3.3: The First and Second Order System Responses and Associated Pole Locations

Learning Objectives

After completing this section, you will be able to:

1. Calculate the first and second order system responses.
2. Describe the relationship between system pole locations and response performances.
3. Calculate system settling time and percentage overshoot for the first and second order systems based upon the model coefficients.

Obtaining system performance measures in general can be challenging. For first and second order systems, most system performance criteria can be found analytically. Fortunately, many physical systems can be simplified into first or second order systems for analyzing and studying system characteristics. Thus, we can study the time responses of first and second order systems and their analytical performance measures in both time and frequency domains.

First, we will address the time response and its performance measures for first order systems. Consider the following standard form of the first order transfer function.

\[G\left(s\right)=\dfrac{K}{Ts+1}\ \label{3.35}\]

Here, \(K\) is the system DC gain (steady-state response \(y_{ss}\) with unit step input) and \(T\) is the system time constant. The unit step input is shown below.

\[ u\left(t\right)=\left\{ \begin{array}{cc} 1 & t\ge 0 \\ 0 & t<0 \end{array} \right.\label{3.36}\]

The Laplace transform of system output with unit step input \(U\left(s\right)=L[u(t)]\) is

\[Y\left(s\right)=G\left(s\right)U\left(s\right)=\dfrac{K}{Ts+1}\cdot \dfrac{1}{s}\ \label{3.37}\]

Using partial fractional expansion, we have the following.

\[Y\left(s\right)=\dfrac{K/T}{s(s+1/T)}=\dfrac{K}{s}-\dfrac{K}{s+1/T}\ \label{3.38}\]

The inverse Laplace transform of the above transfer function leads to the following time response.

\[y\left(t\right)=K\left(1-e^{-t/T}\right),\ t\ge 0 \label{3.39}\]

When \(T>0\), the system is stable. We can calculate the steady-state response as follows.

\[y_{ss}={\mathop{\mathrm{lim}}_{t\to \infty } y(t)\ }=K={\mathop{\mathrm{lim}}_{s\to 0} sY\left(s\right)\ }\ \label{3.40}\]

Figure 3.15 shows the plot of a typical first order system response with \(K=T=1\).

From Figure 3.15, we can see that the system response does not have any overshoot. The maximum response will be achieved as time goes to infinity. As a result, the system does not have any overshoot (or OP = 0%) and peak time \(t_{max}=\infty\); see Figure 3.11 for their definitions. As marked on Figure 3.15, at \(t=T\), the system reaches about 63% of its steady state response \(y_{ss}\). Table 3.3 provides the first order system response with an increment of one time constant.

Table 3.3: First order system response per time constant \(T.\)

Time (second)	\(Ke^{-t/T}\)	\(K(1-e^{-t/T})\)
\(0\)	\(1.0K\)	\(0\)
\(T\)	\(0.3679K\)	\(0.6321K\)
\(2T\)	\(0.1353K\)	\(0.8647K\)
\(3T\ (t_s\left(5\%\right))\)	\(0.0498K\)	\(0.9502K\)
\(4T\ (t_s\left(2\%\right))\)	\(0.0183K\)	\(0.9817K\)
\(5T\)	\(0.0067K\)	\(0.9933K\)

Table 3.3 shows that it takes about \(3T\) time to achieve 95% of the steady-state response, indicating that the 5% settling time \(t_s(5\%)\) is \(3T\) (three times of time constant). Similarly, the 2% settling time \(t_s(2\%)\) is \(4T\) (four times of time constant). The time constant is the time for the step response to reach 63% of the steady state response. As a result, the smaller the time constant \(T\), the faster the system response.

Example 3.9

Consider the following first order system with the transfer function below. Calculate the \(y_{ss}\) and 2% settling time.

\[G\left(s\right)=\dfrac{10}{4s+2} \label{3.41}\]

First, we will convert the system transfer function into the standard form so that the time constant and DC gain can be calculated as below.

\[G\left(s\right)=\dfrac{10}{4s+2}=\dfrac{5}{2s+1} \label{3.42}\]

As a result, the system DC gain is \(K=5\) and time constant \(T=2\). Therefore, we have the following.

\[y_{ss}=K=5,\ t_s\left(2\%\right)=4T=8\mathrm{\ (second)} \label{3.43}\]

Now, let’s study the second order system. The standard form of the second order system can be expressed as follows.

\[G\left(s\right)=\dfrac{Kω^2_n}{s^2+2ζω_ns+ω^2_n} \label{3.44}\]

where \(K\) is the system DC gain, \(ζ\) is the system damping ratio, and \(ω_n\) is the system natural frequency. Now let’s consider the unit step response of the second order system. The Laplace transform of system output is shown below, assuming the underdamped damping ratio (\(0<ζ<1\)).

\[Y\left(s\right)=G\left(s\right)U\left(s\right)=\dfrac{Kω^2_n}{s^2+2ζω_ns+ω^2_n}\cdot \dfrac{1}{s}=\dfrac{Kω^2_n}{{(s+ζω_n)}^2+(1-ζ^2)ω^2_n}\cdot \dfrac{1}{s} \label{3.45}\]

Using fractional expansion leads to

\[Y\left(s\right)=\dfrac{K}{s}-\ \dfrac{K \left(s+ζω_n\right)}{\left(s+ζω_n\right)^2+ω^2_d} - \dfrac{ζ}{\sqrt{1-ζ^2}}\cdot \dfrac{Kω_d}{{(s+ζω_n)}^2+ω^2_d} \label{3.46}\]

where \(ω_d=ω_n\sqrt{1-ζ^2}\) is called the damped natural frequency. The inverse Laplace transform yields the following.

\[y\left(t\right)=K\left\{1-e^{-ζω_nt}{\mathrm{cos} ω_dt\ }-\dfrac{ζ}{\sqrt{1-ζ^2}}e^{-ζω_nt}{\mathrm{sin} ω_dt\ }\right\} \label{3.47}\]

We can reorganize equation \ref{3.47} into the following.

\[y\left(t\right)=K\left\{1-\dfrac{e^{-ζω_nt}}{\sqrt{1-ζ^2}}\left(\sqrt{1-ζ^2}{\mathrm{cos} ω_dt\ }+ζ{\mathrm{sin} ω_dt\ }\right)\right\} \label{3.48}\]

Because \(0<ζ<1\), letting \({\mathrm{sin} φ\ }=\sqrt{1-ζ^2}\) leads to \({\mathrm{cos} φ\ }=ζ\). Substituting both \({\mathrm{sin} φ\ }\) and \({\mathrm{cos} φ\ }\) into equation \ref{3.48} and using the trigonometric sum identity, \({\mathrm{sin} φ\ }{\mathrm{cos} ω_dt\ }+{\mathrm{cos} φ{\mathrm{sin} ω_dt\ }\ }={\mathrm{sin} \left(ω_dt+φ\right),\ }\) leads to the following simplified second order time response with unit step input.

\[y\left(t\right)=K\left\{1-\dfrac{e^{-ζω_nt}}{\sqrt{1-ζ^2}}{\mathrm{sin} \left(ω_dt+φ\right)\ }\right\} \label{3.49}\]

where \(φ={{\mathrm{cos}}^{-1} ζ\ }\). Figure 3.16 plots the system responses for \(K=1\) under the different damping ratio \(0\le ζ\le 2\) with horizontal axis \(ω_nt\). We can see that when \(ζ=0\), the response will not converge to its steady-state. Instead, it keeps oscillating between 0 and 2 because the system is marginally stable and the system response will not converge to its target steady state \(y_{ss}=1\). When \(0<ζ<1\) after a few oscillations, the response will converge to its steady state \(y_{ss}=1\). When \(ζ\ge 1\), the response will monotonically converge to its steady-state \(y_{ss}=1\) without any oscillation.

Note that equation \ref{3.49} can be used for calculating the response for \(0\le ζ<1\), but not for \(ζ\ge 1\). In the case of \(ζ\ge 1\), the response needs to be solved following the same process, but it will be different from equation \ref{3.49}. Note that when \(ζ=1\), the system has two repeated poles \(s_{1,2}=-ω_n\). Furthermore, when \(ζ>1\), the system has two distinguished poles \(s_{1,2}=ω_n\left(-ζ\pm \sqrt{ζ^2-1}\right)\).

From equation \ref{3.49}, we can find that the response of the second order system with \(0<ζ<1\) is bounded by the following equation with plus-sign “\(+\)” for the upper bound and negative-sign “\(-\)” for the lower bound. That is, the oscillated response with \(0<ζ<1\) falls between these two boundary curves.

\[y\left(t\right)=K\left\{1\pm \dfrac{e^{-ζω_nt}}{\sqrt{1-ζ^2}}\right\} \label{3.50}\]

Figure 3.17 plots the boundary curves with different \(ω_n\) values for \(K=1\) and \(ζ=0.4\). The rate of convergence increases as \(ζω_n\) goes up from \(ζω_n=0.4\) to \(ζω_n=1.6\). That is, the system settling time decreases as \(ζω_n\) increases. Notice the case of \(ζω_n=0.4\) (\(ω_n=1)\). Here, the calculated 2% settling time \(t_s\left(2\%\right)\) is \(4/ζω_n=10\) seconds. From Figure 3.17 we can see that the green curve reaches about 0.98 at the 10th second.

From equation \ref{3.49}, the oscillation frequency of the unit step response for the second order system is damped natural frequency \(ω_d=ω_n\sqrt{1-ζ^2}\) instead of its natural frequency \(ω_n\). The response convergence bounds are \(\pm e^{-ζω_nt}\). Comparing these to the convergence response \(e^{-t/T}\) of the first order system, where \(T\) is the time constant, we can see that letting coefficient \(T=1/ζω_n\) makes the convergence rates for the first order and second order systems comparable. Recall that for the first order system, the system response settling times, \(t_s(2\%)\) and \(t_s(5\%)\), are \(4T\) and \(3T\), respectively. As a result, for the second order systems, system response settling times, \(t_s(2\%)\) and \(t_s(5\%)\), are \(4/ζω_n\) and \(3/ζω_n\), respectively.

Next, let’s investigate the system peak time and percentage overshoot, starting with the peak time. Note that peak time candidates can be found by taking the time derivative of \(y(t)\) defined in equation \ref{3.49} and setting the derivative to zero as below.

\[\dot{y}\left(t\right)=K\left\{ζω_n\dfrac{e^{-ζω_nt}}{\sqrt{1-ζ^2}}{\mathrm{sin} \left(ω_dt+φ\right)\ }-ω_d\dfrac{e^{-ζω_nt}}{\sqrt{1-ζ^2}}{\mathrm{cos} \left(ω_dt+φ\right)\ }\right\}=0 \label{3.51}\]

This equation leads to the following.

\[ζ{\mathrm{sin} \left(ω_dt+φ\right)\ }-\sqrt{1-ζ^2}{\mathrm{cos} \left(ω_dt+φ\right)\ }=0 \label{3.52}\]

Using the same definitions of \({\mathrm{sin} φ\ }=\sqrt{1-ζ^2}\) and \({\mathrm{cos} φ\ }=ζ\) as before, and utilizing the trigonometric identity, \({\mathrm{sin} φ\ }{\mathrm{cos} \left(ω_dt+φ\right)\ }+{\mathrm{cos} φ{\mathrm{sin} \left(ω_dt+φ\right)\ }\ }={\mathrm{sin} \left(ω_dt+φ\right),\ }\) we have the following.

\[{\mathrm{cos} φ\ }{\mathrm{sin} \left(ω_dt+φ\right)\ }-{\mathrm{sin} φ\ }{\mathrm{cos} \left(ω_dt+φ\right)\ }={\mathrm{sin} \left(ω_dt+φ-φ\right)\ }={\mathrm{sin} ω_dt\ }=0 \label{3.53}\]

It is clear that when \(\dot{y}\left(t\right)=0\), \(t=nπ/ω_d\) (\(n=0,1,2,\dots\)). From Figure 3.16 and Figure 3.17, the peak time can only occur at \(t_{max}=π/ω_d\). Substituting the peak time into equation \ref{3.49} yields the following.

\[y_{max}=K\left\{1-\dfrac{e^{\dfrac{-ζπ}{\sqrt{1-ζ^2}}}}{\sqrt{1-ζ^2}}{\mathrm{sin} \left(π+φ\right)\ }\right\}=K\left\{1+\dfrac{e^{\dfrac{-ζπ}{\sqrt{1-ζ^2}}}}{\sqrt{1-ζ^2}}{\mathrm{sin} φ\ }\right\}=K\left\{1+e^{\dfrac{-ζπ}{\sqrt{1-ζ^2}}}\right\} \label{3.54}\]

Therefore, the percentage overshoot (PO) can be calculated by the following equation.

\[PO(\%)=\dfrac{y_{max}-y_{ss}}{y_{ss}}100=\dfrac{K\left\{1+e^{\dfrac{-ζπ}{\sqrt{1-ζ^2}}}\right\}-K}{K}100=100e^{\dfrac{-ζπ}{\sqrt{1-ζ^2}}} \label{3.55}\]

Note that the derived percentage overshoot (PO) is a function of the second order system damping ratio \(ζ\). That is, the system damping ratio is the only factor that affects the system PO. Now, we can summarize performance indices of the first and second order systems into Table 3.4.

Table 3.4: Summary of the first and second order system performance measures.

Performance indices	1st Order	2nd Order
Settling time (5%)	\(3T\)	\(\dfrac{3}{{ζω}_n}\)
Settling time (2%)	\(4T\)	\(\dfrac{4}{{ζω}_n}\)
Peak time	\(\infty\)	\(\dfrac{π}{ω_d} = \dfrac{π}{ω_n\sqrt{1-ζ^2}}\)
Percent Overshoot	\(0\)	\(100e^{-\frac{ζπ}{\sqrt{1-ζ^2}}}\)
Peak value	\(K\)	\(K+Ke^{-\frac{ςπ}{\sqrt{1-ζ^2}}}\)

Example 3.10

Consider the closed-loop system shown in Figure 3.18. Calculate \(y_{ss}\), \(y_{max}\), \(t_{max}\), the percentage overshoot (PO), and 2% settling time for \(K=10,\ 25\).

First, we will calculate the closed-loop system transfer function \(G(s)\) from \(U(s)\) to \(Y(s)\).

\[G\left(s\right)=\dfrac{Y(s)}{U(s)}=\dfrac{75K}{s^2+10s+75(K+1)} \label{3.56}\]

For \(K=10,\)

\[ω_n=\sqrt{75\times 11}=28.72, ζ=\dfrac{5}{ω_n}=0.174, \ \sqrt{1-ζ^2}=0.9847 \label{3.57}\]

Now, we are ready to calculate these performance indices as below.

\[y_{ss}=G\left(0\right)=\dfrac{750}{75\times 11}=0.909,\  t_{max}=\dfrac{π}{28.72\times 0.9847}=0.111\mathrm{\ (second)} \label{3.58}\]

\[y_{max}=0.909\left(1+e^{-\dfrac{πζ}{\sqrt{1-ζ^2}}}\right)=1.4316, PO\left(\%\right)=100e^{-\dfrac{πζ}{\sqrt{1-ζ^2}}}=57.4\% \\  t_s\left(2\%\right)=\dfrac{4}{ζω_n}=\dfrac{4}{0.174\times 28.72}=0.8\mathrm{\ (second)} \label{3.60}\]

For \(K=25,\)

\[ω_n=\sqrt{75\times 26}=44.16, ζ=\dfrac{5}{ω_n}=0.113, \sqrt{1-ζ^2}=0.9936 \label{3.61}\]

Then, we will calculate these performance indices as below.

\[y_{ss}=G\left(0\right)=\dfrac{75\times 25}{75\times 26}=0.9615, t_{max}=\dfrac{π}{44.16\times 0.9936}=0.0716\mathrm{\ (second)} \label{3.62}\]

\[y_{max}=0.9615\left(1+e^{-\dfrac{πζ}{\sqrt{1-ζ^2}}}\right)=1.634, PO\left(\%\right)=100e^{-\dfrac{πζ}{\sqrt{1-ζ^2}}}=69.96\% \\(t_s\left(2\%\right)=\dfrac{4}{ζω_n}=\dfrac{4}{0.113\times 44.16}=0.8016\mathrm{\ (second)} \label{3.64}\]

Note that the increased settling time is due mainly to the reduced damping ratio.

Now, we are going to discuss how the closed-loop system pole locations affect system performance indices such as percentage overshoot (PO) and settling time \(t_s\). Consider the second order system defined in equation \ref{3.44}. We can calculate its closed-loop poles by finding the pair of complex conjugate roots of the denominator’s polynomial as follows.

\[s_{1,2}=-ζω_n\pm jω_n\sqrt{1-ζ^2} \label{3.65}\]

Figure 3.19 shows the pole location geometry on the complex plane. Pole \(s_1\) forms a right triangle with three sides of \(ω_n\sqrt{1-ζ^2}\),  \(ζω_n\), and \(ω_n\), where angle \(θ={{\mathrm{cos}}^{-1} ζ\ }\).

As discussed earlier in Table 3.4, the system percentage overshoot (PO) is only a function of damping ratio \(ζ\) (or angle \(θ\) defined in Figure 3.19). The system settling time is inversely proportional to \(ζω_n\). Therefore, we could place (or assign) the closed-loop system pole locations, if possible, to guarantee system performances such as PO and settling time using a certain closed-loop controller. In Example 3.11, we demonstrate how to relate the performance indices to pole locations.

Example 3.11

Find the region of feasible pole locations for a closed-loop second order system with 5% settling time less than \(t_{sm}=3\) (second) and percentage overshoot (PO) less than \({PO}_{max}=4.3\%\). Note that \(t_s\left(5\%\right)=\dfrac{3}{ζω_n} \le t_{sm}\) and \(PO=100e^{-πζ/\sqrt{1-ζ^2}}=100e^{-π/\tan θ} \le PO_{max}\). In this case, we can have the following relationship to meet the performance requirements.

\[ζω_n\ge \dfrac{3}{t_{sm}}=1, \\[4pt] θ \le \tan^{-1} \left(\dfrac{-π}{\ln (0.01 PO_{max})} \right)=θ_{max}={45}^{^\circ } \label{3.66}\]

For both settling time and overshoot requirements of \(t_s(5\%)\le 3\) and \(PO\le 4.3\%\), we can generate feasible pole location maps for settling time and percentage overshoot (PO), as seen in Figure 3.20. The shaded areas are feasible regions of pole locations that satisfy settling time (left) and percentage overshoot (right) requirements, respectively.

Taking the intersection of the two shaded areas in Figure 3.20 leads to the combined feasible region for the closed-loop pole locations, as shown in Figure 3.21. That is, if the closed-loop system poles are located within the shaded area, the second order system response will have a 5% settling time less than 3 seconds and a percentage overshoot less than 4.3%.

In short, we can see that the transient response of the second order system can be characterized by its damping ratio \(ζ\) and undamped natural frequency \(ω_n\). In other words, pole locations are associated with roots of \(s^2+2ζω_ns+ω^2_n\). Delay time and rise time are not as easy to characterize and can be obtained only through simulation investigation. For transient responses of high order systems, normally the only option is computer simulation study. After we find the desired (feasible) pole location region, we can use pole assignment control to locate the closed-loop system poles inside the desired region (see Section 3.5).

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