signals and systems€¦ · outline 1 frequency domain system identiﬁcation introduction choice...

Signals and Systems

Lecture 11: System Identification

Dr. Guillaume Ducard

Fall 2018

based on materials from: Prof. Dr. Raffaello D’Andrea

Institute for Dynamic Systems and Control

ETH Zurich, Switzerland

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Outline

1 Frequency Domain System IdentificationIntroductionChoice of signals inputs

2 Identifying a System based on its Impulse ResponseImpulse response, no noiseImpulse response, with noise

3 Identification Using Sinusoidal InputsExperimental procedureClosed-loop system identificationIdentifying the transfer function

MethodExample

Weighted least squares

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Frequency Domain System IdentificationIdentifying a System based on its Impulse Response

Identification Using Sinusoidal Inputs

Introduction

In this lecture apply tools learned to identify an unknown system.

This process is known as system identification. The key idea:

1 by applying a known input to a system

2 and observing the system’s output,

⇒ a model of the system can be estimated.

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Introduction






In this lecture, we focus on black-box system identification.

Black-box: we do not have a first-principles (i.e. physics-based) model ofthe system. Black-box system identification is a purely data-drivenmodeling tool.

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Introduction








Grey-box: we have a first-principles model, but with unknownparameters. Data is used to identify these parameters. This is covered inthe IDSC lecture System Modeling.

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Introduction








Grey-box: we have a first-principles model, but with unknownparameters. Data is used to identify these parameters. This is covered inthe IDSC lecture System Modeling.

White-box: we have a first-principles model, with no unknownparameters, or where parameters can be measured directly (mass, forexample); no data is required in this case.

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Introduction

This categorization is a simplified view.

In reality : combinations of these system identification methods are usually

applied.

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Introduction



applied.

The objective of black-box system identification is:

1 typically to identify the transfer function H(z) of the system;

2 ⇒ to be used for control design, simulation, etc.

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Introduction



applied.




An intermediate step

1 Often first: estimate the system’s frequency response H(Ω)(like Bode plots) – which can also be used for control designdirectly in the frequency domain (lead/lag compensation, gainand phase margin, Nyquist stability criterion, etc)

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Introduction



applied.




An intermediate step

1 Often first: estimate the system’s frequency response H(Ω)(like Bode plots) – which can also be used for control designdirectly in the frequency domain (lead/lag compensation, gainand phase margin, Nyquist stability criterion, etc)

2 Next, estimate the system’s transfer function of the system,based on the frequency response estimate.

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Introduction

System identification can either be performed in the time domainor frequency domain.

Time domain techniques can be generalized to non-linearsystems; however,

Frequency domain techniques exploit the structure of LTIsystems, and are therefore conceptually easier.

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Introduction




We will only consider frequency domain system identification inthis lecture, where the objective:

1 is to determine H(Ω) from experimental data,2 and then to estimate the transfer function H(z) from the

frequency response estimate.

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Introduction




We will only consider frequency domain system identification inthis lecture, where the objective:

1 is to determine H(Ω) from experimental data,2 and then to estimate the transfer function H(z) from the

frequency response estimate.

⇒ We focus on the black-box identification of causal, linear time-invariantsystems.

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IntroductionChoice of signals inputs

Outline




MethodExample


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Frequency Domain System Identification

We will consider the following system identification problem:

Gue

udy

ydym

Where G is a causal, stable LTI system, and

ue is a known input that we use to excite the system

ud is an unknown process noise, assumed to be white

yd is an unknown measurement noise, assumed to be white

ym, given by ym = Gue + yd + Gud, is a measurement of thesystem’s output, which is corrupted by process noise andmeasurement noise

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Outline




MethodExample


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Input choice for frequency domain system identification

Based on the above problem structure, the task is now to design:

1 a set of inputs ue, such that by analyzing the measuredoutput ym

2 and the known input ue,

⇒ the frequency response of the system H(Ω) can be determinedaccurately, despite inherent process and measurement noise.

We will see later in the lecture how to estimate the system’stransfer function based on its estimated frequency response.

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Input choice for frequency domain system identification

Based on the above problem structure, the task is now to design:

1 a set of inputs ue, such that by analyzing the measuredoutput ym

2 and the known input ue,

⇒ the frequency response of the system H(Ω) can be determinedaccurately, despite inherent process and measurement noise.

We will see later in the lecture how to estimate the system’stransfer function based on its estimated frequency response.

Question: What kind of input signal can be used for identification?

We now introduce two possible inputs:

a unit impulse,

and a sinusoid.

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Impulse response, no noiseImpulse response, with noise

Outline




MethodExample


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Identifying a System based on its Impulse Response

In Lecture 2, we saw that

an LTI system is fully characterized by its impulse response.

ConclusionBy measuring its impulse response, the system can be identified.

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In the absence of process and measurement noise, we have that

ym = G ue .

We then have a simple way of determining freq. resp. H(Ω):1 Let ue[n] = δ[n]

2 Then ym[n] = h[n] and H(Ω) =∞∑

n=0

ym[n]e−jΩn

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In the absence of process and measurement noise, we have that

ym = G ue .

We then have a simple way of determining freq. resp. H(Ω):1 Let ue[n] = δ[n]

2 Then ym[n] = h[n] and H(Ω) =∞∑

n=0

ym[n]e−jΩn

In practicesince most systems have an infinite impulse response

⇒ impossible to measure it entirely,

⇒ we collect N pieces of data, where N is large enough suchthat the impulse response has decayed significantly ( = sothat ym[n] is small enough for n ≥ N).

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We then take the DFT:

Ym[k] =

N−1∑

n=0

ym[n]e−j 2πkN

n.

At the discrete frequency Ωk = 2πk/N where k = 0, 1, . . . , N − 1,(N DFT coefficients).

The frequency response estimate H(Ωk) is defined as:

H (Ωk) := Ym[k] = H (Ωk)−

∞∑

n=N

h[n]e−jΩkn

︸︷︷︸HN (Ωk)

,

where HN (Ωk) is the ‘estimation” error introduced by notmeasuring the entire duration of the impulse response.Remark: Note that HN(Ω) → 0 as N → ∞ since G is stable.

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Example : Estimating frequency response, based on impulse response, without noise

Let the system be given by the transfer function

H(z) =0.9474 + 0.2105z−1 − 0.7368z−2

1− 1.579z−1 + 0.7895z−2,

which has the following frequency response:

π02468

|H(Ω

)|

π

−90

0

90

∠H(Ω

)

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We apply the input ue[n] = δ[n] in the absence of noise, and recordthe system’s output for N = 11, N = 51 and N = 201 samples:

20 40 60 80 100 120 140 160 180 200

−1

0

1

2

y m[n]

ym[n] for N = 201

ym[n] for N = 51

ym[n] for N = 11

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Take DFT of these measurements to estimate system’s frequency responseH(Ω):

π0

2

4

6

8

∣ ∣ ∣H(Ω

)∣ ∣ ∣

H(Ω)

H(Ω) for N = 201

H(Ω) for N = 51

H(Ω) for N = 11

π

−90

0

90

∠H(Ω

)

Conclusion:

1 a larger N gives a higher frequency resolution, since we are dividing theinterval −π to π into N equally spaced frequency intervals.

2 We also note that as N increases, estimation error decreases.G. Ducard 16 / 46




Outline




MethodExample


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With measurement noise

We study effects of measurement noise yd on above estimation procedure.

ym[n] = h[n] + yd[n] for n = 0, 1, . . . , N − 1,

where E(yd[n]) = 0 and E(yd[n]yd[m]) = σ2yδ[n−m].

It follows that H (Ωk) = Ym[k] = H (Ωk)−HN (Ωk) + Yd[k],where E(Yd[k]) = 0 and E(|Yd[k]|

2) = Nσ2y .

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ym[n] = h[n] + yd[n] for n = 0, 1, . . . , N − 1,



2) = Nσ2y .

Thus, H (Ωk)−H (Ωk) = −HN (Ωk) + Yd[k]

E(H (Ωk)−H (Ωk)

)= −HN (Ωk) , est. error approaches zero as

N → ∞.

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ym[n] = h[n] + yd[n] for n = 0, 1, . . . , N − 1,



2) = Nσ2y .

Thus, H (Ωk)−H (Ωk) = −HN (Ωk) + Yd[k]

E(H (Ωk)−H (Ωk)

)= −HN (Ωk) , est. error approaches zero as

N → ∞.

But, mean-square error: E

(∣∣∣H (Ωk)−H (Ωk)∣∣∣2)

= H2N (Ωk) +Nσ2

y

approaches infinity as the length of the sample increases, becauseNσ2

y → ∞ as N → ∞.

Remark : Inclusion of process noise ud has a similar effect, but is morecomplex to derive analytically.

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We now demonstrate the effect of zero-mean process noise ud andmeasurement noise yd with σ2

y = σ2u = 1

3.

Using the identical system, given by

H(z) =0.9474 + 0.2105z−1 − 0.7368z−2

1− 1.579z−1 + 0.7895z−2,

we reapply the input ue[n] = δ[n].We record the system’s output for N = 11, N = 51 and N = 201 samples:

20 40 60 80 100 120 140 160 180 200

−1

0

1

2

ym[n]

ym[n] for N = 201

ym[n] for N = 51

ym[n] for N = 11

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We then take the DFT of these measurements in order to estimate thefrequency response H(Ω) of the system:

π0

5

10

∣ ∣ ∣H(Ω

)∣ ∣ ∣

H(Ω)

H(Ω) for N = 201

H(Ω) for N = 51

H(Ω) for N = 11

π

−90

0

90

∠H(Ω

)

Conclusion: Here we see that the process noise and measurement noise causelarge estimation errors for large N .

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This example highlights the problem of using a unit impulseinput to identify a system:

the energy of the input we apply stays constant,

but the energy of the measurement noise increases as Nincreases (E(|Yd[k]|

2) = Nσ2y .).

To address this problem:

one can increase the input amplitude;however, beyond a certain point, we usually don’t have thisluxury due to saturation, non-linear effects, etc.

An alternative solution is to select a more appropriate inputsignal.

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Experimental procedureClosed-loop system identificationIdentifying the transfer functionWeighted least squares

Outline




MethodExample


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Goal: To identify the frequency response of a system from itsinput-output behavior.An alternative method : is to apply sinusoidal inputs at differentfrequencies .

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Goal: To identify the frequency response of a system from itsinput-output behavior.An alternative method : is to apply sinusoidal inputs at differentfrequencies .

Advantages of this method: more robust to noise !

Recall: the energy of the input and that of the noise bothgrow linearly as a function of N ,

the energy of the noise is spread out across all N frequencies(white noise),

but a pure sinusoid has all energy at one frequency (⇒ better“signal to noise ratio” at every frequency).

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Sinusoidal input, output with noise

Consider the case where measurement noise corrupts the output,such that

ym = G ue + yd.

Now let

ue[n] = ej2πN

ln for n = 0, 1, . . . , NT +N − 1, be a sinusoidwith discrete frequency Ωl = 2πl/N ,

ye = Gue be the ideal output of the system.

In Lecture 6 we saw that, since ej2πN

ln is an eigenfunction of anyLTI system,

ye[n] = H (Ωl)ue[n]− w[n], (1)

where w[n] captures the effect of a causal input and decays to zeroas N → ∞ (transient).

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ye[n] = H (Ωl)ue[n]− w[n],

We now take the DFT of the sequences introduced above forn ≥ NT . Let

Ye[l] =

NT+N−1∑

n=NT

ye[n]e−j

2πN

ln,

Ue[l] =

NT+N−1∑

n=NT

ue[n]e−j

2πN

ln= N,

W [l] =

NT+N−1∑

n=NT

w[n]e−j2πN

ln,

then Ye[l] = H (Ωl)Ue[l]−W [l]. Remark: l is the index for the DT frequency

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We can thus estimate the system’s frequency response as follows. Let

Ym[l] =

NT +N−1∑

n=NT

ym[n]e−j2πN

ln

Yd[l] =

NT +N−1∑

n=NT

yd[n]e−j

2πN

ln,

then, since ym = ye + yd,

Ym[l] = Ye[l] + Yd[[l]] = H (Ωl)Ue[l]−W [l] + Yd[[l]]

Frequency response estimate: H (Ωl)

H (Ωl) :=Ym[l]

Ue[l]= H (Ωl)−

W [l]

N+

Yd[l]

N.

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H (Ωl) :=Ym[l]

Ue[l]= H (Ωl)−

W [l]

N+

Yd[l]

N.

We therefore have1

E(H (Ωl)−H (Ωl)

)= −

W [l]

N,

which approaches zero as NT → ∞ or N → ∞ . Therefore,as N → ∞, we have an unbiased estimate.

2

E

(∣∣∣H (Ωl)−H (Ωl)∣∣∣2)

=W 2[l]

N2+

σ2y

N.

The mean-squared error goes to 0. A similar result holds whenud is included.

Conclusion: A sinusoidal input at different frequencies is a very effective inputfor system identification. G. Ducard 27 / 46




Experimental procedure: summary

Choose NT large enough to let the transient die down (first NT

samples of the output to discard),

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Choose the experiment length N .

A larger N will reduce the influence of noise (as previouslydiscussed),but will increase the duration of each experiment.

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Choose the experiment length N .

A larger N will reduce the influence of noise (as previouslydiscussed),but will increase the duration of each experiment.

Select a frequency of interest by choosing an appropriateinteger l and setting Ωl = 2πl/N .

Note that, because the system is real, we only need to identifythe frequency response for 0 ≤ Ω ≤ π, since H(−Ω) = H∗(Ω).We will thus pick l ∈ [0, N−1

2] if N is odd, or l ∈ [0, N

2] if N is

even.

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Set ue[n] = A cos(Ωln) for n = 0, . . . , NT +N − 1, where A scales the

input.

Choosing a large A increases the signal to noise ratio;however, if A is chosen too large, nonlinearities such assaturation may be excited, thus negatively affecting thefrequency response estimate.

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input.


Calculate

Ym[l] =

NT +N−1∑

n=NT

ym[n]e−jΩln and Ue[l] =

NT +N−1∑

n=NT

ue[n]e−jΩln

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input.


Calculate

Ym[l] =

NT +N−1∑

n=NT

ym[n]e−jΩln and Ue[l] =

NT +N−1∑

n=NT

ue[n]e−jΩln

The frequency response estimate at the frequency Ωl is then given by

H(Ωl) :=Ym[l]

Ue[l].

Repeat this for the frequencies of interest by picking the appropriateinteger l.

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Example:Estimating frequency response using sinusoidal inputs, with noise

We now revisit the previous example using the identical system,given by

H(z) =0.9474 + 0.2105z−1 − 0.7368z−2

1− 1.579z−1 + 0.7895z−2,

and where the system is subject to both the process noise ud andthe measurement noise yd.

1 Compared to the previous example, we increase the noise suchthat the standard deviation is increased by a factor of 5,resulting in σ2

y = σ2u = 25

3 .

2 We perform system identification by applying sinusoidal inputsat N different frequencies, by choosingl = 0, 1, . . . , (N − 1)/2.

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The recorded input ue and output ym for NT = 50, N = 201, A = 1, andl = 7 are as follows:

0 50 100 150 200 250

−5

0

5

n

Amplitude

ue

ym without noise

ym with noise

Observations1. Ideal output of system is: shifted and scaled version of input sinusoid.2. Process noise and measurement noise corrupt the signals and their effectsare noticeable.3. By picking NT = 50, we let the transient decay. G. Ducard 31 / 46




The estimate of the frequency response is as follows:

π0

2

4

6

8

∣ ∣ ∣H(Ω

)∣ ∣ ∣

H(Ω)

H(Ω) for N = 201

H(Ω) for N = 51

H(Ω) for N = 11

π

−90

0

90

∠H(Ω

)

Observations: This method performs very well in the presence of noise. Theabove plot highlights the two main benefits of picking a larger N :

1 the effect of noise is further reduced, since more samples are averaged,

2 a larger N results in a finer frequency resolution, allowing morefrequencies to be tested.

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We now demonstrate what happens if we do not wait for the transient todecay, i.e. we pick NT = 0.

π0

2

4

6

8

∣ ∣ ∣H(Ω

)∣ ∣ ∣

H(Ω)

H(Ω) for N = 201

H(Ω) for N = 51

H(Ω) for N = 11

π

−90

0

90

∠H(Ω

)

We see that, as long as N is much larger than the transient decay time, theeffect of the transient is small. However, for small N the transient significantlyaffects the estimate of the frequency response. G. Ducard 33 / 46




Outline




MethodExample


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Closed-loop system identification

Often, the system to be identified is part of a closed-loop system.

This is the case, for example, if the open-loop system is unstable.⇒ In this case, an appropriate, stabilizing controller C is requiredbefore performing the identification.

Consider the following stable closed-loop system (where G can beunstable)

ucG

C

us ueud yd

ym

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uc

G

C

us ue

ud ydym

Let ue = uc + us. Then, ym = Gue + Gud + yd.

Discussion :

if ue can be measured, the open loop method described in the previoussection can be used to identify the frequency response of the plant.

If G and C have no poles at z = ejΩl , and under some mild assumptionson ud and yd, it can be shown that the bias and mean-squared error ofthe estimate H(Ωl) :=

Ym[l]Ue[l]

, both approach 0 as N → ∞.

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Outline




MethodExample


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Identifying the Transfer Function: Method

After estimating the frequency response at a number of distinct frequencies Ωl,we can identify the transfer function of the system.Given the two design parameters:

A (number of ak coefficients, although a0 is fixed at 1)

and B (number of bk coefficients),

the goal is to identify the unknown parameters: ak and bk of the system’stransfer function, given by

H(z) =

B−1∑k=0

bkz−k

1 +A−1∑k=1

akz−k

,

with frequency response

H(Ω) =

B−1∑k=0

bke−jΩk

1 +A−1∑k=1

ake−jΩk

.

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As a result of the identification experimentswe obtained estimates H(Ωl) of H(Ω) at L frequencies Ωl forl = 0, 1, . . . , L− 1.

Setting H(Ωl) = H(Ωl) at all measurement frequencies, we obtain the systemof equations(1 + a1e

−jΩl+ · · ·+ aA−1e−j(A−1)Ωl

)H(Ωl) =

= b0 + b1e−jΩl + · · ·+ bB−1e

−j(B−1)Ωl for l = 0, 1, . . . , L− 1.

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These equations must hold for both real and imaginary components.We therefore let :

H(Ωl) = Rlejφl and,

using ejθ = cos θ + j sin(θ),

rewrite the system of equations for l = 0, 1, . . . , L− 1 as

Rl cos(φl) + a1Rl cos(φl −Ωl) + · · ·+ aA−1Rl cos(φl − (A− 1)Ωl)

= b0 + b1 cos(Ωl) + · · ·+ bB−1 cos((B − 1)Ωl) Real

Rl sin(φl) + a1Rl sin(φl − Ωl) + · · ·+ aA−1Rl sin(φl − (A− 1)Ωl)

= −b1 sin(Ωl)− · · · − bB−1 sin((B − 1)Ωl) Imaginary

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This system of equations can be converted to the least squaresproblem of minimizing

(FΘ−G)T(FΘ −G),

where

Θ =[a1 a2 . . . aA−1 b0 b1 . . . bB−1

]T

contains the unknown parameters and where

Θ: size (A+B − 1) vector, unknown.

F : size (2L)× (A+B − 1) regression matrix, known.

G: size (2L) output measurement vector, known.

The least-squares solution

Θ∗ = (FTF )−1FTG

yields the estimated transfer function coefficients Θ∗. G. Ducard 41 / 46




Note that there are (A+B − 1) unknown parameters in the vectorΘ, and that there are 2L equations (L frequency measurementswith both real and imaginary components). However, if anyΩl = 0 or π, an equation is lost since the equation correspondingto the imaginary component yields 0 = 0. For the existence of aleast squares solution, the matrix F must have full column rank.We therefore require

2L ≥ A+B − 1 if frequencies Ωl = 0 and Ωl = π were nottested;

2L ≥ A+B if either Ωl = 0 or Ωl = π were tested;

2L ≥ A+B + 1 if both Ωl = 0 and Ωl = π were tested.

In practice this is not a problem as typically L ≫ A+B.

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Identifying the transfer function

In the previous example, we applied the sinusoid method toestimate the frequency response of the system given by

H(z) =0.9474 + 0.2105z−1 − 0.7368z−2

1− 1.579z−1 + 0.7895z−2,

at specified frequencies Ωl. Using the approach described above,we can now identify the transfer function. With A = B = 3, thetransfer function is identified to be:

H(z) =0.9472 + 0.2015z−1 − 0.7223z−2

1− 1.572z−1 + 0.7836z−2, for N = 201;

H(z) =0.9558 + 0.2477z−1 − 0.7693z−2

1− 1.559z−1 + 0.772z−2, for N = 51;

H(z) =0.6371 + 0.2257z−1 − 0.3091z−2

1− 1.5z−1 + 0.778z−2, for N = 11.

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The resulting frequency responses are the following:

π0

2

4

6

8

∣ ∣ ∣H(Ω

)∣ ∣ ∣

H(Ω)

H(Ω) for N = 201

H(Ω) for N = 51

H(Ω) for N = 11

π

−90

0

90

∠H(Ω

)

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Outline




MethodExample


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A useful variant for system identification is weighted least squares: FΘ = G isreplaced by WFΘ = WG, where

W =

w0

w0

w1

w1

. . .

wL

is a (2L× 2L) square, diagonal matrix. Its elements wl are weights whichcapture the confidence in the measurements. We obtain the solution using thesubstitutions F = WF , G = WG:

Θ∗ = (FTF )−1

FTG

= (F TW

TWF )−1

FTW

TWG.

Higher weights result in a better fit at the corresponding frequencies.For example, it is beneficial to use low weights at frequencies where a systemrolls off. There, the measured amplitudes become small compared to the noiseand the measurements are less reliable. G. Ducard 46 / 46

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