Analysis of Discrete Time Control System

Jury stability test (Ogata form)

For $P(z) = a_0z^n + a_1z^{n-1} + \dots + a_n$ with $a_0 > 0$, all roots lie inside the unit circle if and only if:

1. $|a_n| < a_0$
2. $P(1) > 0$
3. $(-1)^nP(-1) > 0$
4. $|b_{n-1}| > |b_0|$, $|c_{n-2}| > |c_0|$, and so on, for the rows of the Jury table

The table entries are

$$b_k = \begin{vmatrix} a_n & a_{n-1-k} \\ a_0 & a_{k+1}\end{vmatrix}, \qquad c_k = \begin{vmatrix} b_{n-1} & b_{n-2-k} \\ b_0 & b_{k+1}\end{vmatrix}$$

Coefficients

$P(z) = 2z^4 + 7z^3 + 10z^2 + 4z + 1$, so $n = 4$, $a_0 = 2$, $a_1 = 7$, $a_2 = 10$, $a_3 = 4$, $a_4 = 1$.

Conditions 1 to 3

• $|a_4| = 1 < a_0 = 2$: satisfied
• $P(1) = 2 + 7 + 10 + 4 + 1 = 24 > 0$: satisfied
• $(-1)^4P(-1) = 2 - 7 + 10 - 4 + 1 = 2 > 0$: satisfied

Jury table

$$\begin{aligned} b_0 &= a_4a_1 - a_0a_3 = 1(7) - 2(4) = -1 \\ b_1 &= a_4a_2 - a_0a_2 = 1(10) - 2(10) = -10 \\ b_2 &= a_4a_3 - a_0a_1 = 1(4) - 2(7) = -10 \\ b_3 &= a_4a_4 - a_0a_0 = 1 - 4 = -3 \end{aligned}$$ $$\begin{aligned} c_0 &= b_3b_1 - b_0b_2 = (-3)(-10) - (-1)(-10) = 20 \\ c_1 &= b_3b_2 - b_0b_1 = (-3)(-10) - (-1)(-10) = 20 \\ c_2 &= b_3b_3 - b_0b_0 = 9 - 1 = 8 \end{aligned}$$

Condition 4

• $|b_3| = 3 > |b_0| = 1$: satisfied
• $|c_2| = 8 > |c_0| = 20$: not satisfied

Conclusion

One Jury condition fails, so the system is unstable.

Check: the roots of $P(z)$ are $z = -1.526 \pm j1.078$ (magnitude 1.868, outside the unit circle) and $z = -0.224 \pm j0.305$ (magnitude 0.379, inside). Two roots lie outside the unit circle, which confirms the result.

Answer: Unstable, because $|c_2| = 8 < |c_0| = 20$.

Step 1: Characteristic equation

For unity feedback, $1 + G(z) = 0$:

$$(z - 0.3679)(z - 1) + K(0.3679z + 0.2642) = 0$$ $$P(z) = z^2 + (0.3679K - 1.3679)z + (0.3679 + 0.2642K) = 0$$

So $n = 2$, $a_0 = 1$, $a_1 = 0.3679K - 1.3679$, $a_2 = 0.3679 + 0.2642K$.

Step 2: Jury conditions for a second-order system

For $n = 2$ there are only three conditions: $|a_2| < a_0$, $P(1) > 0$, and $P(-1) > 0$. No table rows are needed.

Condition 1: $|a_2| < a_0$

$$|0.3679 + 0.2642K| < 1$$ $$\Rightarrow\; -1 < 0.3679 + 0.2642K < 1 \;\Rightarrow\; -5.178 < K < \frac{1 - 0.3679}{0.2642} = 2.3925$$

Condition 2: $P(1) > 0$

$$P(1) = 1 + 0.3679K - 1.3679 + 0.3679 + 0.2642K = 0.6321K > 0 \;\Rightarrow\; K > 0$$

Condition 3: $(-1)^2P(-1) > 0$

$$P(-1) = 1 - 0.3679K + 1.3679 + 0.3679 + 0.2642K = 2.7358 - 0.1037K > 0 \;\Rightarrow\; K < 26.38$$

Step 3: Combine

The intersection of all three is

$$0 < K < 2.3925$$

Step 4: Check at the critical gain

At $K = 2.3925$: $P(z) = z^2 - 0.4877z + 1 = 0$, which gives $z = 0.244 \pm j0.970$ with $|z| = 1.000$. The poles lie on the unit circle, so the system is marginally stable and oscillates.

This $G(z)$ is the ZOH equivalent of $\dfrac{K}{s(s+1)}$ with $T = 1$ s. The continuous system is stable for every $K > 0$, but sampling limits the stable gain to $K < 2.39$.

Answer: The closed-loop system is stable for $0 < K < 2.3925$.

Jury stability test (Ogata form)

For $P(z) = a_0z^n + a_1z^{n-1} + \dots + a_n$ with $a_0 > 0$, all roots lie inside the unit circle if and only if:

1. $|a_n| < a_0$
2. $P(1) > 0$
3. $(-1)^nP(-1) > 0$
4. $|b_{n-1}| > |b_0|$, $|c_{n-2}| > |c_0|$, and so on, for the rows of the Jury table

The table entries are

$$b_k = \begin{vmatrix} a_n & a_{n-1-k} \\ a_0 & a_{k+1}\end{vmatrix}, \qquad c_k = \begin{vmatrix} b_{n-1} & b_{n-2-k} \\ b_0 & b_{k+1}\end{vmatrix}$$

Coefficients

$P(z) = z^3 - 1.1z^2 - 0.1z + 0.2$, so $n = 3$, $a_0 = 1$, $a_1 = -1.1$, $a_2 = -0.1$, $a_3 = 0.2$.

Conditions 1 to 3

• $|a_3| = 0.2 < a_0 = 1$: satisfied
• $P(1) = 1 - 1.1 - 0.1 + 0.2 = 0$: not greater than 0. The equality means there is a root exactly at $z = 1$.
• $(-1)^3P(-1) = -(-1 - 1.1 + 0.1 + 0.2) = 1.8 > 0$: satisfied

Jury table (for completeness)

$$\begin{aligned} b_0 &= a_3a_1 - a_0a_2 = 0.2(-1.1) - 1(-0.1) = -0.12 \\ b_1 &= a_3a_2 - a_0a_1 = 0.2(-0.1) - 1(-1.1) = 1.08 \\ b_2 &= a_3a_3 - a_0a_0 = 0.04 - 1 = -0.96 \end{aligned}$$

• $|b_2| = 0.96 > |b_0| = 0.12$: satisfied

Conclusion

All conditions hold except $P(1) > 0$, which is exactly zero. So one root lies on the unit circle at $z = 1$ and the others lie inside. The system is critically (marginally) stable, not asymptotically stable.

Check by factorising:

$$P(z) = (z - 1)(z^2 - 0.1z - 0.2) = (z - 1)(z - 0.5)(z + 0.4)$$

The roots are $z = 1$, $0.5$ and $-0.4$, which confirms the result.

Answer: Critically (marginally) stable, with a root at $z = 1$ (because $P(1) = 0$).

The s-plane and z-plane are linked by the sampling relation $z = e^{Ts}$, where $T$ is the sampling period. The inverse relation is $s = \dfrac{1}{T}\ln z$.

Mapping $s \to z$

Put $s = \sigma + j\omega$:

$$z = e^{T(\sigma + j\omega)} = e^{\sigma T}e^{j\omega T} \;\Rightarrow\; |z| = e^{\sigma T}, \quad \angle z = \omega T$$

As $\omega$ goes from $-\omega_s/2$ to $\omega_s/2$ (with $\omega_s = 2\pi/T$), the angle $\omega T$ goes from $-\pi$ to $\pi$, covering the circle once. This band is the primary strip. Each complementary strip above or below it maps onto the same z-plane again. So the mapping is many-to-one, and frequencies differing by $\omega_s$ cannot be told apart (aliasing).

   s-plane (primary strip)       z-plane
   jw
   | ws/2 ---------             .---.
   |   LHP  | RHP   ==>       /  in  \  out
 --+--------+---> sigma      | (LHP)  |
   |        |                 \      /
   | -ws/2 ---------            '---'  |z|=1

Mapping $z \to s$

$$s = \frac{1}{T}\ln z = \frac{1}{T}\left[\ln|z| + j\angle z\right]$$

This gives the principal value in the primary strip.

Example: $T = 0.1$ s and $s = -2 + j10$. Then $z = e^{-0.2}\angle 1\text{ rad} = 0.819\angle57.3^\circ$. Going back: $\sigma = \ln(0.819)/0.1 = -2$ and $\omega = 1/0.1 = 10$ rad/s.

Use

The left half-plane maps to the inside of the unit circle. So a discrete system is stable when all its closed-loop poles lie inside $|z| = 1$. Specifications such as $\zeta$, $\omega_n$ and settling time can also be carried from the s-plane to the z-plane for design.

Use $z = e^{Ts}$ with $s = \sigma + j\omega$. Then $|z| = e^{\sigma T}$ and $\angle z = \omega T$. Since $\omega_s = 2\pi/T$, the frequency $\omega = \pm\omega_s/4$ gives the angle

$$\omega T = \pm\frac{\omega_s}{4}\cdot\frac{2\pi}{\omega_s} = \pm\frac{\pi}{2} = \pm90^\circ$$

Mapping of the corner and edge points

Mapping of the edges

• The line $\sigma = -1$ (a-b-c) maps to an arc of radius $e^{-T}$ from $-90^\circ$ to $+90^\circ$.
• The line $\sigma = -2$ (f-e-d) maps to an arc of radius $e^{-2T}$ from $-90^\circ$ to $+90^\circ$.
• The lines $\omega = \pm\omega_s/4$ (c-d and f-a) map to the radial segments along the positive and negative imaginary axes, between the two radii.

Mapped region

The shaded rectangle maps to the right half of an annulus (a half-ring) lying between the circles $|z| = e^{-2T}$ and $|z| = e^{-T}$, for $-90^\circ \le \angle z \le 90^\circ$. It lies entirely inside the unit circle.

          Im z
           |
           c        c: r=e^-T,  +90 deg
           |  .
           d    .   d: r=e^-2T, +90 deg
           | .   .
  ---------+--e---b------- Re z
           | .   .  e: e^-2T  b: e^-T
           f    .
           |  .     f: r=e^-2T, -90 deg
           a        a: r=e^-T,  -90 deg
           |
  Shaded: right half-ring, e^-2T <= |z| <= e^-T

Example with $T = 1$ s: the inner radius is $e^{-2} = 0.135$ and the outer radius is $e^{-1} = 0.368$. So b = 0.368, e = 0.135, c = $j0.368$, d = $j0.135$, a = $-j0.368$, f = $-j0.135$.

Jury stability test (Ogata form)

For $P(z) = a_0z^n + a_1z^{n-1} + \dots + a_n$ with $a_0 > 0$, all roots lie inside the unit circle if and only if:

1. $|a_n| < a_0$
2. $P(1) > 0$
3. $(-1)^nP(-1) > 0$
4. $|b_{n-1}| > |b_0|$, $|c_{n-2}| > |c_0|$, and so on, for the rows of the Jury table

The table entries are

$$b_k = \begin{vmatrix} a_n & a_{n-1-k} \\ a_0 & a_{k+1}\end{vmatrix}, \qquad c_k = \begin{vmatrix} b_{n-1} & b_{n-2-k} \\ b_0 & b_{k+1}\end{vmatrix}$$

Coefficients

$P(z) = z^4 - 1.2z^3 + 0.07z^2 + 0.3z - 0.08$, so $n = 4$, $a_0 = 1$, $a_1 = -1.2$, $a_2 = 0.07$, $a_3 = 0.3$, $a_4 = -0.08$.

Conditions 1 to 3

• $|a_4| = 0.08 < a_0 = 1$: satisfied
• $P(1) = 1 - 1.2 + 0.07 + 0.3 - 0.08 = 0.09 > 0$: satisfied
• $(-1)^4P(-1) = 1 + 1.2 + 0.07 - 0.3 - 0.08 = 1.89 > 0$: satisfied

Jury table

$$\begin{aligned} b_0 &= a_4a_1 - a_0a_3 = (-0.08)(-1.2) - 0.3 = -0.2040 \\ b_1 &= a_4a_2 - a_0a_2 = (-0.08)(0.07) - 0.07 = -0.0756 \\ b_2 &= a_4a_3 - a_0a_1 = (-0.08)(0.3) + 1.2 = 1.1760 \\ b_3 &= a_4^2 - a_0^2 = 0.0064 - 1 = -0.9936 \end{aligned}$$ $$\begin{aligned} c_0 &= b_3b_1 - b_0b_2 = (-0.9936)(-0.0756) - (-0.2040)(1.1760) = 0.3150 \\ c_1 &= b_3b_2 - b_0b_1 = (-0.9936)(1.1760) - (-0.2040)(-0.0756) = -1.1839 \\ c_2 &= b_3^2 - b_0^2 = 0.9872 - 0.0416 = 0.9456 \end{aligned}$$

Condition 4

• $|b_3| = 0.9936 > |b_0| = 0.2040$: satisfied
• $|c_2| = 0.9456 > |c_0| = 0.3150$: satisfied

Conclusion

All the Jury conditions are satisfied, so all the roots lie inside the unit circle and the system is stable.

Check by factorising: $P(z) = (z - 0.8)(z - 0.5)(z + 0.5)(z - 0.4)$. The roots are $0.8$, $0.5$, $-0.5$ and $0.4$, all with $|z| < 1$.

Answer: Stable. All conditions hold, with $|b_3| = 0.9936 > 0.2040$ and $|c_2| = 0.9456 > 0.3150$.

The controller and the ZOH + plant are in cascade with only one sampler in front, so $C(z)/R(z) = G_D(z)G(z)/[1+G_D(z)G(z)]$.

Step 1: Pulse transfer function of controller

Digital PID controller. Using the positional PID form (Ogata), with $K_P = 1$, $K_I = 0.2$, $K_D = 0.2$:

$$\begin{aligned} G_D(z) &= K_P + \frac{K_I}{1-z^{-1}} + K_D(1-z^{-1}) \\ &= \frac{(K_P+K_I+K_D) - (K_P+2K_D)z^{-1} + K_D z^{-2}}{1-z^{-1}} \\ &= \frac{1.4 - 1.4z^{-1} + 0.2z^{-2}}{1-z^{-1}} = \frac{1.4z^2 - 1.4z + 0.2}{z(z-1)} \end{aligned}$$

Step 2: Pulse transfer function of ZOH + plant

$$\begin{aligned} G(z) &= \mathcal{Z}\left[\frac{1-e^{-Ts}}{s}\cdot\frac{1}{s(s+1)}\right] = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] \\ \frac{1}{s^2(s+1)} &= \frac{1}{s^2} - \frac{1}{s} + \frac{1}{s+1} \\ G(z) &= (1-z^{-1})\left[\frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z-e^{-T}}\right] \end{aligned}$$

With $T = 1$ s and $e^{-1} = 0.3679$:

$$G(z) = \frac{0.3679z + 0.2642}{(z-1)(z-0.3679)}$$

Step 3: Open-loop pulse transfer function

$$\begin{aligned} G_D(z)G(z) &= \frac{(1.4z^2-1.4z+0.2)(0.3679z+0.2642)}{z(z-1)^2(z-0.3679)} \\ &= \frac{0.5151z^3 - 0.1451z^2 - 0.2964z + 0.0528}{z^4 - 2.3679z^3 + 1.7358z^2 - 0.3679z} \end{aligned}$$

Step 4: Closed-loop pulse transfer function

Adding numerator and denominator gives the characteristic polynomial:

$$\frac{C(z)}{R(z)} = \frac{G_DG}{1+G_DG} = \frac{0.5151z^3 - 0.1451z^2 - 0.2964z + 0.0528}{z^4 - 1.8528z^3 + 1.5907z^2 - 0.6642z + 0.0528}$$

or, in powers of $z^{-1}$:

$$\frac{C(z)}{R(z)} = \frac{0.5151z^{-1} - 0.1451z^{-2} - 0.2964z^{-3} + 0.0528z^{-4}}{1 - 1.8528z^{-1} + 1.5907z^{-2} - 0.6642z^{-3} + 0.0528z^{-4}}$$

Answer: $\dfrac{C(z)}{R(z)} = \dfrac{0.5151z^3 - 0.1451z^2 - 0.2964z + 0.0528}{z^4 - 1.8528z^3 + 1.5907z^2 - 0.6642z + 0.0528}$

Check: At $z = 1$ the numerator is $0.1264$ and the denominator is $0.1265$, so the ratio is about 1. Because the controller has integral action, the steady-state error for a step input is zero, as expected.

There is one sampler in front of the controller, so the closed-loop pulse transfer function is $C(z)/R(z) = G_D(z)G(z)/[1+G_D(z)G(z)]$, where $G(z)$ is the z-transform of ZOH and plant together.

Step 1: Digital PID controller

Digital PID controller. Using the positional PID form (Ogata), with $K_P = 1$, $K_I = 0.2$, $K_D = 0.2$:

$$\begin{aligned} G_D(z) &= K_P + \frac{K_I}{1-z^{-1}} + K_D(1-z^{-1}) \\ &= \frac{(K_P+K_I+K_D) - (K_P+2K_D)z^{-1} + K_D z^{-2}}{1-z^{-1}} \\ &= \frac{1.4 - 1.4z^{-1} + 0.2z^{-2}}{1-z^{-1}} = \frac{1.4z^2 - 1.4z + 0.2}{z(z-1)} \end{aligned}$$

Here $K_P = K - K_I/2$, $K_I = KT/T_i$ and $K_D = KT_d/T$ are the usual digital PID gains. Trapezoidal integration and a backward difference for the derivative give this form.

Step 2: ZOH + plant $1/(s+1)$

$$\begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+1)}\right] = (1-z^{-1})\left[\frac{z}{z-1} - \frac{z}{z-e^{-T}}\right] \\ &= \frac{1-e^{-T}}{z-e^{-T}} = \frac{0.6321}{z-0.3679} \quad (T = 1) \end{aligned}$$

Step 3: Open-loop pulse transfer function

$$\begin{aligned} G_D(z)G(z) &= \frac{0.6321(1.4z^2-1.4z+0.2)}{z(z-1)(z-0.3679)} \\ &= \frac{0.8850z^2 - 0.8850z + 0.1264}{z^3 - 1.3679z^2 + 0.3679z} \end{aligned}$$

Step 4: Closed-loop pulse transfer function

$$\begin{aligned} \frac{C(z)}{R(z)} &= \frac{0.8850z^2 - 0.8850z + 0.1264}{(z^3 - 1.3679z^2 + 0.3679z) + (0.8850z^2 - 0.8850z + 0.1264)} \\ &= \frac{0.8850z^2 - 0.8850z + 0.1264}{z^3 - 0.4829z^2 - 0.5171z + 0.1264} \end{aligned}$$

In powers of $z^{-1}$:

$$\frac{C(z)}{R(z)} = \frac{0.8850z^{-1} - 0.8850z^{-2} + 0.1264z^{-3}}{1 - 0.4829z^{-1} - 0.5171z^{-2} + 0.1264z^{-3}}$$

Answer: $\dfrac{C(z)}{R(z)} = \dfrac{0.8850z^2 - 0.8850z + 0.1264}{z^3 - 0.4829z^2 - 0.5171z + 0.1264}$

Check: At $z = 1$ both numerator and denominator equal $0.1264$. The DC gain is 1, so there is zero steady-state error to a step input because of the integral term.

A digital PID controller is the discrete-time version of the analog PID law. We obtain it by replacing the integral with a numerical sum (trapezoidal rule) and the derivative with a backward difference. Its pulse transfer function is $G_D(z) = U(z)/E(z)$.

Analog PID law

$$u(t) = K\left[e(t) + \frac{1}{T_i}\int_0^t e(t)\,dt + T_d\frac{de(t)}{dt}\right]$$

Here $K$ is the proportional gain, $T_i$ the integral time and $T_d$ the derivative time.

Discretization (sampling period $T$)

• Integral, trapezoidal rule: $\displaystyle\int_0^{kT} e\,dt \approx T\sum_{h=1}^{k}\frac{e((h-1)T)+e(hT)}{2}$
• Derivative, backward difference: $\displaystyle\frac{de}{dt}\Big|_{kT} \approx \frac{e(kT)-e((k-1)T)}{T}$

$$u(kT) = K\left\{e(kT) + \frac{T}{T_i}\sum_{h=1}^{k}\frac{e((h-1)T)+e(hT)}{2} + \frac{T_d}{T}\big[e(kT)-e((k-1)T)\big]\right\}$$

Taking the z-transform

Let $f(h) = \tfrac12[e((h-1)T)+e(hT)]$, so $F(z) = \tfrac12(1+z^{-1})E(z)$. A running sum transforms as $\mathcal{Z}\left[\sum_{h=0}^{k} f(h)\right] = F(z)/(1-z^{-1})$, with $f(0) = 0$ because $e$ is zero for negative time. Therefore:

$$\begin{aligned} \mathcal{Z}\left[\sum_{h=1}^{k}\frac{e(h-1)+e(h)}{2}\right] &= \frac{1+z^{-1}}{2(1-z^{-1})}E(z) \\ \mathcal{Z}\big[e(k)-e(k-1)\big] &= (1-z^{-1})E(z) \end{aligned}$$ $$U(z) = K\left[1 + \frac{T}{2T_i}\cdot\frac{1+z^{-1}}{1-z^{-1}} + \frac{T_d}{T}(1-z^{-1})\right]E(z)$$

Rearranging into the standard form

Use $\dfrac{1+z^{-1}}{2(1-z^{-1})} = \dfrac{1}{1-z^{-1}} - \dfrac{1}{2}$:

$$\begin{aligned} U(z) &= \left[K - \frac{KT}{2T_i} + \frac{KT}{T_i}\cdot\frac{1}{1-z^{-1}} + \frac{KT_d}{T}(1-z^{-1})\right]E(z) \\ G_D(z) &= \frac{U(z)}{E(z)} = K_P + \frac{K_I}{1-z^{-1}} + K_D(1-z^{-1}) \end{aligned}$$

where

Single-fraction form

$$G_D(z) = \frac{(K_P+K_I+K_D) - (K_P+2K_D)z^{-1} + K_D z^{-2}}{1-z^{-1}}$$

Remarks

• The pole at $z = 1$ comes from the integral action. It makes the steady-state error to a step input zero.
• $K_D(1-z^{-1})$ adds a zero and gives phase lead (anticipation).
• Setting $K_D = 0$ gives the PI controller $K_P + K_I/(1-z^{-1})$. Setting $K_I = 0$ gives the PD controller $K_P + K_D(1-z^{-1})$.

Example: For $K_P = 1$, $K_I = 0.2$, $K_D = 0.2$: $G_D(z) = (1.4 - 1.4z^{-1} + 0.2z^{-2})/(1-z^{-1})$.

The answer depends on whether a sampler separates $G_1$ and $G_2$. With a sampler between them, the two pulse transfer functions multiply. Without one, the two s-domain transfer functions must be multiplied first and then z-transformed.

(a) Sampler between $G_1$ and $G_2$

X(s)  _/  X*  +----+ U(s)  _/  U*  +----+ Y(s)  _/  Y*
 ---o/ o---->| G1 |------o/ o---->| G2 |------o/ o--->
      T      +----+        T      +----+        T

• $U(s) = G_1(s)X^*(s)$, so $U^*(s) = G_1^*(s)X^*(s)$, i.e. $U(z) = G_1(z)X(z)$.
• $Y(s) = G_2(s)U^*(s)$, so $Y(z) = G_2(z)U(z)$.

$$\frac{Y(z)}{X(z)} = G_1(z)\,G_2(z), \qquad G_1(z) = \mathcal{Z}[G_1(s)],\; G_2(z) = \mathcal{Z}[G_2(s)]$$

(b) No sampler between $G_1$ and $G_2$

X(s)  _/  X*  +----+ U(s)  +----+ Y(s)  _/  Y*
 ---o/ o---->| G1 |------->| G2 |------o/ o--->
      T      +----+        +----+        T

• $Y(s) = G_1(s)G_2(s)X^*(s)$, so $Y^*(s) = [G_1G_2]^*(s)\,X^*(s)$.

$$\frac{Y(z)}{X(z)} = G_1G_2(z) = \mathcal{Z}\big[G_1(s)G_2(s)\big]$$

In general, $G_1G_2(z) \neq G_1(z)G_2(z)$.

Example

Take $G_1(s) = \dfrac{1}{s}$ and $G_2(s) = \dfrac{a}{s+a}$.

Case (a):

$$\frac{Y(z)}{X(z)} = \frac{z}{z-1}\cdot\frac{az}{z-e^{-aT}} = \frac{az^2}{(z-1)(z-e^{-aT})}$$

Case (b):

$$\begin{aligned} \frac{a}{s(s+a)} &= \frac{1}{s} - \frac{1}{s+a} \\ \frac{Y(z)}{X(z)} &= \frac{z}{z-1} - \frac{z}{z-e^{-aT}} = \frac{(1-e^{-aT})z}{(z-1)(z-e^{-aT})} \end{aligned}$$

The two results are different. You cannot move or remove a sampler without changing the system.

Absolute stability asks only whether the system is stable or unstable. A closed-loop discrete-time system with pulse transfer function $C(z)/R(z) = G(z)/[1+GH(z)]$ is stable if all roots of the characteristic equation

$$P(z) = 1 + GH(z) = 0$$

lie inside the unit circle $|z| < 1$. A simple pole on $|z| = 1$ makes the system critically stable. Any pole outside the circle, or a repeated pole on it, makes the system unstable.

Factoring a high-order $P(z)$ is tedious, so two direct tests are used:

1. Jury stability test, applied directly in the z-plane.
2. Bilinear transformation + Routh–Hurwitz: substitute $z = (w+1)/(w-1)$. This maps the inside of the unit circle onto the left half of the w-plane, so the ordinary Routh test can be used.

Jury stability test

Write $P(z) = a_0z^n + a_1z^{n-1} + \dots + a_n$ with $a_0 > 0$. The system is stable if and only if all of these hold:

1. $|a_n| < a_0$
2. $P(1) > 0$
3. $(-1)^nP(-1) > 0$
4. In the Jury table, $|b_{n-1}| > |b_0|$, $|c_{n-2}| > |c_0|$, … down to the row with three elements.

The rows of the Jury table are built as:

$$b_k = \begin{vmatrix} a_n & a_{n-1-k} \\ a_0 & a_{k+1} \end{vmatrix},\qquad c_k = \begin{vmatrix} b_{n-1} & b_{n-2-k} \\ b_0 & b_{k+1} \end{vmatrix}$$

Example 1

Test $P(z) = z^3 - 1.3z^2 - 0.08z + 0.24 = 0$. Here $a_0 = 1$, $a_1 = -1.3$, $a_2 = -0.08$, $a_3 = 0.24$.

The condition $P(1) > 0$ fails, so the system is unstable. Factoring confirms this: the roots are $z = 1.2, 0.5, -0.4$, and $z = 1.2$ lies outside the unit circle.

Example 2: stable range of gain

Take $G(z) = \dfrac{K(0.3679z + 0.2642)}{(z-1)(z-0.3679)}$ with unity feedback. The characteristic equation is:

$$z^2 + (0.3679K - 1.3679)z + (0.3679 + 0.2642K) = 0$$

• $|a_2| < 1$: $0.3679 + 0.2642K < 1$, so $K < 2.392$
• $P(1) = 0.6321K > 0$, so $K > 0$
• $P(-1) = 2.7358 - 0.1037K > 0$, so $K < 26.38$

Answer: The system is stable for $0 < K < 2.392$. At $K = 2.392$ it is critically stable and oscillates.