Basics on Digital Signal Processing

z - transform - Digital Filters

Vassilis Anastassopoulos

Electronics Laboratory, Physics Department, University of Patras Outline of the Lecture

1. The z-transform 2. Properties 3. Examples 4. Digital filters 5. Design of IIR and FIR filters 6. Linear phase

2/31 z-Transform

2 N 1 N 1  j nk n X (k)   x(n) e N X (z)   x(n) z n0 n0 Time to frequency Time to z-domain (complex)

Transformation tool is the Transformation tool is complex wave z   e j  e j j j2k / N e  e With amplitude ρ changing(?) with time With amplitude |ejω|=1

The z-transform is more general than the DFT

3/31 z-Transform For z=ejω i.e. ρ=1 we work on the unit circle N 1 X (z)   x(n) z n And the z-transform degenerates n0 into the Fourier transform.

I -z z=R+jI |z|=1

R

The DFT is an expression of the z-transform on the unit circle.

The quantity X(z) must exist with finite value on the unit circle i.e. must posses spectrum with which we can describe a signal or a system.

4/31 z-Transform convergence

We are interested in those values of z for which X(z) converges. This region should contain the unit circle.

I Why is it so?

|a| N 1 N 1 X (z)   x(n) zn   x(n) (1/zn ) n0 n0 R At z=0, X(z) diverges ROC

The values of z for which X(z) diverges are called poles of X(z).

5/31 z-Transform example

Which is the z-transform and the ROC of a discrete time sequence x(n)=an for n0 and a<1 ?

I x(n) 1 a |a| a2 a3 a4 R

n ROC

N 1   X (z)   x(n) zn  an zn  (az1)n n0 n0 n0 -1 1 z which for |az |<1 or |z|>|a| converges to X (z)   1  az1 z  a The pole z=a, is never included in the ROC

6/31 Poles and zeros of X(z)

I Poles: X(z)= Zeros: X(z)=0

For infinite sequences, X(z) converges |a| everywhere outside the circle with radius the pole with maximum value. R

ROC For finite sequences, X(z) converges everywhere except at z=0

x(n) 5  3 3 X(z) x(n)z n  z 1 3z 2 5z 3 3z 4  z 5 1 1  n n

For stable, causal digital systems the region of convergence includes the Unit Circle so that the system possesses spectrum

7/31 z-Transform general form

a  a z1  a z2  a zN k(z  z )(z  z )(z  z ) 0 1 2 N X (z)  1 2 N X (z)  1 2 M b0  b1z  b2z  bN z (z  p1)(z  p2 )(z  pN )

Those values of z (zi) that make the nominator zero are called zeros.

While the poles are the values of z (pi) that make the denominator zero and thus X(z) diverges.

For stable, causal digital systems the region of convergence includes the Unit Circle so that the system possesses spectrum

  H(e jT )  H(z)  h(n)z n  h(n)e jT ze jT   n ze jT n

8/31 Inverse z-Transform

A simple way to evaluate the signal from the X(z) is to perform the division

1 2z 1  z 2 X(z)  1 z 1  0.356z 2

X(z) 1 3z 1  3.6439z 2  2.5756z 3 

The signal is x(0)=1, x(1)=3, x(2)=3.6439, x(3)=2.5756

9/31 z-Transform properties

x1(n)  X1(z) Linearity x2 (n)  X 2 (z)

 ax1(n)  bx2 (n)  aX1(z)  bX 2 (z)

x(n)  X (z) Delay or Shift  x(n  m)  zm X (z)

 Convolution y(n)  h(k)x(n  k)  Y(z)  H(z)X(z) k

10/31 z-Transform and digital systems

N M y(n)  ak x(n  k)  bk y(n  k) k0 k1

x(n) x(n-1) x(n-2) x(n-Ν+1) x(n-Ν) Ts Ts Ts

a0 a1 a2 aΝ-1 aΝ

y(n) N + k ak z k0 H (z)  N bΜ bΜ-1 b2 b1 k 1 bk z k1 T T T y(n-Μ) s y(n-Μ+1) y(n-2) s y(n-1) s

11/31 Digital system z-Transform derivation

N M y(n)  ak x(n  k)  bk y(n  k) z-transform both sides k0 k1   N  M Interchange zn y(n)  zn a x(n  k)  zn b y(n  k)    k   k summations n0 n0 k0 n0 k1

N  M  n n Y (z)  ak z x(n  k)  bk z y(n  k) Replace transforms k0 n0 k1 n0 N M Y (z)  ak X (z  k)  bkY (z  k) Shift property k0 k1 N M k k Common factors Y(z)  ak z X(z)  bk z Y(z) k0 k1 N M k k Y (z)  X (z)ak z Y (z)bk z k0 k1

12/31 Digital system z-Transform derivation

M N k k Common factors Y (z) Y (z)bk z  X (z)ak z k1 k0

M N k k Y (z)(1 bk z )  X (z) ak z x(n) x(n-1) x(n-2) x(n-Ν+1) x(n-Ν)   Ts Ts Ts k1 k0

a0 a1 a2 aΝ-1 aΝ N k ak z y(n) Y (z) +  k0  H(z) Frequency X (z) N 1 b z k  k bΜ bΜ-1 b2 b1 k1

T T T y(n-Μ) s y(n-Μ+1) y(n-2) s y(n-1) s N M y(n)  ak x(n  k)  bk y(n  k) Time k0 k1

13/31 Application

Find the impulse response h(n) and the input-output relationship of the filter described by 1 z 1 H(z)  1 0.5z 1 Examine the stability of the filter and find its frequency response.

Solution -1 Y(z) 1 z 1 x(n) z H(z)    X(z) 1 0.5z 1 1 1 Z1 -1 Y(z)  0.5Y(z)z  X(z)  X(z)z  y(n) y(n)  0.5y(n 1)  x(n)  x(n 1)  -0.5 y(n)  x(n)  x(n 1)  0.5y(n 1) z-1 h(n) is obtained from the terms of the polynomial which results after the division (1 z 1 ) / (1 0.5z 1 )  11.5z 1  0.75z 2  0.375z 3 

14/31 Application

Im 1 z 1 H(z)  Stable 1 0.5z 1 |z|=1 1 Re

Frequency response using MATLAB

5 2

4.5 1.8

4 1.6

3.5 1.4

3 1.2

2.5 1 Phase

Magnitude 2 0.8

1.5 0.6

1 0.4

0.5 0.2

0 0 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 Frequency Frequency

15/31 Frequency on the Unit Circle

π/2 or Fs/4 Im z=ejωt |z|=1 1 z 1 π or F /2 H(z)  s 0 1 0.5z 1 Re

The region from 0 to π corresponds to the region 0-

Fs/2 of the Frequency response

16/31 Digital Filters

•They are characterized by their Impulse Response h(n), their Transfer Function H(z) and their Frequency Response H(ω).

•They can have memory, high accuracy and no drift with time and temperature.

•They can possess linear phase.

•They can be implemented by digital computers.

17/31 Digital Filters - Categories

N M IIR y(n)  a(k)x(n  k)  bk y(n  k) k0 k1 N k ak z x(n) x(n-1) x(n-2) x(n-Ν+1) x(n-Ν)  Ts Ts Ts k0 H(z)  M 1 b z k  k a a a a a k1 0 1 2 N-1 N

y(n) +

bΜ bΜ-1 b2 b1

T T T y(n-Μ) s y(n-Μ+1) y(n-2) s y(n-1) s

18/31 Digital Filters - Categories

N N y(n)  h(k)x(n  k)  a(k)x(n  k) FIR k0 k0 N N k k H(z)  ak z  h(k)z k0 k0

x(n) x(n-1) x(n-2) x(n-Ν+2) x(n-Ν+1) Ts Ts Ts

aΝ-2 ή a ή a0 ή a1 ή a2 ή Ν-1 h(0) h(1) h(2) h(Ν-2) h(Ν-1)

y(n) + •Stable •Linear phase

19/31 Digital Filters - Examples

a  a z 1  a z 2 0 1 2 1.5 1 H(z)  1 2 1 b1z  b2 z 0 1 a0=0.498, a1=0.927, -1 a2=0.498, b1=-0.674

b2=-0.363. 0.5 IIR Phase IIR Magnitude -2 0 -3 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 Frequency Frequency 11 1.5 4 H(z)  h(k)z k  2 k0 1 0 h(1)=h(10)=-0.04506

0.5 FIR Phase

h(2)=h(9)=0.06916 FIR Magnitude -2

h(3)=h(8)=-0.0553 0 -4 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 h(4)=h(7)=-.06342 Frequency Frequency

h(5)=h(4)=0.5789

20/31 IIR filter design

Design using the Bilinear z-transform, BZT

Design using the position of poles and zeros on the unit circle

Fs/4 Im •The frequency response is zero at the points of zeros |z|=1 Fs/2 1 0 or Fs •The frequency response takes a peak at the position of poles. -Fs/2 •In order to have real coefficients of the filter, the poles must appear in pairs. The 3F /4 s same happens for the zeros as well.

21/31 IIR Filter Design - position poles and zeros on the unit circle

Design a band-pass filter with the following specifications Sampling frequency 1000Hz, full rejection at dc and 500 Hz, Narrow pass-band at 250 Hz, 20Hz 3dBs bandwidth.

18 Im 16

14

12 |z|=1

10 1

8 Magnitude Re 6

4

2

0 0 0.1 0.2 0.3 0.4 Frequency

ο Zeros at 0 and Fs/2 or 180 (dc  0, 500/1000 Fs/2). ο ο Poles at 250/1000 Fs/4 or 90 and complex conjugate at -90 The radius of poles should be smaller than 1 (stability)

r  1 (bw / Fs )  1 (20 /1000)  0.937

22/31 IIR Filter Design - position poles and zeros on the unit circle

x(n) Im z-2

|z|=1 y(n) 1 Re 0.87 z-2 y(n)  x(n)  x(n  2)  0.877969y(n  2) r  1 (bw/ Fs )  1 (20/1000)  0.937 (z 1)(z 1) z2 1 1 z 2 H(z)    (z  re j / 2 )(z  re j / 2 ) z2  0.877969 1 0.877969z 2

18 2

16 1.5

14 1

12 0.5 10 0

8 Phase Magnitude -0.5 6

-1 4

2 -1.5

0 -2 0 0.1 0.2 0.3 0.4 0 0.1 0.2 0.3 0.4 Frequency Frequency

23/31 FIR filters

N N N N y(n)  h(k)x(n  k)  a(k)x(n  k) k k   H(z)  ak z  h(k)z k0 k0 k0 k0

x(n) x(n-1) x(n-2) x(n-Ν+2) x(n-Ν+1) Ts Ts Ts

aΝ-2 ή a ή a0 ή a1 ή a2 ή Ν-1 h(0) h(1) h(2) h(Ν-2) h(Ν-1)

y(n) + Design Methods •Stable • Optimal filters • Windows method •Linear phase • Sampling frequency

24/31 Main categories of ideal filters

Low-pass Band-pass High-pass |Η(ω)| |Η(ω)| |Η(ω)|

1 1 1

0 ωc π 2π 0 ωc1 ωc2 π 2π 0 ωc π 2π

Η(ω) Η(ω) j j 0 π 2π -j

0 π 2π Differentiator Hilbert Transformer

25/31 FIR filter design – basic concept

1  h (n)  H ()e jn d  D 2  D

1 c   e jn d  2 c 2 f sin(n )  c c nc

c   / 2 1 sin(n / 2) The filter coefficients result from the Inverse Fourier h(n)  Transform of the Desired Frequency Response. 2 (n / 2)

26/31 FIR filter design – Windows method

|HD(ω)| 0.5 hD(n) 1

...... -π -π/2 0 π/2 π

Ts |H (ω)| 0.5 t ht(n) 1 Gibbs phenomenon

-π 0 π -8 0 8

|W(ω)| 1 w(n) 1 Hamming Window  2n  a  (1 a)cos( ) wR (n)   N -π 0 π -8 8  0

|H(ω)| 0.5 h(n) 1

-π 0 π -8 8

27/31 FIR filter design – Optimal filters

Parks and McClellan method. Distributes Approximation error from the discontinuity all over the frequency band.

28/31 FIR filter design – Comparisons

0

0 -20

-10 -40

-20 -60 Magnitude

-30 -80

-40 0 100 200 300 400 500 Frequency

-50 5 Magnitude

-60

-70 0 Phase -80

-90 0 100 200 300 400 500 -5 Frequency 0 100 200 300 400 500 Frequency

0.4

Windows: 61 Coefficients 0.2 IR

Parks and McClellan method: 46 0 and equiripple -0.2 0 5 10 15 20 25 30 35 40 45 Time

29/31 FIR filters – Linear phase

Linear phase is strictly related with the symmetry of the Impulse Response ()  a ()  b  a

Condition h(n)  h(N  n 1) 6 H()   h(k)e jkT  k0  h(0)  h(1)e jT  h(2)e j2T  h(3)e j3T  h(4)e j4T  h(5)e j5T  h(6)e j6T  e j3T h(0)e j3T  h(1)e j2T  h(2)e jT  h(3)  h(4)e jT  h(5)e j2T  h(6)e j3T   e j3T h(0)(e j3T  e j3T )  h(1)(e j2T  e j2T )  h(2)(e jT  e jT )  h(3)  e j3T 2h(0)cos(3T )  2h(1)cos(2T )  2h(2)cos(T )  h(3)

The phase is introduced by the term e-j3ωT and equals θ(ω)=3ωΤ=ωΤ (7-1)/2

30/31 Linear phase - Same time delay

180o

3*180o

5*180o

If you shift in time one signal, then you have to shift the other signals the same amount of time in order the final wave remains unchanged. For faster signals the same time interval means larger phase difference. Proportional to the frequency of the signals. -αω

31/31