP and Digital Filters (2015-6724) Transforms: 3 1 / 15

DFT Problems


DFT ProblemsDCT


DCT Properties

IDCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform

MDCT (ModifiedDCT)

IMDCT & TDAC

Summary

MATLAB routines


For processing 1-D or 2-D signals (especially coding), a common method isto divide the signal into frames and then apply an invertible transform toeach frame that compresses the information into few coefficients.

The DFT has some problems when used for this purpose:

N real x[n] N complex X[k] : 2 real, N2 1 conjugate pairs

DFT the DTFT of a periodic signal formed by replicating x[n] . Spurious frequency components from boundary discontinuity.

N=20f=0.08

The Discrete Cosine Transform (DCT) overcomes these problems.

DCT


DFT Problems

DCTDCT of sine waveDCT/DFTEquivalence

DCT Properties

IDCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform

MDCT (ModifiedDCT)

IMDCT & TDAC

Summary

MATLAB routines


To form the Discrete Cosine Transform (DCT), replicate x[0 : N 1] but inreverse order and insert a zero between each pair of samples:

0 12 23

y[r]

Take the DFT of length 4N real symmetric sequenceResult is real, symmetric and anti-periodic: only need first N values

012

23

Y[k]

Forward DCT: XC [k] =N1

n=0 x[n] cos2pi(2n+1)k

4N for k = 0 : N 1

Compare DFT: XF [k] =N1

n=0 x[n] expj2pi(4n+0)k

4N

DCT of sine wave


DFT Problems

DCT


DCT Properties

IDCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform

MDCT (ModifiedDCT)

IMDCT & TDAC

Summary

MATLAB routines


DCT: XC [k] =N1


4N

f = mN

f 6= mN

x[n]N=20f=0.10

N=20f=0.08

|XF [k]|

|XC [k]|

DFT: RealComplex; Freq range [0, 1]; Poorly localizedunless f = m

N; |XF [k]| k

1 for Nf < k N2DCT: RealReal; Freq range [0, 0.5]; Well localized f ;

|XC [k]| k2 for 2Nf < k < N

DCT/DFT Equivalence


DFT Problems

DCT

DCT of sine wave

DCT/DFTEquivalence

DCT Properties

IDCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform

MDCT (ModifiedDCT)

IMDCT & TDAC

Summary

MATLAB routines


XC [k] =N1


4N0 12 23

y[r]

Define y[r] =

0 r even

x[r12

]r odd, 1 r 2N 1

x[4N1r

2

]r odd, 2N + 1 r 4N 1

YF [k] =4N1

r=0 y[r]Wkr4N where WM = e

j2piM

(i)=2N1

n=0 y[2n+ 1]W(2n+1)k4N

(ii)=N1

n=0 y[2n+ 1]W(2n+1)k4N

+N1

m=0 y[4N 2m 1]W(4N2m1)k4N

(iii)=N1

n=0 x[n]W(2n+1)k4N +

N1m=0 x[m]W

(2m+1)k4N

= 2N1


4N = 2XC [k]

(i) odd r only: r = 2n+ 1, (ii) reverse order for n N : m = 2N 1 n

(iii) substitute y definition & W 4Nk4N = ej2pi 4Nk

4N 1

DCT Properties


DFT Problems

DCT


DCT PropertiesIDCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform

MDCT (ModifiedDCT)

IMDCT & TDAC

Summary

MATLAB routines


Definition: X[k] =N1


4N Linear: x[n] + y[n] X[k] + Y [k] DFT ConvolutionMultiplication property does not hold / Symmetric: X[k] = X[k] since cosk = cos+k Anti-periodic: X[k + 2N ] = X[k] because:

cos ( + ) = cos 2(2n+ 1)(k + 2N) = 2(2n+ 1)k + 8Nn+ 4NX[N ] = 0 since X[N ] = X[N ] = X[N + 2N ]

Periodic: X[k + 4N ] = X[k + 2N ] = X[k]DCT basis functions:

0:3:

1: 4:

2: 5:

IDCT


DFT Problems

DCT


DCT Properties

IDCTEnergy Conservation

Energy Compaction

Frame-based coding

Lapped Transform

MDCT (ModifiedDCT)

IMDCT & TDAC

Summary

MATLAB routines


y[r] = 14N4N1

k=0 Y [k]Wrk4N =

12N

4N1k=0 X[k]W

rk4N

x[n] = y[2n+ 1] = 12N4N1

k=0 X[k]W(2n+1)k4N [Y [k] = 2X[k]]

(i)= 12N

2N1k=0 X[k]W

(2n+1)k4N

12N2N1

l=0 X[l]W(2n+1)(l+2N)4N

012

23

Y[k]

(ii)= 1

N

2N1k=0 X[k]W

(2n+1)k4N W

ba = e

j 2piba

(iii)= 1

NX[0] + 1

N

N1k=1 X[k]W

(2n+1)k4N +

1NX[N ]W

(2n+1)N4N

+ 1N

N1r=1 X[2N r]W

(2n+1)(2Nr)4N

(iv)= 1

NX[0] + 1

N

N1k=1 X[k]W

(2n+1)k4N

+ 1N

N1r=1 X[r]W

(2n+1)r+2N4N

x[n] = 1NX[0] + 2

N

N1k=1 X[k] cos

2pi(2n+1)k4N = Inverse DCT

(i) k = l + 2N for k 2N and X[k + 2N ] = X[k]

(ii) (2n+1)(l+2N)4N =(2n+1)l

4N + n+12

(iii) k = 2N r for k > N (iv) X[N ] = 0 and X[2N r] = X[r]

Energy Conservation


DFT Problems

DCT


DCT Properties

IDCT

EnergyConservation

Energy Compaction

Frame-based coding

Lapped Transform

MDCT (ModifiedDCT)

IMDCT & TDAC

Summary

MATLAB routines


DCT: X[k] =N1


4N

IDCT: x[n] = 1NX[0] + 2

N

N1k=1 X[k] cos

2pi(2n+1)k4N

rep

0 12 23

y[r]

DFT 0

1223

Y[k]

2

Energy: E =N1

n=0 |x[n]|2: E 2E 8NE 0.5NE

E = 1N|X|2 [0] + 2

N

N1n=1 |X|

2 [n]

Orthogonal DCT (preserves energy)

Define: c[k] =

2kN

c[0] =

1N, c[k 6= 0] =

2N

ODCT: X[k] = c[k]N1


4N

IODCT: x[n] =N1

k=0 c[k]X[k] cos2pi(2n+1)k

4N

Note: MATLAB dct() calculates the ODCT

Energy Compaction


DFT Problems

DCT


DCT Properties

IDCT

Energy Conservation

EnergyCompaction

Frame-based coding

Lapped Transform

MDCT (ModifiedDCT)

IMDCT & TDAC

Summary

MATLAB routines


If consecutive x[n] are positively correlated, DCT concentrates energy in afew X[k] and decorrelates them.

Example: Markov Process: x[n] = x[n 1] +1 2u[n]

where u[n] is i.i.d. unit Gaussian.Then

x2[n]

= 1 and x[n]x[n 1] = .

Covariance of vector x is Si,j =xxH

i,j

= |ij|.

Suppose ODCT of x is Cx and DFT is Fx.

Covariance of Cx isCxxHCH

= CSCH (similarly FSFH)

Diagonal elements give mean coefficient energy.

5 10 15 20 25 30

50

60

70

80

90

100DCT

DFT

=0.9N=32

No of coefficients

C

u

m

u

l

a

t

i

v

e

e

n

e

r

g

y

(

%

)

Used in MPEG and JPEG (superseded byJPEG2000 using wavelets)

Used in speech recognition to decorrelatespectral coeficients: DCT of log spectrum

Energy compaction good for coding (low-valued coefficients can be set to 0)Decorrelation good for coding and for probability modelling

Frame-based coding


DFT Problems

DCT


DCT Properties

IDCT

Energy Conservation

Energy Compaction

Frame-basedcoding

Lapped Transform

MDCT (ModifiedDCT)

IMDCT & TDAC

Summary

MATLAB routines


Divide continuous signalinto frames

Apply DCT to each frame

Encode DCT

e.g. keep only 30 X[k]

Apply IDCT y[n]

x[n]

X[k] k=30/220

y[n]y[n]-x[n]

Problem: Coding may create discontinuities at frame boundariese.g. JPEG, MPEG use 8 8 pixel blocks

8.3 kB (PNG) 1.6 kB (JPEG) 0.5 kB (JPEG)

Lapped Transform


DFT Problems

DCT


DCT Properties

IDCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped TransformMDCT (ModifiedDCT)

IMDCT & TDAC

Summary

MATLAB routines


Modified Discrete Cosine Transform (MDCT): overlapping frames 2N long

x[0 : 2N 1]MDCT X0[0 : N 1]IMDCT y0[0 : 2N 1]

x[N : 3N 1]MDCT X1[N : 2N 1]IMDCT y1[N : 3N 1]

x[2N : 4N 1]MDCT X2[2N : 3N 1]IMDCT y2[2N : 3N 1]

y[n] = y0[n] + y1[n] + y2[n]

X0[k]

y0[n]X1[k]

y1[n]

X2[k]

y2[n] y[n]

y[n]-x[n] = error

x[n]

0 N

N

2N

2N

3N

3N

4N

4N

MDCT: 2N N coefficients, IMDCT: N 2N samplesAdd yi[n] together to get y[n]. Only two non-zero terms far any n.Errors cancel exactly: Time-domain alias cancellation (TDAC)

MDCT (Modified DCT)


DFT Problems

DCT


DCT Properties

IDCT

Energy Conservation

Energy Compaction

Frame-based coding

Lapped Transform

MDCT (ModifiedDCT)

IMDCT & TDAC

Summary

MATLAB routines


MDCT: XM [k] =2N1

n=0 x[n] cos2pi(2n+1+N)(2k+1)

8N

IMDCT: yA,B [n] =1N

N1k=0 XM [k] cos

2pi(2n+1+N)(2k+1)8N

MDCT from a DCT of length 2N :

Split x[n] into four N2 -vectors:x =

[a b c d

]

Form the N -vectoru =

[dc a

b

]

where c is in reverse order

Form v =[u u

]

Take 2N DCT: v VC

Symmetry VC [2k] = 0

Set XM [k] = VC [2k + 1]

a b c da

-b 1peaks sharpened.Pole at z = p gives peak bandwidth 2 |log |p|| 2 (1 |p|)For pole near unit circle, decrease bandwidth by 2 log

IIR Impulse response examples

5: Filters

Difference Equations

FIR Filters

FIR Symmetries

IIR FrequencyResponse

Negating z

Cubing z

Scaling z

IIR Impulseresponse examples

IIR Impulse response

Other BIBOresponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase

Linear Phase Filters

Summary

MATLAB routines

DSP and Digital Filters (2015-6724) Filters: 5 9 / 19

To find the impulse response of B(z)A(z) , use partial fractions:

Example 1: M < N

B(z)A(z) =

13.4z1

10.3z10.4z2 =3

1+0.5z1 +2

10.8z1

h[n] = (3 (0.5)n 2 (0.8)n) u[n]Example 2: M N

B(z)A(z) =

25.7z1+0.2z2+0.8z3

10.3z10.4z2

= 1 2z1 + 13.4z110.3z10.4z2= 1 2z1 + 31+0.5z1 + 210.8z1

h[n] = [n] 2[n 1] + (3 (0.5)n 2 (0.8)n) u[n]Example 3: repeated pole at z = 0.5

B(z)A(z) =

35z11.3z2

1+0.3z10.55z20.2z3

= 21+0.5z1 +3

(1+0.5z1)2+ 210.8z1

h[n] = (2 (0.5)n + 3(n+ 1) (0.5)n 2 (0.8)n)u[n]


5: Filters


FIR Filters

FIR Symmetries


Negating z

Cubing z

Scaling z

IIR Impulse responseexamples

IIR Impulseresponse

Other BIBOresponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


H(z) = B(z)A(z) we want h[n], the inverse z-Transform of H(z).

(1) Assume M < N else, do a long division: H(z) = D(z) + B(z)A(z)

(2) Factorize A(z) =Np

i=1

(1 piz1

)Ki (Np= # of distinct poles)(3) Then B(z)

A(z) =Np

i=1

Kik=1

ri,k

(1piz1)k (Ki= multiplicity of pi)

where ri,k =(pi)

kKi

(Kik)!dKik

d(z1)Kik

(B(z)A(z) (1 piz1)Ki

)z=pi

For the common case Ki = 1 i (e.g. all poles distinct):B(z)A(z) =

Npi=1

ri1piz1

where ri =B(z)A(z) (1 piz1)

z=pi

(4) The impulse response is given by

h[n] =Np

i=1

Kik=1C

n+k1k1 ri,kp

ni for n 0

where C =!

!()! is a binomial coefficient Cn+k1k1 nk1

For the common case Ki = 1 i: h[n] =Np

i=1 ripni for n 0

[Impulse Response: Proof of (3)]

DSP and Digital Filters (2015-6724) Filters: 5 note 1 of slide 10

[You do not need to memorise this proof]

We assume thatB(z)A(z)

=Np

j=1

Kjl=1

rj,l

(1pjz1)l and we want to determine ri,k

(1) Multiply both sides by(1 piz

1)Ki :

B(z)A(z)

(1 piz

1)Ki =Npj=1Kjl=1 rj,l(1piz1)Ki(1pjz1)l

now separate out the term with j = i:

B(z)A(z)

(1 piz

1)Ki =Ki

l=1 ri,l(1 piz

1)Kil +j 6=iKjl=1 rj,l(1piz1)Ki(1pjz1)l

(2) DifferentiateKi k times with respect to z1 : d

Kik

d(z1)Kikand evaluate at z = pi

1 k Ki Ki k < Ki so we are differentiating < Ki times.

Hence all terms in

j 6=i

Kjl=1

rj,l(1piz1)Ki

(1pjz1)l will still include a factor

(1 piz

1)in the

numerator and will equal zero at z = pi.

Similarly, terms inKi

l=1 ri,l(1 piz

1)Kil with l < k will also vanish.

Also, terms inKi

l=1 ri,l(1 piz

1)Kil with l > k will also differentiate to zero.

The only term remaining is: (p)Kik (Ki k)!ri,k from which the result follows.

[Similarly(dn

dxn(xm)

)x=0

is zero unless m = n in which case it equals m!]

[Impulse Response: Proof of (4)]


[You do not need to memorise this proof]

We use induction on k to show that(1 pz1

)k=

n=0 Cn+k1k1 p

nzn

(1) Suppose the formula is true for k and differentiate both sides with respect to z1:

dd(z1)

{(1 pz1

)k=

n=0 Cn+k1k1 p

nzn}

This gives:

pk(1 pz1

)(k+1)=

n=0 nCn+k1k1 p

nz(n1) =

n=1 nCn+k1k1 p

nz(n1)

where we have omitted the term with n = 0 because it equals zero.

Now substitute, m = n 1:

pk(1 pz1

)(k+1)=

m=0(m+ 1)Cm+kk1 p

m+1zm

(1 pz1

)(k+1)=

m=0m+1k

Cm+kk1 p

mzm =

m=0 Cm+kk

pmzm

which proves the result for k + 1.

(2) To start the induction at k = 1:(1 pz1

)1=

n=0 Cn0 p

nzn =

n=0 pnzn

This is true from the standard geometric progression formula. (Note that Cn0 = 1n).

Other BIBO responses

5: Filters


FIR Filters

FIR Symmetries


Negating z

Cubing z

Scaling z



Other BIBOresponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


All LTI systems described by difference equations have h[n] decayingexponentially or exponentially times a power of n.

It is perfectly possible to have BIBO stable LTI systems whose impulseresponses decay more slowly than this.

Example: h[n] = 11+n2 for n 0BIBO stable since

|h[n]| = 2.077...

Stability Triangle

5: Filters


FIR Filters

FIR Symmetries


Negating z

Cubing z

Scaling z



Other BIBOresponses

Stability TriangleLow-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


Not normally easy to tell if poles lie within|z| = 1, but easy test for a 2nd order filter:Suppose A(z) = 1 + a[1]z1 + a[2]z2

The roots are p1,2 =a[1]

a[1]24a[2]

2

Want the conditions that |p1,2| < K-2K 0 2K

0

PQ

RSK

2

-K2

a[1]

Case 1: Real roots:(P) a[1]2 4a[2](Q) 2K < a[1]

a[1]2 4a[2] < 2K

a[1]2 4a[2] < 2K a[1]

a[1]2 4a[2] < 4K2 4a[1]K + a[1]2 a[2] > K2 a[1]K

Case 2: Complex roots:(R) a[1]2 < 4a[2]

(S) |p1,2|2 = a[2] < K2

Hence coefficients must lie within a stability triangle.

Low-pass filter

5: Filters


FIR Filters

FIR Symmetries


Negating z

Cubing z

Scaling z



Other BIBOresponses

Stability Triangle

Low-pass filterAllpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


1st order low pass filter: extremely commony[n] = (1 p)x[n] + py[n 1] H(z) = 1p1pz1

Impulse response:h[n] = (1 p)pn = (1 p)en

where = 1 ln p is the time constant in samples.

Magnitude response:H(ej) = 1p

12p cos+p2

Low-pass filter with DC gain of unity.

3 dB frequency is 3dB = cos1(1 (1p)22p

) 2 1p1+p 1

Compare continuous time: HC(j) =1

1+j

Indistinguishable for low but H(ej) is periodic, HC(j) is not

-1 0 1-1

-0.5

0

0.5

1

(z)

p=0.80

0.01 0.1

-30

-20

-10

0

H(j)

HC(j)

1/ 2pi

(rad/sample)

Allpass filters

5: Filters


FIR Filters

FIR Symmetries


Negating z

Cubing z

Scaling z



Other BIBOresponses

Stability Triangle

Low-pass filter

Allpass filtersGroup Delay

Minimum Phase


Summary

MATLAB routines


If a[n] is real and a[n] = b[M n] then we have an allpass filterH(ej) =

Mr=0 b[r]e

jr

Mr=0 b[r]e

j(Mr) = ejM

Mr=0 b[r]e

jr

Mr=0 b[r]e

jr

The two sums are complex conjugates they have the same magnitudeHence

H(ej) = 1 allpassHowever phase is not constant: H(ej) = M + 2B(ej)

1st order allpass: H(z) = p+z1

1pz1 = p 1p1z1

1pz1

Pole at p and zero at p1: reflected in unit circle

Constant distance ratio:ej p = |p|

ej 1p

0 1 2 300.20.40.60.8

1

0 1 2 3

-3

-2

-1

0

-1 0 1-1

-0.5

0

0.5

1

(z)

In an allpass filter, the zeros are the poles reflected in the unit circle.

Group Delay

5: Filters


FIR Filters

FIR Symmetries


Negating z

Cubing z

Scaling z



Other BIBOresponses

Stability Triangle

Low-pass filter

Allpass filters

Group DelayMinimum Phase


Summary

MATLAB routines


Group delay: H(ej) = dH(ej)

d= delay of the modulation envelope.

Trick to get at phase: lnH(ej) = lnH(ej)+ jH(ej)

H =d((lnH(ej)))

d=

(1

H(ej)dH(ej)

d

)=

(zH(z)

dHdz

)z=ej

H(ej) =

n=0 h[n]ejn= F (h[n]) [F = DTFT]

dH(ej)d

=

n=0jnh[n]ejn= jF (nh[n])

H = (

1H(ej)

dH(ej)d

)=

(jF(nh[n])F(h[n])

)=

(F(nh[n])F(h[n])

)

Example: H(z) = 11pz1 H = [1 p]= (pej

1pej

)

-1 0 1-1

-0.5

0

0.5

1

(z)

0 1 2 3

-0.8

-0.6

-0.4

-0.2

0p=0.80

0 1 2 3

0

1

2

3 p=0.80

H

Average group delay (over ) = (# poles # zeros) within the unit circleZeros on the unit circle count

[Group Delay Properties]


The group delay of a filter H(z) at a frequency , measured in seconds or samples, gives the time delay

of the envelope of a modulated sine wave at a frequency . It is defined as H(ej) =

dH(ej)d

.

For example, H(z) = zk defines a filter that delays its input by k samples and we can calculate thegroup delay by evaluating

H(ej) =

dH(ej)

d=

d

d

(ejk

)=

d

d(k) = k

which tells us that this filter has a constant group delay of k samples that is independent of .

The average value of H equals the total change in H(ej) as goes from pi to +pi divided by

2pi. If you imagine an elastic string connecting a pole or zero to the point z = ej, you can see that as goes from pi to +pi the string will wind once around the pole or zero if it is inside the unit circle butnot if it is outside. Thus, the total change in H(ej) is equal to 2pi times the number of poles insidethe unit circle minus the number of zeros inside the unit circle. A zero that is exactly on the unit circlecounts 1

2since there is a sudden discontinuity of pi in H(ej) as passes through the zero position.

When you multiply or divide complex numbers, their phases add or subtract, so it follows that whenyou multiply or divide transfer functions their group delays will add or subtract. Thus, for example,

the group delay of an IIR filter, H(z) =B(z)A(z)

, is given by H = B A. This means too that we

can determine the group delay of a factorized transfer function by summing the group delays of theindividual factors.

[Group Delay from h[n] or H(z)]


The slide shows how to determine the group delay, H , from either the impulse response, h[n], orthe transfer function, H(z). We start by using a trick that is very common: if you want to get atthe magnitude and phase of a complex number separately, you can do so by taking its natural log:ln(rej

)= ln |r| + j or, in general, lnH = ln |H| + jH. By rearranging this equation, we get

H = (lnH) where ( ) denotes taking the imaginary part of a complex number. Using this, we canwrite

H =d((lnH(ej)

))d

=

(d(lnH(ej)

)d

)=

(1

H(ej)

dH(ej)

d

). (1)

By going back to the definition of the DTFT, we find that H(ej) = F (h[n]) anddH(ej)

d=

jF (nh[n]) where F ( ) denotes the DTFT. Substituting these expressions into the above equationgives us a formula for H in terms of the impulse response h[n].

H =

(F (nh[n])

F (h[n])

)(2)

In order to express H in terms of z, we first note that if z = ej then dz

d= jz. By substituting

z = ej into equation (1), we get

H =

(1

H(z)

dH(z)

d

)=

(1

H(z)

dH(z)

dz

dz

d

)=

(jz

H(z)

dH(z)

dz

)=

(z

H(z)

dH(z)

dz

)z=ej

.

[Group Delay Example]


As an example, suppose we want to determine the group delay of : H(z) = 11pz1

. As noted above,

if H(z) =B(z)A(z)

, then H = B A. In this case B = 0 so H = [1 p].

Using equation (2) gives H = (

F([0 p])F([1 p])

)since nh[n] = [0 1] [1 p].

Applying the definition of the DTFT, we get

H =

(pej

1 pej

)=

(p

ej p

)=

(p(ej p

))(ej p) (ej p)

=p cos p2

1 2p cos + p2

As demonstrated above, the average value of H is zero for this filter because there is one pole and onezero inside the unit circle.

Minimum Phase

5: Filters


FIR Filters

FIR Symmetries


Negating z

Cubing z

Scaling z



Other BIBOresponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum PhaseLinear Phase Filters

Summary

MATLAB routines


Average group delay (over ) = (# poles # zeros) within the unit circle

zeros on the unit circle count Reflecting an interior zero to the exteriormultiplies

H(ej) by a constant butincreases average group delay by 1 sample. 0 1 2 3

0

1

2

3

4

-1 0 1

-1-0.5

00.5

1

(z)

0 1 2 3

-10

-5

0

-1 0 1

-1-0.5

00.5

1

(z)

0 1 2 3-10

0

10

20

30

H

A filter with all zeros inside the unit circle is a minimum phase filter: Lowest possible group delay for a given magnitude response Energy in h[n] is concentrated towards n = 0


5: Filters


FIR Filters

FIR Symmetries


Negating z

Cubing z

Scaling z



Other BIBOresponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase

Linear PhaseFilters

Summary

MATLAB routines


The phase of a linear phase filter is: H(ej) = 0 Equivalently constant group delay: H = dH(e

j)d

=

A filter has linear phase iff h[n] is symmetric or antisymmetric:h[n] = h[M n] n or else h[n] = h[M n] n

M can be even ( mid point) or odd ( mid point)Proof :2H(ej) =

M0 h[n]e

jn +M

0 h[M n]ej(Mn)= ej

M2

M0 h[n]e

j(nM2 ) + h[M n]ej(nM2 )

h[n] symmetric:

2H(ej) = 2ejM2

M0 h[n] cos

(n M2

)

h[n] anti-symmetric:

2H(ej) = 2jejM2 M0 h[n] sin(n M2

)

= 2ej(pi2 +

M2 )M

0 h[n] sin(n M2

)

Summary

5: Filters


FIR Filters

FIR Symmetries


Negating z

Cubing z

Scaling z



Other BIBOresponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase




Useful filters have difference equations: Freq response determined by pole/zero positions N M zeros at origin (or M N poles) Geometric construction of |H(ej)|

Pole bandwidth 2 |log |p|| 2 (1 |p|) Stable if poles have |p| < 1

Allpass filter: a[n] = b[M n] Reflecting a zero in unit circle leaves |H(ej)| unchanged

Group delay: H(ej

)= dH(ej)

dsamples

Symmetrical h[n] H(ej

)= M2

Average H over = (# poles # zeros) within the unit circle Minimum phase if zeros have |q| 1

Lowest possible group delay for given |H(ej)| Linear phase = Constant group Delay = symmetric/antisymmetric h[n]

For further details see Mitra: 6, 7.

MATLAB routines

5: Filters


FIR Filters

FIR Symmetries


Negating z

Cubing z

Scaling z



Other BIBOresponses

Stability Triangle

Low-pass filter

Allpass filters

Group Delay

Minimum Phase


Summary

MATLAB routines


filter filter a signalimpz Impulse response

residuez partial fraction expansiongrpdelay Group Delayfreqz Calculate filter frequency response

6: Window Filter Design

6: Window FilterDesign

Inverse DTFT

Rectangular window

Dirichlet Kernel

Window relationships

Common Windows

Uncertainty principle

Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines

DSP and Digital Filters (2015-6729) Windows: 6 1 / 12

Inverse DTFT


Inverse DTFTRectangular window

Dirichlet Kernel


Common Windows


Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


For any BIBO stable filter, H(ej) is the DTFT of h[n]

H(ej) = h[n]e

jn h[n] = 12

H(ej)ejnd

If we know H(ej) exactly, the IDTFT gives the ideal h[n]

Example: Ideal Lowpass filter

H(ej) =

{1 || 0

0 || > 0 h[n] = sin0n

n

-2 0 20

0.5

1

20

0

2pi/0

Note: Width in is 20, width in n is20: product is 4 always

Sadly h[n] is infinite and non-causal. Solution: multiply h[n] by a window

Rectangular window


Inverse DTFT

Rectangularwindow

Dirichlet Kernel


Common Windows


Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


Truncate to M2 to make finite; h1[n] is now of length M + 1

MSE Optimality:Define mean square error (MSE) in frequency domain

E = 12

H(ej)H1(ej)2 d= 12

H(ej)M2M

2

h1[n]ejn

2 dMinimum E is when h1[n] = (h[n])Proof: Differentiate w.r.t. h1[r] and set to zero

However: 9% overshoot at a discontinuity even for large n.

0

h1[n]M=14

0 1 2 30

0.5

1

M=14M=28

Normal to delay by M2 to make causal. Multiplies H(ej) by ej

M

2.

Dirichlet Kernel


Inverse DTFT

Rectangular window

Dirichlet KernelWindow relationships

Common Windows


Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


Truncation Multiply h[n] by a rectangular window, w[n] = M2nM

2

Circular Convolution HM+1(ej) = 12H(e

j)W (ej)

W (ej) =M

2

M2

ejn(i)= 1 + 2

0.5M1 cos (n)

(ii)= sin 0.5(M+1)sin 0.5

Proof: (i) ej(n) + ej(+n) = 2 cos (n) (ii) Sum geom. progression

Effect: convolve ideal freq response with Dirichlet kernel (aliassed sinc)

-2 0 2

0

0.5

1

-2 0 2

0

0.5

1

4pi/(M+1)

-2 0 2

0

0.5

1

-2 0 2

0

0.5

1 M=14

Provided that 4M+1 < 20 M + 1 >

20:

Passband ripple: 4M+1 , stopband

2M+1

Transition pk-to-pk: 4M+1

Transition Gradient: d|H|d

=0

M+12

[Dirichlet Kernel]

DSP and Digital Filters (2015-6729) Windows: 6 note 1 of slide 4

Other properties of W (ej):

The DTFT of a symmetric rectangular window of length M + 1 is W (ej) =sin 0.5(M+1)

sin 0.5. For

small x we can approximate sinx x; the error is < 1% for x < 0.25. So, for < 0.5,we haveW (ej) 21 sin 0.5(M + 1).

The peak value is at = 0 and equals M + 1;this means that the peak gradient of HM+1(ej) will

be M+12pi

.

The minimum value of W (ej) is approximately equal to the minimuum of 21 sin 0.5(M + 1)which is when sin 0.5(M + 1) = 1 i.e. = 1.5pi

0.5(M+1)= 3pi

M+1. Hence minW (ej)

min 21 sin 0.5(M + 1) = M+11.5pi

.

Passband and Stopband ripple:

The ripple in W (ej) =sin 0.5(+1)sin 0.5

has a period of = 2pi0.5(+1)

= 4piM+1

and this gives rise to

ripple with this period in both the passband and stopband of HM+1(ej).

However the stopband ripple takes the value of HM+1(ej) alternately positive and negative. If you

plot the magnitude response,HM+1(ej) then this ripple will be full-wave rectified and will double in

frequency so its period will now be 2piM+1

.



Inverse DTFT

Rectangular window

Dirichlet Kernel

Windowrelationships

Common Windows


Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


When you multiply an impulse response by a window M + 1 longHM+1(e

j) = 12H(ej)W (ej)

-2 0 20

0.5

1

-2 0 2

0

10

20

M=20

-2 0 20

0.5

1

1

(a) passband gain w[0]; peak w[0]2 +0.52

mainlobe

W (ej)drectangular window: passband gain = 1; peak gain = 1.09

(b) transition bandwidth, = width of the main lobetransition amplitude, H = integral of main lobe2rectangular window: = 4

M+1 , H 1.18

(c) stopband gain is an integral over oscillating sidelobes of W (ej)rect window:

minH(ej) = 0.09 minW (ej) = M+11.5(d) features narrower than the main lobe will be broadened and

attenuated

Common Windows


Inverse DTFT

Rectangular window

Dirichlet Kernel


Common WindowsUncertainty principle

Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


Rectangular: w[n] 1dont use

Hanning: 0.5 + 0.5c1ck = cos

2knM+1

rapid sidelobe decay

Hamming: 0.54 + 0.46c1best peak sidelobe

Blackman-Harris 3-term:0.42 + 0.5c1 + 0.08c2

best peak sidelobe

Kaiser:I0

(

1( 2nM )

2

)

I0()

controls width v sidelobesGood compromise:Width v sidelobe v decay

0 1 2 3

-50

0-13 dB6.27/(M+1)

0 1 2 3

-50

0-31 dB

12.56/(M+1)

0 1 2 3

-50

0

-40 dB12.56/(M+1)

0 1 2 3

-50

0

-70 dB

18.87/(M+1)

0 1 2 3

-50

0

-40 dB13.25/(M+1)

= 5.3

0 1 2 3

-50

0

-70 dB

21.27/(M+1) = 9.5



Inverse DTFT

Rectangular window

Dirichlet Kernel


Common Windows

Uncertaintyprinciple

Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


CTFT uncertainty principle:(

t2|x(t)|2dt|x(t)|2dt

) 12

( 2|X(j)|2d|X(j)|2d

) 12

12

The first term measures the width of x(t) around t = 0.

It is like if |x(t)|2was a zero-mean probability distribution.

The second term is similarly the width of X(j) in frequency.A signal cannot be concentrated in both time and frequency.

So a short window cannot have a narrow main lobe.

Proof:Assume

|x(t)|

2dt = 1

|X(j)|

2d = 2 [Parseval]

Set v(t) = dxdt V (j) = jX(j) [by parts]

Nowtxdx

dtdt= 12 tx

2t=

12x

2dt = 12 [by parts]

So 14 = txdx

dtdt2 ( t2x2dt) ( dx

dt

2 dt) [Schwartz]=(

t2x2dt) (

|v(t)|2 dt)=(

t2x2dt) (

12

|V (j)|2 d

)=(

t2x2dt) (

12

2 |X(j)|

2d)

No exact equivalent for DTFT/DFT but a similar effect is true

[Uncertainty Principle Proof Steps]

DSP and Digital Filters (2015-6729) Windows: 6 note 1 of slide 7

(1) Suppose v(t) = dxdt

. Then integrating the CTFT definition by parts w.r.t. t gives

X(j) =

x(t)ejtdt =

[1j

x(t)ejt]

+ 1j

dx(t)dt

ejtdt = 0 + 1j

V (j)

(2) Since ddt

(12x2

)= x dx

dt, we can apply integration by parts to get

tx dx

dtdt =

[t 1

2x2

]

t=

dtdt 1

2x2dt = 1

2

x2dt = 1

2 1 = 1

2

It follows that

tx dxdt

dt

2 = ( 12 )2 = 14 which we will use below.(3) The Cauchy-Schwarz inequality is that in a complex inner product space

|u v|2 (u u) (v v). For the inner-product space of real-valued square-integrable functions,

this becomes

u(t)v(t)dt2

u2(t)dt

v2(t)dt. We apply this with u(t) = tx(t)

and v(t) =dx(t)dt

to get

14=

tx dxdt

dt

2 ( t2x2dt)( (

dxdt

)2dt

)=

(t2x2dt

) (v2(t)dt

)

(4) From Parsevals theorem for the CTFT,v2(t)dt = 1

2pi

|V (j|2 d. From step (1), we can

substitute V (j) = jX(j) to obtainv2(t)dt = 1

2pi

2 |X(j|2 d. Making this substitution

in (3) gives14

(t2x2dt

) (v2(t)dt

)=

(t2x2dt

) (12pi

2 |X(j|2 d

)

Order Estimation


Inverse DTFT

Rectangular window

Dirichlet Kernel


Common Windows


Order EstimationExample Design

Frequency sampling

Summary

MATLAB routines


Several formulae estimate the required order of a filter, M .

E.g. for lowpass filter

Estimated order is

M 5.64.3 log10()21

820 log10 2.2

Required M increases as either thetransition width, 2 1, or the gaintolerances and get smaller.Only approximate.

Example:Transition band: f1 = 1.8 kHz, f2 = 2.0 kHz, fs = 12 kHz,.

1 =2f1fs

= 0.943, 2 =2f2fs

= 1.047

Ripple: 20 log10 (1 + ) = 0.1 dB, 20 log10 = 35 dB

= 100.1

20 1 = 0.0116, = 1035

20 = 0.0178

M 5.64.3 log

10(2104)1.0470.943 =

10.250.105 = 98 or

3582.2 = 117

Example Design


Inverse DTFT

Rectangular window

Dirichlet Kernel


Common Windows


Order Estimation

Example DesignFrequency sampling

Summary

MATLAB routines


Specifications:Bandpass: 1 = 0.5, 2 = 1Transition bandwidth: = 0.1Ripple: = = 0.02

20 log10 = 34 dB20 log10 (1 + ) = 0.17 dB

Order:M 5.64.3 log10()

21= 92

Ideal Impulse Response:Difference of two lowpass filtersh[n] = sin2n

n sin1n

n

Kaiser Window: = 2.5

0

M=92

0 1 2 30

0.5

1

0 1 2 30

0.5

1 M=92 = 2.5

0 1 2 3-60

-40

-20

0 M=92 = 2.5

Frequency sampling


Inverse DTFT

Rectangular window

Dirichlet Kernel


Common Windows


Order Estimation

Example Design

Frequencysampling

Summary

MATLAB routines


Take M + 1 uniform samples of H(ej); take IDFT to obtain h[n]

Advantage:exact match at sample points

Disadvantage:poor intermediate approximation if spectrum is varying rapidly

Solutions:(1) make the filter transitions smooth over width(2) oversample and do least squares fit (cant use IDFT)(3) use non-uniform points with more near transition (cant use IDFT)

-2 0 20

0.5

1 M+1=93

0 1 2 3

0

0.5

1 M+1=93

Summary


Inverse DTFT

Rectangular window

Dirichlet Kernel


Common Windows


Order Estimation

Example Design

Frequency sampling



Make an FIR filter by windowing the IDTFT of the ideal response

Ideal lowpass has h[n] = sin0nn

Add/subtract lowpass filters to make any piecewise constantresponse

Ideal filter response is with the DTFT of the window Rectangular window (W (z) =Dirichlet kernel) has 13 dB

sidelobes and is always a bad idea Hamming, Blackman-Harris are good Kaiser good with trading off main lobe width v. sidelobes

Uncertainty principle: cannot be concentrated in both time andfrequency

Frequency sampling: IDFT of uniform frequency samples: not so great

For further details see Mitra: 7, 10.

MATLAB routines


Inverse DTFT

Rectangular window

Dirichlet Kernel


Common Windows


Order Estimation

Example Design

Frequency sampling

Summary

MATLAB routines


diric(x,n) Dirichlet kernel: sin 0.5nxsin 0.5xhanninghammingkaiser

Window functions(Note periodic option)

kaiserord Estimate required filter order and

7: Optimal FIR filters

7: Optimal FIRfilters

Optimal Filters

Alternation TheoremChebyshevPolynomials

Maximal ErrorLocationsRemez ExchangeAlgorithm

Determine Polynomial

Example Design

FIR Pros and Cons

Summary

MATLAB routines

DSP and Digital Filters (2015-6729) Optimal FIR: 7 1 / 11

Optimal Filters


Optimal FiltersAlternation TheoremChebyshevPolynomials



Example Design

FIR Pros and Cons

Summary

MATLAB routines


We restrict ourselves to zero-phase filters of odd length M + 1, symmetricaround h[0], i.e. h[n] = h[n].

H() = H(ej) =M

2

M2

h[n]ejn= h[0] + 2M

2

1 h[n] cosn

H() is real but not necessarily positive (unlikeH(ej)).

Weighted error: e() = s()(H() d()

)where d() is the target.

Choose s() to control the error variation with .

Example: lowpass filter

d() =

{1 0 1

0 2

s() =

{1 0 1

1 2

e() = 1 when H() lies at the edge of the specification.

Minimax criterion: h[n] = argminh[n]max |e()|: minimize max error

Alternation Theorem


Optimal Filters

AlternationTheorem

ChebyshevPolynomials



Example Design

FIR Pros and Cons

Summary

MATLAB routines


Want to find the best fit line: with the smallest maximal error.

Best fit line always attains themaximal error three times withalternate signs

2 4 6 82

4

6

8

Proof:Assume the first maximal deviation from the line is negative as shown.There must be an equally large positive deviation; or else just move the linedownwards to reduce the maximal deviation.This must be followed by another maximal negative deviation; or else youcan rotate the line and reduce the deviations.

Alternation Theorem:A polynomial fit of degree n to a bounded set of points is minimax if andonly if it attains its maximal error at n+ 2 points with alternating signs.There may be additional maximal error points.Fitting to a continuous function is the same as to an infinite number ofpoints.

Chebyshev Polynomials


Optimal Filters

Alternation Theorem

ChebyshevPolynomials



Example Design

FIR Pros and Cons

Summary

MATLAB routines


H() = H(ej) = h[0] + 2M

2

1 h[n] cosn

But cosn = Tn(cos): Chebyshev polynomial of 1st kind

cos 2 = 2 cos2 1 = T2(cos) T2(x) = 2x2 1

cos 3 = 4 cos3 3 cos = T3(cos) T3(x) = 4x3 3x

Recurrence Relation:Tn+1(x) = 2xTn(x) Tn1(x) with T0(x) = 1, T1(x) = x

Proof: cos (n + ) + cos (n ) = 2 cos cosn

So H() is an M2 order polynomial in cos: alternation theorem applies.

Example: Symmetric lowpass filter of orderM = 4H(z) = 0.1766z2 + 0.4015z + 0.2124 + 0.4015z1 + 0.1766z2

Maximal Error Locations


Optimal Filters


Maximal ErrorLocations

Remez ExchangeAlgorithm


Example Design

FIR Pros and Cons

Summary

MATLAB routines


Maximal error locations occur either at bandedges or when dH

d= 0

H() = h[0] + 2M

2

1 h[n] cosn= P (cos)

where P (x) is a polynomial of order M2 . 0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

|

H

|

M=18

dHd

= P (cos) sin

= 0 at = 0, and at most M2 1 zeros of polynomial P(x).

With two bands, we have at most M2 + 3 maximal error frequencies.

We require M2 + 2 of alternating signs for the optimal fit.

Only three possibilities exist (try them all):

(a) = 0 + two band edges + all(M2 1

)zeros of P (x).

(b) = + two band edges + all(M2 1

)zeros of P (x).

(c) = {0 and } + two band edges +(M2 2

)zeros of P (x).

Remez Exchange Algorithm


Optimal Filters


Maximal ErrorLocations

Remez ExchangeAlgorithm


Example Design

FIR Pros and Cons

Summary

MATLAB routines


1. Guess the positions of the M2 + 2 maximal error frequencies and givealternating signs to the errors (e.g. choose evenly spaced ).

2. Determine the error magnitude, , and the M2 + 1 coefficients ofthe polynomial that passes through the maximal error locations.

3. Find the local maxima of the error function by evaluatinge() = s()

(H() d()

)on a dense set of .

4. Update the maximal error frequencies to be an alternating subset ofthe local maxima + band edges + {0 and/or }.If maximum error is > , go back to step 2. (typically 15 iterations)

5. Evaluate H() on M + 1 evenly spaced and do an IDFT to get h[n].

Remex Step 2: Determine Polynomial


Optimal Filters



DeterminePolynomial

Example Design

FIR Pros and Cons

Summary

MATLAB routines


For each extremal frequency, i for 1 i M2 + 2

d(i) = H(i) +(1)is(i)

= h[0] + 2M

2

n=1 h[n] cosni +(1)is(i)

Method 1: (Computation time M3)Solve M2 + 2 equations in

M2 + 2 unknowns for h[n] + .

In step 3, evaluate H() = h[0] + 2M

2

n=1 h[n] cosni

Method 2: Dont calculate h[n] explicitly (Computation time M2)Multiply equations by ci =

j 6=i

1cosicosj

and add:M2+2

i=1 ci

(h[0] + 2

M2

n=1 h[n] cosn +(1)is(i)

)=M

2+2

i=1 cid(i)

All terms involving h[n] sum to zero leavingM2+2

i=1(1)icis(i)

=M

2+2

i=1 cid(i)

Solve for then calculate the H(i) then use Lagrange interpolation:

H() = P (cos) =M

2+2

i=1 H(i)

j 6=icoscosjcosicosj(

M2 + 1

)-polynomial going through all the H(i) [actually order

M2 ]

Example Design


Optimal Filters




Example DesignFIR Pros and Cons

Summary

MATLAB routines


Filter Specifications:Bandpass = [0.5, 1], Transition widths: = 0.2Stopband Attenuation: 25 dB and 15 dBPassband Ripple: 0.3 dB

Determine gain tolerances for each band:25 dB = 0.056, 0.3 dB = 1 0.034, 15 dB = 0.178

Predicted order: M = 36M2 + 2 extremal frequencies are distributed between the bandsFilter meets specs ,; clearer on a decibel scale

Most zeros are on the unit circle + three reciprocal pairsReciprocal pairs give a linear phase shift

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

|

H

|

M=36

0 0.5 1 1.5 2 2.5 3-30

-25

-20

-15

-10

-5

0

|

H

|

(

d

B

)

M=36

-1 0 1 2

-1

-0.5

0

0.5

1

FIR Pros and Cons


Optimal Filters




Example Design

FIR Pros and ConsSummary

MATLAB routines


Can have linear phase no envelope distortion, all frequencies have the same delay , symmetric or antisymmetric: h[n] = h[n]n or h[n]n antisymmetric filters have H(ej0) = H(ej) = 0 symmetry means you only need M2 + 1 multiplications

to implement the filter.

Always stable ,

Low coefficient sensitivity ,

Optimal design method fast and robust ,

Normally needs higher order than an IIR filter / Filter order M dBatten3.5 where is the most rapid transition

Filtering complexity M fs dBatten3.5 fs =

dBatten3.5 f

2s

f2s for a given specification in unscaled units.

Summary


Optimal Filters




Example Design

FIR Pros and Cons



Optimal Filters: minimax error criterion

use weight function, s(), to allow different errorsin different frequency bands

symmetric filter has zeros on unit circle or in reciprocal pairs Response of symmetric filter is a polynomial in cos Alternation Theorem: M2 + 2 maximal errors with alternating signs

Remez Exchange Algorithm (also known as Parks-McLellan Algorithm)

multiple constant-gain bands separated by transition regions very robust, works for filters with M > 1000 Efficient: computation M2

can go mad in the transition regions

Modified version works on arbitrary gain function

Does not always converge

For further details see Mitra: 10.

MATLAB routines


Optimal Filters




Example Design

FIR Pros and Cons

Summary

MATLAB routines


firpm optimal FIR filter designfirpmord estimate require order for firpmcfirpm arbitrary-response filter designremez [obsolete] optimal FIR filter design

8: IIR Filter Transformations

8: IIR FilterTransformations

Continuous TimeFilters

Bilinear Mapping

Continuous TimeFiltersMapping Poles andZerosSpectralTransformationsConstantinidesTransformations

Impulse Invariance

Summary

MATLAB routines

DSP and Digital Filters (2015-6707) IIR Transformations: 8 1 / 10

Continuous Time Filters

8: IIR FilterTransformations


Bilinear Mapping


Impulse Invariance

Summary

MATLAB routines


Classical continuous-time filters optimize tradeoff:passband ripple v stopband ripple v transition widthThere are explicit formulae for pole/zero positions.

Butterworth: G2() =H(j)2 = 11+2N

Monotonic G() = 1 12

2N + 384N +

Maximally flat: 2N 1 derivatives are zero

Chebyshev: G2() = 11+2T 2

N()

where polynomial TN (cosx) = cosNx passband equiripple + very flat at

Inverse Chebyshev: G2() = 11+(2T 2N (1))

1

stopband equiripple + very flat at 0

Elliptic: [no nice formula] Very steep + equiripple in pass and stop bands

0.1 0.2 0.5 1 2 5 100

0.2

0.4

0.6

0.8

1 N=5

Frequency (rad/s)

|

H

|

0.1 0.2 0.5 1 2 5 100

0.2

0.4

0.6

0.8

1

Frequency (rad/s)

|

H

|

0.1 0.2 0.5 1 2 5 100

0.2

0.4

0.6

0.8

1

Frequency (rad/s)

|

H

|

0.1 0.2 0.5 1 2 5 100

0.2

0.4

0.6

0.8

1

Frequency (rad/s)

|

H

|

Bilinear Mapping

8: IIR FilterTransformationsContinuous TimeFilters

Bilinear MappingContinuous TimeFiltersMapping Poles andZerosSpectralTransformationsConstantinidesTransformations

Impulse Invariance

Summary

MATLAB routines


Change variable: z = +ss s =

z1z+1 : a one-to-one invertible mapping

axis (s) axis (z)

axis (s) Unit circle (z)

Proof: z = ejs = ej1ej+1 =

ej 2 ej

2

ej 2 +ej

2= j tan 2= j

Left half plane(s) inside of unit circle (z)

Proof: s = x+ jy |z|2 = |(+x)+jy|2

|(x)jy|2

= 2+2x+x2+y2

22x+x2+y2 = 1 +4x

(x)2+y2

x < 0 |z| < 1

Unit circle (s) axis (z)

-4 -2 0 2 4-4

-3

-2

-1

0

1

2

3

4s-plane

= 1

-2 -1 0 1 2-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5z-plane

= 1

0 2 4 6 8 100

0.5

1

1.5

2

2.5

3

/

Continuous Time Filters


Bilinear Mapping


Mapping Poles andZerosSpectralTransformationsConstantinidesTransformations

Impulse Invariance

Summary

MATLAB routines


Take H(s) = 1s2+0.2s+4 and choose = 1

Substitute: s = z1z+1 [extra zeros at z = 1]

H(z) = 1( z1z+1 )

2+0.2 z1

z+1+4

= (z+1)2

(z1)2+0.2(z1)(z+1)+4(z+1)2

= z2+2z+1

5.2z2+6z+4.8 = 0.191+2z1+z2

1+1.15z1+0.92z2

Frequency response is identical (both magnitude andphase) but with a distorted frequency axis:

Frequency mapping: = 2 tan1

=[ 2 3 4 5

] =

[1.6 2.2 2.5 2.65 2.75

]

1 2 3 4 5 60

0.5

1

1.5

2

2.5

Frequency (rad/s)

|

H

|

0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

(rad/sample)

|

H

|

=1

0 2 4 6 8 100

0.5

1

1.5

2

2.5

3

/

Choosing : Set = 0tan 1

20

to map 0 0

Set = 2fs =2T

to map low frequencies to themselves

Mapping Poles and Zeros


Bilinear Mapping


Mapping Poles andZeros

SpectralTransformationsConstantinidesTransformations

Impulse Invariance

Summary

MATLAB routines


Alternative method: H(s) = 1s2+0.2s+4

Find the poles and zeros: ps = 0.1 2jMap using z = 1+s1s pz = 0.58 0.77j

After the transformation we will always end up withthe same number of poles as zeros:Add extra poles or zeros at z = 1

H(z) = g (1+z1)2

(1+(0.580.77j)z1)(1+(0.58+0.77j)z1)

= g 1+2z1+z2

1+1.15z1+0.92z2

Choose overall scale factor, g, to give the same gainat any convenient pair of mapped frequencies:

At 0 = 0 s0 = 0H(s0)

= 0.25 0 = 2 tan

1 0

= 0 z0 = ej0 = 1

|H(z0)| = g 4

3.08 = 0.25 g = 0.19

H(z) = 0.19 1+2z1+z2

1+1.15z1+0.92z2

1 2 3 4 5 60

0.5

1

1.5

2

2.5

Frequency (rad/s)

|

H

|

-3 -2 -1 0 1 2

-2

-1

0

1

2

s

-1 0 1

-1

-0.5

0

0.5

1

z

=1

0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

(rad/sample)

|

H

|

=1

Spectral Transformations


Bilinear Mapping

Continuous TimeFiltersMapping Poles andZeros

SpectralTransformations

ConstantinidesTransformations

Impulse Invariance

Summary

MATLAB routines


We can transform the z-plane to change the cutofffrequency by substituting

z = z1z z =z+1+z

Frequency Mapping:

If z = ej, then z = z 1+z1

1+z has modulus 1since the numerator and denominator arecomplex conjugates.Hence the unit circle is preserved.

ej = ej+

1+ej

Some algebra gives: tan 2 =(1+1

)tan 2

Equivalent to:z s = z1

z+1 s =11+s z =

1+s1s

Lowpass Filter example:Inverse Chebyshev

0 =2 = 1.57

=0.6 0 = 0.49

-2 -1 0 1 2-2

-1

0

1

2z

-2 -1 0 1 2-2

-1

0

1

2z^

= 0.6

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

(rad/s)

|

H

|

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

^ (rad/s)

|

H

^

|

= 0.6

^ = 0.49

Constantinides Transformations


Transform any lowpass filter with cutoff frequency 0 to:

Target Substitute Parameters

Lowpass < 1

z1 = z1

1z1 =sin(012 )sin(0+12 )

Highpass > 1

z1 = z1+

1+z1 =cos(0+12 )cos(012 )

Bandpass1 < < 2

z1 = (1)2z1+(+1)z2

(+1)2z1+(1)z2 =cos( 2+12 )cos( 212 )

= cot(21

2

)tan

(02

)Bandstop

1 2

z1 = (1)2z1+(+1)z2

(+1)2z1+(1)z2 =cos( 2+12 )cos( 212 )

= tan(21

2

)tan

(02

)Bandpass and bandstop transformations are quadratic and so will double the order:

-3 -2 -1 0 1 2 30

0.5

1

(rad/s)

|

H

|

Lowpass

-3 -2 -1 0 1 2 30

0.5

1

^ (rad/s)

|

H

^

|

Bandpass

Impulse Invariance


Bilinear Mapping


Impulse InvarianceSummary

MATLAB routines


Bilinear transform works well for a lowpass filter but the non-linearcompression of the frequency distorts any other response.

Alternative method: H(s)L

1

h(t)sample h[n] = T h(nT )

Z H(z)

Express H(s) as a sum of partial fractions H(s) =N

i=1gi

spi

Impulse response is h(t) = u(t)N

i=1 giepit

Digital filter H(z)T

=N

i=1gi

1epiT z1has identical impulse response

Poles of H(z) are pi = epiT (where T = 1

fsis sampling period)

Zeros do not map in a simple way

Properties:, Impulse response correct. , No distortion of frequency axis./ Frequency response is aliased.

Example: Standard telephone filter - 300 to 3400 Hz bandpass

0 5 10 15 20 250

0.5

1

Frequency (krad/s)

|

H

|

Analog Filter

0 0.5 1 1.5 2 2.5 30

0.5

1

(rad/sample)

|

H

|

Bilinear (fs = 8 kHz)

Matched at 3.4 kHz

0 0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1

(rad/sample)

|

H

|

Impulse Invariance (fs = 8 kHz)

Summary


Bilinear Mapping


Impulse Invariance



Classical filters have optimal tradeoffs in continuous time domain Order transition width pass ripple stop ripple Monotonic passband and/or stopband

Bilinear mapping Exact preservation of frequency response (mag + phase) non-linear frequency axis distortion can choose to map 0 0 for one specific frequency

Spectral transformations lowpass lowpass, highpass, bandpass or bandstop bandpass and bandstop double the filter order

Impulse Invariance Aliassing distortion of frequency response preserves frequency axis and impulse response

For further details see Mitra: 9.

MATLAB routines


Bilinear Mapping


Impulse Invariance

Summary

MATLAB routines


bilinear Bilinear mappingimpinvar Impulse invariancebutter

butterordAnalog or digitalButterworth filter

cheby1cheby1ord

Analog or digitalChebyshev filter

cheby2cheby2ord

Analog or digitalInverse Chebyshev filter

ellipellipord

Analog or digitalElliptic filter

9: Optimal IIR Design

9: Optimal IIRDesign

Error choices

Linear Least Squares

Frequency Sampling

Iterative Solution

Newton-Raphson

Magnitude-onlySpecification

Hilbert RelationsMagnitude PhaseRelation

Summary

MATLAB routines

DSP and Digital Filters (2015-6729) Optimal IIR: 9 1 / 11

Error choices

9: Optimal IIRDesign

Error choicesLinear Least Squares

Frequency Sampling

Iterative Solution

Newton-Raphson

Magnitude-onlySpecification

Hilbert RelationsMagnitude PhaseRelation

Summary

MATLAB routines

DSP and Digital Filters (2015-6729) Optimal IIR: 9 2 / 11

We want to find a filter H(ej) = B(ej)

A(ej) that approximates a target

response D(). Assume A is order N and B is order M .

Two possible error measures:

Solution Error: ES() = WS()(B(ej)A(ej) D()

)

Equation E

dsp & digital filters

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