ELEG–305: Digital Signal Processing Lecture 9: Inverse Systems and Deconvolution; Ideal Sampling and Reconstruction Kenneth E. Barner Department of Electrical and Computer Engineering University of Delaware

Fall 2008

K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Fall 2008

1 / 22

Outline 1

Review of Previous Lecture

2

Lecture Objectives

3

Inverse Systems and Deconvolution Inverse Systems Recursive Solution Min/Max/Mixed-Phase Systems Decomposition of Nonminimum–Phase Pole–Zero Systems

4

Sampling and Reconstruction of Signals Discrete–Time Processing of Continuous–Time Signals

K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Fall 2008

2 / 22

Review of Previous Lecture

Review of Previous Lecture Filter Design Through Pole–Zero Placement – Poles increase response and zeroes decrease response (magnitude) Lowpass, Highpass, and Bandpass Filters – Pass signal content at the appropriate frequencies; Lowpass-to-Highpass filter transformation Digital Resonators – Two–pole bandpass filters with a pair of complex–conjugate poles near the unit circle Notch filters – Systems with one or more nulls in their frequency response Comb Filters – Notch filter with nulls spaced periodically across the spectrum All–Pass Filter – Constant magnitude response systems

K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Fall 2008

3 / 22

Lecture Objectives

Lecture Objectives

Objective Determine the inverse of LTI Systems with closed form expressions; Derive recursive inverse solutions for systems without closed forms; Define Min/Max/Mixed–Phase systems; Decompose nonminimum phase systems into minimum phase and all–pass components; Review ideal D/A and A/D systems Reading Chapters 5 and 6 (5.6-6.2); Next lecture, Sampling, Quantization, and Reconstruction (Chapter 6)

K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Fall 2008

4 / 22

Inverse Systems and Deconvolution

Inverse Systems

Objective I: Develop equalizers, or inverse systems, that invert communication channel distortions. Similarly, deconvolution inverts the convolution affects of a systems. Objective II: Determine the characteristics of an unknown system through system identification Identity system

x(n)

y(n)

T

T-1

Direct system

w(n) = x(n)

Inverse system

The cascade of a system and its inverse is the identity system w(n) = T −1 [y(n)] = T −1 [T [x(n)]] = x(n) Note: Not all systems are invertible, e.g., y(n) = ax 2 (n) K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Inverse Systems and Deconvolution

Fall 2008

5 / 22

Inverse Systems

Denote the impulse response of the inverse system as h I (n) w(n) = hI (n) ∗ h(n) ∗ x(n) = x(n) =⇒ h(n) ∗ hI (n) = δ(n) =⇒ H(z)HI (z) = 1 =⇒ HI (z) =

1 H(z)

If H(z) has a rational system function H(z) =

B(z) A(z)

=⇒

HI (z) =

A(z) B(z)

Result: The zeroes of H(z) became the poles of the inverse system, and vice versa Question: Does this raise any stability constraints? K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Fall 2008

6 / 22

Inverse Systems and Deconvolution

Inverse Systems

Example Determine the inverse of the system with impulse response h(n) = δ(n) − 0.5δ(n − 1) H(z) = 1 − 0.5z −1 , ROC: |z| > 0 1 z ⇒ HI (z) = = z − 0.5 1 − 0.5z −1 Two possible inverse systems: ROC: |z| > 0.5, causal hI (n) = (0.5)n u(n) ROC: |z| < 0.5, anticausal hI (n) = −(0.5)n u(−n − 1) K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Inverse Systems and Deconvolution

Fall 2008

7 / 22

Recursive Solution

Problem: Suppose no closed form of H(z) exists Solution: Use recursion and assume h I (n) is causal n 

h(k)hI (n − k) = δ(n)

k =0

1 h(0) n  h(k)hI (n − k) = 0 [for n ≥ 1] =⇒ h(0)hI (n) + [let n = 0] =⇒ hI (0) =

k =1

n =⇒ hI (n) = −

k =1 h(k)hI (n

− k)

h(0)

Note: Recursion uses known samples, i.e., h(n) and h I (m) for m < n Method Problems: h(0) = 0 causes problems – solved by delaying to first nonzero sample; Bigger problem – roundoff errors accumulate K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Fall 2008

8 / 22

Inverse Systems and Deconvolution

Min/Max/Mixed-Phase Systems

Consider two systems 1 1 H1 (z) = 1 + z −1 and H2 (z) = + z −1 2 2 Since H2 (z) = z −1 H1 (z −1 ), the systems have reciprocal zeroes and  5 + cos ω |H1 (ω)| = |H2 (ω)| = 4 The phase responses, however, are unique Θ1 (ω) = −ω + tan−1

1 2

sin ω + cos ω

K. E. Barner (Univ. of Delaware)

and Θ2 (ω) = −ω + tan−1

ELEG–305: Digital Signal Processing

Inverse Systems and Deconvolution

sin ω 2 + cos ω

Fall 2008

9 / 22

Min/Max/Mixed-Phase Systems

Observations on H1 (z): H1 (z) = 1 + 12 z −1 has a pole inside the unit circle (z = − 12 ) H1 (z) has minimum phase change, i.e., Θ 1 (π) − Θ1 (0) = 0 Generalization: when all system zeroes are inside the unit circle, the net zero contribution to phase is 0 radians – minimum phase systems Observations on H2 (z): H2 (z) =

1 2

+ z −1 has a pole outside the unit circle (z = −2)

H2 (z) has maximum phase change, i.e., Θ 2 (π) − Θ2 (0) = π Generalization: when all system zeroes are outside the unit circle, the net zero contribution to phase is Mπ radians (M = # of zeroes) – maximum phase systems

K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Fall 2008

10 / 22

Inverse Systems and Deconvolution

Min/Max/Mixed-Phase Systems

Min/Max/Mixed-Phase Systems Definition (Minimum–Phase System) H(z) is minimum–phase if all its poles and zeros of are inside the unit circle. Definition (Maximum–Phase System) H(z) is maximum–phase if all its poles are inside the unit circle and all its zeros are outside the unit circle. Definition (Mixed–Phase System) H(z) is mixed–phase if all poles of H(z) are inside the unit circle and some, but not all, of its zeros are outside the unit circle. Note: Pole restriction inside the unit circle is required for stability of causal systems K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Inverse Systems and Deconvolution

Fall 2008

11 / 22

Decomposition of Nonminimum–Phase Pole–Zero Systems

Recall H(z) =

B(z) A(z)

=⇒

H −1 (z) =

A(z) B(z)

Observations: H(z) minimum–phase =⇒ H −1 (z) stable and minimum–phase H(z) maximum or mixed phase =⇒ H −1 (z) unstable Theorem (Decomposition of Nonminimum–Phase Pole–Zero Systems) Any nonminimum–phase pole-zero system can be expressed as H(z) = Hmin (z)Hap (z) where Hmin (z) is a minimum-phase system and Hap (z) is an all-pass system. Note: Magnitude response of H(z) completely captured by H min (z) K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Fall 2008

12 / 22

Inverse Systems and Deconvolution

Decomposition of Nonminimum–Phase Pole–Zero Systems

To prove, consider a causal and stable system H(z) =

B(z) A(z)

Factor B(z) into minimum & maximum phase components B(z) = B1 (z)   

B2 (z)   

min. phase max. phase

⇒ B2 (z −1 ) is min.–phase (zeroes reflected inside unit circle). Define Hmin (z) =

B1 (z)B2 (z −1 ) A(z)

⇒ H(z) =

B1 (z)B2 (z −1 ) A(z)

and Hap (z) =



B2 (z) B2 (z −1 )

B2 (z) B2 (z −1 )

= Hmin (z)Hap (z)

Note: Hap (z) is a stable, maximum–phase, all–pass system K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Inverse Systems and Deconvolution

Fall 2008

13 / 22

Decomposition of Nonminimum–Phase Pole–Zero Systems

Example Let H(z) =

1 − 3.5z −1 + 6z −2 + 4z −3 1 − √1 z −1 + 14 z −2 2

Plot the P/Z diagram, determine phase characteristics, plot magnitude and phase, and express as H(z) = H min (z)Hap (z) Use MATLAB. Thus » B=[1,-3.5,6,4] » A=[1,-1/sqrt(2),0.25] » abs(roots(A)) ans = 0.5000 0.5000

K. E. Barner (Univ. of Delaware)

» angle(roots(A))/pi ans = 0.2500 -0.2500 Thus p1,2 = 12 e±jπ/4

ELEG–305: Digital Signal Processing

Fall 2008

14 / 22

Inverse Systems and Deconvolution

Decomposition of Nonminimum–Phase Pole–Zero Systems

» abs(roots(B)) » angle(roots(B))/pi ans = ans = 2.8284 0.2500 2.8284 -0.2500 0.5000 1.0000 √ ±jπ/4 and z3 = − 12 Thus z1,2 = 2 2e » zplane(B,A) 2 1.5

Imaginary Part

1 0.5 0 −0.5 −1 −1.5 −2 −2

−1

0

1

2

3

Real Part

K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Inverse Systems and Deconvolution

H(z) =

Fall 2008

15 / 22

Decomposition of Nonminimum–Phase Pole–Zero Systems

1 − 3.5z −1 + 6z −2 + 4z −3 1 − √1 z −1 + 14 z −2 2

=

(1 − z1 z −1 )(1 − z2 z −1 )(1 − z3 z −1 ) B(z) = A(z) (1 − p1 z −1 )(1 − p2 z −1 )

√ where z1,2 = 2 2e ±jπ/4 and z3 = − 12 and p1,2 = 12 e±jπ/4 . Let B(z) = B1 (z)B2 (z), where B1 (z) = (1 − z3 z −1 )   

and B2 (z) = (1 − z1 z −1 )(1 − z2 z −1 )   

min. phase

max. phase

Then H(z) = Hmin (z)Hap (z), where Hmin (z) = = min = =⇒ z1,2

1 √ e±jπ/4 2 2

K. E. Barner (Univ. of Delaware)

B1 (z)B2 (z −1 ) A(z) (1 − z3 z −1 )(1 − z1 z)(1 − z2 z) (1 − p1 z −1 )(1 − p2 z −1 )

min = 1 e ±jπ/4 and z3min = − 12 and p1,2 2 ELEG–305: Digital Signal Processing

Fall 2008

16 / 22

Inverse Systems and Deconvolution

Decomposition of Nonminimum–Phase Pole–Zero Systems

Hmin (z) P/Z Diagram 1

» Zmin=[sqrt(2)/4*exp(j*pi/4), sqrt(2)/4*exp(-j*pi/4), -0.5] » Bmin=poly(Zmin); » zplane(Bmin,A) Converts roots to polynomial coefficients and plots P/Z diagram. Next Hap (z) =

0.8 0.6

Imaginary Part

0.4 0.2 0 −0.2 −0.4 −0.6 −0.8 −1 −1

−0.5

0 Real Part

0.5

1

Hap (z) P/Z Diagram

B2 (z) B2 (z −1 )

2 1.5 1 Imaginary Part

» temp=roots(B); » ZB2=temp(1:2); »zplane(poly(ZB2),poly(1./ZB2))

0.5 0 −0.5 −1 −1.5 −2 −2

−1

0

1

2

3

Real Part

K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Inverse Systems and Deconvolution

Plot magnitude & phase » W=-pi:pi/100:pi; » H=freqz(B,A,W); » plot(W/pi,abs(H)) » plot(W/pi,angle(H)) » H=freqz(Bmin,A,W); » plot(W/pi,abs(H)) » plot(W/pi,angle(H)) » H=freqz(poly(ZB2), poly(1./ZB2),W); » plot(W/pi,abs(H)) » plot(W/pi,angle(H)) Top to bottom, H, H min , Hap

Fall 2008

17 / 22

Decomposition of Nonminimum–Phase Pole–Zero Systems

|H(ω)|

Θ(ω)

16

4

14

3 2

12

1 10 0 8 −1 6

−2

4

2 −1

−3

−0.5

0

0.5

1

−4 −1

2

0.8

1.8

0.6

1.6

0

0.5

1

−0.5

0

0.5

1

−0.5

0

0.5

1

0.4

1.4

0.2

1.2

0

1

−0.2

0.8

−0.4

0.6

−0.6

0.4 −1

−0.5

−0.5

0

0.5

1

8

−0.8 −1

4 3

8

2 1

8

0 −1

8

−2 −3

8 −1

K. E. Barner (Univ. of Delaware)

−0.5

0

ELEG–305: Digital Signal Processing

0.5

1

−4 −1

Fall 2008

18 / 22

Sampling and Reconstruction of Signals

Discrete–Time Processing of Continuous–Time Signals

Discrete-Time Processing of Continuous-Time Signals

Questions: How do we design the prefilter, A/D converter, D/A converter? Approach: Ideal cases are reviewed first, followed by the introduction & analysis of practical approaches

K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Sampling and Reconstruction of Signals

Fall 2008

19 / 22

Discrete–Time Processing of Continuous–Time Signals

Ideal A/D Converter Case

x(n) = xa (t)|t=nT = xa (nT ) ∞ 1  Xa (F − kFs ) X (F ) = T

[Time domain] [Frequency domain]

k =−∞

where Fs = 1/T , the value of which should be chosen to satisfy the Sampling Theorem K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Fall 2008

20 / 22

Sampling and Reconstruction of Signals

Discrete–Time Processing of Continuous–Time Signals

Ideal D/A Converter Case

ya (t) =

∞ 

y(n)ga (t − nT )

[Time domain]

n=−∞

Ya (F ) = Ga (F )Y (F )

[Frequency domain]

For perfect reconstruction the Ideal Interpolator is used  sin(πt/T ) F T , |F | ≤ Fs /2 ←→ Ga (F ) = ga (t) = 0, otherwise πt/T K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Fall 2008

21 / 22

Lecture Summary

Lecture Summary LTI System Inverse – closed form case H(z) =

B(z) A(z)

=⇒

HI (z) =

A(z) B(z)

Note P/Z location restrictions for stability Recursive inverse solution for systems without closed forms 1 h(0) n h(k)hI (n − k) [for n ≥ 1] =⇒ hI (n) = − k =1 h(0) [let n = 0] =⇒ hI (0) =

Min/Max/Mixed–Phase systems; H(z) = Hmin (z)Hap (z) decomposition; Ideal D/A and A/D systems Next Lecture – Sampling, Quantization, and Reconstruction (Chapter 6) K. E. Barner (Univ. of Delaware)

ELEG–305: Digital Signal Processing

Fall 2008

22 / 22