University of Warwick: amr summer School 4th-6th



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University of Warwick: AMR Summer School

4th-6th July, 2016

Structural Identifiability Analysis

Dr Mike Chappell,

School of Engineering, University of Warwick Coventry CV4 7AL

Examples Sheet: Solutions

1. For the given system we have the following:

and u(t) = (t). Therefore the transfer function for this model is given by:



Using the uniqueness of the coefficients of powers of s in the numerator and denominator (the moment invariants, i) it is seen that the following are unique:



.

Since c2 and d1 are both known we see that is unique for the given input-output behaviour and so k12 is globally identifiable.

From the second moment invariant we have and so substituting into the third moment invariant gives:

Thus for generic parameter vectors and there are two solutions for k2e (and hence k1e). Hence both k2e and k1e are locally identifiable.

Since two of the three unknown parameters are locally identifiable and the remaining one is globally identifiable the model is structurally locally identifiable.

If, in addition, we can assume that then the second moment invariant implies that . Hence, in this case, both k2e and k1e are globally identifiable and so the model becomes structurally globally identifiable.



2. Consider the model given by:

The first Taylor series coefficient is given by:



and so the parameter c is unique for the output – it is globally identifiable.

The second coefficient is given by:

and so 1 = k1(1 – k2) is also unique. The third coefficient is given by:



and so 2 = k1(1 – 2k2) is also unique. Dividing these two expressions gives



and so k2 is unique. Since k1(1 – k2) this means that k1 is unique.

Summarising: All of the parameters k1, k2 and c are unique for the given output and hence globally identifiable. Therefore the model is structurally globally identifiable.

For the given measured values:



and


Dividing second equation by first gives



.

Substituting into the first equation gives:




3. (i) For the given system to be at steady state:

From the second equation either or .

Substituting into the first equation yields a contradiction since a > 0.

Substituting into the first equation yields:



Therefore the steady state is given by .

(ii) To determine the linearisation (for arbitrary a and b) first determine the corresponding Jacobian matrices (from initial bookwork):

There is no input and so the B matrix is zero. Therefore the linearisation is given by:



For the given values for a and b the characteristic equation is given by:



Therefore, the eigenvalues of the system are given by . Since the real parts are both negative this is a stable focus (when a = 2 and b = 3).

(iii) Now suppose that the system has observation given by

so that the linearised system is a state space model with matrix A given by Part (ii) and C matrix given by C = (1 0). Applying the Observability Rank Criterion we have:



This Observability matrix has full rank provided a > 0. Thus, the linearisation is observable for any values of a, b > 0.



4. Taking Laplace transforms of the system equations:

Alternatively, writing the system in the standard form first the transfer function is given by:



Using the uniqueness of the coefficients of powers of s in the numerator and denominator of the transfer function (the moment invariants, i) it is seen that the following are unique:



.

Since c12 is known and we see that b1 is unique and so globally identifiable.

Note that and so we only need consider the last two moment invariants.

From the third moment invariant we have and so substituting into the fourth moment invariant gives:



Thus for generic parameter vectors and there are two solutions for k2e (and hence k12). Hence both k2e and k12 are locally identifiable.

Since two of the three unknown parameters are locally identifiable and the remaining one is globally identifiable the model is structurally locally identifiable.

The given measured transfer function is given by:



and so the moment invariants are given by:



Using the above:







Hence the there are two sets of parameters:



b1 = 10, k12 = 5 and k2e = 2 or b1 = 10, k12 = 2 and k2e = 5.
5. For the given system:

We determine the Taylor series coefficients and check uniqueness:



Therefore c1 is unique for the given output and hence globally identifiable.



Therefore c1a21 is unique for the given output and, since c1 is globally identifiable, a21 is globally identifiable.



Since c1a21 we have that (a21 + a12) is also unique. Since a21 is globally identifiable we have that a12 is also globally identifiable.



Since c1, a12 and a21 are globally identifiable, and hence unique, we see from this coefficient that a02 is also unique and therefore globally identifiable.

Since all parameters are globally identifiable the model is structurally globally identifiable.

Using the coefficients determined in Part (ii) with the measured values we have the following:



(l-1)

(s-1)

(s-1)


6. (i) For the given system to be at steady state:

From the first equation and subtituting for in the second equation yields



The numerator of this expression then gives



So or b-c and the steady states are or .

(ii) To determine the linearisations first determine the corresponding Jacobian matrices (from initial bookwork):

There is no input and so the B matrix is zero.

Therefore the linearisations are given by:

At



The A matrix has eigenvalues so both real and negative implies a stable steady state.

At

The A matrix has eigenvalues, so one is positive, the other negative which implies an unstable steady state.

(iii) For the given initial observation condition, need to verify the ORC for each linearisation.

Note that the observation is given by



So and hence need to check the rank of .

For the linearisation at , , so

Which has rank 2 (det=-3) so the ORC is satisfied.

For the linearisation at , , so

Which has rank 2 (det=3/4) so the ORC is again satisfied.


7. For the given system:

We determine the Taylor series coefficients and check uniqueness:



Therefore c is unique for the given output and hence globally identifiable.



Therefore β – α is unique for the given output since c is globally identifiable.



Since c (β – α) is unique, we have that (β – 3α) is also unique. But this is just (β – α) – 2α, and so α is globally identifiable.

Therefore (since α + β is unique from the second coefficient) β is also globally identifiable.

Since all parameters are globally identifiable there is only one parameter vector that gives rise to each (generic) output – therefore, the model is structurally globally identifiable.



Using the coefficients determined in Part (iii) with the measured values we have the following:

(l-1)

(1)

(2)

Substracting Eq (1) from Eq (2) gives 2α = 1 and so α = 0.5 (s-1). Therefore β = 2.5 (s-1 μmol-2).

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