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absolute stability for quantized system -凯发k8网页登录

this example uses:

this example shows how to enforce absolute stability when a linear time-invariant system is in feedback interconnection with a static nonlinearity that belongs to a conic sector.

feedback connection

consider the feedback connection as shown in figure 1.

figure 1: feedback connection

$g$ is a linear time invariant system, and $n(y)$ is a static nonlinearity that belongs to a conic sector $[\alpha,\beta]$ (where $\alpha<\beta$); that is,

$$ \alpha y^2<~yn(y)<~\beta y^2$$

for this example, $g$ is the following discrete-time system.

a = [0.9995, 0.0100, 0.0001;
    -0.0020, 0.9995, 0.0106;
          0,      0, 0.9978];
b = [0, 0.002, 0.04]';
c = [2.3948, 0.3303, 2.2726];
d = 0;
g = ss(a,b,c,d,0.01);

sector bounded nonlinearity

in this example, the nonlinearity $n(y)$ is the logarithmic quantizer, which is defined as follows:

$$n(y) = \left\{ \begin{array}{ll} \rho^j, & \mbox{if~ $ \frac{1 \rho}{2}\rho^j < y \leq \frac{1 \rho}{2\rho}\rho^j$};\\0, & \mbox{if ~$y = 0$}; \\ -n(-y), & \mbox{if~ $y < 0$} \end{array} \right. $$

where, $j\in \{0,\pm1,\pm2,\dots \}$. this quantizer belongs to a sector bound $[\frac{2\rho}{1 \rho},\frac{2}{1 \rho}]$. for example, if $\rho = 0.1$, then the quantizer belongs to the conic sector [0.1818,1.8182].

% quantizer parameter
rho = 0.1;
% lower bound
alpha = 2*rho/(1 rho)
% upper bound
beta = 2/(1 rho)

alpha =
    0.1818
beta =
    1.8182

plot the sector bounds for the quantizer.

plotsectorbound(rho)

$\rho$ represents the quantization density, where $0<\rho<1$. if $\rho$ is larger, then the quantized value is more accurate. for more details about this quantizer, see [1].

conic sector condition for absolute stability

the conic sector matrix for the quantizer is given by

$$ q = \left(\begin{array}{cc} 1 & -\frac{\alpha \beta}{2} \\ -\frac{\alpha \beta}{2} & \alpha\beta \end{array}\right). $$

to guarantee stability of the feedback connection in figure 1, the linear system $g$ needs to satisfy

$$\int_0^t \left(\begin{array}{c} u(t)\\-y(t)\end {array} \right)^t q \left(\begin{array}{c} u(t)\\-y(t)\end {array} \right) > 0$$

where, $u$ and $y$ are the input and output of $g$, respectively.

this condition can be verified by checking if the sector index, $r$, is less than 1.

define the conic sector matrix for a quantizer with $\rho = 0.1$.

q = [1,-(alpha beta)/2;-(alpha beta)/2,alpha*beta];

get the sector index for q and g.

r = getsectorindex([1;-g],-q)

r =
    1.8247

since $r>1$, the closed-loop system is not stable. to see this instability, use the following simulink model.

mdl = 'dtquantization';
open_system(mdl)

run the simulink model.

sim(mdl)
open_system('dtquantization/output')

from the output trajectory, it can be seen that the closed-loop system is not stable. this is because the quantizer with $\rho = 0.1$ is too coarse.

increase the quantization density by letting $\rho = 0.25$. the quantizer belongs to the conic sector [0.4,1.6].

% quantizer parameter
rho = 0.25;
% lower bound
alpha = 2*rho/(1 rho)
% upper bound
beta = 2/(1 rho)

alpha =
    0.4000
beta =
    1.6000

plot the sector bounds for the quantizer.

plotsectorbound(rho)

define the conic sector matrix for a quantizer with $\rho = 0.25$.

q = [1,-(alpha beta)/2;-(alpha beta)/2,alpha*beta];

get the sector index for q and g.

r = getsectorindex([1;-g],-q)

r =
    0.9702

the quantizer with $\rho = 0.25$ satisfies the conic sector condition for stability of the feedback connection since $r<1$.

run the simulink model with $\rho = 0.25$.

sim(mdl)
open_system('dtquantization/output')

as indicated by the sector index, the closed-loop system is stable.

reference

[1] m. fu and l. xie,"the sector bound approach to quantized feedback control," ieee transactions on automatic control 50(11), 2005, 1698-1711.

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