Z-Transform ENGI 4559 Signal Processing for Software Engineers

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Presentation transcript:

Z-Transform ENGI 4559 Signal Processing for Software Engineers Dr. Richard Khoury Fall 2009

Z-Transform: Overview ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform: Overview Definition Properties Contour integration Inverse Z-transform ROC Causality Stability Relationship with the frequency domain One-sided Z-transform State-space analysis

ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform We’re well familiar with signals in the time or space domain We’ve learned about signals in the frequency domain, and we can convert back and forth with the Fourier transform equations There is also the Z-domain Z-transform More general than Fourier transform

Z-Transform The Z-transform equation is ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform The Z-transform equation is We can see a special case if r = 1 It’s the DTFT! Z-domain and frequency domain are related

Z-Transform Sequence converges in some domain of Z ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Sequence converges in some domain of Z That domain is the region of convergence (ROC) Right-sided sequence if s(n) ≠ 0 for all n ≥ n0 Left-sided sequence if s(n) ≠ 0 for all n ≤ n0

ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Example State if the following signal is left- or right-defined, and compute its Z-Transform and ROC From the definition of u(n), we know that It’s a right-defined signal

Z-Transform Example Computing the Z-transform ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Example Computing the Z-transform Recall: infinite geometric series

Z-Transform Example The ROC is the region where |z| > |β| ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Example The ROC is the region where |z| > |β| z = β is a pole of S(z), and the circle is a pole circle The ROC is outside the pole circle for right-sided sequences, inside for left-sided sequences

Z-Transform Exercise Compute the Z-Transform and ROC of s(0) ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Exercise Compute the Z-Transform and ROC of s(0)

ENGI 4559 © Dr. Richard Khoury, 2009 Poles and Zeros A very important class of Z-transform are those for which S(z) is a rational function If a0 ≠ 0 and b0 ≠ 0 This function has M zeros at z = z0, z1, …, zM and N poles at z = p0, p1, …, pN

Poles and Zeros We can represent them in a pole-zero plot ENGI 4559 © Dr. Richard Khoury, 2009 Poles and Zeros We can represent them in a pole-zero plot This pole is outside the ROC

Z-Transform Properties ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Properties Assume the pair With a ROC Rs ROC bounded in Where ri and ro can be zero or infinity

Z-Transform Properties ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Properties Linearity The ROC can be a ring, when it is the sum of a left-sided and right-sided sequence

Z-Transform Properties ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Properties Time-shifting Multiplication by exponential Differentiation in the Z domain

Z-Transform Properties ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Properties Time reversal Convolution in time Complex conjugation

Z-Transform Properties ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Properties Multiplication in time Where C1 is the common region of the ROC of S(v) and H(z/v) Or C2 is the common region of the ROC of H(v) and S(z/v) is the contour integration

Z-Transform Properties ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Properties Parseval’s Theorem Where C1 is the common region of the ROC of S(v) and H*(1/v*) Or C2 is the common region of the ROC of H* (v) and S(1/v) is the contour integration The energy of a signal is

ENGI 4559 © Dr. Richard Khoury, 2009 Contour Integration Contour integration is a method of computing the integral of a region in the complex plane by following a closed contour inside the region There are multiple ways of evaluating this integral We will use Cauchy’s Residue Theorem

Cauchy’s Residue Theorem ENGI 4559 © Dr. Richard Khoury, 2009 Cauchy’s Residue Theorem Assume that there is a simple (non-repeated) pole inside C, at a point p0 Then F(v) contains a term (v – p0) in the denominator The residue is computed by multiplying F(v) by (v – p0) and evaluating at v = p0 Pole = singularity, division by zero

Cauchy’s Residue Theorem ENGI 4559 © Dr. Richard Khoury, 2009 Cauchy’s Residue Theorem Assume that there is a repeated Nth-order pole inside C, at a point p0 Then F(v) contains a term (v – p0)N in the denominator The residue is computed by multiplying F(v) by (v – p0)N, taking the N–1th derivative, evaluating at v = p0 , and normalizing The previous equation was the special case at N = 1

21 ENGI 4559 © Dr. Richard Khoury, 2009 Cauchy Exercise Compute this contour integral given that the contour C is given to be the circle |v| = 2

22 ENGI 4559 © Dr. Richard Khoury, 2009 Cauchy Exercise

23 ENGI 4559 © Dr. Richard Khoury, 2009 Cauchy Exercise

24 ENGI 4559 © Dr. Richard Khoury, 2009 Cauchy Exercise

25 ENGI 4559 © Dr. Richard Khoury, 2009 Cauchy Exercise

ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Example Find the Z-transform of the product of s(n) and h(n) First we compute the Z-transform of each

Z-Transform Example Next, recall the multiplication in time property ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Example Next, recall the multiplication in time property Fill in the equation

Z-Transform Example So, where is C1? And what are the poles? ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Example So, where is C1? And what are the poles?

Z-Transform Example What poles are inside C1? Only v = 2 ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Example What poles are inside C1? Only v = 2

Z-Transform Example Evaluate the equation ENGI 4559 © Dr. Richard Khoury, 2009 Z-Transform Example Evaluate the equation

Inverse Z-Transform Where C is inside the ROC of S(z) ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform Where C is inside the ROC of S(z) With a substitution σ = z-1 Where is the reciprocal of C

Inverse Z-Transform Example ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform Example Using our first equation, we have: For n > 0, this is easy; zn does not create any poles The only pole is at z = a, which is inside C

Inverse Z-Transform Example ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform Example For n ≤ -1, we have more poles We will avoid this using our second equation Numerator does not contribute to poles Only pole at σ = 1/a, outside of

Inverse Z-Transform Example ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform Example Putting it together:

Inverse Z-Transform There is a simpler way of computing it ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform There is a simpler way of computing it Start by dividing by z And compute the partial fraction expansion

ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform Compute each coefficient… it’s the residue at each pole! Multiply by z Compute the inverse Z-transform of each term

Inverse Z-Transform Example ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform Example Factorize denominator and divide by z

Inverse Z-Transform Example ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform Example Compute the coefficients

Inverse Z-Transform Example ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform Example These are all common Z-transform pairs Putting it all together

Inverse Z-Transform The previous formula works for simple poles ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform The previous formula works for simple poles A similar one applies for multiple poles

ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform This method only holds if the denominator contains positive powers of z If that is not the case, multiply to express the ratio with positive powers of z Example:

Inverse Z-Transform Exercise ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform Exercise

Inverse Z-Transform Exercise ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform Exercise

Inverse Z-Transform Exercise ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform Exercise

Inverse Z-Transform Exercise ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform Exercise

Inverse Z-Transform Exercise ENGI 4559 © Dr. Richard Khoury, 2009 Inverse Z-Transform Exercise

Region of Convergence We already know that z is a complex number ENGI 4559 © Dr. Richard Khoury, 2009 Region of Convergence We already know that z is a complex number Can take any value on the complex plane ROC is the region of the complex plane for which the infinite sum converges Four possible shapes Nonexistent Outside outermost pole circle Inside innermost pole circle Bounded between two pole circles

Region of Convergence ROC is nonexistent if summation doesn’t converge ENGI 4559 © Dr. Richard Khoury, 2009 Region of Convergence ROC is nonexistent if summation doesn’t converge There are no values of z for which this converges No ROC

ENGI 4559 © Dr. Richard Khoury, 2009 Region of Convergence ROC outside outermost pole circle if function is right-sided s(n) ≠ 0 for all n ≥ n0 Converges for all values of |z| > |β|

ENGI 4559 © Dr. Richard Khoury, 2009 Region of Convergence ROC inside innermost pole circle if function is left-sided s(n) ≠ 0 for all n ≤ n0 Converges for all values of |z| < |β|

ENGI 4559 © Dr. Richard Khoury, 2009 Region of Convergence ROC bounded between two pole circles if the function is neither right- nor left-sided Converges for all values of |α| < |z| < |β|

ROC Exercise Find the region of convergence of these functions 52 ENGI 4559 © Dr. Richard Khoury, 2009 ROC Exercise Find the region of convergence of these functions

ROC and Causality Recall our definition of causality ENGI 4559 © Dr. Richard Khoury, 2009 ROC and Causality Recall our definition of causality The output s(n) at sample n0 does not depend on the input f(n) at n > n0 The output at sample n0 can only depend on the input at n0 and on the input and output at n < n0 And consider a function Will be causal if h(i) = 0 for all i < 0 s(n) is right-sided! ROC outside outermost pole circle

ROC and Causality Will be anticausal if h(i) = 0 for all i ≥ 0 ENGI 4559 © Dr. Richard Khoury, 2009 ROC and Causality Will be anticausal if h(i) = 0 for all i ≥ 0 s(n) is left-sided ROC inside innermost pole circle Will be non-causal (acausal) if h(i) has non-zero values both sides of i = 0 s(n) is neither left- nor right-sided ROC bounded between two pole circles

ROC and Stability Recall our definition of stability ENGI 4559 © Dr. Richard Khoury, 2009 ROC and Stability Recall our definition of stability A system is BIBO stable if and only if its impulse response is absolutely summable Let’s look at the Z-transform of h(n) Its ROC is the region where it converges

ROC and Stability ROC is: BIBO stability is: ENGI 4559 © Dr. Richard Khoury, 2009 ROC and Stability ROC is: BIBO stability is: They’re the same at r = 1! System is stable if its ROC includes the unit circle

ROC, Causality and Stability ENGI 4559 © Dr. Richard Khoury, 2009 ROC, Causality and Stability System is stable if its ROC includes the unit circle ROC of a causal system is outside the outermost pole circle Causal system will be stable if all poles are inside the unit circle ROC of an anticausal system is inside innermost pole circle Anticausal system will be stable if all poles are outside the unit circle

ENGI 4559 © Dr. Richard Khoury, 2009 Relationship Example A function with only one real pole… there’s only one!

ENGI 4559 © Dr. Richard Khoury, 2009 Unit Circle We’ve mentioned another important feature of the z-domain unit circle It’s the link between the Z-transform and the DTFT The Fourier frequency domain is on the unit circle of the Z-domain

Z-Domain and Frequency Domain ENGI 4559 © Dr. Richard Khoury, 2009 Z-Domain and Frequency Domain Consider a simple function: the 11-point running sum There are 11 zeros at There are 11 poles, ten at z = 0 and one at z = 1 z = 1 is both a pole and a zero! Cancels out: there are 10 zeros and 10 poles

Z-Domain and Frequency Domain ENGI 4559 © Dr. Richard Khoury, 2009 Z-Domain and Frequency Domain Now compute the DTFT There are 10 zeros, at Those are the phase angle of the zeros on the unit circle in the Z-domain!

Z-Domain and Frequency Domain ENGI 4559 © Dr. Richard Khoury, 2009 Z-Domain and Frequency Domain If we plot S(z) in the Z-domain, and highlight the unit circle It’s the frequency response!

Z-Domain Frequency Filtering ENGI 4559 © Dr. Richard Khoury, 2009 Z-Domain Frequency Filtering The 11-point running sum filters out 10 sinusoid frequencies Corresponding to the 10 zeros at

Z-Domain Frequency Filtering ENGI 4559 © Dr. Richard Khoury, 2009 Z-Domain Frequency Filtering We can generalize this to an L-point running sum filter to target the frequencies we want This filters out L-1 frequencies at the zeros:

One-Sided Z-Transform ENGI 4559 © Dr. Richard Khoury, 2009 One-Sided Z-Transform One-sided (unilateral) Z-transform The Z-transform we’ve seen so far is the bilateral Z-transform

One-Sided Z-Transform ENGI 4559 © Dr. Richard Khoury, 2009 One-Sided Z-Transform It’s a special case of the bilateral Z-transform Linear system with finite number of initial conditions It’s the most common form of the Z-transform in signal processing Systems typically start at n = 0, with known initial conditions, so we don’t need to compute the values from -∞ to -1 It’s right-sided and causal

One-Sided Z-Transform Example ENGI 4559 © Dr. Richard Khoury, 2009 One-Sided Z-Transform Example Compute the unilateral Z-transform of both sides

One-Sided Z-Transform Example ENGI 4559 © Dr. Richard Khoury, 2009 One-Sided Z-Transform Example We isolate the system output on one side Zero input response Zero state response

ENGI 4559 © Dr. Richard Khoury, 2009 State-Space Analysis The state of a system at time n0 is all the information needed, in addition to the input signal for n ≥ n0, in order to uniquely compute the output of the system at time n ≥ n0 Composed of state variables For continuous-time systems, it could be the charge stored in the capacitors For discrete-time systems, it’s the time-delay elements of the system

State-Space Analysis Example ENGI 4559 © Dr. Richard Khoury, 2009 State-Space Analysis Example Consider a system with this equation D

State-Space Analysis Example ENGI 4559 © Dr. Richard Khoury, 2009 State-Space Analysis Example D The state will be composed of three state variables

State-Space Analysis Example ENGI 4559 © Dr. Richard Khoury, 2009 State-Space Analysis Example If we integrate these variables in the equation of the system And we could put all this information in matrix form

State-Space Analysis This is the state-space description of the system ENGI 4559 © Dr. Richard Khoury, 2009 State-Space Analysis This is the state-space description of the system State equation Dynamic (memory) subsystem Static (memoryless) subsystem Output equation

ENGI 4559 © Dr. Richard Khoury, 2009 State-Space Analysis More general formulation of the state-space description State equation Output equation

State-Space Analysis {A,b,c,d} is the structure of the filter ENGI 4559 © Dr. Richard Khoury, 2009 State-Space Analysis {A,b,c,d} is the structure of the filter Any system: whose input, output and states are related by state-space equations at n ≥ n0 and where {A,b,c,d} have arbitrary but fixed values is a linear time-invariant system If even one of the values in {A,b,c,d} depends on time, the system is time-variant

Solving State-Space Equations ENGI 4559 © Dr. Richard Khoury, 2009 Solving State-Space Equations So we can represent any linear time-invariant system with two simple equations The next step is to solve the equations, to get the transfer function of the system Can be done easily in the Z-domain

Solving State-Space Equations ENGI 4559 © Dr. Richard Khoury, 2009 Solving State-Space Equations Take the unilateral z-transform of the state equation Where: And v(0) is the initial values of the state variables But we’re looking for the transfer function, which assumes zero initial conditions

Solving State-Space Equations ENGI 4559 © Dr. Richard Khoury, 2009 Solving State-Space Equations Take the unilateral z-transform of the output equation And by using our pervious result

Solving State-Space Equations ENGI 4559 © Dr. Richard Khoury, 2009 Solving State-Space Equations The transfer function is the ratio of the output to the input, in the Z-domain and with zero initial conditions What’s more, we can find that Where adj is the adjoint matrix and det is the determinant

Solving State-Space Equations ENGI 4559 © Dr. Richard Khoury, 2009 Solving State-Space Equations The poles of the system are the zeros of det(zI – A) det(zI – A) is the characteristic polynomial of A The roots of the polynomial are the eigenvalues of A

ENGI 4559 © Dr. Richard Khoury, 2009 State-Space Exercise Determine the transfer function H(z) and the impulse response h(n) of this system

ENGI 4559 © Dr. Richard Khoury, 2009 State-Space Exercise

ENGI 4559 © Dr. Richard Khoury, 2009 State-Space Exercise

ENGI 4559 © Dr. Richard Khoury, 2009 State-Space Exercise

ENGI 4559 © Dr. Richard Khoury, 2009 State-Space Exercise

2D Extensions 1D Z-transform: 2D Z-transform: ENGI 4559 © Dr. Richard Khoury, 2009 2D Extensions 1D Z-transform: 2D Z-transform:

2D Extensions 1D ROC is the region of z where the sum converges ENGI 4559 © Dr. Richard Khoury, 2009 2D Extensions 1D ROC is the region of z where the sum converges In 2D, we have two z and two sums Each one converges in a certain region The ROC is the intersection of the two

ENGI 4559 © Dr. Richard Khoury, 2009 2D Z-Transform Example y x

2D Poles and Zeros Consider a 2D system Compute the transfer function ENGI 4559 © Dr. Richard Khoury, 2009 2D Poles and Zeros Consider a 2D system Compute the transfer function

2D Poles and Zeros The poles are the points where ENGI 4559 © Dr. Richard Khoury, 2009 2D Poles and Zeros The poles are the points where The roots are the points where They are called nonessential singularities of the first kind Points that are both poles are roots are nonessential singularities of the second kind

2D Stability Recall in 1D Equal at r = 1 BIBO stability: ROC: ENGI 4559 © Dr. Richard Khoury, 2009 2D Stability Recall in 1D BIBO stability: ROC: Equal at r = 1 System is stable if ROC includes unit circle All poles are inside unit circle

2D Stability BIBO stability in 2D: In the Z-domain: ENGI 4559 © Dr. Richard Khoury, 2009 2D Stability BIBO stability in 2D: In the Z-domain: They’re the same at r1 = 1 and r2 = 1 (unit bidisk) System is stable if all poles are inside the unit bidisk

2D Stability Huang’s stability test 1: 2: 3 option 1: 3 option 2: 93 ENGI 4559 © Dr. Richard Khoury, 2009 2D Stability Huang’s stability test Three conditions that must be satisfied for a system to be stable Given a system 1: 2: 3 option 1: 3 option 2: 93

2D Example Consider the system with the transfer function ENGI 4559 © Dr. Richard Khoury, 2009 2D Example Consider the system with the transfer function There are infinite zeros There are infinite poles

2D Example There is one nonessential singularity of the second kind ENGI 4559 © Dr. Richard Khoury, 2009 2D Example There is one nonessential singularity of the second kind is both a pole and a zero The system is not stable Pole (1,1) is on the unit bidisk There are infinitely many poles outside the unit bidisk

96 ENGI 4559 © Dr. Richard Khoury, 2009 2D Exercise Compute the Z-transform of this signal, determine its ROC, find its poles and zeros, and determine whether or not it is stable.

97 ENGI 4559 © Dr. Richard Khoury, 2009 2D Exercise

98 ENGI 4559 © Dr. Richard Khoury, 2009 2D Exercise

99 ENGI 4559 © Dr. Richard Khoury, 2009 2D Exercise

100 ENGI 4559 © Dr. Richard Khoury, 2009 2D Exercise

101 ENGI 4559 © Dr. Richard Khoury, 2009 2D Exercise

Summary Z-Transform Inverse Z-Transform One-sided Z-Transform ENGI 4559 © Dr. Richard Khoury, 2009 Summary Z-Transform Inverse Z-Transform One-sided Z-Transform

Summary Multiple ways of computing the inverse Z-Transform ENGI 4559 © Dr. Richard Khoury, 2009 Summary Multiple ways of computing the inverse Z-Transform Residue equations, with or without substitution Factorization

Summary Poles and zeros ROC, and its different shapes ENGI 4559 © Dr. Richard Khoury, 2009 Summary Poles and zeros ROC, and its different shapes Relationship between the ROC shape and causality Relationship between the ROC, the unit circle, the poles and stability

Summary Relationship between the Z-transform and the DTFT ENGI 4559 © Dr. Richard Khoury, 2009 Summary Relationship between the Z-transform and the DTFT The unit circle in the Z-domain is the Fourier frequency response

Summary State-Space Analysis 2D Z-transform What is a state ENGI 4559 © Dr. Richard Khoury, 2009 Summary State-Space Analysis What is a state What is a state variable What is the state space description How to solve the equations in the Z-domain to get the transfer function of the system 2D Z-transform 2D ROC 2D poles and zeros 2D stability conditions and the unit bidisk

Textbook Material Readings Recommended problems ENGI 4559 © Dr. Richard Khoury, 2009 Textbook Material Readings Chapter 3, excluding 3.6, 3.8, and 3.9 Recommended problems 1, 2, 3, 4, 5, 6, 7, 9, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 42, 43, 44, 45, 46, 51, 55, 56, 57, 58, 59