System Identification Tips

7/29/2019 System Identification Tips

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ECE 621, Project # 2 Question 1

Qn:1 Driving System with Sinusiods

a)Input Signal DefinitionA general function for the input chosen is shown below:

For the three cases analyzed, the parameters were varied.

b)Results with one sinusoidParameter Value

0.01

1.00

0.00

0.00

5.00

NA

NA

0.00

NA

NA

For this analysis, the Batch LS algorithm was chosen, as directed. A 1000 sample sinusoidal signal was

chosen for this analysis. Figure 1 below shows the response of the system. Note that the first 200

samples have been discarded for the system identification analysis, as directed.


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Figure 1 1-Sine Response

Results

True value Estimate with

1 Sinusoid

Condition Number

Of Data Matrix

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

0.7867

-4.0551

1.5249

-76.0945

504.1890

-336.1260

4.2589e+21

With one sinusoid, the system was not identified.

0 100 200 300 400 500 600 700 800 900-1

-0.5

0

0.5

1

Input

0 100 200 300 400 500 600 700 800 900-200

-100

0

100

200

Output

0 100 200 300 400 500 600 700 800 900-1

-0.5

0

0.5

1

InputconsideredforI/D

Samples

0 100 200 300 400 500 600 700 800 900-200

-100

0

100

200

OutputconsideredforI/D


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c) Results with two sinusoidsParameter Value

0.01

1.00 1.00

0.00

5.00

2.00

NA

0.00

NA

For this analysis, the Batch LS algorithm was chosen, as directed. A 1000 sample sinusoidal signal waschosen for this analysis. Figure 1 below shows the response of the system. Note that the first 200



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Results


2 Sinusoids

Condition Number

Of Data Matrix

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

-3.1668

3.7274

-1.5170

140.4014

-307.9648

185.5163

6.5278e+20

Even with 2 sinusoids, the system was not identified.

0 100 200 300 400 500 600 700 800 900-2

-1

0

1

2

Input

0 100 200 300 400 500 600 700 800 900-400

-200

0

200

400

Output

0 100 200 300 400 500 600 700 800 900-2

-1

0

1

2


Samples

0 100 200 300 400 500 600 700 800 900-400

-200

0

200

400



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d)Results with three sinusoidsParameter Value

0.01

1.00 1.00

1.00

5.00

2.00

1.00

0.00

For this analysis, the Batch LS algorithm was chosen, as directed. A 1000 sample sinusoidal signal waschosen for this analysis. Figure 1 below shows the response of the system. Note that the first 200



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Results


3 Sinusoids

Condition Number

Of Data Matrix

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

2.7145e+12

With a combination of three sinusoids, the system was identified.

0 100 200 300 400 500 600 700 800 900-4

-3

-2

-1

0

1

2

3

4

Input

0 100 200 300 400 500 600 700 800 900-1000

-800

-600

-400

-200

0

200

400

600

Output

0 100 200 300 400 500 600 700 800 900

0


Samples

0 100 200 300 400 500 600 700 800 900-1000

0



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e)ConclusionThe persistency of the input signal used is clearly seen to affect the ability of the algorithm to identify

the system. The weak/strong persistency order (PE) of a sinusoidal signal is shown to be 2. In order that

the Data Matrix has full rank, (in this case 6, since there are 6 parameters to be identified), it means thatthe input sequence must have a PSE of at least 6.

Input TypeOrder

of P.E.Identifiability

Condition

Number of QComment

1 Sinusoid 2 < 6, Not Identifiable 4.2589e+21

Does not meet the Identifiability

requirement. This is seen from the large

condition number.

2 Sinusoids 4 < 6, Not Identifiable 6.5278e+20



condition number.

3 Sinusoids 6 >= 6, Identifiable 2.7145e+12

Meets the Identifiability requirement.The condition number is still large due to

the sinusoidal input, however the matrix

is invertible


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g)Code% ECE 621, Systems Identification% Project # 2, Qn 01%% Author:%

% Date: 2012.10.30%

clear all; clc;close allhidden;

% Display true system parameters for comparisonp1 = 0.80;p2 = 0.68;p3 = 0.50;b0 = 5.00;b1 = 2.00;b2 = 3.00;

a1 = - (p1 + p2 + p3);a2 = p1*p2 + p2*p3 + p1*p3;a3 = - p1*p2*p3;disp ('Theta matrix from System');disp([a1;a2;a3;b0;b1;b2]);

clear all;

% Batch estimation% Drive system with random input and get output vector.t = (0:1:999);u = 0 + 1*sin(2*pi*5*0.01*t);u = u + 0*sin(2*pi*2*0.01*t + pi/6);

u = u + 0*sin(2*pi*1*0.01*t - pi/12);y = g_proj_1(u);

% Length of input/output vectornum = length(u);

% Samples to discarddiscard = 200;

% Length of theta_hat vector = m + nm = 3;n = 3;

% Initialize Phi Matrixphi = zeros(num-discard+1, m+n);

% Define regressor phi'% phi_t = [-y(t-1) -y(t-2) -y(t-3) u(t-1) u(t-2) u(t-3)]

for i = discard : num

i1 = i-1;


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i2 = i-2;i3 = i-3;

if i10)y1 = y(i1); y2 = y(i2); y3 = y(i3);u1 = u(i1); u2 = u(i2); u3 = u(i3);

end

phi(i,:) = [-y1 -y2 -y3 u1 u2 u3];

end

% Form Q Matrixq = phi'*phi;disp(['Condition Number of Data Matrix = ', num2str(cond(q))]);

% Batch compute theta_hat with PHI matrixtheta_hat = q\(phi'*y);disp ('Theta hat matrix from System Identification');disp(theta_hat);

figure;

subplot(211);[AX,H1,H2] = plotyy(t,u,t,y); grid on;set(AX(1), 'xlim', [0 t(num)]); set(AX(2), 'xlim', [0 t(num)]);ax1y = get(AX(1), 'ylim'); ax2y = get(AX(2), 'ylim');set(get(AX(1),'ylabel'), 'string', 'Input');set(get(AX(2),'ylabel'), 'string', 'Output');

clear AX;

subplot(212);[AX,H1,H2] = plotyy(t(discard:num), u(discard:num),t(discard:num),y(discard:num)); grid on;set(AX(1), 'xlim', [0 t(num)]); set(AX(2), 'xlim', [0 t(num)]);

set(AX(1), 'ylim', ax1y); set(AX(2), 'ylim', ax2y);set(get(AX(1),'ylabel'), 'string', 'Input considered for I/D');set(get(AX(2),'ylabel'), 'string', 'Output considered for I/D');xlabel('Samples');


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Qn:2 Driving System with Symmetric

Square waveforms

h)Results with Period = 6 samples

Figure 4 Response with 6 sample symmetric square wave

The first 200 samples (208s) of the input and output signals were discarded and only the rest was fed

into the identifier.

0 5 10 15 20 25 30 35-1

-0.5

0

0.5

1

Input

0 5 10 15 20 25 30 35-20

0

20

40

60

Output

Samples

180 190 200 210 220 230 240 250

-1

-0.5

0

0.5

1


Time (s)

180 190 200 210 220 230 240 250-20

-10

0

10

20



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Results


6 sample sym. Sqr. wave

Condition Number

Of Data Matrix

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

4.6946

-9.1376

-11.7038

-106.7417

-7.0509

98.1983

4.0311e+17

With a 6 sample symmetric square waveform, the system was not identified.

0 50 100 150 200 250 300 350 400 450 500

-1

-0.5

0

0.5

1

Input

0 50 100 150 200 250 300 350 400 450 500-20

0

20

40

60

Output

0 50 100 150 200 250 300 350 400 450 500

-1

-0.5

0

0.5

1


Time (s)

0 50 100 150 200 250 300 350 400 450 500-20

-10

0

10

20

Outputconsidered

forI/D


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i) Results with Period = 12 samples


The first 200 samples (104s) of the input and output signals were discarded and only the rest was fed

into the identifier.

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

-0.5

0

0.5

1

Input

0 2 4 6 8 10 12 14 16 18 20-100

-50

0

50

100

150

Output

Samples

70 80 90 100 110 120 130 140

-1

-0.5

0

0.5

1


Time (s)

70 80 90 100 110 120 130 140-100

-50

0

50

100



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Results

True value Estimate with12 sample sym. Sqr. wave Condition NumberOf Data Matrix

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

374614.1391

With a 12 sample symmetric square waveform, the system was identified correctly.

0 50 100 150 200 250 300 350 400 450 500

-1

0

1

I

nput

0 50 100 150 200 250 300 350 400 450 500-200

0

200

Output

0 50 100 150 200 250 300 350 400 450 500

-1

-0.5

0

0.5

1

InputconsideredforI/

D

Time (s)

0 50 100 150 200 250 300 350 400 450 500-100

-50

0

50

100

OutputconsideredforI

/D


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j) Results with Period = 20 samples


The first 200 samples (63s) of the input and output signals were discarded and only the rest was fed intothe identifier.

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

-0.5

0

0.5

1

Input

0 2 4 6 8 10 12 14 16 18 20-200

-100

0

100

200

300

Output

Sam les

45 50 55 60 65 70 75 80 85 90

-1

-0.5

0

0.5

1


Time (s)

45 50 55 60 65 70 75 80 85 90-200

-100

0

100

200



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Results

True value Estimate with20 sample sym. Sqr. wave

Condition NumberOf Data Matrix

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

800462.8466

With a 20 sample symmetric square waveform as well, the system was identified correctly.

0 50 100 150 200 250 300 350 400 450 500

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Input

0 50 100 150 200 250 300 350 400 450 500-200

-150

-100

-50

0

50

100

150

200

250

300

Output

0 50 100 150 200 250 300 350 400 450 500

-1

-0.5

0

0.5

1


Time (s)

0 50 100 150 200 250 300 350 400 450 500-200

-100

0

100

200



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k)ConclusionThe persistency of the input signal used is clearly seen to affect the ability of the algorithm to identify

the system. The weak/strong persistency order (PE) of a symmetric square wave is known to be function

of the number of samples per period. For a symmetric square wave with M samples per period, theorder of persistency is M/2.

In order that the Data Matrix has full rank, (in this case 6, since there are 6 parameters to be identified),

it means that the input sequence must have a PSE of at least 6. Hence the input symmetric square wave

shall have at least 12 samples per period.

Input TypeOrder

of P.E.Identifiability

Condition

Number of QComment

6 sample

symmetric

sq. wave

3< 6, Not

Identifiable4.0311e+17



condition number.12 sample

symmetric

sq. wave

6 >= 6, Identifiable 374614.1 Meets the Identifiability requirement.

20 sample

symmetric

sq. wave

10 >= 6, Identifiable 800462.8 Meets the Identifiability requirement.


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m) Code.% ECE 621, Systems Identification% Project # 2, Qn 02%% Author:

%% Date: 2012.10.30%


% Display true system parameters for comparisonp1 = 0.80;p2 = 0.68;p3 = 0.50;b0 = 5.00;b1 = 2.00;

b2 = 3.00;a1 = - (p1 + p2 + p3);a2 = p1*p2 + p2*p3 + p1*p3;a3 = - p1*p2*p3;disp ('Theta matrix from System');disp([a1;a2;a3;b0;b1;b2]);

clear all;

% Batch estimation% Drive system with random input and get output vector.

nmp = 11; % Number of periodssmp = 20; % Samples per periodt = linspace(1e-10,nmp*2*pi,smp*nmp);% Symmetric Square Waveu = (2*(sin(t)>0)-1);

y = g_proj_1(u);

figure;subplot(211);stem(t(1:3*smp),u(1:3*smp)); grid on; ylabel('Input');subplot(212);stem(t(1:3*smp),y(1:3*smp)); grid on; ylabel('Output');xlabel('Samples');

% Length of input/output vector

num = length(u);




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i1 = i-1;

i2 = i-2;i3 = i-3;

if i10)y1 = y(i1); y2 = y(i2); y3 = y(i3);u1 = u(i1); u2 = u(i2); u3 = u(i3);

end

phi(i,:) = [-y1 -y2 -y3 u1 u2 u3];

end


% Batch compute theta_hat with PHI matrixtheta_hat = q\(phi'*y);disp ('Theta hat matrix from System Identification');disp(theta_hat);figure;subplot(211);[AX,H1,H2] = plotyy(t,u,t,y); grid on;set(AX(1), 'xlim', [0 t(num)]); set(AX(2), 'xlim', [0 t(num)]);set(AX(1), 'ylim', [-1.1 1.1]);set(get(AX(1),'ylabel'), 'string', 'Input');set(get(AX(2),'ylabel'), 'string', 'Output');

subplot(212);

[AX,H1,H2] = plotyy(t(discard:num), u(discard:num),t(discard:num),y(discard:num)); grid on;set(AX(1), 'xlim', [0 t(num)]); set(AX(2), 'xlim', [0 t(num)]);set(AX(1), 'ylim', [-1.1 1.1]);set(get(AX(1),'ylabel'), 'string', 'Input considered for I/D');set(get(AX(2),'ylabel'), 'string', 'Output considered for I/D');xlabel('Time (s)');


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Qn:3 Over Parametrization

n)True ModelThe true model for the system can be defined as follows:

An expression for y(t) in discrete time domain can then be defined as follows:

where:

and with the following parameters:

p1 = 0.80;p2 = 0.68;p3 = 0.50;b0 = 5.00;b1 = 2.00;b2 = 3.00;

Parameter True value

a0a1a2b0b1b2

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000


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o)Input Used for IdentificationA symmetric square wave with 20 samples per period was chosen for system identification, for all cases

discussed here. This should be sufficient, since such a signal has an order of persistence of 10, and can

be used to estimate up to 10 parameters. Even if an extra parameter is added to both the numerator

and the denominator, the total number of parameters to be estimated would only be 8 in this case.

Hence the chosen signal has sufficient order of P.E..

0 5 10 15 20 25 30 35-1

-0.5

0

0.5

1

Input

0 5 10 15 20 25 30 35-200

-100

0

100

200

300

Output

Samples

0 10 20 30 40 50 60 70

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Input

0 10 20 30 40 50 60 70-200

-150

-100

-50

0

50

100

150

200

250

300

Output

0 10 20 30 40 50 60 70

-1

-0.5

0

0.5

1


Time (s)

0 10 20 30 40 50 60 70-200

-100

0

100

200



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p)Extra Input TermIf the System ID model were modeled with an extra input term, the regressor can be expressed as:

[ ]

Results


extra i/p parameter

Condition Number

Of Data Matrix

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

0.0000

1.4112e+07

It can be seen that in this case, the least squares identifier was able to correctly identify the system and

the excess parameter. The excess parameter b3 was identified as 0, as expected.

q)Extra Output TermIf the System ID model were modeled with an extra output term, the regressor can be expressed as:

[ ]

Results


extra o/p parameter

Condition Number

Of Data Matrix

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

-1.9800

1.2840

-0.2720

-0.0000

5.0000

2.0000

3.0000

2711800.233

It can be seen that in this case, the least squares identifier was able to correctly identify the system and

the excess parameter. The excess parameter a3 was identified as 0, as expected.


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r) Extra Term in Numerator and DenominatorIf the System ID model were modeled with an extra term in the numerator and the denominator,, the

regressor can be expressed as:

[ ]

Results


extra i/p parameter

Condition Number

Of Data Matrix

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

7.2433

-16.9782

11.5708

-2.5087

5.0000

48.1167

21.4467

27.6700

8.4400e+17

It can be seen that in this case, the least squares identifier was not able to identify the system. Several

attempts were made to evaluate the possibility of complete system identification with the

overparametrization:

Increased Samples per period to upto 50 sampleso No effect: System was not identified

Increased total number of samples considered for identificationo No effect: System was not identified

Changed input to a pseudorandom signalo No effect: System was not identified. But estimates slightly better

True value Estimate Cond(Q)

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

-2.7020

2.7135

-1.1990

0.1964

5.0000

-1.6100

1.5560

-2.1660

5.2243e+16

s) ConclusionOverparametrization introduces limitation to estimation of an ARMA system only if excess parameters

are introduced both at the input and at the autoregressive terms. This is because it creates linear

dependencies in the regressor, if extra input terms and output terms are present together.


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t) Code: (+1 param in numerator)% ECE 621, Systems Identification% Project # 2, Qn 03 a (+1 param in numerator)%% Author:%

% Date: 2012.10.30%




clear all;



y = g_proj_1(u);

% This will assume that the system is an ARMA model with a form as defined% as below:% y(t) = bo u(t-1) + b1 u(t-2) + b2 u(t-3) - a1 y(t-1) - a2 y(t-2) - a3 y(t-3)% Goal of the batch estimator will be to find b0, b1, b2, a1, a2, & a3





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% Define regressor phi'% Add param in numerator (b4)

% phi_t = [-y(t-1) -y(t-2) -y(t-3) u(t-1) u(t-2) u(t-3) u(t-4)]


% indices will be +ve as long as discard >=4% here discard = 200 samples

i1 = i-1;i2 = i-2;i3 = i-3;i4 = i-4;

y1 = y(i1); y2 = y(i2); y3 = y(i3);u1 = u(i1); u2 = u(i2); u3 = u(i3); u4 = u(i4);

phi(i,:) = [-y1 -y2 -y3 u1 u2 u3 u4];

end




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u)Code: (+1 param in denominator)% ECE 621, Systems Identification% Project # 2, Qn 03 b (+1 param in denominator)%% Author:%

% Date: 2012.10.30%




clear all;



y = g_proj_1(u);






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i1 = i-1;i2 = i-2;i3 = i-3;i4 = i-4;

y1 = y(i1); y2 = y(i2); y3 = y(i3); y4 = y(i4);u1 = u(i1); u2 = u(i2); u3 = u(i3);

phi(i,:) = [-y1 -y2 -y3 -y4 u1 u2 u3];

end




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v)Code: (+1 param in numerator & denominator)% ECE 621, Systems Identification% Project # 2, Qn 03 c (+1 param in num and den)%% Author:%

% Date: 2012.10.30%




clear all;


nmp = 50; % Number of periodssmp = 50; % Samples per periodt = linspace(1e-10,nmp*2*pi,smp*nmp);% Symmetric Square Waveu = (2*(sin(t)>0)-1);clear u; load rand_uy = g_proj_1(u);






28/44







i1 = i-1;i2 = i-2;i3 = i-3;i4 = i-4;

y1 = y(i1); y2 = y(i2); y3 = y(i3); y4 = y(i4);u1 = u(i1); u2 = u(i2); u3 = u(i3); u4 = u(i4);

phi(i,:) = [-y1 -y2 -y3 -y4 u1 u2 u3 u4];

end




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Qn:4 Effect of Feedback

The system is modeled as the following per the problem description:

This can be simplified to:

For the analysis, a pseudorandom sequence was used for measurement noise v(t). For the sake of

comparison, the same measurement noise sequence was used for all the cases analyzed. The following

table shows the list of cases that have been analysed.

Controller

Type

Controller

Reference r(t)

Measurement

Noise

P 0 random

P random random

PI 0 random

PI random random

v(t)

r(t)

y(t)

d(t)

y0(t)

Plant

Controller

v(t)

r(t)

y(t)

u(t)

d(t)

y0(t)


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w)Case 1: P Controller, r(t) = random

Results


P Controller, r(t) = random

Condition Number

Of Data Matrix-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0001

42903.3689

With a measurement noise magnitude of 0.000001 (Noise Power Level of 7.083e-6 compared to Output

Power), the system was identified to adequate precision, in the presence of a pseudorandom reference

signal. The following table shows the effect of measurement noise on the estimates with a P controller.

100 200 300 400 500 600 700 800-2

-1

0

1

2x 10

-3

PlantInput(u)

Time (s)

100 200 300 400 500 600 700 800-0.2

-0.1

0

0.1

0.2

PlantMeasu

redOutput(y)

100 200 300 400 500 600 700 800-2

-1

0

1

2x 10

-3

PlantInput(u)

Time (s)

100 200 300 400 500 600 700 800-1

-0.5

0

0.5

1

ControllerRefernce(r)

750 755 760 765 770 775 780 785 790 795 800-0.1

-0.05

0

0.05

0.1

PlantTrueOutput(y0

)Zoom In on Plant Output (True vs Measured), Noise Level :7.0833e-006

Time (s)

750 755 760 765 770 775 780 785 790 795 800-0.1

-0.05

0

0.05

0.1

PlantMeasuredOutput

(y)


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Table 1 Comparison of Results with a P Controller for various Measurement Noise Levels

Noise Amplitude Estimate RMS Estimate Error Cond(Q) Noise to Output Ratio

1.00E-10

-1.9800

2.40E-09 47020.04436 6.88853E-10

1.2840

-0.2720

5.0000

2.00003.0000

1.00E-09

-1.9800

2.40E-08 47020.04436 6.88853E-09

1.2840

-0.2720

5.0000

2.0000

3.0000

1.00E-08

-1.9800

0.0000002 47020.04431 6.88853E-08

1.2840

-0.2720

5.0000

2.0000

3.0000

1.00E-07

-1.9800

0.0000024 47020.0438 6.88853E-07

1.2840-0.2720

5.0000

2.0000

3.0000

1.00E-06

-1.9800

0.0000245 47020.03626 6.88853E-06

1.2840

-0.2720

5.0000

2.0000

3.0000

1.00E-05

-1.9800

0.0002843 47019.70911 6.88852E-05

1.2839

-0.2720

5.0000

2.00023.0001

0.0001

-1.9790

0.0068166 46991.29833 0.000688852

1.2821

-0.2711

4.9997

2.0056

3.0032

0.001

-1.9096

0.4328803 44408.34346 0.006888419

1.1566

-0.2115

4.9958

2.3511

3.1981

0.01

-1.1454

5.2706347 17059.58527 0.068871096

-0.17140.3887

5.0249

6.1491

5.7042

0.1

-0.4081

14.7885249 14540.0083 0.684050624

-0.2804

-0.1290

5.2693

10.3454

15.0018


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x)Case 2: P Controller, r(t) = 0

Results


P Controller, r(t) = 0

Condition Number

Of Data Matrix-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

-0.0407

-0.0886

-0.3442

33.2457

140.6121

-8.3892

1.4493e+021

With r(t) = 0, the system was not identified. This is because of the following:

Controller input d(t) is:

Hence Plant Input u(t):

Since the regressor is the following:

there is linear dependence between u and y. Hence the data matrix is not invertible.

100 200 300 400 500 600 700 800-2

-1

0

1

2x 10

-9

PlantInput(u)

Time (s)

100 200 300 400 500 600 700 800-1

-0.5

0

0.5

1x 10

-6

PlantMeasu

redOutput(y)

100 200 300 400 500 600 700 800-2

-1

0

1

2x 10

-9

PlantInput(u)

Time (s)

100 200 300 400 500 600 700 800-1

-0.5

0

0.5

1


750 755 760 765 770 775 780 785 790 795 800-1

-0.5

0

0.5

1

x 10-7

PlantTrueOutput(y0

)Zoom In on Plant Output (True vs Measured), Noise Level :6.308

Time (s)

750 755 760 765 770 775 780 785 790 795 800-1

-0.5

0

0.5

1

x 10-6

PlantMeasuredOutput

(y)


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y)Case 3: PI Controller, r(t) = randomFor a PI controller, the block diagram is represented as before, but the coeffients A(z) and B(z) will be

different:

where:

An expression for y(t) can hence be obtained as follows:

With a measurement noise magnitude of 0.000001 (Noise Power Level of 7.083e-6 compared to Output

Power), the system was identified to adequate precision, in the presence of a pseudorandom reference

signal.

Results


PI Controller, r(t) = random

Condition Number

Of Data Matrix

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0000

-1.9800

1.2840

-0.2720

5.0000

2.0000

3.0001

92529.8786

v(t)

r(t)

y(t)

d(t)

y0(t)


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The following table shows the effect of measurement noise on the estimates with a PI controller.

100 200 300 400 500 600 700 800-2

-1

0

1

2x 10

-3

PlantInput(u)

Time (s)

100 200 300 400 500 600 700 800-0.2

-0.1

0

0.1

0.2

PlantMeasuredOutput(y)

100 200 300 400 500 600 700 800-2

-1

0

1

2x 10

-3

PlantInput(u)

Time (s)

100 200 300 400 500 600 700 800-1

-0.5

0

0.5

1


750 755 760 765 770 775 780 785 790 795 800-0.2

-0.1

0

0.1

0.2

PlantTrueOutput(y0

)Zoom In on Plant Output (True vs Measured), Noise Level :4.5487e-006

Time (s)

750 755 760 765 770 775 780 785 790 795 800-0.2

-0.1

0

0.1

0.2

PlantMeasuredO

utput(y)


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Table 2 Comparison of Results with a PI Controller for various Measurement Noise Levels

Noise Amplitude Estimate RMS Estimate Error Cond(Q) Noise to Output Ratio

1.00E-10

-1.9800

2.22E-09 97123.60802 4.51E-10

1.2840

-0.2720

5.0000

2.00003.0000

1.00E-09

-1.9800

2.20E-08 97123.60803 4.51E-09

1.2840

-0.2720

5.0000

2.0000

3.0000

1.00E-08

-1.9800

0.0000002 97123.60809 4.51E-08

1.2840

-0.2720

5.0000

2.0000

3.0000

1.00E-07

-1.9800

0.0000022 97123.60867 4.51E-07

1.2840-0.2720

5.0000

2.0000

3.0000

1.00E-06

-1.9800

0.0000224 97123.61018 4.51E-06

1.2840

-0.2720

5.0000

2.0000

3.0000

1.00E-05

-1.9800

0.0002580 97123.19356 4.51E-05

1.2839

-0.2720

5.0000

2.00023.0001

0.0001

-1.9791

0.0059687 97075.89533 0.000451

1.2824

-0.2713

4.9998

2.0049

3.0029

0.001

-1.9208

0.3759261 92591.31712 0.004508

1.1775

-0.2223

5.0038

2.3037

3.1782

0.01

-1.1632

5.3322007 35372.40991 0.045086

-0.14550.3732

5.1671

6.2032

5.7588

0.1

-0.4577

17.1558106 27309.19736 0.449590

-0.3035

-0.1117

7.3639

12.0901

16.4930


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z)Case 4: P Controller, r(t) = 0With r(t) = 0 , the LS identifier will not be able to identify the parameters due to the linear dependency

introduced in the regressor by the PI controller. Since r(t) = 0, the plant input u(t) is:

or

Since the regressor is the following:

there is linear dependence between u and y. Hence the data matrix is not invertible.

Results


PI Controller, r(t) = 0

Condition Number

Of Data Matrix

-1.9800

1.2840-0.2720

5.0000

2.0000

3.0000

-0.0133

0.4766-1.6467

-183.7949

669.0858

-41.7854

8.7032e+022

100 200 300 400 500 600 700 800-2

-1

0

1

2x 10

-9

PlantInput

(u)

Time (s)

100 200 300 400 500 600 700 800-1

-0.5

0

0.5

1x 10

-6

PlantMeasuredO

utput(y)

100 200 300 400 500 600 700 800-2

-1

0

1

2x 10

-9

PlantInput(u)

Time (s)

100 200 300 400 500 600 700 800-1

-0.5

0

0.5

1


750 755 760 765 770 775 780 785 790 795 800-2

-1

0

1

2x 10

-7

PlantTrueOutput(y0

)Zoom In on Plant Output (True vs Measured), Noise Level :4.1584

Time (s)

750 755 760 765 770 775 780 785 790 795 800-1

-0.5

0

0.5

1x 10

-6

PlantMeasuredOutput(y)


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aa) ConclusionIt was seen that feedback introduces linear dependencies in the regressor, and hence led to singularities

in the data matrix inversion calculations, if the controller reference signal was set to 0. If the controller

reference signal was non-zero, and independent of u(t) and y(t), the system identifier did not have any

issues estimating the parameters.

Furthermore, the introduction of measurement noise was shown to significantly impact the parameter

estimates. As shown in Table 1 and Table 2, the noise levels had to be significantly reduced to yield

parameter estimates close to truth.

Table 3 Comparison of Results

Controller

Type

Controller

Reference r(t)

Measurement

Noise

Condition Number

Of Data MatrixEstimate

P 0 random 1.4493e+021 Not IdentifiedP random random 42903.3689 Identified

PI 0 random 8.7032e+022 Not Identified

PI random random 92529.8786 Identified


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bb) Code: (P Controller)% ECE 621, Systems Identification% Project # 2, Qn 04%% Author:%

% Date: 2012.11.02%




clear all;

% Batch estimation

load rand_rload rand_v

% Length of input/output vectornum = 800;

r(num+1:end)=[];v(num+1:end)=[];

rmax = 0.0;vmax = 0.000001;

rmean = -mean(r);vmean = -mean(v);

t = (1:num);

r = rmean*rmax + rmax*r;v = vmean*vmax + vmax*v;

for i = 1:length(r)


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i1 = i-1;i2 = i-2;i3 = i-3;

if i10)y01 = y0(i1); y02 = y0(i2); y03 = y0(i3);u1 = u(i1); u2 = u(i2); u3 = u(i3);r1 = r(i1);

d1 = d(i1);y1 = y(i1);end

d(i) = r(i) - y1;

kp =0.0023;u(i) = kp*d(i);

y0curr = [y03 y02 y01];ucurr = [u3 u2 u1];

% Compute current outputy0(i) = g_proj_1_current(ucurr, y0curr);

y(i) = y0(i) + v(i);

end

y = y';





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phi(i,:) = [-y(i-1) -y(i-2) -y(i-3) u(i-1) u(i-2) u(i-3)];

end



pnum = [t(1) t(end)];

figure;

subplot(311);[AX,H1,H2] = plotyy(t,u,t,y); grid on;set(H2,'LineStyle','-.', 'Linewidth',1.2)set(AX(1), 'xlim', pnum); set(AX(2), 'xlim', pnum);set(get(AX(1),'ylabel'), 'string', 'Plant Input (u)');set(get(AX(2),'ylabel'), 'string', 'Plant Measured Output (y)');xlabel('Time (s)');

subplot(312);[AX,H1,H2] = plotyy(t,u,t,r); grid on;set(H2,'LineStyle','-.', 'Linewidth',1.2)set(AX(1), 'xlim', pnum); set(AX(2), 'xlim', pnum);set(get(AX(1),'ylabel'), 'string', 'Plant Input (u)');set(get(AX(2),'ylabel'), 'string', 'Controller Refernce (r)');xlabel('Time (s)');

subplot(313);[AX,H1,H2] = plotyy(t(num-50:num),y0(num-50:num),t(num-50:num),y(num-50:num)); grid on;set(H2,'LineStyle','.')set(get(AX(1),'ylabel'), 'string', 'Plant True Output (y_0)');set(get(AX(2),'ylabel'), 'string', 'Plant Measured Output (y)');ttt = ['Zoom In on Plant Output (True vs Measured), Noise Level :',num2str(std(v)/std(y0))];title(ttt)xlabel('Time (s)');size_figs_4PPT


41/44


cc) Code: (PI Controller)% ECE 621, Systems Identification% Project # 2, Qn 04%% Author:%

% Date: 2012.11.02%




clear all;

% Batch estimation

load rand_rload rand_v

% Length of input/output vectornum = 800;

r(num+1:end)=[];v(num+1:end)=[];

rmax = 0.0;vmax = 0.000001;

rmean = -mean(r);vmean = -mean(v);

t = (1:num);

r = rmean*rmax + rmax*r;v = vmean*vmax + vmax*v;

for i = 1:length(r)

i1 = i-1;i2 = i-2;i3 = i-3;


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if i10)y01 = y0(i1); y02 = y0(i2); y03 = y0(i3);u1 = u(i1); u2 = u(i2); u3 = u(i3);r1 = r(i1);d1 = d(i1);y1 = y(i1);

end

d(i) = r(i) - y1;

dcurr = [d1 d(i)];

u(i) = pi_control_current(dcurr, u1);

y0curr = [y03 y02 y01];ucurr = [u3 u2 u1];

% Compute current outputy0(i) = g_proj_1_current(ucurr, y0curr);

y(i) = y0(i) + v(i);

end

y = y';





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phi(i,:) = [-y(i-1) -y(i-2) -y(i-3) u(i-1) u(i-2) u(i-3)];

end



pnum = [t(1) t(end)];

figure;

subplot(311);[AX,H1,H2] = plotyy(t,u,t,y); grid on;set(H2,'LineStyle','-.', 'Linewidth',1.2)set(AX(1), 'xlim', pnum); set(AX(2), 'xlim', pnum);set(get(AX(1),'ylabel'), 'string', 'Plant Input (u)');set(get(AX(2),'ylabel'), 'string', 'Plant Measured Output (y)');xlabel('Time (s)');

subplot(312);[AX,H1,H2] = plotyy(t,u,t,r); grid on;set(H2,'LineStyle','-.', 'Linewidth',1.2)set(AX(1), 'xlim', pnum); set(AX(2), 'xlim', pnum);set(get(AX(1),'ylabel'), 'string', 'Plant Input (u)');set(get(AX(2),'ylabel'), 'string', 'Controller Refernce (r)');xlabel('Time (s)');

subplot(313);[AX,H1,H2] = plotyy(t(num-50:num),y0(num-50:num),t(num-50:num),y(num-50:num)); grid on;set(H2,'LineStyle','.')set(get(AX(1),'ylabel'), 'string', 'Plant True Output (y_0)');set(get(AX(2),'ylabel'), 'string', 'Plant Measured Output (y)');ttt = ['Zoom In on Plant Output (True vs Measured), Noise Level :',num2str(std(v)/std(y0))];title(ttt)xlabel('Time (s)');size_figs_4PPT


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dd) Code: (Functions)function y = g_proj_1_current(u, yold)

p1 = 0.80;p2 = 0.68;

p3 = 0.50;b0 = 5.00;b1 = 2.00;b2 = 3.00;

a1 = - (p1 + p2 + p3);a2 = p1*p2 + p2*p3 + p1*p3;a3 = - p1*p2*p3;

il = length(u);ol = length(yold);

y = b0*u(il) + b1*u(il-1) + b2*u(il-2) - a1*yold(ol) - a2*yold(ol-1) -a3*yold(ol-2);

function u = pi_control_current(d, uold)

il = length(d);ol = length(uold);

kp =0.0023;ki =0.00023;

if (il ~=2 || ol ~=1 )disp(['i/p vector has length ', int2str(il)]);disp(['o/p vector has length ', int2str(ol)]);error ('i/p vector should have length of 2, & prev o/p vector should have

length of 1');end

u = uold + (kp + ki)*d(il) - kp*d(il-1);

System Identification Tips

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