8. z-transforms1 agenda r1. difference equations r2. z-transforms r3. discrete fourier transform r4....

8. Z-transforms 1

Agenda

1. Difference equations2. Z-transforms3. Discrete Fourier transform4. Power spectral density

8. Z-transforms 2

1. Difference equations

DefinitionExamplesSample-date systemsResponse of discrete LTI system


8. Z-transforms 3

Definition

Difference equations are equations that relate the state of the system at discrete points.

Difference equations normally assume that there is a single, fixed time between samples

Difference equations are easily handled by computers and even by spreadsheets

Difference equations are a special-case of sampled-data systems


8. Z-transforms 4

Examples (1 of 4)

unitdelay

2

x[n] y[n]

y[n-1]

y[n] = 2 y[n-1] + x[n]y[n] - 2 y[n-1] = x[n]

A first-order, linear, time-invariant (LTI) difference equation

Example 1


8. Z-transforms 5

Examples (2 of 4)

unitdelay2

x[n] y[n]

q[n-1]

y[n] = q[n] + 4 q[n-1]q[n] = 2 q[n-1] + x[n]

4

q[n]

4 q[n-1] + q[n] = y[n] -2 q[n-1] + q[n] = x[n]

q[n-1] = 1/6(-x[n] + y[n]) q[n] = 1/3(2 x[n] + y[n])q[n-1] = 1/3(2 x[n-1] + y[n-1])

y[n] - 2 y[n-1] = x[n] + 4 x[n-1]

Example 2


8. Z-transforms 6

Examples (3 of 4)

unitdelay

2

x[n] y[n]

4

unitdelay

y[n] = 4 y[n-1] + 2 y[n-2] + x[n]

Example 3

y[n-1]y[n-2]


8. Z-transforms 7

Examples (4 of 4)

unitdelay

x[n] y[n]

1/3

unitdelay

y[n] = 1/3 (x[n] + x[n-1] + x[n-2])

1/3

1/3Example 4

x[n-1] x[n-2]


8. Z-transforms 8

Sample-data systems

A system that receives information intermittently

Sampling may be periodic, aperiodic, or random

State information is not available between samples

sampler

computerhold

circuitplant

x(t) e(t)

-+

y(t)


8. Z-transforms 9

Response of discrete LTI systemImpulse response• h[n] = T{[n]}

Arbitrary input

• x[n] = x[k] [n-k]Response to arbitrary input

• y[n] = T {x[n]} = T { x[k] [n-k] }

• = x[k] h[n-k]

• = convolution sum

k=-

k=-

k=-


8. Z-transforms 10

2. Z-transform

IntroductionZ transformationDefinitionTransformsExamples

2. Z-transforms

8. Z-transforms 11

IntroductionThe z-transform is the discrete-time

counterpart of the Laplace transformThe Laplace transform converts integro-

differential equations into algebraic equations.

The z-transform converts difference equations into algebraic equations

Properties of z-transforms parallel those of Laplace transforms, but there are important distinctions

2. Z-transforms

8. Z-transforms 12

Z transformation

lineardifference equation

timedomainsolution

z transformedequation

z transformsolution

time domain

summing

algebra

z transforminverse z transform

z domain orcomplex frequency domain

2. Z-transforms

8. Z-transforms 13

Definition

X(z) = x[n] z -n

where z is a complex number

n=-

2. Z-transforms

8. Z-transforms 14

Transforms (1 of 4)

Unit impulse sequence, [n]

X(z) = [n] z -n = 1n=-

Unit impulse

2. Z-transforms

8. Z-transforms 15

Transforms (2 of 4)

Unit step sequence, u[n]

X(z) = u[n] z -n

= z -n

= 1/(1 - z-1)

= z/(z + 1)

n=-

0

Unit step

2. Z-transforms

8. Z-transforms 16

Transforms (3 of 4)

Unit step sequence times an, an u[n]

X(z) = u[n] an z -n

= an z -n

= 1/(1 - a z-1)

= z/(z - a)

n=-

0

an

2. Z-transforms

8. Z-transforms 17

Transforms (4 of 4)

x1[n] x2[n]

x[n - m]

zon x[n]

z[-n]

x[n]

X1(z) ± X2(z)

z-m X(z)

X(z/zo)

X(1/z)

z/(z-1) X(z)

Linearity

Time shift

Multiplication by zon

Time reversal

Accumulation k=-

n

Properties

8. Z-transforms 18

Examples

y[n] = 2 y[n-1] + x[n]• Y(z)/X(z) = 1/(1-2 z-1)

y[n] - 2 y[n-1] = x[n] + 4 x[n-1]• Y(z)/X(z) = (1+ 4 z-1) /(1-2 z-1)

y[n] = 4 y[n-1] + 2 y[n-2] + x[n]• Y(z)/X(z) = 1 /(1- 4 z-1 - 2 z-2 )

2. Z-transforms

8. Z-transforms 19

3. Discrete Fourier transform

DFTAliasingSampling theoremFrequency, resolution, and lengthLinear convolutionCircular convolutionBlock filteringFilteringWindowing


8. Z-transforms 20

Inverse DFT

DFT

DFT (1 of 4)

X[k] = x [n] e -j(2/N)kn = x [n] WNkn

n=0

N-1

x[n] = 1/N X[k] e j(2/N)nk = 1/N X[k] WN-kn

k=0

N-1

n=0

N-1

k=0

N-1

WN = e -j(2/N)

Where


8. Z-transforms 21

Matrix format

DFT (2 of 4)

X = DN x

1 1 1 1

1 WN1 WN

2 … WN(N-1)

1 WN2 WN

4 … WN2(N-1)

.

.

.

1 WN(N-1)(N-1) WN

(N-1)(N-1) … WN(N-1)(N-1)

X[0]

X[1]

X[2]

.

.

.

X[N-1]

=

x[0]

x[1]

x[2]

.

.

.

x[N-1]


8. Z-transforms 22

DFT (3 of 4)

e -j(2/4)0*0 e -j(2/4)1*0 e -j(2/4)2*0 e -j(2/4)3*0

e -j(2/4)0*1 e -j(2/4)1*1 e -j(2/4)2*1 e -j(2/4)3*1

e -j(2/4)0*2 e -j(2/4)1*2 e -j(2/4)2*2 e -j(2/4)3*2

e -j(2/4)0*3 e -j(2/4)1*3 e -j(2/4)2*3 e -j(2/4)3*3)

X[0]X[1]X[2]X[3]

=

x[0]x[1]x[2]x[3]

1 1 1 11 -j -1 j1 -1 1 -11 j -1 -j

X[0]X[1]X[2]X[3]

=

1100

=

21 - j01 + j

Example DFT


8. Z-transforms 23

DFT (4 of 4)

e +j(2/4)0*0 e +j(2/4)1*0 e +j(2/4)2*0 e +j(2/4)3*0

e +j(2/4)0*1 e +j(2/4)1*1 e +j(2/4)2*1 e +j(2/4)3*1

e +j(2/4)0*2 e +j(2/4)1*2 e +j(2/4)2*2 e +j(2/4)3*2

e +j(2/4)0*3 e +j(2/4)1*3 e +j(2/4)2*3 e +j(2/4)3*3)

x[0]x[1]x[2]x[3]

=1/4

X[0]X[1]X[2]X[3]

1 1 1 11 +j -1 -j1 -1 1 -11 -j -1 j

x[0]x[1]x[2]x[3]

=1/4

21 - j01 + j

=

1100

Example DFT (continued)


8. Z-transforms 24

Aliasing

The condition in which a continuous-time sinusoid of higher frequency acquires the identity of a lower-frequency sinusoid


8. Z-transforms 25

Sampling theorem

The signal x(t) must be bandlimited such that its spectrum contains no frequencies above fmax.

The sampling rate fs must be chosen to be 2 fmax

2 fmax is the Nyquist rate

fs/2 is the Nyquist frequency or folding frequency


8. Z-transforms 26

Frequency, resolution, & length (1 of 4)

f = 1/To = fs/N = 1/(NT) = 1/To

f = frequency resolution

fs = sampling frequency

To = record length

T = sampling interval

N = number of samples

Equations


8. Z-transforms 27


Relationship of record length• An analog signal having a 0.2 second record

length is sampled at a frequency of 2.5 kHz Maximum frequency that can be present in

this sample if there is no aliasing

• fmax = Nyquist frequency fs/2 = 1250 Hz

Frequency resolution of a DFT for this signalf = 1/To = 1/0.2 = 5 Hz

Example


8. Z-transforms 28


Record length• N = To/T = 0.2/(1/2500) = 500 samples

Frequencies present• f = 0, 5, 10, … , 1250, -1245, -1240, … , -5

Example (continued)


8. Z-transforms 29


N T To f fmax

N 2T 2 To 0.5 f 0.5 fmax

2N 0.5 T To f 2 fmax

2N T 2 To 0.5 f fmax

Effect of changing numbers


8. Z-transforms 30

Linear convolution (1 of 7)

h[n]x[n] y[n]

y[n] are determined by convolving x[n] and h[n]

Determining y[n] from x[n] and h[n]


8. Z-transforms 31

Convolution by multiplication

Introduction to Signal Processing

Sophacles J. Orfanidis page 147

Introduction to Signal Processing

Sophacles J. Orfanidis page 147


h[n] = [1 2 -1 1]x[n] = [1 1 2 1 2 2 1 1 ]y[n] = [1 3 3 5 3 7 4 3 3 0 1]

11=4+8-1

1 1 2 1 2 2 1 11 2 -1 1

1 1 2 1 2 2 1 1 2 2 4 2 4 4 2 2 -1 -1 -2 -1 -2 -2 -1 -1 1 1 2 1 2 2 1 1

1 3 3 5 3 7 4 3 3 0 1


8. Z-transforms 32


[1 2 -1 1][1 1 2 2 1 2 1 1 ] 1

3353743301

[1 1 2 2 1 2 1 1 ][1 1 2 2 1 2 1 1 ]

[1 1 2 2 1 2 1 1 ][1 1 2 2 1 2 1 1 ]

[1 1 2 2 1 2 1 1 ][1 1 2 2 1 2 1 1 ]

[1 1 2 2 1 2 1 1][1 1 2 2 1 2 1 1 ]

[1 1 2 2 1 2 1 1 ][1 1 2 2 1 2 1 1 ]

time

Convolution by filtering in time


8. Z-transforms 33


[1 1 2 1 2 2 1 1 ][1 -1 2 1] 1

3353743301

[1 -1 2 1][1 -1 2 1]

[1 -1 2 1][1 -1 2 1]

[1 -1 2 1][1 -1 2 1]

[1 -1 2 1][1 -1 2 1]

[1 -1 2 1][1 -1 2 1]

filter slideConvolution by sliding filter


8. Z-transforms 34


h[n] = [1 2 -1 1] x[n] = [1 1 2 1 2 2 1 1 ]

Need to have sequences of same length, and length needs to be 4 + 8 -1 = 11 to avoid circular convolution

h[n] = [1 2 -1 1 0 0 0 0 0 0 0 0 0 0 0 0] x[n] = [1 1 2 1 2 2 1 1 0 0 0 0 0 0 0 0]

Convolution by DFT


8. Z-transforms 35


X[k] = 11 1.3244 - 7.6583j-1.7071 - 0.2929j 1.2168 - 0.8210j - j-0.6310 - 0.5784j-0.2929 + 1.7071j 2.0898 + 0.5843j 1 2.0898 - 0.5843j-0.2929 - 1.7071j-0.6310 + 0.5784j + j 1.2168 + 0.8210j-1.7071 + 0.2929j 1.3244 + 7.6583j

H[k] = 3 2.5233 - 0.9821j 1.7071 - 1.1213j 1.5486 - 0.7580j 2 - j 1.8656 - 2.1722j 0.2929 - 3.1213j-1.9375 - 2.3964j-3-1.9375 + 2.3964j 0.2929 + 3.1213j 1.8656 + 2.1722j 2 + j 1.5486 + 0.7580j 1.7071 + 1.1213j 2.5233 + 0.9821j

Y[k]=X[k] H[k] = 33-4.1796 - 20.6253j-3.2426 + 1.4142j 1.2620 - 2.1937j -1 - 2j-2.4335 + 0.2916j 5.2426 + 1.4141j-2.6488 - 6.1400j-3-2.6488 + 6.1400j 5.2426 - 1.4141j-2.4335 - 0.2916j-1 + 2j 1.2620 + 2.1937j-3.2426 - 1.4142j-4.1796 +20.6253j

y[n]=1335 374330100000


8. Z-transforms 36


MATLAB code for convolution by DFT

x = [1,1,2,1,2,2,1,1,0,0,0,0,0,0,0,0]h = [1,2,-1,1,0,0,0,0,0,0,0,0,0,0,0,0]X = fft(x)H = fft(h)Y = X.*Hy = ifft(Y)

Convolution by DFT (continued)


8. Z-transforms 37

Circular convolution (1 of 4)

yc[n] = x[m] h[ <n-m>N]m=0

N-1

x[n] = [1 1 2 0], h[n] = [1 2 -1 1]

yc[0] = [1 1 2 0] [ 1 1 -1 2] = 0yc[1] = [1 1 2 0] [ 2 1 1 -1] = 5yc[2] = [1 1 2 0] [-1 2 1 1] = 3yc[3] = [1 1 2 0] [ 1 -1 2 1] = 4

Definition and example


8. Z-transforms 38


yc[0]yc[1]yc[2]...yc[N-1]

h[0] h[N-1] h[N-2] … h[1]h[1] h[0] h[N-1] … h[2]h[2] h[1] h[0] … h[3]. . . … .. . . … .. . . … .h[N-1] h[N-2] h[N-3] … h[0]

x[0]x[1]x[2]...x[N-1]

=

Matrix format


8. Z-transforms 39


h[n] = [1 2 -1 1] x[n] = [1 1 2 1]

1 1 -1 2 2 1 1 -1-1 2 1 1 1 -1 2 1

1121

2445

1 1 2 11 2 -1 11 1 2 1 2 2 4 2 -1 -1 -2 -1 1 1 2 11 3 3 5 1 1 1

=

Linear Circular 1 3 3 51 1 12 4 4 5

Example

8. Z-transforms 40


Circular convolution is what DFTs compute naturally when convolving two sequences of length N using a DFT of length N

The matrix format shows a formal way of computing circular convolution

A simpler way to compute circular convolution is to compute the linear convolution and then wrap the part of the linear convolution that is longer than N

Explanation


8. Z-transforms 41

Block filtering (1 of 9)

In many applications, the data is too long to handle in one step

In these cases, filtering may be done in blocks

There are two methods:• Overlap add• Overlap save

Reason for block filtering


8. Z-transforms 42


Given• Signal x is of length L• Filter h of length M

Then

• The length of convolution Ly = L + M - 1

• The size of the DFT N Ly to ensure that the circular convolution gives the same results as the linear convolution

To use a DFT of length N, we must pad x and h with zeros to make them of length N also

Lengths


8. Z-transforms 43

Block filtering (3 of 9)Overlap add

block xo block x1 block x2

L M

L M

L M

yo =

y1 =

y2 =

x =

L L L

ytemp

ytemp

ytemp

n=0 n=L n=2L n=3L


8. Z-transforms 44


h[n] = [1 2 -1 1] x[n] = [1 1 2 1 2 2 1 1 ]y[n] = [1 3 3 5 3 7 4 3 3 0 1]

x1[n] = [1 1 2], x2[n] = [1 2 2], x3[n] = [1 1]

1 2 -1 11 1 21 3 3 4 -1 2

1 2 -1 11 2 21 4 5 3 0 2

1 2 -1 11 11 3 1 0 1

1 3 3 4 -1 2 1 4 5 3 0 2 1 3 1 0 11 3 3 5 3 7 4 3 3 0 1

Divide input sequence into 3 parts

Overlap-add (continued)


8. Z-transforms 45


x = [1,1, 2, 0,0,0,0,0]h = [1,2,-1,1, 0,0,0,0]X = fft(x)H = fft(h)Y = X.*Hy = ifft(Y)

y[n] = [1 3 3 4 -1 2 0 0]

MATLAB code for convolution by DFT


8. Z-transforms 46

Block filtering (6 of 9)Overlap save

x = M MM

block xo block x1 block x2 block x3

N-MM

N-MM

N-MM

N-MM

yo=

y1=

y2=

y3=

n=0 n=N n=2N n=3N3. Discrete Fourier transform

8. Z-transforms 47


h[n] = [1 -1 -1 1]x[n] = [1,1,1,1,3,3,3,3,1,1,1,2,2,2,2,1,1,1,1]y[n] = [1,0,-1,0,2,0,-2,0,-2,0,2,1,0,-1,0,-1,0,1,0,-1,0,1]

Divide input sequence into sequences of length 8that overlap by 3

But first, attach 8-3 = 5 zeros to front of sequence

x[n] = [0,0,0,0,0,1,1,1,1,3,3,3,3,1,1,1,2,2,2,2,1,1,1,1]

Overlap save (continued)


8. Z-transforms 48


Form sequences[0,0,0,0,0,1,1,1][1,1,1,1,3,3,3,3][3,3,3,1,1,1,2,2][1,2,2,2,2,1,1,1][1,1,1,1,0,0,0,0]

Form circular convolutions[X,X,X,0,0,1,0,-1][X,X,X,0,2,0,-2,0][X,X,X,-2,0,2,1,0][X,X,X,-1,0,-1,0,1][X,X,X,0,-1,0,1,0]

y[n] = [1,0,-1,0,2,0,-2,0,-2,0,2,1,0,-1,0,-1,0,1,0,-1,0,1]

Overlap save (continued)


8. Z-transforms 49


Cost• Linear convolution = (M+1) (N-M)

• DFT = N (log2 N +1)

• Ratio = log2 N /M for N>>M>>1

• Overlap-add and overlap-save have about the same cost

Cost


8. Z-transforms 50

Filtering (1 of 11)

Filter -- A filter is an algorithm that converts in input sequence into an output sequence

Linear, constant-coefficient difference equation (LCCDE)

y[n] = b[k] x[n-k] - a[k] y[n-k]k=0

q

k=1

p

Definition


8. Z-transforms 51

Filtering (2 of 11)

Impulse response, h[n] -- h[n] is found by solving the LCDDE for x[n] = [n]

Impulse response


8. Z-transforms 52

Filtering (3 of 11)

Infinite impulse response (IIR) filter -- A FIR is a filter in which the length of h[n] is infinite. It corresponds to an LCCDE in which not all the values of a[n-k] are zero• y[n] = -0.5 y[n-1] + 0.5 x[n]

IIR advantages• Useful for implementing analog filters in

digital format• Sometimes has fewer coefficients

IIR


8. Z-transforms 53

Filtering (4 of 11)

Finite impulse response (FIR) filter -- A FIR is a filter in which the length of h[n] is finite. It corresponds to an LCCDE in which all the values of a[n-k] are zero• y[n] = 0.25 x[n] + 0.50 x[n-1] + 0.25 x[n-2]

FIR advantages• Simple• Always stable• Linear phase

FIR


8. Z-transforms 54

Filtering (5 of 11)

Filters may be of many types• Low-pass• High-pass• All-pass• Band-pass

Types of filters


8. Z-transforms 55

Filtering (6 of 11)

Ideal filter• Magnitude -- constant across pass-band• Phase -- linear across pass-band

Reason for ideal filter• The envelope of a signal will not be

distorted when passing through a filter with linear phase

Ideal filter


8. Z-transforms 56

Filtering (7 of 11)

c-c

M

stop-band stop-bandpass-band

-

c-c

-

ideal low-pass filter

constant magnitude

linear phaseP


8. Z-transforms 57

Filtering (8 of 11)

Four types of FIRs with linear phase• Type 1 -- N=odd, real and symmetric• Type 2 -- N=even, real and symmetric• Type 3 -- N=odd, real and antisymmetric• Type 4 -- N=even, real and antisymmetric

Four types of FIRs with linear phase


8. Z-transforms 58

Filtering (9 of 11)

Linear phase -- An LTI system has linear phase if • H(e j ) = |H (e j )| e -j

Generalized linear phase -- An LTI has generalized linear phase if • H(e j ) = |H (e j )| e -j ( - )

Linear phase


8. Z-transforms 59

Filtering (10 of 11)

type 1 type 2

type 3 type 4

Types (continued)


8. Z-transforms 60

Filtering (11 of 11)

The design of an FIR involves finding the coefficients of h[n] that result in a frequency response that satisfies a set of filter design specifications

Design


8. Z-transforms 61

Windowing (1 of 4)

Windowing -- The process of reducing the length of the original signal because we can take DFTs of unlimited size

Purpose• Reduces the frequency resolution• Introduces spurious frequency

components into the spectrum -- called frequency leakage

Equation

• xL [n] = x[n] w[n]

Definitions


8. Z-transforms 62

Windowing (2 of 4)

0 5 10 15 20 25 30 350

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Rectangular window


8. Z-transforms 63

Windowing (3 of 4)

0 5 10 15 20 25 30 35 400

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Hanning window


8. Z-transforms 64

Windowing (4 of 4)

0 5 10 15 20 25 30 350

2

4

6

8

10

12

14

16

0 5 10 15 20 25 30 350

1

2

3

4

5

6

7

8

9

Rectangular window withsidelobe at -13 dB

Hamming window withsidelobe at -40 dB

Comparison of effect of rectangular & Hamming windows


8. Z-transforms 65

4. Power spectral density

Power spectrum (PS)Power spectral density (PSD)ExperimentUse of PSDEffect of linear system on PSD


8. Z-transforms 66

Power spectrum (PS)

x2 = 1/2 Cn C*n

n=1

The mean square value of a periodic time function is the sum of the mean square value of the individual harmonic components present

The power spectrum G(fn) = 1/2 Cn C*n is the contribution to the mean square error in the frequency interval f

x2 = G(fn) n=1

8. Z-transforms 67

Power spectral density (PSD)

The power spectral density S(fn) is the power spectrum divided by the frequency interval f

S(fn) = G(fn) / f

x2 = S(fn) f n=1


8. Z-transforms 68

Experiment

shaker accelerometer amplifier filterRMSmeter

f

420-580 Hz480-520 Hz495-505 Hz

f

160 Hz40 Hz10 Hz

meter

8g4g2g

G(fn)

64g2

16g2

4g2

S(fn)/ f

0.40g2/Hz0.40g2/Hz0.40g2/Hz

Theory of Vibration with Applications, William T. Thompson


8. Z-transforms 69

Use of PSD

G12 G2

2 G32

Grms = SQRT(G12 + G2

2 + G32 )

f

PSD (g2/Hz)


8. Z-transforms 70

Effect of linear system on PSD

H(j)

Syy()= H(j) H*(j) Sxx () = |H(j)|2 Sxx ()

Where Sxx () = power spectrum of inputSyy () = power spectrum of outputH(j) = transfer function with s = j

x y


8. z-transforms1 agenda r1. difference equations r2. z-transforms r3. discrete fourier transform r4....

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