Observer Design by Matching Coefficients

PROBLEM: A time-scaled model for the body’s blood glucose level is shown in Eq. (12.105). The output is the deviation in glucose concentration from its mean value in mg/100 ml, and the input is the intravenous glucose injection rate in g/kg/hr (Milhorn,1966).
$G(s)=\frac{407(s+0.916)}{(s+1.27)(s+2.69)}$
Design an observer for the phase variables with a transient response described by ζ = 0.7 and $\omega _{n}=100$ .

Question

Observer Design by Matching Coefficients

PROBLEM: A time-scaled model for the body’s blood glucose level is shown in Eq. (12.105). The output is the deviation in glucose concentration from its mean value in mg/100 ml, and the input is the intravenous glucose injection rate in g/kg/hr (Milhorn,1966).

[latex]G(s)=\frac{407(s+0.916)}{(s+1.27)(s+2.69)}[/latex]

Design an observer for the phase variables with a transient response described by ζ = 0.7 and [latex] \omega _{n}=100[/latex].

Accepted Answer

We can first model the plant in phase-variable form. The result is shown in Figure 12.20(a).

For the plant,

[latex]A=\left[\begin{matrix} 0 & 1 \ -3.42 & -3.96 \end{matrix} ight] ;[/latex] [latex]C=\left[\begin{matrix} 372.81 & 407 \end{matrix} ight] [/latex]

Calculation of the observability matrix, [latex]O_{M} =\left[\begin {matrix} C & CA \end{matrix} ight] ^{T} [/latex], shows that the plant is observable andwe can proceed with the design. Next find the characteristic equation of the observer. First we have

[latex]A-LC= \left[\begin{matrix} 0 & 1 \ -3.42 & -3.96 \end{matrix} ight]-\left[\begin{matrix} l_{1} \ l_{2} \end{matrix} ight]\left[\begin{matrix} 372.81 & 407 \end{matrix} ight][/latex]

[latex]=\left[\begin{matrix} -372.81l_{1} & (1-407l_{1}) \ -(3.42+372.81l_{2}) & -(3.96+407l_{2}) \end{matrix} ight][/latex]

Now evaluate [latex]\left[\lambda I-(A-LC) ight] =0[/latex] in order to obtain the characteristic equation:

[latex]\det \left[\lambda I-(A-LC) ight]=\det \left[\begin{matrix} (\lambda +372.81l_{1}) & -(1-407l_{1}) \ (3.42+372.81l_{2}) & (\lambda +3.96+407l_{2}) \end{matrix} ight][/latex]

[latex]=\lambda ^{2}+(3.96+372.81l_{1}+407l_{2})\lambda +(3.42+84.39l_{1}+372.81l_{2})[/latex]

=0

From the problem statement, we want ζ = 0.7 and [latex] \omega _{n}=100[/latex]. Thus,

[latex]\lambda ^{2}+140\lambda +10,000=0[/latex]

Comparing the coefficients of Eqs. (12.108)

[latex]\det \left[\lambda I-(A-LC) ight]=\det \left[\begin{matrix} (\lambda +372.81l_{1}) & -(1-407l_{1}) \ (3.42+372.81l_{2}) & (\lambda +3.96+407l_{2}) \end{matrix} ight][/latex]

[latex]=\lambda ^{2}+(3.96+372.81l_{1}+407l_{2})\lambda +(3.42+84.39l_{1}+372.81l_{2})[/latex]

=0

and (12.109),

[latex]\lambda ^{2}+140\lambda +10,000=0[/latex]

we find the values [latex]l_{1}[/latex] and [latex]l_{2}[/latex] to be
-38.397 and 35.506, respectively. Using Eq. (12.60),

[latex]\dot{\hat{x} }=A\hat{x} +Bu+L(y-\hat{y} )[/latex]

[latex]\hat{y}=C\hat{x}[/latex]

where

[latex]A=\left[\begin{matrix} 0 & 1 \ -3.42 & -3.96 \end{matrix} ight] ; B= \left[\begin{matrix} 0 \ 1 \end{matrix} ight] ; C=\left[\begin{matrix} 372.81 & 407 \end{matrix} ight] ;[/latex]

[latex]L=\left[\begin{matrix} -38.397 \ 35.506 \end{matrix} ight][/latex]

the observer is implemented and shown in Figure 12.20(b).

Question 12.9: Observer Design by Matching Coefficients PROBLEM: A time-sca...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions