Lecture2 Small

8/13/2019 Lecture2 Small

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Z-Transform

Z-Transform

Recall theZ-Transform definition:

y(z) :=

k=

y(k)zk,

wherez C. This has an associated region of convergence: r0< |z|< R0.

Example: y(k) =

0 fork


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Transfer functions

Transfer Functions

Applying Z-transforms to the difference equations gives:

y(z) =

k=

y(k)zk

=

k=

(a1y(k 1). . . any(k n) +b0u(k) . . . +bmu(k m)) zk,

= a1

k=

y(k 1)zk . . . an

k=

y(k n)zk +b0

k=

u(k)zk . . . +bm

k=

u(k m)zk,

= a1z1y(z). . . anz

ny(z) +b0u(z) . . . bmzmu(z),

Rearranging gives,1 +a1z

1 + +anzn

y(z) =

b0+b1z1 + +bmz

m

u(z).

From this we get the transfer function,

P(z) =y(z)

u(z)=

b0+b1z1 + +bmz

m

1 +a1z1 + +anzn =

b0zn +b1z

n1 + +bmznm

zn +a1zn1 + +an=

b(z)

a(z).

Roy Smith: ECE 147b2: 4

Z-Transform

Example (continued)

y(z) =k=0

eakTzk =k=0

eaTz1

k

= 1

1 eaTz1, for eaT


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Unit pulse response

Unit pulse response

Note that ifu(k) is the unit pulse then,

u(z) =

k=

u(k)zk = 1z0 = 1.

So the system pulse response is given by,

y(k) = Z1{y(z)}= Z1{P(z)u(z)}= Z1{P(z)} ,

which is the inverseZ-transform of the transfer function. (cf. continuous-time).

First order example:

P(z) = b1z

z a1, and so the pulse response is y(k) = Z1{P(z)}= b1 a

k1.

We will use this example to look at the various behaviors possible.


Transfer functions

Pole/Zero locations

P(z) = b(z)

a(z)

The pole and zero positions determine thetypical plant responses.

The poles are the roots ofa(z) = 0.

The zeros are the roots ofb(z) = 0.

Unit pulse response: u(k) =

0 fork 0

This is used to characterise LTI discrete-time systems (cf. impulse response)

!

4 !

3 !

2 !

1 0 1 2 3 4 5

0

0.2

0.4

0.6

0.8

1

1.2

unit pulse

Index: k



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Pole positions

First order example continued: (a1< 0)

1< a1< 0(inside the unit circle)

0 2 4 6 8 10 12 14 16 18 20!4

!2

0

2

4

10 20

y(k)

!1 < a1 < 0

Index: k

a1< 1(outside the unit circle)

0 2 4 6 8 10 12 14 16 18 20!4

!2

2

4

5 10 15 20

y(k)

a1 < !1

Index: k

We can see that for |a1|< 1 the responses decay. If|a1|> 1 the responses blow up.Ifa1 is negative the responses alternate in sign.


Pole positions

First order example continued

The examples are illustrated forb1= 1.5.

0< a1< 1(inside the unit circle)

0 5 10 15 20

1.0

2.0

3.0

Index: k

y(k)

0 < a1 1(outside the unit circle)

0 5 10 15 20

1.0

2.0

3.0

y(k)

a1 > 1

Index: k



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Pole positions

Generic second order system

For example, consider the pulse response:

y(k) = rk cos k, fork 0, withr >0.

Which corresponds to

P(z) = 1 r cos z1

1 2r cos z1 +r2z2 =

z(z r cos )

z2 2r cos z+r2.

Pole/zero pattern r


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Pole positions

Oscillatory responses:Pole angle () determines how oscillatory the response will be.

Periodic responses (with periodN): cos(k) = cos((k+N)) orN=2

.

Examples

= /4 implies

N= 8 samples/oscillation

= /2 impliesN= 4 samples/oscillation

1.0

0

1.0

5 10 15 20

y(k) pole angle = 45 degrees

Index: k

1.0

0

1.0

5 10 15 20

y(k)

Index: k

pole angle = 90 degrees

Roy Smith: ECE 147b 2: 12

Pole positions

Decay rates

Ifr


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Stability

BIBO Stability (sufficiency)

Assume thatu(l) is bounded: |u(l)| M < , for all l.

Now look at,

|y(k)| =

l=

p(k l)u(l)

l=

|p(k l)| |u(l)|.

As |u(l)| M,

|y(k)| M

l=

|p(k l)|.

So if

l=

|p(l)|< ,

then the system is BIBO stable.


BIBO stability

Discrete-time System Stability

P(z) has pulse responsep(k),

p(k) = Z1{P(z)} .

The output,y(k), is given by the discrete-timeconvolution,

y(k) =

l=

p(k l)u(l).

P(z)u(z)y(z)

BIBO stability

If|u(k)|< , under what conditions is |y(k)|< ?



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Stability

BIBO Stability (necessity)

If

l=

|p(l)|= , is the system necessarily unstable?

Does there exist a particularu(k) which makes|y(l)|= for some particularl?

Consider: u(l) =

p(l)

|p(l)| ifp(l) = 0,

0 ifp(l) = 0

.

This is a sequence made up of 0, 1, and -1. It is bounded (M= 1).

Now look aty(0).

y(0) =

l=

p(l)u(l) =

l=

(p(l))2

|p(l)| =

l=

|p(l)| =

l=

|p(l)| = .

Unbounded!


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Lecture2 Small