# 7.4: Stability of Sampled-Data Systems

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### Stability Region in the Complex Plane

In continuous-time control system design, $$s=j\omega$$ defines the stability boundary.

The $$z$$-plane stability boundary is obtained from the transform: $$z=e^{j\omega T} =1\angle \omega T$$; it maps the $$j\omega$$-axis to the unit circle and the open left-half plane to the inside of the unit circle: $$|z|<1$$.

Thus, the sampled-data system is stable if and only if the pulse characteristic polynomial $$\Delta (z)$$ has its roots inside the unit circle, i.e., $$|z_\rm i |<1.$$where $$z_\rm i$$ is the root of $$\Delta (z)$$.

The analytical conditions for a $$z$$-domain polynomial, $$A\left(z\right)$$, to have its roots inside the unit circle are given by the Schur–Cohn stability test. When applied to real polynomials, the Schur–Cohn test results in a criterion similar to the Ruth’s test, and is known as Jury’s stability test.

### Jury’s Stability Test

Assume that the $$n$$th order $$z$$-domain polynomial to be investigated is given as:

$A(z)=a_{0} z^{n} +a_{1} z^{n-1} +\ldots +a_{n-1} z+a_{n} ;a_{0} >0. \nonumber$

For stability determination, the Jury’s table is built as follows:

$\left|\begin{array}{c} {\begin{array}{ccc} {a_{n} } & {a_{n-1} } & {\begin{array}{cc} {\ldots } & {a_{0} } \end{array}} \\ {a_{0} } & {a_{1} } & {\begin{array}{cc} {\ldots } & {a_{n} } \end{array}} \end{array}} \\ {\begin{array}{ccc} {b_{n-1} } & {b_{n-2} } & {\begin{array}{cc} {\ldots } & {b_{0} } \end{array}} \\ {b_{0} } & {b_{1} } & {\begin{array}{cc} {\ldots } & {b_{n-1} } \end{array}} \end{array}} \\ {\begin{array}{ccc} {c_{n-2} } & {c_{n-1} } & {\begin{array}{cc} {\ldots } & {c_{0} } \end{array}} \\ {c_{0} } & {c_{1} } & {\begin{array}{cc} {\ldots } & {c_{n-2} } \end{array}} \end{array}} \\ {\vdots } \end{array}\right. \nonumber$

where the first two rows reflect the coefficients of the polynomial; the coefficients of the third and subsequent rows are computed as:

$b_{k} =\left|\begin{array}{cc} {a_{n} } & {a_{n-1-k} } \\ {a_{0} } & {a_{k+1} } \end{array}\right|,\; \; k=0,\ldots ,n-1 \nonumber$

$c_{k} =\left|\begin{array}{cc} {b_{n-1} } & {b_{n-2-k} } \\ {b_{0} } & {b_{k+1} } \end{array}\right|,\; \; k=0,\ldots ,n-2 \nonumber$

and so on. The necessary conditions for polynomial stability are:

$A(1)>0,(-1)^{n} A(-1)>0. \nonumber$

The sufficient conditions for stability, given by the Jury’s test, are:

$a_{0} >|a_{n} |,\; |b_{n-1} |>|b_{0} |,\; |c_{n-2} |>|c_{0} |,\ldots \rm (n-1) constraints. \nonumber$

#### Second-Order Polynomial

Let $$A(z)=z^{2} +a_{1} z+a_{2}$$; then, the Jury’s table is given as:

$\left|\begin{array}{c} {\begin{array}{ccc} {a_{2} } & {a_{1} } & {1} \\ {1} & {a_{1} } & {a_{2} } \end{array}} \\ {\begin{array}{ccc} {b_{1} } & {b_{0} } & {} \\ {b_{0} } & {b_{1} } & {} \end{array}} \end{array}\right. \nonumber$

The resulting necessary conditions are: $$1+a_{1} +a_{2} >0,\; \; 1-a_{1} +a_{2} >0.$$

The sufficient conditions are: $$|a_{2} |<1,|1+a_{2} |>|a_{1} |$$.

### Stability Determination through Bilinear Transform

The bilinear transform (BLT) defines a linear map between $$s$$-domain and $$z$$-domain.

The BLT is based on the first-order Pade’ approximation of $$z=e^{sT}$$, given as:

$z=\frac{e^{sT/2} }{e^{-sT/2} } \cong \frac{1+sT/2}{1-sT/2} ,\; \; s=\frac{2}{T} \frac{z-1}{z+1} \nonumber$

Since $$T$$ has no impact on stability determination, we may use $$T=2$$ for simplicity. Further, to differentiate from continuous-time systems, a new complex variable, $$w$$, is introduced. Thus $$z=\frac{1+w}{1-w} ,\; \; w=\frac{z-1}{z+1}$$

Let $$\Delta (z)$$ represent the polynomial to be investigated; application of BLT to $$\Delta (z)$$ returns a polynomial, $$\Delta (w)$$, whose stability is determined through the application of Hurwitz criterion (Sec. 2.5).

#### Second-Order Polynomial

Let $$\Delta (z)=z^{2} +a_{1} z+a_{2}$$; then, application of BLT, ignoring the denominator term, results in:

$\Delta (w)=\Delta (z)|_{z=\frac{1+w}{1-w} } =(1-a_{1} +a_{2} )w^{2} +2(1-a_{2} )w+1+a_{1} +a_{2} \nonumber$

Application of the Hurwitz criterion results in the following stability conditions for $$\Delta (z)$$:

$a_{2} +a_{1} +1>0, a_{2} -a_{1} +1>0, 1-a_{2} >0 \nonumber$

The above conditions are similar those obtained from the application of Jury’s stability test.

### Stability of the Closed-Loop System

Let the pulse transfer function be given as: $$G\left(z\right)=\frac{n\left(z\right)}{d\left(z\right)}$$; then, for a static controller, the closed-loop pulse characteristic polynomial is given as: $$\mathit{\Delta}\left(z\right)=d\left(z\right)+Kn(z)$$.

The stability of the closed-loop characteristic polynomial can be determined by applying Jury’s stability test. Alternatively, BLT can be used to determine stability through the application of Routh’s test to the transformed polynomial, $$\mathit{\Delta}(w)$$.

##### Example $$\PageIndex{1}$$

Let $$G(s)=\frac{1}{s+1}$$, $$T=0.2\rm s$$; then, the pulse transfer function is given as: $$G(z)=\frac{0.181}{z-0.819}$$.

The closed-loop characteristic polynomial is: $$\mathit{\Delta}\left(z\right)=z-0.819+0.181K$$.

The polynomial is stable for $$\left|0.819+0.181K\right|<1$$, or $$-1<K<10$$ for stability.

##### Example $$\PageIndex{2}$$

Let $$\Delta (z)=z^{2} +z+K$$; then, $$\Delta (w)=\Delta (z)|_{z=\frac{1+w}{1-w} } =Kw^{2} +2(1-K)w+1+K$$.

The application of the Hurwitz criteria to $$\Delta (w)$$ reveals $$0<K<1$$ for stability.

##### Example $$\PageIndex{3}$$

Let $$G(s)=\frac{1}{s(s+1)} ,\; \; T=0.2s$$; then, the pulse transfer function is given as: $$G(z)=\frac{0.0187z+0.0175}{(z-1)(z-0.181)}$$.

The characteristic polynomial is:

$\Delta (z)=z^{2} +(0.0187K-1.819)z+0.0175K+0.819. \nonumber$

The $$w$$-polynomial obtained through BLT is given as:

$\Delta (w)=(3.637-0.0012K)w^{2} +(0.363-0.035K)w+0.036K \nonumber$

The application of the Hurwitz criteria to $$\Delta (w)$$ gives $$0<K<10.34$$ for stability.

This page titled 7.4: Stability of Sampled-Data Systems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Kamran Iqbal.