Chapter 15 Observability II
15.1 Examples
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Example 15.1. Consider the second order scalar differential equation
where
, the state is . HenceWe can see from the definition that this system is observable. If
for all , then by differentiating we have that for all and therefore for all . In particular and therefore the only unobservable state is zero.
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Example 15.2. Consider the second order scalar differential equation
where
, the state is . HenceWe can show from the definition that this system is observable, but this is slightly more complicated than it was for Example 15.1. Assume that
for all , i.e. that for all . Then for all . From the differential equation we then obtain that for all . Hence for all and in particular . We see that zero is the only state which is unobservable and therefore that the system is observable.We can also consider the Kalman observability matrix. We have
so that
which has determinant
, which is nonzero, and therefore is invertible and therefore is injective. Hence we have observability.The Hautus observability matrix is
Picking the second and last rows gives the matrix
which is upper-triangular with nonzero elements on the diagonal and is therefore invertible (no matter what
is). It follows that the Hautus observability matrix is injective for all and from this we see that the system is observable.If
, then is asymptotically stable and we can consider the infinite-time observability Gramian. The observation Lyapunov equation is (using that is symmetric)which is
The top-left corner gives that
. The bottom-right corner then gives that . The off-diagonal entry then gives . HenceSince
is invertible, we see that the system is observable.
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Example 15.4. We use the Kalman observability matrix to show that the state-output system with
is observable. We have that the Kalman observability matrix is
Developing by the first row, we see that the determinant of this matrix is 1. Therefore the Kalman observability matrix is invertible and therefore injective. It follows that the system is observable.
15.2 Case study: control of a tape drive*
Since we measure the whole state, it is clear that the state is observable from the measurement. To make tape drives smaller, manufacturers are looking into removing the sensor which measure the tension in the tape. This would leave only the two velocity measurements. We then have
The Kalman observability matrix is (since
We have
so that
which has rank 3 (the first three rows form a lower triangular matrix with nonzero elements on the diagonal). Therefore the Kalman observability matrix is injective (note that we didn’t even need the
If we measure only the tension in the tape, then
and we have
whose determinant (e.g. developing by the first row) is
is in the nullspace of the Kalman observability matrix and therefore is in the unobservable subspace. This means that if the two mass-damper systems are identical, then if the velocities of the two reels are the same, we cannot determine what that common velocity is from measurement of the tension in the tape.
In concordance with Theorem 11.3 we consider observability of
The third row of
Therefore the Kalman observability matrix includes all rows in the following matrix (we pick all rows from
and since this is lower-triangular with nonzero entries on the diagonal this matrix is invertible and therefore the Kalman observability matrix is injective and the pair
15.3 Case study: a suspension system*
If we measure the whole state, then we certainly have observability. We consider the case where we only measure the suspension stroke and the difference of the velocity of the two masses. This gives
We then have
and
and
The Kalman observability matrix therefore is
Omitting duplicate rows gives the matrix
Omitting the last row gives the matrix
We develop the determinant by the first row:
and the last row
We therefore see that the Kalman observability matrix is injective and therefore we obtain observability with only these two measurements.
15.4 Case study: a suspension system (optimal fixed structure)*
We consider the fixed structure suspension system. One way of choosing the constants
where
where
with
Defining
Setting the partial derivative with respect to
Solving this gives
Substituting this back into the equation for
The minimum value of
We give the impulse responses in Figure 15.1a and the Bode magnitude diagrams in Figure 15.1b. Comparing these to Figures 7.3 and 6.7, we see that the output is far less oscillatory and the peaks in the Bode diagram are far less pronounced.