CS6501: Basics

Z-Transforms and Transfer Functions

(Tools for analyzing the dynamics of systems)

Spring 2015

CS6501: Basics

Outline

• Signals and Systems• Z-Transforms

– What are Z-Transforms– What are inverse Z-Transforms– How to infer properties of a signal from

its Z-transform

• Transfer Functions– What are Transfer Functions– How to infer properties of a system from

its Transfer Function

CS6501: Basics

Important

• Z-transforms and transfer functions enable you to analyze signals and systems (general techniques)

– With or without a controller!!!

CS6501: Basics

Signals

• The signals we are studying – Discrete Signals– A discrete signal takes value at each

non-negative time instance

0 2 4 6 8 10 120

2

4

6

8

10

12

14

16

18

CS6501: Basics

Example of a System

0 2 4 6 8 10 120

2

4

6

8

10

12

14

16

18

0 2 4 6 8 10 120

2

4

6

8

10

12

14

16

18

raw readings from a noisy temperature sensor- Input Signal

smooth temperature values after filtering- Output Signal

Filter

33)u(k2)u(k1)u(k

y(k)

A (SISO) system takes an input signal, manipulates it and gives a corresponding output signal.

CS6501: Basics

Control System

Controller TargetSystem

Transducer

ReferenceInput

Controlerror

ControlInput

MeasuredOutput

TransducerOutput

CS6501: Basics

Common Signals

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

-1 0 1 2 3 40

1

2

3

4

0 5 10 150

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-1 0 1 2 3 4 5 6 7 8 90

1

2

3

4

5

6

a=1.2

0 2 4 6 8 10 12 14 16 18-1

-0.5

0

0.5

1

sin(k*pi/6)

0 2 4 6 8 10 12 14 16 18-1

-0.5

0

0.5

1

cos(k*pi/6)

0 2 4 6 8 10 12 14 16 18-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

u(k)=c os(k*pi/6)*0.9k

impulse

delayed impulse

step

ramp

exponential

|a|<1|a|>1

sine

cosine

exponentiallymodulatedcosine/sine

(ak)

CS6501: Basics

Other Signals – (arbitrary)

• From a temperature sensor• From an acoustic sensor

CS6501: Basics

Z-Transform of a Signal

u(k)Z-1

u(0)

u(1)

u(2)

u(3)

u(4)

…

0k

kzu(k)U(z)

U(z)Z

u(0) · z0

+u(1) · z-1

+u(2) · z-2

+u(3) · z-3

+u(4) · z-4

…

CS6501: Basics

Z-Transform – Cont’d

• Mapping from a discrete signal to a function of z– Many Z-Transforms have this form:

• Helps intuitively derive the signal properties– Does it converge?– To which value does it converge? – How fast does it converges to the value?

m

0j

jj

n

0i

ii

zb

zaU(z)

Rational Function of z

CS6501: Basics

Z Transform of Unit Impulse Signaluimpulse(k) Uimpulse(z)Z

Z-1

u(0) = 1

u(1) = 0

u(2) = 0

u(3) = 0

u(4) = 0

…

1 · z0

+0 · z-1

+0 · z-2

+0 · z-3

+0 · z-4

…

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

1(z)Uimpulse

CS6501: Basics

Delayed Unit Impulse Signal

udelay(k) Udelay(z)ZZ-1

u(0) = 0

u(1) = 1

u(2) = 0

u(3) = 0

u(4) = 0

…

0 · z0

+1 · z-1

+0 · z-2

+0 · z-3

+0 · z-4

…

1delay z(z)U

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

CS6501: Basics

Z-Transform of Unit Step Signalustep(k) Ustep(z)Z

Z-1

u(0) = 1

u(1) = 1

u(2) = 1

u(3) = 1

u(4) = 1

…

1 · z0

+1 · z-1

+1 · z-2

+1 · z-3

+1 · z-4

…

...zzz1(z)U 321step

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

CS6501: Basics

Unit Step Signal - continued

,n 1,|a|

a1

a1

a1

)a...aaa)(1(1a...aa1

1n

n2n2

A little bit of math …

assuming

a1

1a1

a1lim

a1

)a...aaa)(1(1lim...aa1

1n

n22

n

n

1-321

step z-11

...zzz1(z)U

CS6501: Basics

Z-Transform of Exponential Signaluexp(k) Uexp(z)Z

Z-1

u(0) = 1

u(1) = a

u(2) = a2

u(3) = a3

u(4) = a4

…

1 · z0

+a · z-1

+a2 · z-2

+a3 · z-3

+a4 · z-4

…

1-

33221exp

az-11

...zazaaz1(z)U

-1 0 1 2 3 4 5 6 7 8 90

1

2

3

4

5

6

a=1.2

Remember this!

CS6501: Basics

What about an arbitrary signal?

• Apply z-transform to any signal

• Just use the z-transform formula

• May not have a convenient formula that summarizes the signal

CS6501: Basics

LTI Systems

• Linear, Time Invariant (LTI) System– Many systems we analyze or design are

or can be approximated by LTI systems– We have a well-established theory for

LTI system analysis and design

• Example - A simple moving average– y(k)=[u(k-1)+u(k-2)+u(k-3)]/3

3-MAu(k) y(k)

CS6501: Basics

What does “Linear” mean exactly?• Scaling

• Superposition

3-MAu(k) y(k)

3-MAλu(k) λy(k)

3-MAu1(k) y1(k)

3-MAu2(k) y2(k)

3-MAu1(k)+u2(k) y1(k)+y2(k)

CS6501: Basics

Time Invariance

3-MAu(k) y(k)

3-MAu’(k)=u(k-n) y’(k)=y(k-n)

Idiom:u(k-n) is u(k) delayed by n time units!

CS6501: Basics

Reality Check

• Typically speaking, are computing systems linear? Why/why not?– Consider saturation …– Assume RT for one job alone in system is X

• Is RT for 3 jobs 3X?

• Typically speaking, are computing systems time-invariant? Why/why not?– Resource allocations are in different states at

different times

CS6501: Basics

Unit Impulse Response

3-MAuimpulse(k) yimpulse(k)

-1 0 1 2 3 4 5 6 7 8 9 100

0.5

1

Claim: If we know yimpulse(k), we can obtain y(k) corresponing to ANY input u(k)!

yimpulse(k) contains ALL information about the input-output relationship of an LTI system.

-1 0 1 2 3 4 5 6 70

0.5

1

33)u(k2)u(k1)u(k

y(k)

Key Points

• Impulse Response – input impluse - basis of convolution (time domain)

• Frequency Response – input sine wave – basis of DFT

CS6501: Basics

CS6501: Basics

An Example: 3-MA

-1 0 1 2 3 4 5 6 70

0.5

1

3MAuimpulse(k) yimpulse(k)

-1 0 1 2 3 4 5 6 70

0.5

1

3MAu (k) y (k) ?

-1 0 1 2 3 4 5 6 70

0.5

1

-1 0 1 2 3 4 5 6 70

0.5

1

-1 0 1 2 3 4 5 6 70

0.5

1

6 x

-1 0 1 2 3 4 5 6 70

1

2

3

4

5

6

7

8

9

9 x

3 x

u(k) =

+

+

+…

uimpulse(k)

uimpulse(k-1)

uimpulse(k-2)

Input: Scaled and delayed

CS6501: Basics

-1 0 1 2 3 4 5 6 70

0.5

1

-1 0 1 2 3 4 5 6 70

0.5

1

An Example: 3-MA

-1 0 1 2 3 4 5 6 70

0.5

1

3MAuimpulse(k) yimpulse(k)

-1 0 1 2 3 4 5 6 70

0.5

1

3MAu (k) y (k) ?

6 x

9 x

3 x

+

+

+…

yimpulse(k-1)

yimpulse(k-2)

-1 0 1 2 3 4 5 6 70

1

2

3

4

5

6

7

8

9

y(k) =

-1 0 1 2 3 4 5 6 70

0.5

1 yimpulse(k)

CS6501: Basics

Convolution

• y(5)= u(0) · yimpulse(k)

+ u(1) · yimpulse(k-1)

+ u(2) · yimpulse(k-2)

+ u(3) · yimpulse(k-3)

+ u(4) · yimpulse(k-4)

-1 0 1 2 3 4 5 6 70

0.5

1

-1 0 1 2 3 4 5 6 70

0.5

1

u(0) x

u(1) x

u(2) x

+

+

+…

yimpulse(k-1)

yimpulse(k-2)

-1 0 1 2 3 4 5 6 70

1

2

3

4

5

6

7

8

9

y(k) =

-1 0 1 2 3 4 5 6 70

0.5

1 yimpulse(k)

(k)yu(k)

i)](ky[u(i)y(k)

impulse

1k

0iimpulse

*

CS6501: Basics

Important Theorem

u(k) *(convolution)

v(k) = y(k)

U(z) V(z) Y(z)=·(multiplication)

Z Z-1 Z Z-1 Z Z-1

Time Domain

Z Domain

CS6501: Basics

Z-Transform/Inverse Z-Transform

LTI: yimpuse(k)=0.3k-1u (k)=0.7k y (k)?

0 5 10 150

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

*(convolution)

10.7z11

Z

=

·(multiplication)

= )0.7z)(10.3z(1z

11

1

0 5 10 150

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Z-1

1

-1

0.3z1z

Z Transfer Function

0 2 4 6 8 10 12 14 16 180

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

CS6501: Basics

Delay the Unit Step Signal

-1 0 1 2 3 40

0.5

1

ustep (k) udelayed(k) udstep(k)*(convolution)

=

y(k)=u(k-1)

1z11

Z Transfer Function

·(multiplication)

Zz-1 =

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

1

-1

z1z

Z

LTI: yimpuse(k)=udelayed(k)

u (k) y (k)

CS6501: Basics

Delayed Unit Step Signal – Cont’d

udstep(k) Udstep(z)ZZ-1

u(0) = 0

u(1) = 1

u(2) = 1

u(3) = 1

u(4) = 1

…

0 · z0

+1 · z-1

+1 · z-2

+1 · z-3

+1 · z-4

…

1z1

z-1z

...zzz(z)U

1-

1-

321dstep

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

Remember this!

CS6501: Basics

Signals in Computer Systems

-1 0 1 2 3 40

0.5

1

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

0 1 2 3 4 5 6 7 8 9 100

2

4

6

8

10

12

Spike, one-time fluctuation in input/output, or disturbance

Change of reference value

Multiple changes of reference value

CS6501: Basics

Transfer Function

• Z-transforms can be used to describe signals

• They can also be used to describe systems (called a transfer function)

• G(z) = Y(z)/U(z) or Y(z) = G(z)U(z)

• Output/InputU(k) Y(k)

CS6501: Basics

Transfer Function

• Transfer function provides a much more intuitive way to understand the input-output relationship, or system characteristics of an LTI system– Stability– Accuracy– Settling time– Overshoot

CS6501: Basics

An LTI System – Discrete Integrator

LTI: yimpuse(k)=udstep(k)

u (k) y (k)

ustep(k) udstep(k) uramp(k)*(convolution)

=

y(k)=y(k-1)+u(k-1)

Z-1Transfer Function

·(multiplication)

Z

1

-1

z1z

-1 0 1 2 3 40

0.5

1

Y(k)=u(k-1)+u(k-2)+…+u(1)+u(0)

-1 0 1 2 3 40

0.5

1

-1 0 1 2 3 40

1

2

3

4

Z

1z11

= 21

-1

)z(1z

CS6501: Basics

Inverse Z-Transform

• Table Lookup – if the Z-Transform looks familiar, look it up in the Z-Transform table!

• Long Division• Partial Fraction Expansion

u(k) U(z)ZZ-1?

2k 3

(k)2u(k)3uu(k) rampstep

(z)2U(z)3U

)z(12z

z13

U(z)

rampstep

21

1

1

Z-1?

CS6501: Basics

Long Division

• Sort both nominator and denominator with descending order of z first

21

1

z2z1z3

U(z)

• u(0), u(1), u(2), u(3), …, are coefficients of the z terms in the answer above (remember, a list of signal values is encoded with z terms: 3,5,7,9…)

CS6501: Basics

Partial Fraction Expansion

• Many Z-transforms of interest can be expressed as division of polynomials of z

m

0j

jj

n

0i

ii

zb

zaU(z)

May be trickier:complex rootduplicate root

)p)...(zp)(zp(zb

zb...zbzbb

m21m

mm

2210

m

1j j

j0 pz

ccU(z)

,1kjdexpp p(k)u

j

m

1jdexppimpulse0 (k)u(k)ucu(k)

j

where k>0

1j

1

zp1

cz

j

CS6501: Basics

An Example

86zz1414z3z

U(z) 2

2

4z

c2z

ccU(z) 21

0

0k,4c2c

0k,cu(k) 1k

21k

1

0

(z-2)(z-4)

U1(z)=c0 Z-1 u1(k)=c0*uimpulse(k)

Z-1

Z-1 u2(k)=c1*2k-1, k>02z

c(z)U 1

2

4zc

(z)U 23

u3(k)=c2*4k-1, k>0

c0? c1? c2?

CS6501: Basics

Get The Constants!

86zz1414z3z

U(z) 2

2

4z

c2z

ccU(z) 21

0

(z-2)(z-4)

,4z

c2z

ccU(z) 21

0

,z ,cU(z) 0 3

86zz1414z3z

limc 2

2

z0

,c2z4)(zc

4)c(z4)U(z)-(zK(z) 21

0

3|2z

1414z3zcK(4) 4z

2

2

CS6501: Basics

An Example – Complete Solution

386zz1414z3z

limU(z)limc 2

2

zz0

4-z1414z3z

86zz1414z3z

2)(z(z)U

2

2

2

2

2-z

1414z3z

86zz

1414z3z4)(z(z)U

2

2

2

3

86zz1414z3z

U(z) 2

2

4z

c2z

ccU(z) 21

0

14-2

1421423(2)Uc

2

21

32-4

1441443(4)Uc

2

32

4z3

2z1

3U(z)

0k,432

0k3,u(k) 1k1k

CS6501: Basics

Solving Difference Equations

m)u(kb...1)u(kbn)y(ka...1)y(kay(k) m1n1

U(z)zb...U(z)zbY(z)za...Y(z)zaY(z) mm

11

nn

11

U(z)za...za1

zb...zbY(z) n

n1

1

mm

11

...y(k)

Z

Z-1Transfer Function

CS6501: Basics

Signal Characteristics from Z-

Transform • If U(z) is a rational function, and

• Then Y(z) is a rational function, too

• Poles are more important – determine key characteristics of y(k)

m)u(kb...1)u(kbn)y(ka...1)y(kay(k) m1n1

m

1jj

n

1ii

)p(z

)z(z

D(z)N(z)

Y(z)

zeros

poles

CS6501: Basics

Determine Properties of System• Most properties only require knowledge of

roots of denominator– SASO properties

• Denominator is called the characteristic polynomial

• Roots of denominator may have complex poles– Represented in rectangular or polar coordinates

CS6501: Basics

Why are poles important?

m

1j j

j0m

1jj

n

1ii

pz

cc

)p(z

)z(z

D(z)N(z)

Y(z)

m

1j

1-kjjimpulse0 pc(k)ucY(k)

Z-1

Z domain

Time domain

poles

components

CS6501: Basics

Various pole values (1)

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

1.5

2

2.5

-1 0 1 2 3 4 5 6 7 8 90

0.2

0.4

0.6

0.8

1

-1 0 1 2 3 4 5 6 7 8 90

0.2

0.4

0.6

0.8

1

-1 0 1 2 3 4 5 6 7 8 9-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

-1 0 1 2 3 4 5 6 7 8 9-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1 0 1 2 3 4 5 6 7 8 9-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

p=1.1

p=1

p=0.9

p=-1.1

p=-1

p=-0.9

CS6501: Basics

Various pole values (2)

-1 0 1 2 3 4 5 6 7 8 90

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-1 0 1 2 3 4 5 6 7 8 90

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-1 0 1 2 3 4 5 6 7 8 90

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

p=0.9

p=0.6

p=0.3

-1 0 1 2 3 4 5 6 7 8 9-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1 0 1 2 3 4 5 6 7 8 9-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1 0 1 2 3 4 5 6 7 8 9-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

p=-0.9

p=-0.6

p=-0.3

CS6501: Basics

Conclusion for Real Poles

• If and only if all poles’ absolute values are smaller than 1, y(k) converges to 0

• The smaller the poles are, the faster the corresponding component in y(k) converges

• A negative pole’s corresponding component is oscillating, while a positive pole’s corresponding component is monotonic

CS6501: Basics

How fast does y(k) converge?

• U(k)=ak, consider u(k)≈0 when the absolute value of u(k) is smaller than or equal to 2% of u(0)’s absolute value

|a|ln4

k

3.912ln0.02|a|kln

0.02|a| k

110.36

4|0.7|ln

4k

0.7a

RememberThis!

0 2 4 6 8 10 120

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

y(k)=0.7k

y(11)=0.0198

CS6501: Basics

Property - Settling Time (k units of time)

• Use Formula for k

|a|ln

4k

a is the magnitude of the (dominant) pole

CS6501: Basics

Example

10.8z11

U(z)

U(z)0.6zY(z)0.4zY(z) 11

0.4z0.6

U(z)Y(z)

G(z)

)0.8z-)(10.4z-(10.6z

U(z)0.4z1

0.6zY(z) 1-1-

-1

1

1

LTI: y(k)=0.4y(k-1)+0.6u(k-1)u (k)=0.8k y (k)?

ZZ

1k1k 0.8b0.4ay(k)

Z-1

SettlingTime

CS6501: Basics

When There Are Complex Poles …

U(z)za...za1

zb...zbY(z) n

n1

1

mm

11

c)...bz(az2

2a4acbb

z2

0,4acb2 )

2a4acbb

)(z2a

4acbba(zcbzaz

222

0,4acb2 )

2ab4acib

)(z2a

b4aciba(zcbzaz

222

If

If

Or in polar coordinates,

)irr)(zirra(zcbzaz2 θθθθ sincossincos

CS6501: Basics

What If Poles Are Complex• If Y(z)=N(z)/D(z), and coefficients of both

D(z) and N(z) are all real numbers, if p is a pole, then p’s complex conjugate must also be a pole– Complex poles appear in pairs

l

1j22

j

j0

l

1j j

j0

r)z(2rz)rdz(zbzr

pz

cc

irrzc'

irrzc

pz

ccY(z)

θ

θθ

θθθθ

cos

cossin

sincossincos

coskθdrsinkθbrpc(k)ucy(k) kkm

1j

1-kjjimpulse0

Z-1

Time domain

CS6501: Basics

An Example

0 2 4 6 8 10 12 14 16 18 20-1

-0.5

0

0.5

1

1.5

2

)3

kπcos(0.8)

3kπ

sin(0.82y(k)

0.640.8zzzz

Y(z)

kk

2

2

Z-Domain: Complex Poles

Time-Domain:Exponentially Modulated Sin/Cos

CS6501: Basics

Poles on Complex Plane

CS6501: Basics

Observations

• Using poles to characterize a signal– The smaller is |r|, the faster the signal converges

• |r| < 1, converge• |r| > 1, does not converge, unbounded• |r|=1?

– When the angle increase from 0 to pi, the frequency of oscillation increases

• Extremes – 0, does not oscillate, pi, oscillate at the maximum frequency

CS6501: Basics

Change Angles

0.9-0.9 Re

Im

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

CS6501: Basics0 2 4 6 8 10 12 14

-6

-4

-2

0

2

4

6

8

10

12

Changing Absolute Value

Im

Re

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10 12 14-3

-2

-1

0

1

2

3

4

CS6501: Basics

Conclusion for Complex Poles

• A complex pole appears in pair with its complex conjugate

• The Z-1-transform generates a combination of exponentially modulated sin and cos terms

• The exponential base is the absolute value of the complex pole

• The frequency of the sinusoid is the angle of the complex pole (divided by 2π)

CS6501: Basics

Steady-State Analysis• If a signal finally converges, what value

does it converge to?• When it does not converge

– Any |pj| is greater than 1– Any |r| is greater than or equal to 1

• When it does converge– If all |pj|’s and |r|’s are smaller than 1, it

converges to 0– If only one pj is 1, then the signal converges to cj

• If more than one real pole is 1, the signal does not converge

kdrkbr kk cossin

m

1j

1-kjjimpulse0 pc(k)ucy(k)

CS6501: Basics

An Example (one pole is = 1)

kk 0.9)(30.52u(k)0.9z

3z0.5zz

1z2z

U(z)

0 10 20 30 40 50 60-1

0

1

2

3

4

5

6

converge to 2

CS6501: Basics

Final Value Theorem

• Enable us to decide whether a system has a steady state error (yss-rss)

CS6501: Basics

Final Value Theorem

1

Theorem: If all of the poles of (1 ) ( ) lie within the unit circle, then

lim ( ) lim ( 1) ( )k z

z Y z

y k z Y z

````````````````````````````

2

1 1

0.11 0.11( )

1.6 0.6 ( 1)( 0.6)

0.11( 1) ( ) | | 0.275

0.6z z

z zY z

z z z z

zz Y z

z

0 5 10 15-0.35

-0.3

-0.25

-0.2

-0.15

-0.1

-0.05

0

k

y(k)

If any pole of (z-1)Y(z) lies out of or ON the unit circle, y(k) does not converge!

CS6501: Basics

What Can We Infer from TF?

• Almost everything we want to know– Stability– Steady-State– Transients

• Settling time• Overshoot

– …

CS6501: Basics

0 5 10 15 200

0.2

0.4

0.6

0.8

1

Bounded Signals

-5

0

5a=0.4

-5

0

5a=0.9

-5

0

5a=1.2

0 5 10-5

0

5a=-0.4

0 5 10-5

0

5a=-0.9

0 5 10-5

0

5a=-1.2

0 10 20 30 40 50 60 70-1

-0.5

0

0.5

1 0 2 4 6 8-1

-0.5

0

0.5

1

CS6501: Basics

BIBO Stability

• Bounded Input Bounded Output Stability– If the Input is bounded, we want the

Output is bounded, too– If the Input is unbounded, it’s okay for

the Output to be unbounded

• For some computing systems, the output is intrinsically bounded (constrained), but limit cycle may happen

CS6501: Basics

Limit Cycle

Solution: make sure the system works in a linearized operating region

Output constrained, But oscillating – Bad!

Imagine CPU utilization Constantly switching from 1 to 0, 0 to 1, …

CS6501: Basics

Example of Stability

10.8z11

U(z)

U(z)0.6zY(z)0.4zY(z) 11

0.4z0.6

U(z)Y(z)

G(z)

)0.8z-)(10.4z-(10.6z

U(z)0.4z1

0.6zY(z) 1-1-

-1

1

1

LTI: y(k)=0.4y(k-1)+0.6u(k-1)u (k)=0.8k y (k)?

ZZ

BIBO? – only one pole at 0.4, so BIBO!

CS6501: Basics

Steady State Gain

yss

CS6501: Basics

Steady-State Gain – Cont’d

• Which value does the output converge to when the input is an unit step signal?– First of all, it has to converge

G(1)

zG(z)lim1z

z1)G(z)(zlim

1)Y(z)(zlimy(k)limy

1z

1z

1zkss

Final Value Theorem

Unit Step Input

CS6501: Basics

More General Case

m)u(kb...1)u(kbn)y(ka...1)y(kay(k) m1n1

U(z)za...za1

zb...zbY(z) n

n1

1

mm

11

n1

m1ss a...a1

b...by

Z

z=1

Transfer Function

Recall y(ss) = G(1) (that is, when z = 1)

CS6501: Basics

Example of Steady State Gain

1z11

U(z)

U(z)0.6zY(z)0.4zY(z) 11

0.4z0.6

U(z)Y(z)

G(z)

)z-)(10.4z-(10.6z

U(z)0.4z1

0.6zY(z) 1-1-

-1

1

1

LTI: y(k)=0.4y(k-1)+0.6u(k-1)u (k)=1 y (k)?

ZZ

Yss? G(1)=1, so yss=1

CS6501: Basics

System Order

• System Order = Number of Poles• The higher the system order is, the

more complex the system behavior is

• Some poles are more important than others– Why?

– If |pi|<|pj|,|pi/pj|k-1 approaches 0 when k is large (pi

k-1 converges faster than pjk-1)

m

1j

1-kjjimpulse0 pc(k)ucy(k)

CS6501: Basics

Overshoot and Settling Time

• If not all poles are positive real numbers, overshoot may happen– Easy to figure out when the system is

first order– For higher order systems, approximation

to first order systems works under certain conditions

• Settling time– First order system– Higher order systems

|p|ln4

ks

CS6501: Basics

How fast does y(k) converge?

• U(k)=ak, consider u(k)≈0 when the absolute value of u(k) is smaller than or equal to 2% of u(0)’s absolute value

|p|ln4

k

3.912ln0.02|p|kln

0.02|p| k

110.36

4|0.7|ln

4k

0.7p

RememberThis!

0 2 4 6 8 10 120

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

y(k)=0.7k

y(11)=0.0198

CS6501: Basics

Examples: Positive Pole

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

0.9z0.1

0.1)-0.9)(z(z0.09

0.3)-0.9)(z(z0.07

0.1)-0.3)(z-0.9)(z(z0.063

DominantPole: 0.9

CS6501: Basics

Examples: Negative Pole

0 5 10 15 20 25 300

0.5

1

1.5

2

0 5 10 15 20 25 300

0.5

1

1.5

2

0 5 10 15 20 25 300

0.5

1

1.5

2

0 5 10 15 20 25 300

0.5

1

1.5

2

0.9z1.9

0.1)-0.9)(z(z1.71

0.3)-0.9)(z(z1.33

0.1)-0.3)(z-0.9)(z(z1.197

DominantPole: -0.9

CS6501: Basics

Dominant Pole

• We can approximate a high-order system with a first-order system with the dominant pole of the high-order system– IF the dominant pole DOES exist– Can give a pretty good estimation of

settling time– Can give a reasonable estimate of the

maximum overshoot

• Some high-order systems do not have dominant pole!

CS6501: Basics

Dominant Pole – Cont’d

• If there is a dominant pole, it must be the pole with the maximum magnitude– The largest pole should have at least

twice the magnitude of the other poles!

• If the dominant pole is real (p’), the high-order system can be approximated by a first-order system

p'z)p'G(1)(1

(z)G'

|p'|ln4

ks

CS6501: Basics

Summary• Signals/Systems

– An LTI system can be specified by• Difference equation• Unit impulse response• Transfer function

• Characterize a signal with Z-transform– Z-domain (poles) -> Time domain (convergence,

etc.)

• Characterize a system with Transfer function– BIBO stability– Steady-State Gain– Transients: overshoot, settling time

• If there exists a dominant pole

CS6501: Basics

Extra Slides

• z-transforms of sin and cos and exponentially modulated sine

CS6501: Basics

sin? cos?

0 2 4 6 8 10 12 14 16 18-1

-0.5

0

0.5

1

sin(k*pi/6)

0 2 4 6 8 10 12 14 16 18-1

-0.5

0

0.5

1

cos(k*pi/6)

CS6501: Basics

From Exponential to Trigonometric

isincose i

isincos)isin()cos(e

1-

33221exp

az-11

...zazaaz1(z)U

?

Euler Formula:

Z[cos(kθ)]?Z[sin(kθ)]?

2ee

cosii

2iee

sinii

CS6501: Basics

Z-Transform of sin/cos

ikeu(k) 1-i ze-11

U(z)

2ee

)cos(ku(k)ikik

Time Domain Z-Transform

2iee

)sin(ku(k)ikik

-ikeu(k)1-i- ze-1

1U(z)

21

1

2121

1

1111

1i1-i

zz2cos1zcos1

)z(sin)zcos(1zcos1

)/2zisinzcos1

1zisinzcos1

1(

)/2ze1

1ze-1

1(U(z)

21-

-1

zz2cos-1zsin

U(z)

CS6501: Basics

Exponentially Modulated sin/cos

2)(ae)(ae

)cos(ka(k)ukiki

kexpcos

2i

)(ae)(ae)sin(ka(k)u

kikik

expsin

221-

-1

zazcos2a-1zsina

U(z)

0 2 4 6 8 10 12 14 16 18-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

u(k)=c os(k*pi/6)*0.9k

221-

-1

zazcos2a-1zsina

U(z)

A damped oscillating signal – a typical output of a second order system

CS6501: Basics

Are these BIBO?

Unity y(k+1) = 1

P Controller y(k+1) = KP u(k)

Integrator y(k+1) = y(k) + u(k)

I Controller y(k+1) = y(k) + KI u(k)

M/M/1/K y(k+1) = 0.49y(k) + 0.033u(k)

Mystery y(k+1) = -1.3y(k) + 2.3u(k)

CS6501: Basics

Better Way to Decide BIBO or NOTTheorem:

A system G(z) is BIBO stable iff all the poles of G(z) are inside the unit circle.

System Time domain Eq Transfer Function Poles

Unity y(k+1) = 1 G(z) = 1 N/A

P Controller

y(k+1) = KP u(k) G(z) = KP N/A

Integrator y(k+1) = y(k) + u(k) G(z) = 1/(z-1) z=1

I Controller

y(k+1) = y(k) + KI u(k) G(z) = KI/(z-1) z=1

M/M/1/K y(k+1) = 0.49y(k) + 0.033u(k)

G(z) = 0.033/(z-0.49)

z = 0.49

Mystery y(k+1) = -1.3y(k) + 2.3u(k)

G(z) = 2.3/(z+1.3) z = -1.3

CS6501: Basics

No Dominant Pole

0 5 10 15 20 25 30 35 40 45 50-3

-2

-1

0

1

2

3

4

5Step Response

Time (sec)

Am

plit

ud

e

pole=-0.9

pole=-0.7

poles=-0.9, -0.7

CS6501: Basics

Why do we need Z-Transform?• A signal can be characterized with

its Z-transform (poles, final value …)• In an LTI system, Z-transform Y(z) is

the multiplication of Z-transform U(z) and the transfer function

• The LTI system can be characterized by the transfer function, or the Z-transform of the unit impulse response