3.2: System Stability and Its Performance Measures

Learning Objectives

After completing this section, you will be able to:

1. Provide definitions of bounded-input/bounded-output and asymptotical stability.
2. Use Routh-Hurwitz criterion to determine system stability.
3. Describe step response performance criteria, including settling time, percentage overshoot, and rise time.

System stability is essential for mechatronic systems. It determines whether the system response will be bounded or not, which is the basic requirement for a mechatronic system. In this section, we will introduce two stability criteria: bounded-input and bounded-output (BIBO) stability and asymptotically stability. In addition, we will provide the condition making these two stability criteria equivalent. We will start with BIBO stability.

Definition 3.1

A dynamic system is bounded-input and bounded-output (BIBO) stable if for any bounded control input \({\mathop{\mathrm{max}}_{t\ge 0} \left|u(t)\right|<γ\ }\), where \(γ>0\) is a positive number, the generated system output \(y(t)\) is also bounded such that \({\mathop{\mathrm{max}}_{t\ge 0} \left|y(t)\right|<μ\ }\) with zero initial condition, where \(μ>0\) is a positive number.

Figure 3.3 illustrates BIBO stability

Definition 3.2

A dynamic system is asymptotically stable if for any given system initial condition with zero control input, that is, \(u\left(t\right)=0\)), the system output \(y(t)\) converges to zero as time goes to infinity.

Figure 3.4 demonstrates asymptotical stability.

Remark 3.2 [1]

For linear time-invariant (LTI) systems defined by a transfer function \(G(s)\), the BIBO stability is equivalent to the asymptotical stability.

Now, consider an LTI system represented by the transfer function defined below.

\[G\left(s\right)=\dfrac{m(s)}{d(s)} \label{3.22}\]

where both \(m\left(s\right)\) and \(d(s)\) are polynomials with order of \(d(s)\) greater than or equal to that of \(m\left(s\right).\) That is, \(G(s)\) is proper. Let \(s_i\) (\(i=1,\ 2,\ \dots ,\ n\)) be roots of the denominator’s polynomial \(d(s)\) defined in equation \ref{3.22}, also called system poles of \(G(s)\); here \(n\) is the polynomial order of \(d(s)\). Then \(G(s)\) is BIBO and asymptotically stable if \(Re\left(s_i\right)<0\) for \(i=1,\ 2,\ \dots ,\ n\). Theorem 3.3 defines whether system \(G(s)\) is asymptotically (or BIBO) stable, marginally stable, or unstable.

Theorem 3.3 [1]

For a linear time-invariant (LTI) system defined by a transfer function \(G\left(s\right)=m(s)/d(s)\), let \(s_i\) (\(i=1,\ 2,\ \dots ,\ n\)) be roots of the denominator’s polynomial \(d(s)\), where \(n\) is the order of \(d(s)\). Then \(G(s)\) is asymptotically (or BIBO) stable if \(Re\left(s_i\right)<0\) for \(i=1,\ 2,\ \dots ,\ n\); marginally stable if \(Re\left(s_i\right)\le 0\) for \(i=1,\ 2,\ \dots ,\ n\) with simple roots (no repeated ones) for these with \(Re\left(s_i\right)=0\); and unstable if it is neither stable nor marginally stable.

Figure 3.5 shows the system stable and unstable region of pole locations on the complex plane (\(s\)-plane). Here, if all system poles are located at the open left half plane (OLHP), the system is asymptotically (or BIBO) stable. If all system poles are located at the closed left half plane (CLHP) without any repeated poles on the on the imaginary (\(jω\)) axis, the system is marginally stable. Finally, if there is any system pole located at the open right half plane (ORHP) or repeated pole on the imaginary (\(jω\)) axis, the system is unstable. In other words, if the system is neither stable nor marginally stable, the system is unstable.

Example 3.5

Show if system transfer functions listed below are stable, marginally stable, or unstable.

\[G_1\left(s\right)=\dfrac{1}{s^2+1}, G_2\left(s\right)=\dfrac{1}{s^2+s+1}, \ G_3\left(s\right)=\dfrac{1}{{\left(s^2+1\right)}^2} \label{3.23}\]

For \(G_1\left(s\right)\), there are two poles, \(s_{1,2}=\pm j\). Because there is a pair of (nonrepeating) poles on the imaginary axis, system \(G_1\left(s\right)\) is marginally stable. For \(G_2\left(s\right)\), there is a pair of poles located at the open left half plane (\(s_{1,2}=-0.5\pm j\sqrt{3}\)), and system \(G_2\left(s\right)\) is asymptotically stable. For \(G_3\left(s\right)\), there are two pairs of repeated poles (\(s_{1,2}=\pm j\)) on the imaginary axis, and system \(G_3\left(s\right)\) is unstable.

To investigate the stability of a mechatronic system described by a rational proper transfer function, the system poles of \(G(s)\) can be calculated manually or using a MATLAB command [16], especially for a high-order system. Manually calculating system poles for high-order systems is challenging. Decades ago, the Routh-Hurwitz criterion [1] was developed to determine whether the roots of a polynomial are in the open left-half plane without calculating such roots. Even though nowadays software such as MATLAB can easily be used to calculate the roots of a polynomial with a fairly high degree, it turns out that the Routh-Hurwitz criterion is still useful in the analysis of feedback control systems. It can be used to determine the range of system parameters with guaranteed system stability. The Routh-Hurwitz criterion is established based on the Routh table defined below. Consider the denominator’s polynomial \(d(s)\) of a transfer function \(G(s)\).

\[d\left(s\right)=a_ns^n+a_{n-1}s^{n-1}+\cdots +a_2s^2+a_1s+a_0 \label{3.24}\]

Assume that \(a_0\neq 0\). Note that if \(a_0=0\), the polynomial can be reorganized into the following equation, with \({\hat{a}}_0\neq 0\).

\[d\left(s\right)=s^m\left({\hat{a}}_{n-m}s^{n-m}+{\hat{a}}_{n-m-1}s^{n-m-1}+\cdots +{\hat{a}}_1s+{\hat{a}}_0\right)=s^m\hat{d}(s) \label{3.25}\]

Here, \(\hat{d}(s)\) can be used for constructing the Routh table. Note that a Routh table can be formed for a given \(d\left(s\right)\) following the rules defined in Figure 3.6, where coefficients \(a_i\) (\(i=0,1,\ 2,\ \dots ,\ n\)) are from the polynomial \(d(s)\) defined in \ref{3.24}.

Theorem 3.4 [1]

The Routh-Hurwitz criterion states that the number of roots in the open right half-plane (ORHP) is equal to the number of sign changes in the first column of the Routh table.

Therefore, based on the Routh-Hurwitz criterion, the dynamic system \(G\left(s\right)\) is stable if there is no sign change in the first column of the Routh table, where coefficients of 0 in the first column of Routh table need to be handled differently.

Example 3.6

Find the range of control gain \(K\) that stabilizes the closed-loop system shown in Figure 3.7.

First, let’s calculate the closed-loop transfer function \(G(s)\) as below.

\[G\left(s\right)=\dfrac{n(s)}{d(s)}=\dfrac{2K}{s^3+4s^2+5s+2(K+1)} \label{3.26}\]

Based upon the denominator’s polynomial \(d\left(s\right)\) of the closed-loop transfer function, we can form the Routh table shown in Figure 3.8. To have no sign change in the first column of the Routh table, control gain \(K\) needs to be greater than \(-1\) and less than \(9\).

Remark 3.3

Even though Figure 3.6 provides ways to calculate coefficients for the Routh table, there are two special cases that need to be handled. First, if \(0\) appears in the first column of a nonzero row in the Routh table, replace it with a small positive number \(ε>0\). Then check for very small \(ε\) to see if there is any sign change. If so, \(d(s)\) has some roots in the right half plan (RHP) (same rule in Theorem 3.4 [1] applies).

Second, if a zero row appears in the Routh table, \(d(s)\) has roots either on the imaginary axis or in RHP. In this case, the zero row in the Routh table should be replaced by coefficients obtained by taking the derivative of the auxiliary polynomial defined by the row above; see Example 3.7. The pole locations on the imaginary or RHP can be calculated by finding roots of the auxiliary polynomial.

Examples 3.7 and 3.8 demonstrate how to deal with these two cases—with zero in the first column of a nonzero row and when a zero row appears in the Routh table.

Example 3.7

Consider the system with the following \(d(s)\).

\[d\left(s\right)=s^5+2s^4+2s^3+4s^2+11s+10 \label{3.27}\]

The Routh table in Figure 3.9 can be formed based on the rules addressed in Figure 3.6 and in Remark 3.3.

For the \(s^3\) row in Figure 3.9, the first entry is \(0\). A small positive number \(ε\) is used to replace the \(0\) and the calculation continues, resulting where the first entry of \(s^2\) row is \(c_1=4-12/ε.\) For a very small \(ε>0\), note that \(c_1=4-12/ε<0\). As a result, there are two sign changes, leading to two poles in the open right half plane. Using MATLAB, we can obtain the following five pole locations

\[s_{1,2}=0.8950\pm 1.4561j, s_{3,4}=-1.2407\pm 1.0375j, \ s_5=-1.3087 \label{3.28}\]

which confirms that there are two poles at ORLP.

Example 3.8

Consider the system with the following \(d(s)\).

\[d\left(s\right)=s^5+s^4+2s^3+2s^2+s+1=(s+1){\left(s^2+1\right)}^2 \label{3.29}\]

The Routh table in Figure 3.10 can be formed based on the rules addressed in Figure 3.6 and in Remark 3.3.

For the zero \(s^3\) row in Figure 3.10, we replace the two zeros by coefficients of derivative auxiliary polynomial in the line above (row of \(s^4)\). The associated auxiliary polynomial is \(s^4+2s^2+1\) with its derivative of \(4s^3+4s\). Therefore, the zero coefficients in \(s^3\) are replaced by \([4\ 4]\) as shown. For the zero \(s^1\) row (even though there is only one zero) in the figure, we replace the zero in row \(s^1\) by coefficients of the derivative auxiliary polynomial in the line above (row of \(s^2)\). The associated auxiliary polynomial is \(s^2+1\) with its derivative of \(2s\). Therefore, the zero coefficients in \(s^1\) are replaced by \(2\) as shown in Figure 3.10. Note that the auxiliary polynomial, \(s^4+2s^2+1={\left(s^2+1\right)}^2\), leads to two repeated pairs of complex poles \(s_{1,2}=s_{3,4}=\pm j\). From \ref{3.29} there are the following five poles for \(d\left(s\right)\)

\[s_{1,2}=s_{3,4}=\pm j, \ s_5=-1.0 \label{3.30}\]

which confirms that there are two pairs of poles at imaginary axis. Hence, this system is unstable.

Remark 3.4

There are a few direct methods to justify whether the denominator’s polynomial of a closed-loop transfer function is stable or not. For the first and second order transfer functions where \(d\left(s\right)=a_1s+a_0\) and d\(\left(s\right)=a_2s^2+a_1s+a_0\), all system poles are located at open left half plane (OLHP)— if and only if all coefficients (\(a_1,\ a_0\) for the first order system and \(a_2,a_{1,}a_0\) for the second order) have the same sign and are not equal to zero, and, as a result, the associated system is asymptotically stable. For the \(d(s)\) defined in equation \ref{3.24}, the necessary condition for the system to be asymptotically stable is that all coefficients (\(a_n,\dots ,a_{1,}a_0)\ \)of \(d(s)\) have the same sign. These conditions can be used to find out if the system is stable or unstable.

There is also a stability measure for nonlinear systems described by the following nonlinear differential state equation; see reference [2].

\[\dot{x}=f(x,u)\ \label{3.31}\]

where \(x\) is the system state vector, \(u\) is the system input vector, and \(f(\cdot ,\cdot )\) is a nonlinear function of state and control. Lyapunov stability criterion can be used in this case; see details in [2].

The rest of this section discusses system performance criteria in both time and frequency domains. First, we will address the performance measures for a stable system excited by a step input. Table 3.2 shows the detailed specification of a step response.

Table 3.2: Performance measure definitions for a time-domain step response.

Performance Measures	Definition
Delay time \(t_d\)	Time to reach 50% of \(y_{ss}\) (steady state value)
Peak time \(t_{max}\)	Time to reach the peak response value
Percentage overshoot (PO)	Ratio of max overshoot \(\dfrac{δy_{max}}{y_{ss}}100\%\) (see Figure 3.11 for \(δy_{max}\))
Rise time \(t_r\)	Time to rise from 10% \(y_{ss}\) to 90% \(y_{ss}\)
Settling time (\(5\%\))	Time to reach 5% error band and remain with 5% error band
Settling time (2\(\%\))	Time to reach 2% error band and remain with 2% error band
Steady state error \({SS}_{err}\)	\({SS}_{err}=1-y_{ss}\)

Both Figure 3.11 and Table 3.2 define seven performance measures for a stable system excited by a unit step input, where the response speed is measured by delay time (\(t_d\)), rise time (\(t_r\)), and settling (\(t_s\)); the system’s relative stability is measured by system percentage overshoot (PO). In general, fast system response leads to large overshoot. A large percentage overshoot provides small stability margin. In the control design, one needs to take a tradeoff to balance response speed and relative stability. That is, it is not practical to have extremely fast response without reducing system stability margin. To have a good stability margin, we need to keep the system percent overshoot (PO) below 15%. In Section 3.3, we will address the relationship between stability margin and closed-loop pole locations.

Next we will discuss the frequency domain stability margin measures in terms of system gain and phase margins. Before defining the system gain and phase margins, consider the concept of loop transfer function for a typical closed-loop system shown in Figure 3.12, where \(G\left(s\right)\) is the system plant model, \(H(s)\) the system sensor dynamics, and \(C(s)\) system closed-loop controller.

Define the system loop transfer function \(L(s)\) as the following.

\[L(s)=C\left(s\right)G\left(s\right)H(s)\ \label{3.32}\]

That is, the system loop transfer function is the product of all transfer functions over the feedback loop.

In this case, the system’s relative stability can be measured by its loop transfer function gain and phase margins. In order to demonstrate and define these margins, consider the following loop transfer function.

\[L\left(s\right)=C\left(s\right)G\left(s\right)H\left(s\right)=\dfrac{10}{s}\cdot \dfrac{50}{s+50}\cdot \dfrac{5}{s+5}=\dfrac{2500}{s^3+55s^2+250s}\ \label{3.33}\]

We can plot the Bode diagram based on the definition provided before Figure 3.2. Here, Figure 3.13 is the Bode diagram of \(L(s)\) defined in equation \ref{3.33}. It is plotted using MATLAB with system gain and phase margins marked on the plot as “Gain Margin” and “Phase Margin” (see Definition 3.3).

Definition 3.3

A dynamic system gain margin and phase margin are defined by its Bode plot of the loop transfer function \(L(s)\). The gain margin is the negative dB value of the magnitude plot at \(ω_p\) (i.e., \(20{log}_{10}\left|L(jω_p)\right|\)), where \(ω_p\) is the frequency when the phase plot crosses the \(-{180}^{{}^\circ }\) line. The phase margin is \({180}^{{}^\circ }\) plus the phase angle value at \(ω_g\) (i.e., \({180}^{{}^\circ }+\angle L(jω_p)\)), where \(ω_g\) is the frequency when the gain plot crosses the 0 dB line.

Figure 3.13 displays the Bode plot for the loop transfer function defined by equation \ref{3.33}. The gain and phase margins are marked on both gain and phase plots. The phase crossover frequency \(ω_p\) is the frequency when the phase plot crosses the \(-{180}^{{}^\circ }\) line whereas the gain crossover frequency \(ω_g\) is the frequency when the gain plot crosses the 0 dB line.

From reference [1], the closed-loop system is stable if both gain and phase margin are greater than zero. That is, as the gain margin (or phase margin) reduces down to zero, the system stability margin also goes to zero. To have a good stability margin for a closed-loop system, the system gain margin should be greater than 6 dB and the phase margin should be greater than \({45}^{{}^\circ }\). From Figure 3.13, we can find that the gain margin for \(L\mathrm{(}s\mathrm{)}\) defined in equation \ref{3.33} is 15.4 dB and the phase margin is \({45}^{{}^\circ }\).

Note that the controller \(C\left(s\right)=10/s\) defined in equation \ref{3.33} is an integral controller; it is a PID control with nonzero integral gain of only \(10\). We can add a proportional gain to improve the system stability margin. For example, let \(C\left(s\right)=(5s+10)/s\). Then \(L(s)\) becomes the following.

\[L\left(s\right)=C\left(s\right)G\left(s\right)H\left(s\right)=\dfrac{5s+10}{s}\cdot \dfrac{50}{s+50}\cdot \dfrac{5}{s+5}=\dfrac{1225s+2500}{s^3+55s^2+250s}\ \label{3.34}\]

Figure 3.14 shows the associated Bode plot. In this case, with \(ω_p=\infty\), the gain margin is also infinity from the Bode plot shown in Figure 3.14, and the phase margin is about \({105}^{{}^\circ }.\ \) Using the PI (proportional-integral) control can significantly improve closed-loop system stability margin. Note that Bode plot analysis can help tune the PID gains to achieve improved stability margin.

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