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

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Outline

• 1. Motivation of Z-transform
• 2. Z-transform definition and implications to LTI

systems

• 3. Examples and theorems

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• 1. Motivation of Z-transform

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Impulse Response Models for LTI Systems

◮ Impulse response of an LTI system initially at

rest is given by g(n):

Time Invariant System

{g(n)} {δ(n)}

◮ 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

• peration, ∗: {y(n)} = {u(n)} ∗ {g(n)}

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An example of output calculation

{y(n)} =

∞

• k=−∞

u(k){g(n − 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

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MCQ

The symbol 1(n) denotes

• 1. Unit step response
• 2. Unit step sequence
• 3. How can a constant be a function of n?

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Another Example

◮ Take a system with g(n) = 0.8n1(n) ◮ Excite it with u(n) = 0.5n1(n) ◮ Determine the output y(n) = u(n) ∗ g(n) ◮ What does this mean? ◮ Same as, y(n) = ∞

k=−∞ u(k)g(n − k)

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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

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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

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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

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Polynomial calculation

◮ z a position marker - coeff. of z−i at ith 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

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• 2. Z-transform definition and implications

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Definition of Z-transform

Z-transform of a sequence {u(n)}, denoted by U(z), is defined as: U(z) =

∞

• n=−∞

u(n)z−n where z is such that there is absolute convergence. That is, z should be chosen so as to satisfy

∞

• n=−∞

|u(n)z−n| < ∞

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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

∞

• n=−∞

|g(n)| < ∞

◮ That is, ∞

n=−∞ |g(n)| < ∞ ⇔ BIBO Stability

◮ Don’t care about what unbounded input does...

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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

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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

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Examples of stability

Is the following system stable? G1(z) = (z + 0.5) (z − 0.5)(z + 0.75) How about G2(z) = (z + 0.5) (z − 0.5)(z + 2)

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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

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Examples of causality

Is the following system causal? G3(z) = (z + 0.5)(z + 0.25) (z + 0.75) How about G4(z) = (z + 0.5) (z + 3)(z + 2)

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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

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Recap of a benefit of Z-transform

Instead of convolution, can take Z-transform, multiply and then invert!

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• 3. Examples and theorems

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Example

◮ u1(n) = an1(n) ◮ U1(z) = ∞

n=0 anz−n

◮ = ∞

n=0 (az−1)n

◮ = 1 + (az−1) + (az−1)2 + · · · ◮ If |az−1| < 1, the sum converges to ◮ =

1 1 − az−1 = z z − a

◮ We write, an1(n) ↔

z z − a, |z| > |a|

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Example: Z-transform of Convolution

Determine the step response of a system with impulse response g(n) = 0.5n1(n) using the Z-transform approach.

G(z) =

∞

• j=−∞

g(j)z−j =

∞

• j=0

0.5jz−j = [1 + 0.5z−1 + (0.5z−1)2 + · · · ] = 1 1 − 0.5z−1

where we have assumed |z| > 0.5. Also have

U(z) =

∞

• j=−∞

g(j)z−j =

∞

• j=0

z−j = [1 + z−1 + z−2 + · · · ] = 1 (1 − z−1), |z| > 1

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Example: Z-transform of Convolution

Y(z) = G(z)U(z) = 1 1 − z−1 1 1 − 0.5z−1, |z| > 1 Carry out partial fraction expansion. It is easy to check that the above expression is equivalent to Y(z) = 2 1 − z−1 − 1 1 − 0.5z−1, |z| > 1 On inverting this expression, we obtain y(n) = 2 × 1(n) − 0.5n1(n) On simplifying, y(n) = (2 − 0.5n)1(n)

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Example 1 - Linearity

Find the Z-transform of u1(n) = δ(n) − 3δ(n − 2): U1(z) =

∞

• n=−∞

δ(n)z−n − 3

∞

• n=−∞

δ(n − 2)z−n On simplifying, U1(z) = 1 − 3z−2 ∀z−1 finite

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Z-transform - Shifting

Z [u(n + d)] =

∞

• n=−∞

u(n + d)z−n = zd

∞

• n=−∞

u(n + d)z−(n+d) = zdU(z)

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Z-transform - Shifting

Example: If {u(n)} ↔ U(z), then {u(n + 3)} ↔ z3U(z) {u(n − 2)} ↔ z−2U(z)

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Reference

Reference:

• K. M. Moudgalya, Digital Control, Wiley,

Chichester, 2007 Also, New Delhi, 2009

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What we learnt today

◮ Motivation of Z-transform ◮ Definition and implications to LTI systems ◮ Examples ◮ Theorems

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Thank you

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