discrete systems & z-transforms · f(t) cne 0 f t e dt t c t t jn t n / 2 / 2 ( ) 0 1 f t dt t...

1

Discrete Systems & Z-Transforms

© 2014 School of Information Technology and Electrical Engineering at The University of Queensland

TexPoint fonts used in EMF.

Read the TexPoint manual before you delete this box.: AAAAA

http://elec3004.org

Lecture Schedule: Week Date Lecture Title

1 2-Mar Introduction

3-Mar Systems Overview

2 9-Mar Signals as Vectors & Systems as Maps

10-Mar [Signals]

3 16-Mar Sampling & Data Acquisition & Antialiasing Filters

17-Mar [Sampling]

4 23-Mar System Analysis & Convolution

24-Mar [Convolution & FT]

5 30-Mar Discrete Systems & Z-Transforms 31-Mar [Z-Transforms]

6 13-Apr Frequency Response & Filter Analysis

14-Apr [Filters]

7 20-Apr Digital Filters

21-Apr [Digital Filters]

8 27-Apr Introduction to Digital Control

28-Apr [Feedback]

9 4-May Digital Control Design

5-May [Digitial Control]

10 11-May Stability of Digital Systems

12-May [Stability]

11 18-May State-Space

19-May Controllability & Observability

12 25-May PID Control & System Identification

26-May Digitial Control System Hardware

13 31-May Applications in Industry & Information Theory & Communications

2-Jun Summary and Course Review

30 March 2015 - ELEC 3004: Systems 2

http://itee.uq.edu.au/~metr4202/

http://itee.uq.edu.au/~metr4202/

http://creativecommons.org/licenses/by-nc-sa/3.0/au/deed.en_US

http://elec3004.com/

http://elec3004.com/

2

Convolution ℱ: Fourier Series

(Periodic functions)

ℒ ℱ: (𝜉 = 𝜎 + 𝑖𝜏)

(ℝ ℂ) ℂ: Poles & Zeros DFFT Z-Transform

Lecture Overview

ODE

ℒ: Laplace (s)

Transfer functions

Cascade of LCC ODE

Convolution

Z-Transform

• Course So Far:

• Lecture(s):


For c(τ)= :

1. Keep the function f (τ) fixed

2. Flip (invert) the function g(τ) about the vertical axis (τ=0)

= this is g(-τ)

3. Shift this frame (g(-τ)) along τ (horizontal axis) by t0.

= this is g(t0 -τ)

For c(t0):

4. c(t0) = the area under the product of f (τ) and g(t0 -τ)

5. Repeat this procedure, shifting the frame by different values

(positive and negative) to obtain c(t) for all values of t.

Graphical Understanding of Convolution


3

Graphical Understanding of Convolution (Ex)


Complex Numbers and Phasors

Y

X

Re j

R

Rcos( )

Rsin( )

Re ( cos , sin )

cos sin

(cos sin )

j R R

R jR

R j

Positive Frequency

component


4

Complex Numbers and Phasors

Y

X

Re j

R

R cos( )

Rsin( )

Re ( cos( ), sin( ))

cos( ) sin( )

(cos sin )

j R R

R jR

R j

Negative frequency

component


Positive and Negative Frequencies • Frequency is the derivative of phase

more nuanced than : 1

𝜏= 𝑟𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒

• Hence both positive and negative frequencies are possible.

• Compare – velocity vs speed

– frequency vs repetition rate


5

• Q: What is negative frequency?

• A: A mathematical convenience

• Trigonometrical FS – periodic signal is made up from

– sum 0 to of sine and cosines ‘harmonics’

• Complex Fourier Series & the Fourier Transform – use exp(jωt) instead of cos(ωt) and sin(ωt)

– signal is sum from 0 to of exp(jωt)

– same as sum - to of exp(-jωt)

– which is more compact (i.e., less chalk!)

Negative Frequency


Rotating wheel and peg

Top

View

Front

View

Need both top and front

view to determine rotation

Another way to see Aliasing Too!


6

Frequency Response

Fourier Series Fourier Transforms


Typical Linear Processors • Convolution h(n,k)=h(n-k)

• Cross Correlation h(n,k)=h(n+k)

• Auto Correlation h(n,k)=x(k-n)

• Cosine Transform h(n,k)=

• Sine Transform h(n,k)=

• Fourier Transform h(n,k)=

cos2

Nnk

sin2

Nnk

exp jN

nk2


7

• Signal measured (or known) as a function of an independent

variable – e.g., time: y = f(t)

• However, this independent variable may not be the most

appropriate/informative – e.g., frequency: Y = f(w)

• Therefore, need to transform from one domain to the other – e.g., time frequency

– As used by the human ear (and eye)

Transform Analysis

Signal processing uses Fourier, Laplace, & z transforms etc


Sinusoids and Linear Systems

If

or

x t A t( ) cos( ) 0 0

x n A nt( ) cos( ) 0 0

then in steady state

h(t) or h(n)

x(t) or x(n) y(t) or y(n)

y t AC t( ) ( )cos( ( )) 0 0 0 0

y n AC T nt T( ) ( )cos( ( )) 0 0 0 0


8

• The pair of numbers C(ω0) and q(ω0) are the complex gain of

the system at the frequency ω0 .

• They are respectively, the magnitude response and the phase

response at the frequency ω0 .

Sinusoids and Linear Systems

y t AC t( ) ( )cos( ( )) 0 0 0 0

y n AC T nt T( ) ( )cos( ( )) 0 0 0 0


• Why probe system with sinusoids?

• Sinusoids are eigenfunctions of linear systems???

• What the hell does that mean?

• Sinusoid in implies sinusoid out

• Only need to know phase and magnitude (two parameters) to

fully describe output rather than whole waveform – sine + sine = sine

– derivative of sine = sine (phase shifted - cos)

– integral of sine = sine (-cos)

• Sinusoids maintain orthogonality after sampling (not true of

most orthogonal sets)

Why Use Sinusoids?


9

Frequency Response



Fourier Series

• Deal with continuous-time periodic signals.

• Discrete frequency spectra.

A Periodic Signal

T 2T 3T

t

f(t)

Source: URI ELE436


10

Two Forms for Fourier Series

T

ntb

T

nta

atf

n

n

n

n

2sin

2cos

2)(

11

0Sinusoidal

Form

Complex

Form:

n

tjn

nectf 0)( dtetfT

cT

T

tjn

n

2/

2/

0)(1

dttfT

aT

T2/

2/0 )(

2tdtntf

Ta

T

Tn 0

2/

2/cos)(

2

tdtntfT

bT

Tn 0

2/

2/sin)(

2

Source: URI ELE436


• Any finite power, periodic, signal x(t) – period T

• can be represented as () summation of – sine and cosine waves

• Called: Trigonometrical Fourier Series

Fourier Series

00 0

1

( ) cos( ) sin( )2

n n

n

Ax t A n t B n t

• Fundamental frequency ω0=2/T rad/s or 1/T Hz

• DC (average) value A0 /2


11

0 1 2 3 4 5 6 -2

-1

0

1

2

time (t)

y =

f(t

)

0 1 2 3 4 5 6 7 8 0

0.5

1

1.5

frequency (f)

Am

plit

ude

Frequency representation (spectrum) shows signal contains:

• 2Hz and 5Hz components (sinewaves) of equal amplitude


• An & Bn calculated from the signal, x(t) – called: Fourier coefficients

Fourier Series Coefficients

,3,2,1)sin()(2

,2,1,0)cos()(2

2/

2/

0

2/

2/

0

ndttnwtxT

B

ndttnwtxT

A

T

T

n

T

T

n

Note: Limits of integration can vary,

provided they cover one period


12

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -1.5

-1

-0.5

0

0.5

1

1.5

time (s)

signal (black), approximation (red) and harmonic (green)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -1.5

-1

-0.5

0

0.5

1

1.5

time (s)

harm

onic


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -1.5

-1

-0.5

0

0.5

1

1.5

time (s)


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -1.5

-1

-0.5

0

0.5

1

1.5

time (s)

harm

onic


13

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -1.5

-1

-0.5

0

0.5

1

1.5

time (s)


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -1.5

-1

-0.5

0

0.5

1

1.5

time (s)

harm

onic


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -1.5

-1

-0.5

0

0.5

1

1.5

time (s)


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -1.5

-1

-0.5

0

0.5

1

1.5

time (s)

harm

onic


14

Example: Square wave

))cos(1(2

cos11coscoscos

)sin()sin()sin()(

0sinsin

)cos()cos()cos()(

).2(

;21,1

;10,1

)(

2

1

1

0

2

1

1

0

2

0

2

1

1

0

2

1

1

0

2

0

nn

B

n

n

nnn

n

n

tn

n

tnB

dttndttndttntxB

n

tn

n

tnA

dttndttndttntxA

tx

t

t

tx

n

n

n

n

n

periodic! i.e., x(t + 2) = x(t)

No cos terms as sin(n) = 0 n

x(t) has odd symmetry

Sin terms only

cos(2n) = 1 n


Example: Square wave

etc

harmonic)(fifth )5sin(5

4

harmonic)(fourth 0

harmonic) (third)3sin(3

4

harmonic) (second0

al)(fundament)sin(4

x(t)

gives, terms theExpanding

)sin())cos(1(2

)(

is, seriesFourier ricTrigonomet Therefore,

1

t

t

t

tnnn

txn

• Only odd harmonics;

• In proportion

1,1/3,1/5,1/7,…

• Higher harmonics

contribute less;

• Therefore, converges


15

How to Deal with Aperiodic Signal?

A Periodic Signal

T

t

f(t)

If T, what happens? Source: URI ELE436


tjn

n

nT ectf 0)(

Fourier Integral

n

tjnT

T

jn

T edefT

002/

2/)(

1

dtetfT

cT

T

tjn

Tn

2/

2/

0)(1

T

20

2

1 0

T

n

tjnT

T

jn

T edef 00

0

2/

2/)(

2

1

Let T

20

0 dT

n

tjnT

T

jn

T edef 002/

2/)(

2

1

dedef tjj

T )(2

1

Source: URI ELE436


16

dedeftf tjj)(2

1)(

Fourier Integral

F(j)

dtetfjF tj

)()(

dejFtf tj)(2

1)( Synthesis

Analysis

Source: URI ELE436


Fourier Series vs. Fourier Integral

n

tjn

nectf 0)(Fourier

Series:

Fourier

Integral:

dtetfT

cT

T

tjn

Tn

2/

2/

0)(1

dtetfjF tj

)()(

dejFtf tj)(2

1)(

Period Function

Discrete Spectra

Non-Period

Function

Continuous Spectra

Source: URI ELE436


17

Complex Fourier Series (CFS)

• Also called Exponential

Fourier series

– As it uses Euler’s relation

• FS as a Complex phasor

summation

j

tjnwtjnwtnw

tjnwtjnwtnw

twjAtwAtjwA

2

)exp()exp()sin(

2

)exp()exp()cos(

implies,which

)sin()cos()exp(

000

000

000

n

n tjnwXtx )exp()( 0Where Xn are the

CFS coefficients


Complex Fourier Coefficients

• Again, Xn calculated from

x(t)

• Only one set of coefficients,

Xn

– but, generally they are

complex

2/

2/

0 )exp()(1

T

T

n dttjnwtxT

X

Remember: fundamental ω0 = 2/T !


18

Relationships

• There is a simple

relationship between

– trigonometrical and

– complex Fourier

coefficients,

.0,2

;0,2

2

00

njBA

njBA

X

AX

nn

nn

n

Therefore, can calculate simplest form and convert

Constrained to be

symmetrical, i.e.,

complex conjugate

*

nn XX


Example: Complex FS

• Consider the pulse train

signal

• Has complex Fourier series:

).(

;2

,0

;2

0,

)(

Ttx

Tt

tA

tx

T

2exp

2exp

exp1

exp)(1

00

0

2

2

0

2

2

0

jnjn

Tjn

A

dttjnAT

dttjntxT

X

T

T

n

A

Note: n is the

ind. variable

Note: ×

by / …


19

Example: Complex FS

• Which using Euler’s identity

reduces to:

Tw

nwT

A

nw

nw

T

AX n

2

)2(sa2

)2sin(

0

0

0

0

sin2sincossincos

sincossincos

expexp

jjj

jj

jj

Note: letting 2

0

n

Note:

cos(-) = cos(): even

sin(-) = -sin(): odd


Frequency Response



20

• A Fourier Transform is an integral transform that re-expresses

a function in terms of different sine waves of varying

amplitudes, wavelengths, and phases.

• When you let these three waves interfere with each other you

get your original wave function!

Fourier Transform

1-D Example:

Source: Tufts Uni Sykes Group


Fourier Series • What we have produced is a processor to calculate one

coefficient of the complex Fourier Series

• Fourier Series Coefficients = Heterodyne and average over

observation interval T

CT

h t e dtk

jT

ktT

1

2

0

( )


21

• If we change the limits of integration to the entire real line,

remove the division by T, and make the frequency variable

continuous, we get the Fourier Transform

ELEC 3004: Systems 30 March 2015 - 57

Fourier Transform

C h t e dtj t( ) ( )

Fourier Transform (is not the Fourier Series per se)

Fourier

Series

Discrete

Fourier

Transform

Continuous

Fourier

Transform

Fourier

Transform

Continuous

Time

Discrete

Time

Per

iodic

A

per

iodic

Source: URI ELE436


22

• Fourier series – Only applicable to periodic signals

• Real world signals are rarely periodic

• Develop Fourier transform by – Examining a periodic signal

– Extending the period to infinity

Fourier Transform


• Problem: as T , Xn 0 – i.e., Fourier coefficients vanish!

• Solution: re-define coefficients – Xn’ = T × Xn

• As T – (harmonic frequency) nω0 ω (continuous freq.)

– (discrete spectrum) Xn’ X(ω) (continuous spect.)

– ω0 (fundamental freq.) reduces dω (differential) • Summation becomes integration

Fourier Transform


23

Fourier Transform Pair

dtetfjF tj

)()(

dejFtf tj)(2

1)( Synthesis

Analysis

Fourier Transform:

Inverse Fourier Transform:

Source: URI ELE436


dtetfjF tj

)()(

Continuous Spectra

)()()( jjFjFjF IR

)(|)(| jejF FR(j)

FI(j)

()

Magnitude

Phase

Source: URI ELE436


24

-10 -8 -6 -4 -2 0 2 4 6 8 10 0

0.2

0.4

0.6

0.8

1

Pulse width = 1

x(t

)

time (t)

-20 -5 0 5 20 -0.5

0

0.5

1

X(w

)

Angular frequency (w)

rect(t)

sinc(w/2)

Time limited

Infinite bandwidth


-10 -8 -6 -4 -2 0 2 4 6 8 10 0

0.2

0.4

0.6

0.8

1

Pulse width = 2

x(t

)

time (t)

-10 -5 0 5 10 -0.5

0

0.5

1

1.5

2

X(w

)


rect(t/2)

2 sinc(w/)

Parseval’s Theorem


25

-10 -8 -6 -4 -2 0 2 4 6 8 10 0

0.2

0.4

0.6

0.8

1

Pulse width = 4

x(t

)

time (t)

-10 -5 0 5 10 -1

0

1

2

3

4

X(w

)


rect(t/4)

4 sinc(2w/)


-10 -8 -6 -4 -2 0 2 4 6 8 10 0

0.2

0.4

0.6

0.8

1

Pulse width = 8

x(t

)

time (t)

-10 -5 0 5 10 -2

0

2

4

6

8

X(w

)


rect(t/8)

8 sinc(4w/)


26

-40 -30 -20 -10 0 10 20 30 40 -0.25

0

0.5

1

Symmetry: F{sinc(t/2)} = 2 rect(-w)

x(t

)

time (t)

-5 -4 -3 -2 -1 0 1 2 3 4 5 0

2

X(w

)


2 rect(-w)

sinc(t/2)

Infinite time

Finite bandwidth

‘Ideal’ Lowpass filter 30 March 2015 - ELEC 3004: Systems 70

• Linearity – F {a x(t) + b y(t)} = a X(ω) + b Y(ω)

• Time and frequency scaling – F {x(at)} = 1/a X(ω/a)

– broader in time narrower in frequency • and vice versa

• Symmetry (duality) – 2x(-ω) = X(t) exp(-jωt)dt

• i.e., Fourier transform ‘pairs’

Properties of Fourier Transform

Time limited signal limited has infinite bandwidth;

Signal of finite bandwidth has infinite time support


27


• if…

• x(t) is real

• x(t) is real and even

• x(t) is real and odd

• Then…

• X(-ω) = X(ω)*

– {X(ω)} is even

– {X(ω)} is odd

– |X(ω)| is even

– X(ω) is odd

• X(ω) is real and even

• X(ω) is imaginary and odd


Fourier Transforms

-15 -10 -5 0 5 10 15

-1

-0.5

0

0.5

1

t

x(t

)

-ω0 0

ω0

0

1

2

3

ω

X(w

)

x(t) = cos(ω0t)

(real & even)

X(w) = [(ω-ω0)

+ (ω+ω0)]

(real and even)

1

0 0

1| | exp

2X x t F X x t j t

Note: cos(ω0t) has energy! But is dual of (w – ω0)


28

Fourier Transforms

x(t) = sin(ω0t)

(real and odd)

X(w) = j[(ω+ω0)

- (ω-ω0)]

(imaginary & odd)

-15 -10 -5 0 5 10 15

-1

-0.5

0

0.5

1

t

x(t

)

-ω0 0

ω0 3 4

-4

-2

0

2

4

{X

(w)}

Note: sin & cos have same Mag spectrum, Phase is only difference


• Time Shift – F {x(t - )} = exp(-j w)X(w)

• time shift phase shift

• Convolution and multiplication – F {x(t) y(t)} = X(ω) Y(ω)

• i.e., implement convolution in Fourier domain

– F {x(t) y(t)} = 1/2 (X(ω) Y(ω)) • i.e., Fourier interpretation of multiplication (e.g., frequency modulation)



29

-15 -10 -5 0 5 10 15 0

0.2

0.4

0.6

0.8

1

1.2

Frequency (rad/s)

|exp

(-j

)|

-15 -10 -5 0 5 10 15 -200

-100

0

100

200

Frequency (rad/s)

ph

ase

(d

egre

es)

Magnitude and Phase of exp(-j w)

modifies phase only

as

cos2 +

sin2 =1


0

0.5

1

x(t) = rect(t)

0

0.5

1

h(t) = rect(t)

-1 0 1 2 0

x(t)*h(t) = tri(t)

-0.5

0

0.5

1 X(w) = sinc(w/2)

-0.5

0

0.5

1 H(w) = sinc(w/2)

-20 -10 0 10 20 0

0.5

1 X(w)H(w) = sinc

2 (w/2)

F

F

F

ZOH

ZOH

FOH


30

• Differentiation in time

• Integration in time

More properties of the FT

n

n

dF x t j X

dt

dF x t j X

dt

Differentiation ×ω (Note: HPF & DC x zero)

1

0

t

F x t dt X Xj

Integration /ω + DC offset (LPF & opposite of differentiation)


More Fourier Transforms

ω

F(w) = (2/t)(ω - 2n/t)

t

f(t) =(t - nt) = (t)

… … … …

F

ω

F(w) = (2/t) (ω - n/t)

t

f(t) = (t - 2nt)

… … … …

F

Impulse train, ‘comb’ or ‘Shah’ function


31

More Fourier Transforms

t

f(t) = (t)

ω

F(ω) = 1

F

ω

F(ω) = 2(ω)

t

f(t) = 1

F

Limit of previous as t and t 0 respectively

Note: f(t) = 1 has energy! But is dual of (t)

Note: u(t) also has energy! But F{u(t)} = F{(t)} i.e., apply integration property


• If F{x(t)} = X(w) – F{x(2t)} =?

– F{x(t/4)} =?

• F{(t)} = ?

• F{1} = ?

“Pop Quiz”: Questions


32

… No Worries LectopiaLand !

• If F{x(t)} = X(ω) – F{x(2t)} = 1/2X(ω/2)

• narrower in t broader in freq

– F{x(t/4)} = 4X(4ω) • broader in t narrower in freq (but increased amplitude)

• F{(t)} = 1 – i.e. flat spectrum (all frequencies equally)

• F{1} = (ω) – i.e. impulse at DC only

Pop Quiz: Answers!


• Represents (usually finite energy) signals – as sum of cosine waves

• at all possible frequencies

• |X(ω)|dω/2 is amplitude of cosine wave – i.e., in frequency band ω to ω + dω

• X(ω) is phase shift of cosine wave

• Also represents finite power, periodic signals – Using (ω)

• Distribution with frequency of – both magnitude & phase

– called a Frequency spectrum (continuous)

Interpretation of Fourier Transform


33

Fourier Image Examples

Lena Bridge


Fourier Magnitude and Phase

20*log10(abs(fft(Lena))) angle(fft(Lena))

0 10 20 30 40 50 60 70 80 90 1000

0.5

1

Frequency

Magnitude

1/f

‘random’

range()

Bridge spectra look similar


34

Magnitude and Phase Only

ifft(abs(fft(Lena)) + angle(0)) ifft(abs(fft(Bridge)) + angle(fft(Lena)))

Lena magnitude only Lena phase + bridge magnitude

Note: titles are illustrative only and are not the actual Matlab commands used!


FFT Fourier Transform (not just yet – we’ll come back to this)


35

• For a discrete time sequence we define two classes of

Fourier Transforms:

• the DTFT (Discrete Time FT) for sequences having

infinite duration,

• the DFT (Discrete FT) for sequences having finite

duration.

Fourier Analysis of Discrete Time Signals


• Given a sequence x(n) having infinite duration, we define the

DTFT as follows:

The Discrete Time Fourier Transform (DTFT)

X DTFT x n x n e

x n IDTFT X X e d

j n

n

j n

( ) ( ) ( )

( ) ( ) ( )

1

2

x n( )

nN 1

X ( )

discrete time continuous

frequency

….. …..


36

Observations:

• The DTFT is periodic with period ;

• The frequency is the digital frequency and therefore it is limited to

the interval

)(X 2

Recall that the digital frequency is a normalized frequency relative to

the sampling frequency, defined as

sF

F 2

0

0

2/sF2/sFsF sF F

22

one period of )(X


Example

0 1N

1

][nx

n

2/sin

2/sin

1

1)(

2/)1(

1

0

Ne

e

eeX

Nj

j

NjN

n

nj

DTFT

since


37

Fast Fourier Transform Algorithms • Consider DTFT

• Basic idea is to split the sum into 2 subsequences of length N/2

and continue all the way down until you have N/2

subsequences of length 2

Log2(8)

N


Radix-2 FFT Algorithms - Two point FFT • We assume N=2^m

– This is called Radix-2 FFT Algorithms

• Let’s take a simple example where only two points are given n=0, n=1; N=2

Source: http://www.cmlab.csie.ntu.edu.tw/cml/dsp/training/coding/transform/fft.html

y0

y1

y0

Butterfly FFT

Advantage: Less

computationally

intensive: O(N/2*log(N))


38

• First break x[n] into even and odd

• Let n=2m for even and n=2m+1 for odd

• Even and odd parts are both DFT of a N/2 point sequence

• Break up the size N/2 subsequent in half by letting 2mm

• The first subsequence here is the term x[0], x[4], …

• The second subsequent is x[2], x[6], …

1

1)2sin()2cos(

)]12[(]2[

2/

2

2/

2/

2/2/

2/

2/

2/

2

12/

0

2/

12/

0

2/

N

N

jN

N

m

N

N

N

m

N

Nm

N

mk

N

mk

N

N

m

mk

N

k

N

N

m

mk

N

W

jeW

WWWW

WW

mxWWmxW

General FFT Algorithm


Example

1

1)2sin()2cos(

)]12[(]2[][

2/

2

2/

2/

2/2/

2/

2/

2/

2

12/

0

2/

12/

0

2/

N

N

jN

N

m

N

N

N

m

N

Nm

N

mk

N

mk

N

N

m

mk

N

k

N

N

m

mk

N

W

jeW

WWWW

WW

mxWWmxWkX

Let’s take a simple example where only two points are given n=0, n=1; N=2

]1[]0[]1[]0[)]1[(]0[]1[

]1[]0[)]1[(]0[]0[

1

1

0

0

1.0

1

1

1

0

0

1.0

1

0

0

0.0

1

0

1

0

0

0.0

1

xxxWxxWWxWkX

xxxWWxWkX

mm

mm

Same result


39

FFT Algorithms - Four point FFT

First find even and odd parts and then combine them:

The general form:


FFT Algorithms - 8 point FFT

http://www.engineeringproductivitytools.com/stuff/T0001/PT07.HTM

Applet:

http://www.falstad.com/fourier/directions.html


40

Poles and Zeros


Poles and Zeros

Source: Boyd, EE102,5-12


41

Poles and Zeros



Pole Zero Plot



42

Partial Fraction Expansion



Partial Fraction Expansion Example



43

How to Handle the Digitization?

(z-Transforms)


The z-transform

• The discrete equivalent is the z-Transform†:

𝒵 𝑓 𝑘 = 𝑓(𝑘)𝑧−𝑘∞

𝑘=0

= 𝐹 𝑧

and

𝒵 𝑓 𝑘 − 1 = 𝑧−1𝐹 𝑧

Convenient!

†This is not an approximation, but approximations are easier to derive

F(z) y(k) x(k)


44

The z-Transform

• It is defined by:

Or in the Laplace domain:

𝑧 = 𝑒𝑠𝑇

• Thus: or

• I.E., It’s a discrete version of the Laplace:

𝑓 𝑘𝑇 = 𝑒−𝑎𝑘𝑇 ⇒ 𝒵 𝑓 𝑘 =𝑧

𝑧 − 𝑒−𝑎𝑇


The z-transform • In practice, you’ll use look-up tables or computer tools (ie. Matlab)

to find the z-transform of your functions

𝑭(𝒔) F(kt) 𝑭(𝒛)

1

𝑠

1 𝑧

𝑧 − 1

1

𝑠2

𝑘𝑇 𝑇𝑧

𝑧 − 1 2

1

𝑠 + 𝑎

𝑒−𝑎𝑘𝑇 𝑧

𝑧 − 𝑒−𝑎𝑇

1

𝑠 + 𝑎 2

𝑘𝑇𝑒−𝑎𝑘𝑇 𝑧𝑇𝑒−𝑎𝑇

𝑧 − 𝑒−𝑎𝑇 2

1

𝑠2 + 𝑎2

sin (𝑎𝑘𝑇) 𝑧 sin𝑎𝑇

𝑧2− 2cos𝑎𝑇 𝑧 + 1


45

ℒ(ZOH)=??? : What is it?

• Lathi

• Franklin, Powell, Workman

• Franklin, Powell, Emani-Naeini

• Dorf & Bishop

• Oxford Discrete Systems:

(Mark Cannon)

• MIT 6.002 (Russ Tedrake)

• Matlab

Proof!

• Wikipedia


• Assume that the signal x(t) is zero for t<0, then the output

h(t) is related to x(t) as follows:

Zero-order-hold (ZOH)

x(t) x(kT) h(t) Zero-order

Hold Sampler


46

• Recall the Laplace Transforms (ℒ) of:

• Thus the ℒ of h(t) becomes:

Transfer function of Zero-order-hold (ZOH)


… Continuing the ℒ of h(t) …

Thus, giving the transfer function as:

Transfer function of Zero-order-hold (ZOH)

𝓩


discrete systems & z-transforms · f(t) cne 0 f t e dt t c t t jn t n / 2 / 2 ( ) 0 1 f t dt t...

Documents