Lecture 33 Z-Transform Process Control Prof. Kannan M. Moudgalya - PowerPoint PPT Presentation

Lecture 33 Z-Transform Process Control Prof. Kannan M. Moudgalya IIT Bombay Wednesday, 6 November 2013 1/30 Process Control Z-Transform

Outline 1. Motivation of Z-transform 2. Z-transform definition and implications to LTI systems 3. Examples and theorems 2/30 Process Control Z-Transform

1. Motivation of Z-transform 3/30 Process Control Z-Transform

Impulse Response Models for LTI Systems ◮ Impulse response of an LTI system initially at rest is given by g(n): { δ ( n ) } { g ( n ) } Time Invariant System ◮ Let the output of such a system be y(n) for an arbitrary input u(n) ◮ The output is given by { y(n) } = � ∞ k= −∞ u(k) { g(n − k) } ◮ This can be represented using the convolution operation, ∗ : { y(n) } = { u(n) } ∗ { g(n) } 4/30 Process Control Z-Transform

An example of output calculation ∞ � { y(n) } = u(k) { g(n − k) } k= −∞ { g(n) } = { 1 , 2 , 3 } , { u(n) } = { 4 , 5 , 6 } g, u start at n = 0. They are zero for n < 0. y(0) = u(0)g(0) = 4 y(1) = u(0)g(1) + u(1)g(0) = 13 y(2) = u(0)g(2) + u(1)g(1) + u(2)g(0) = 28 y(3) = u(1)g(2) + u(2)g(1) = 27 y(4) = u(2)g(2) = 18 All other terms that don’t appear above are 0 5/30 Process Control Z-Transform

MCQ The symbol 1(n) denotes 1. Unit step response 2. Unit step sequence 3. How can a constant be a function of n? 6/30 Process Control Z-Transform

Another Example ◮ Take a system with g(n) = 0 . 8 n 1(n) ◮ Excite it with u(n) = 0 . 5 n 1(n) ◮ Determine the output y(n) = u(n) ∗ g(n) ◮ What does this mean? ◮ Same as, y(n) = � ∞ k= −∞ u(k)g(n − k) 7/30 Process Control Z-Transform

Causality of LTI Systems ◮ If output depends only on past inputs, called causal ◮ If output depends on future inputs, not causal ◮ Can we say anything about g(n) for LTI causal systems? ◮ Initial state is zero ◮ No input until n = 0 - impulse input ◮ So, impulse response can begin only from n = 0 ◮ For LTI causal systems, g(n) = 0 for n < 0 8/30 Process Control Z-Transform

Convolution is cumbersome ◮ In general, we will work with infinite number of terms ◮ To calculate every term of y(n), we need to sum up infinite number of terms ◮ We need to calculate y(n) for infinite values of n ◮ Extremely tedious ◮ Propose a convenient method in the next slide 9/30 Process Control Z-Transform

Polynomial Calculation ≡ Convolution Also carryout multiplication: (u(0) + u(1)z − 1 + u(2)z − 2 ) × (g(0) + g(1)z − 1 + g(2)z − 2 ) = u(0)g(0)+ (u(0)g(1) + u(1)g(0))z − 1 + (u(0)g(2) + u(1)g(1) + u(2)g(0))z − 2 + (u(1)g(2) + u(2)g(1))z − 3 + u(2)g(2)z − 4 10/30 Process Control Z-Transform

Polynomial calculation ◮ z a position marker - coeff. of z − i at i th instant ◮ u(0) + u(1)z − 1 + u(2)z − 2 - a way of representing a sequence with three terms: { u(0) , u(1) , u(2) } ◮ If we can represent this polynomial ◮ e.g. u(0) + u(1)z − 1 + u(2)z − 2 + · · · in a compact way, ◮ it will simplify the convolution calculation 11/30 Process Control Z-Transform

2. Z-transform definition and implications 12/30 Process Control Z-Transform

Definition of Z-transform Z-transform of a sequence { u(n) } , denoted by U(z), is defined as: ∞ � u(n)z − n U(z) = n= −∞ where z is such that there is absolute convergence. That is, z should be chosen so as to satisfy ∞ � | u(n)z − n | < ∞ n= −∞ 13/30 Process Control Z-Transform

External (BIBO) Stability of LTI Systems ◮ If every Bounded Input produces Bounded Output, ◮ system is externally stable ◮ equivalently, system is BIBO stable ◮ Necessary and sufficient condition for BIBO stability of LTI systems is ∞ � | g(n) | < ∞ n= −∞ ◮ That is, � ∞ n= −∞ | g(n) | < ∞ ⇔ BIBO Stability ◮ Don’t care about what unbounded input does... 14/30 Process Control Z-Transform

Transfer function ◮ Z-transform of g(n), denoted by G(z), is called the transfer function ◮ That is, g(n) ↔ G(z) ◮ Poles and zeros are defined for G(z) just as in continuous time systems 15/30 Process Control Z-Transform

Important properties of transfer functions: Stability Given a causal, BIBO stable system with impulse response g(n) ◮ Z-transform of g(n), namely G(z), will have poles inside unit circle 16/30 Process Control Z-Transform

Examples of stability Is the following system stable? (z + 0 . 5) G 1 (z) = (z − 0 . 5)(z + 0 . 75) How about (z + 0 . 5) G 2 (z) = (z − 0 . 5)(z + 2) 17/30 Process Control Z-Transform

Important properties of transfer functions: Causality ◮ Let g(n) be the impulse response of an LTI causal system ◮ Let G(z) be the Z-transform of g(n). ◮ G(z) = N(z) D(z) with ◮ N(z) is a polynomial of degree n ◮ D(Z) is a polynomial of degree m ◮ n ≤ m 18/30 Process Control Z-Transform

Examples of causality Is the following system causal? G 3 (z) = (z + 0 . 5)(z + 0 . 25) (z + 0 . 75) How about (z + 0 . 5) G 4 (z) = (z + 3)(z + 2) 19/30 Process Control Z-Transform

Recall: Motivation for Z-transform (u(0) + u(1)z − 1 + u(2)z − 2 ) × (g(0) + g(1)z − 1 + g(2)z − 2 ) = u(0)g(0)+ (u(0)g(1) + u(1)g(0))z − 1 + (u(0)g(2) + u(1)g(1) + u(2)g(0))z − 2 + (u(1)g(2) + u(2)g(1))z − 3 + u(2)g(2)z − 4 That is, u(n) ∗ g(n) ↔ U(z)G(z) Instead of convolution, work with Z-transforms 20/30 Process Control Z-Transform

Recap of a benefit of Z-transform Instead of convolution, can take Z-transform, multiply and then invert! 21/30 Process Control Z-Transform

3. Examples and theorems 22/30 Process Control Z-Transform

Example ◮ u 1 (n) = a n 1(n) ◮ U 1 (z) = � ∞ n=0 a n z − n ◮ = � ∞ n=0 (az − 1 ) n ◮ = 1 + (az − 1 ) + (az − 1 ) 2 + · · · ◮ If | az − 1 | < 1, the sum converges to 1 z ◮ = 1 − az − 1 = z − a z ◮ We write, a n 1(n) ↔ z − a , | z | > | a | 23/30 Process Control Z-Transform

Example: Z-transform of Convolution Determine the step response of a system with impulse response g(n) = 0 . 5 n 1(n) using the Z-transform approach. ∞ ∞ g(j)z − j = � � 0 . 5 j z − j G(z) = j= −∞ j=0 1 = [1 + 0 . 5z − 1 + (0 . 5z − 1 ) 2 + · · · ] = 1 − 0 . 5z − 1 where we have assumed | z | > 0 . 5. Also have ∞ ∞ g(j)z − j = � � z − j U(z) = j= −∞ j=0 1 = [1 + z − 1 + z − 2 + · · · ] = (1 − z − 1 ) , | z | > 1 24/30 Process Control Z-Transform

Example: Z-transform of Convolution 1 1 Y(z) = G(z)U(z) = | z | > 1 1 − 0 . 5z − 1 , 1 − z − 1 Carry out partial fraction expansion. It is easy to check that the above expression is equivalent to 2 1 Y(z) = 1 − z − 1 − 1 − 0 . 5z − 1 , | z | > 1 On inverting this expression, we obtain y(n) = 2 × 1(n) − 0 . 5 n 1(n) On simplifying, y(n) = (2 − 0 . 5 n )1(n) 25/30 Process Control Z-Transform

Example 1 - Linearity Find the Z-transform of u 1 (n) = δ (n) − 3 δ (n − 2): ∞ ∞ δ (n)z − n − 3 � � δ (n − 2)z − n U 1 (z) = n= −∞ n= −∞ On simplifying, ∀ z − 1 finite U 1 (z) = 1 − 3z − 2 26/30 Process Control Z-Transform

Z-transform - Shifting ∞ � u(n + d)z − n Z [u(n + d)] = n= −∞ ∞ = z d � u(n + d)z − (n+d) n= −∞ = z d U(z) 27/30 Process Control Z-Transform

Z-transform - Shifting Example: If { u(n) } ↔ U(z) , then { u(n + 3) } ↔ z 3 U(z) { u(n − 2) } ↔ z − 2 U(z) 28/30 Process Control Z-Transform

Reference Reference: K. M. Moudgalya, Digital Control, Wiley, Chichester, 2007 Also, New Delhi, 2009 29/30 Process Control Z-Transform

What we learnt today ◮ Motivation of Z-transform ◮ Definition and implications to LTI systems ◮ Examples ◮ Theorems 30/30 Process Control Z-Transform

Thank you 31/30 Process Control Z-Transform