Second-order performance - System Analysis and Control

In this section, we dive deeper into the analysis of second-order systems. We will define quantitative notions of performance, and relate them to the canonical parameters we already learned about, such as natural frequency and damping ratio.

Performance of second-order systems¶

Let’s start with the step response (30), which we derived in the previous section:

y(t) = K \left(1 - \frac{e^{-\zeta \omega_n t}}{\sqrt{1 - \zeta^2}} \sin(\omega_d t + \phi)\right)

(1)

We can make several observations about this function:

Initial value: At $t=0$ , we have $y(0) = K \bigl(1 - \frac{1}{\sqrt{1 - \zeta^2}} \sin(\phi)\bigr)$ . Recall that $\cos\phi = \zeta$ , so $\sin(\phi) = \sqrt{1 - \zeta^2}$ . Therefore, $y(0) = 0$ . This makes sense, since the system starts at rest and the step input is applied at $t=0$ .
Steady-state value: As $t \to \infty$ , the exponential term $e^{-\zeta \omega_n t}$ goes to zero, so $y(t) \to K$ . This means that the system will eventually settle at the value $K$ . This is similar to what we observed for first-order systems, where the steady-state value is also determined by the DC gain.
Exponential envelope: Since the sine term can only be between -1 and 1, the exponential $e^{-\zeta \omega_n t}$ defines an envelope that bounds the oscillations of the response:
$K \left(1 - \frac{e^{-\zeta \omega_n t}}{\sqrt{1 - \zeta^2}}\right) \leq y(t) \leq K \left(1 + \frac{e^{-\zeta \omega_n t}}{\sqrt{1 - \zeta^2}}\right)$
(2)

This is enough information to make a first pass and sketch the step response. Here is what a step response with $\zeta = 0.25$ looks like:

Step response of a second-order system with \zeta=0.25 showing peak time, settling time, and maximum overshoot. — Figure 1:Step response of a second-order system with $\zeta=0.25$ showing peak time, settling time, and maximum overshoot.

When it comes to performance, we are often interested in the following quantities:

Peak time $t_p$ : the time at which the response reaches its maximum value. Generally, we want this to be as small as possible, since it indicates how quickly the system responds to the step input.
Percent overshoot $M_p$ : the amount by which the response overshoots its steady-state value, expressed as a percentage. We typically want this to be small, since large overshoots are often undesirable in practice.
Settling time $t_s$ : the time it takes for the response to get sufficiently close to its steady-state value and stay there. This is similar to notion of settling time for first-order systems.

We will now go into more detail about each of these performance metrics, and how they relate to the parameters of the second-order system.

Peak time¶

The peak time $t_p$ is the time at which the response reaches its maximum value. To find this, we can take the time derivative of the step response (1) and set it equal to zero:

\begin{aligned} \frac{\dd y}{\dd t} &= K \left( \frac{e^{-\zeta \omega_n t}}{\sqrt{1 - \zeta^2}} \left( \zeta \omega_n \sin(\omega_d t + \phi) - \omega_d \cos(\omega_d t + \phi) \right) \right) = 0 \\ &\implies \zeta \omega_n \sin(\omega_d t + \phi) - \omega_d \cos(\omega_d t + \phi) = 0 \\ &\implies \omega_n\left( \zeta \sin(\omega_d t + \phi) - \sqrt{1 - \zeta^2} \cos(\omega_d t + \phi) \right) = 0 \\ &\implies \sin(\omega_d t + \phi)\cos\phi - \cos(\omega_d t + \phi)\sin\phi = 0 \\ &\implies \sin(\omega_d t + \phi - \phi) = 0 \\ &\implies \sin(\omega_d t) = 0 \\ &\implies t = \frac{n\pi}{\omega_d}, \quad n = 0, 1, 2, \ldots \end{aligned}

(3)

This tells us that the response reaches its maximum value at integer multiples of $\pi/\omega_d$ . In particular, the slope is zero at $t=0$ , and the peak time is given by setting $n=1$ .

\boxed{t_p = \frac{\pi}{\omega_d} = \frac{\pi}{\omega_n \sqrt{1 - \zeta^2}}}

(4)

Peak and envelope

The peak time is the time at which the response reaches its maximum value, which does not correspond to where the response touches the exponential envelope! The response touches the envelope when $\sin(\omega_d t + \phi) = \pm 1$ , which happens when

t = \frac{n\pi}{\omega_d} + \frac{\frac{\pi}{2} - \phi}{\omega_d}

(5)

for $n=0,1,2,\ldots$ . This occurs slightly after each point of zero slope, which can also be deduced by looking at the plot of the step response.

Also notice that the exponential envelope does not pass through zero. It goes slightly negative at $t=0$ , which is what allows the response to have a horizontal slope at $t=0$ but then touch the envelope at a slightly later time.

Percent overshoot¶

The percent overshoot $M_p$ is the amount by which the response overshoots its steady-state value, expressed as a percentage. Mathematically, the formula is:

M_p = \frac{y(t_p) - K}{K} \times 100\%

(6)

Substituting the formula for the step response (1) and the formula for $t_p$ (4), we get:

\begin{aligned} M_p &= \frac{K \left(1 - \frac{e^{-\zeta \omega_n t_p}}{\sqrt{1 - \zeta^2}} \sin(\omega_d t_p + \phi)\right) - K}{K} \times 100\% \\ &= \left( - \frac{e^{-\zeta \omega_n t_p}}{\sqrt{1 - \zeta^2}} \sin(\omega_d t_p + \phi) \right) \times 100\% \\ &= \left( - \frac{e^{-\zeta \omega_n t_p}}{\sqrt{1 - \zeta^2}} \sin(\pi + \phi) \right) \times 100\% \\ &= \left( - \frac{e^{-\zeta \omega_n t_p}}{\sqrt{1 - \zeta^2}} (-\sin(\phi)) \right) \times 100\% \\ &= \left( \frac{e^{-\zeta \omega_n t_p}}{\sqrt{1 - \zeta^2}} \sin(\phi) \right) \times 100\% \\ &= \left( \frac{e^{-\zeta \omega_n t_p}}{\sqrt{1 - \zeta^2}} \sqrt{1 - \zeta^2} \right) \times 100\% \\ &= e^{-\zeta \omega_n t_p} \times 100\% \\ &= e^{-\frac{\zeta \pi}{\sqrt{1 - \zeta^2}}} \times 100\% \end{aligned}

(7)

We can also express this formula in terms of the pole angle $\phi$ using the fact that:

\frac{\sqrt{1-\zeta^2}}{\zeta} = \frac{\sin\phi}{\cos\phi} = \tan\phi

(8)

Therefore, we obtain the formulas:

\boxed{M_p = e^{-\frac{\zeta \pi}{\sqrt{1 - \zeta^2}}} \times 100\% = e^{-\pi / \tan\phi} \times 100\%}

(9)

The percent overshoot only depends on the damping ratio $\zeta$ (or equivalently, the pole angle $\phi$ ). However, it can be cumbersome to calculate due to the complexity of the formula. Instead, we can simply plot $M_p$ as a function of $\zeta$ to see how the overshoot changes with damping to get a better intuition.

Relationship between percent overshoot M_p and damping ratio \zeta for a second-order system. The percent overshoot decreases as the damping ratio increases, and approaches zero as \zeta \to 1. — Figure 2:Relationship between percent overshoot $M_p$ and damping ratio $\zeta$ for a second-order system. The percent overshoot decreases as the damping ratio increases, and approaches zero as $\zeta \to 1$ .

For example, we see from Figure 2 that when $\zeta = 0.25$ , the percent overshoot is around 45%, which matches what we see in the step response Figure 1.

Settling time¶

To find the settling time, we need to determine when the response gets sufficiently close to its steady-state value $K$ and stays there. This is difficult to determine exactly, so instead we will use the exponential envelope to find an upper bound on the settling time. As with first-order systems, we will use the 2% criterion.

The exponential envelope decays like $e^{-\zeta \omega_n t}$ . If we compare that to a standard exponential decay $e^{-t/\tau}$ , we can see that the time constant $\tau$ of the envelope is given by:

\tau = \frac{1}{\zeta \omega_n}

(10)

This means that the envelope will decay to 2% of its initial value after approximately $4\tau$ seconds. Therefore, we can use the following formula to get an upper bound on the settling time:

\boxed{t_s \approx \frac{4}{\zeta \omega_n}}

(11)

Sketching the step response¶

Now that we have a bit more information, we can make a more detailed sketch of the step response. Here are the main steps to follow:

Sketching steps

Canonical form. Put the system in canonical form (7) and identify the parameters $K$ , $\omega_n$ , and $\zeta$ . Ensure that $0<\zeta<1$ (underdamped case). Also, calculate the damped frequency $\omega_d$ using the formula $\omega_d = \omega_n \sqrt{1 - \zeta^2}$ .
Steady-state value. Let $2K$ be the top of the $y$ -axis. Draw a horizontal line at $y=K$ to indicate the steady-state value.
Time scale. Use the formula (10) to calculate the time constant of the exponential envelope. Make the $x$ -axis go up to about $4\tau$ , since this is roughly how long it takes for the response to settle.
Exponential envelope. Sketch the exponential envelope. It should start a bit below zero and a bit above $2K$ at $t=0$ , and then approach $K$ as $t \to \infty$ . After about $4\tau$ , the envelope should be very close to $K$ .
Peak time. Use the formula (4) to calculate the peak time $t_p$ . Make vertical lines at $\{t_p, 2t_p, 3t_p, \dots\}$ to indicate where the response has zero slope. The first vertical line at $t_p$ is the peak response.
Sketch! Starting at $t=0$ , sketch the response. It should start at zero with zero slope, and then oscillate between the upper and lower bounds defined by the exponential envelope. The first peak should occur at $t_p$ , and subsequent valleys and peaks should occur at multiples of $t_p$ . The response should eventually settle at $K$ as $t \to \infty$ .
Verify percent overshoot. Use the formula (9) to calculate the percent overshoot $M_p$ . Check that the first peak of your sketch is approximately $M_p$ percent above $K$ .

Example: response sketch¶

Consider the following spring-mass-damper system

with $m=1$ kg, $k=10$ N/m, and $b=1$ Ns/m. Let’s sketch the step response of this system. In other words, let’s plot $x(t)$ with unit step force $f(t) = H(t) \cdot 1$ N.

Canonical form. The transfer function of this system is given by:

G(s) = \frac{X(s)}{F(s)} = \frac{1}{s^2 + s + 10}

(12)

We can calculate the parameters:

\begin{aligned} \omega_n &= \sqrt{10} \approx 3.16\,\text{rad/s}\\ \zeta &= \tfrac{1}{2\sqrt{10}} \approx 0.158\quad\textsf{(underdamped)}\\ K &= \tfrac{1}{10} = 0.1 \\ \omega_d &= \omega_n \sqrt{1 - \zeta^2} \approx 3.12\,\text{rad/s} \\ \phi &= \arccos(\zeta) \approx 1.412\,\text{rad} \approx 80.9^\circ \end{aligned}

(13)

Therefore, the step response is given by the formula:

\begin{aligned} x(t) &= K \left( 1 - \frac{1}{\sqrt{1-\zeta^2}} e^{-\zeta\omega_n} \sin( \omega_d t + \phi) \right) \\ &= 0.1 \left( 1 - 1.01 e^{-0.5t}\sin(3.12t + 1.412) \right) \end{aligned}

(14)

Performance parameters. We can calculate the performance parameters:

\begin{aligned} t_p &= \frac{\pi}{\omega_d} \approx 1.01\,\text{s} \\ \tau &\approx \frac{1}{\zeta \omega_n} \approx 2.00\,\text{s} \\ t_s &\approx 4\tau \approx 8.00\,\text{s} \\ M_p &= e^{-\frac{\zeta \pi}{\sqrt{1 - \zeta^2}}} \times 100\% \approx 60.5\% \end{aligned}

(15)

Detailed sketch. We can now make our detailed sketch using the steps outlined above. We can verify that the first peak of our sketch is approximately 60.5% above the steady-state value, which matches our calculation of $M_p$ .

Step response sketch for the spring-mass-damper system with m=1 kg, k=10 N/m, and b=1 Ns/m. The response has a peak time of approximately 1 second, a settling time of approximately 8 seconds, and a percent overshoot of approximately 60.5%. — Figure 4:Step response sketch for the spring-mass-damper system with $m=1$ kg, $k=10$ N/m, and $b=1$ Ns/m. The response has a peak time of approximately 1 second, a settling time of approximately 8 seconds, and a percent overshoot of approximately 60.5%.

Pole locations and performance¶

Let’s recap the performance measures we defined so far:

t_p = \frac{\pi}{\omega_d}, \qquad M_p = e^{-\pi/\tan\phi} \times 100\%, \qquad t_s \approx \frac{4}{\zeta \omega_n}

(16)

Let’s also recall that for an undamped second order system, the poles are located at

s = -\zeta \omega_n \pm j \omega_d

(17)

This allows us to infer performance directly from the pole location!

The peak time $t_p$ is inversely proportional to the imaginary part of the poles. Poles with larger imaginary part have smaller peak times (faster response).
The percent overshoot $M_p$ is determined by the angle $\phi$ of the poles. Poles with smaller angle (closer to the real axis) will have smaller percent overshoot.
The settling time $t_s$ is inversely proportional to the real part of the poles. Therefore, poles with larger negative real part will have smaller settling times (faster settling).

Example: admissible pole locations¶

Sometimes, we will be given performance specifications and asked to determine where the poles of the system can be located to meet those specifications. For example, suppose we have a second-order system and we would like:

a peak time of at most 2 second,
a percent overshoot of at most 20%, and
a settling time of at most 3 seconds.

Let’s see where the poles of such a system can be located. We can use the formulas for $t_p$ , $M_p$ , and $t_s$ to derive inequalities that the parameters $\zeta$ and $\omega_n$ must satisfy, and then translate those into constraints on the pole locations in the complex plane.

$t_p \leq 2$ implies $\frac{\pi}{\omega_d} \leq 2$ , which means $\omega_d \geq \frac{\pi}{2}$ . So the imaginary part of the poles must be at least $\frac{\pi}{2} \approx 1.57$ .
$M_p \leq 20\%$ implies (see Figure 2) that $\zeta > 0.45$ . Therefore the pole angle satisfies $\phi < \arccos(0.45) \approx 63^\circ$ , which means the poles must be located within 63 degrees of the negative real axis.
$t_s \leq 3$ implies $\frac{4}{\zeta \omega_n} \leq 3$ , which means $\zeta \omega_n \geq \frac{4}{3}$ . So the real part of the poles must be at least $\frac{4}{3} \approx 1.33$ .

We can combine these constraints and plot them on the complex plane to see where the poles can be located to meet the performance specifications.

Figure 5:The shaded region indicates where the poles of a second-order system can be located to meet the performance specifications of peak time at most 2 seconds, percent overshoot at most 20%, and settling time at most 3 seconds.

Note that there are two regions of admissible pole locations, one in the upper half-plane and one in the lower half-plane. This is because the poles of a second-order system come in complex conjugate pairs, so if one pole is located at $s = -\zeta \omega_n + j \omega_d$ , then the other pole must be located at $s = -\zeta \omega_n - j \omega_d$ .

Example: moving poles and performance¶

We can also examine what happens to the step response as we move the poles around in the complex plane.

If we move the poles horizontally (i.e., change the real part), then damped frequency and peak time remain the same.
If we move the poles vertically (i.e., change the imaginary part), then the settling time remains approximately constant.
If we move the poles radially away from the origin, then the percent overshoot remains approximately constant.

Here is a figure that illustrates all three cases.

Figure 6:Figure showing how the step response changes as the poles of a second-order system move around in the complex plane, either horizontally, vertically, or radially away from the origin.

Interactive applet¶

Using the applet below (click the icon in the lower-right corner to fullscreen it), you can interactively explore how the step response of a second-order system changes as you change the parameters $K$ , $\zeta$ , and $\omega_n$ . You can also see how $t_p$ , $M_p$ , and $t_s$ change as you adjust the parameters, and how the pole locations move around.

Test your knowledge¶

Solution to Exercise 1 #

The poles of $G$ are the roots of the denominator. So if we want the poles to be located at $s_{1,2} = -4\pm 2j$ , then the denominator must be equal to $(s - (-4 + 2j))(s - (-4 - 2j))$ . We can expand this to get:
$(s + 4 - 2j)(s + 4 + 2j) = s^2 + 8s + 20$
(19)
Therefore, we have $Q = 8$ and $R = 20$ . To find $P$ , we can use the fact that in canonical form, $P = K\omega_n^2$ and $R = \omega_n^2$ . Therefore, $K = P/R$ . We want $K=0.5$ and we have $R=20$ , so we get $P = 0.5 \cdot 20 = 10$ . Therefore, the transfer function is:
$G(s) = \frac{10}{s^2 + 8s + 20}$
(20)
If we double the value of $R$ , then the denominator becomes $s^2 + 8s + 40$ . The poles of the system are now the roots of this new denominator, which can be calculated using the quadratic formula:
$\begin{aligned} s &= \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 40}}{2 \cdot 1} \\ &= \frac{-8 \pm \sqrt{64 - 160}}{2} \\ &= \frac{-8 \pm \sqrt{-96}}{2} \\ &= -4 \pm 4.9j \end{aligned}$
(21)

Solution to Exercise 2 #

We derived in an earlier example that

\omega_n = \sqrt{k}, \quad \zeta = \frac{b}{2 \sqrt{k}}, \quad K = \frac{1}{k}

(23)

Based on the step response, we see that the steady-state value is $K=0.04$ . Therefore, we find $k=\frac{1}{K} = 25$ .

We can also see that the height of the first peak is around 0.05, so the percent overshoot is around:

M_p = \frac{0.05 - 0.04}{0.04} \times 100\% = 25\%

(24)

Looking this up on Figure 2, we find that $\zeta \approx 0.4$ . Therefore, we can solve for $b$ :

b = 2 \zeta \sqrt{k} = 2 \cdot 0.4 \cdot \sqrt{25} = 4

(25)

Therefore, our estimates for the parameters are:

\boxed{k = 25, \quad b = 4}

(26)

We could have estimated other performance metrics from the plot, such as peak time or settling time, and then used those to find $\zeta$ and ultimately $b$ . However, both of these approaches would have been more complicated. Also, the percent overshoot is much easier to estimate from this sort of plot than the settling time!