Higher-order systems - System Analysis and Control

In recent sections, we studied the transient response of first and second-order systems in detail. However, many systems of interest are higher-order, meaning that their transfer functions have order greater than 2. In this section, we will discuss how to analyze the transient response of higher-order systems and introduce the concept of dominant poles.

Before we can analyze higher-order systems, we need to introduce a powerful tool called the final value theorem, which allows us to determine the steady-state behavior of a system (of any order) directly from its transfer function.

Final value theorem¶

The final value theorem (FVT) is a powerful tool that allows us to determine the steady-state behavior of a system directly from its transfer function.

Sketch of proof

The rough steps are as follows:

\begin{aligned} \lim_{t\to\infty} y(t) &= \lim_{t\to\infty} y(t) - y(0) + y(0) \\ &= \Bigl[ y(t) \Bigr]_{t=0}^{t=\infty} + y(0) \\ &= \int_0^\infty \dot{y}(t)\, \dd t + y(0) \\ &= \left( \lim_{s\to 0} \int_0^\infty \dot{y}(t) e^{-st}\, \dd t \right) + y(0) \\ &= \lim_{s\to 0} \Lap\{\dot{y}(t)\} + y(0) \\ &= \lim_{s\to 0} \left( sY(s) - y(0) \right) + y(0) \\ &= \lim_{s\to 0} sY(s) \end{aligned}

(2)

The validity of the various steps (convergence of integrals, existence of limits, etc.) is guaranteed by the assumptions of the theorem.

The most common way to apply the FVT is find the steady-state response of a system to a step input. For example, if we have a system with transfer function $G(s)$ and we apply a unit step input $u(t) = H(t)$ , then the output in the Laplace domain is:

Y(s) = G(s)U(s) = \frac{1}{s}G(s)

(3)

Therefore, the steady-state response is:

\lim_{t\to\infty} y(t) = \lim_{s\to 0} sY(s) = \lim_{s\to 0} G(s) = G(0)

(4)

So the steady-state response is simply the value of the transfer function at $s = 0$ . Let’s verify this property on a couple familiar examples.

Example: first-order system¶

Consider a first-order system in canonical form:

G(s) = \frac{K}{\tau s + 1}

(5)

Using the FVT, the steady-state response is:

G(0) = \frac{K}{1} = K

(6)

This matches our intuition (and the step response sketch we made earlier); the steady-state response of a first-order system to a step input is equal to the DC gain $K$ .

Example: second-order system¶

Consider a second-order system in canonical form:

G(s) = \frac{K\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}

(7)

Using the FVT, the steady-state response is:

G(0) = \frac{K\omega_n^2}{\omega_n^2} = K

(8)

Again, this matches our intuition; the steady-state response of a second-order system to a step input is equal to the DC gain $K$ .

Example: first-order with P-control¶

Recall the closed-loop transfer function of a first-order system $\frac{1}{\tau s + 1}$ in unity feedback with proportional controller $k_p$ :

G(s) = \frac{k_p}{\tau s + 1 + k_p}

(9)

Using the FVT, the steady-state response is:

G(0) = \frac{k_p}{1 + k_p}

(10)

which is what we found earlier by using the canonical form.

Example: second-order with P-control¶

Recall the transfer function of a second-order system $\frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}$ in unity feedback with proportional gain $k_p$ :

G(s) = \frac{k_p\omega_n^2}{s^2 + 2\zeta\omega_n s + (1+k_p)\omega_n^2}

(11)

Using the FVT, the steady-state response is:

G(0) = \frac{k_p\omega_n^2}{(1+k_p)\omega_n^2} = \frac{k_p}{1 + k_p}

(12)

Again, this matches what we found earlier by using the canonical form.

When does the FVT not apply?

If the assumptions of the FVT are not satisfied, then we obtain nonsense results.

Consider $G(s) = \frac{1}{s^2+1}$ . This system has poles at $\pm j$ , so the FVT does not apply. If we try to apply it anyway, we get $G(0) = 1$ . However, the step response of this system is $y(t) = 1 - \cos(t)$ , which just oscillates forever and does not converge to a finite value as $t\to\infty$ .
Consider $G(s) = \frac{1}{s-1}$ . This system has a pole in the right half plane, so the FVT does not apply. If we try to apply it anyway, we get $G(0) = -1$ . However, the step response of this system is $y(t) = e^t - 1$ , which diverges as $t\to\infty$ .

There is an analogous result to the FVT called the Initial Value Theorem (IVT) that applies to the limit as $t\to 0$ instead of $t\to\infty$ . This isn’t as useful for us, but we include it for completeness:

Sketch of proof

Start from the Laplace transform definition:

Y(s)=\int_{0}^{\infty} y(t)\,e^{-st}\,\dd t.

(14)

Multiply by $s$ and change variables $\tau = st$ (so $t=\tau/s$ and $\dd t=\dd\tau/s$ ):

sY(s) = s\int_{0}^{\infty} y(t)e^{-st}\,\dd t = \int_{0}^{\infty} y(\tau/s)\,e^{-\tau}\,\dd\tau.

(15)

Now take the limit $s\to\infty$ . For each fixed $\tau$ , we have $\tau/s\to 0^+$ , hence $y(\tau/s)\to y(0^+)$ . Under the stated assumptions (which guarantee the needed limit/interchange steps), this yields

\begin{aligned} \lim_{s\to\infty} sY(s) &= \int_{0}^{\infty} \left(\lim_{s\to\infty} y(\tau/s)\right)e^{-\tau}\,\dd\tau \\ &= \int_{0}^{\infty} y(0^+)\,e^{-\tau}\,\dd\tau \\ &= y(0^+)\int_{0}^{\infty} e^{-\tau}\,\dd\tau \\ &= y(0^+) \end{aligned}

(16)

Therefore, $\displaystyle\lim_{t\to 0^+} y(t)=y(0^+)=\lim_{s\to\infty} sY(s)$ .

Higher-order systems¶

Any transfer function can be expressed as a sum of first-order and second-order terms via PFE. By superposition, the response to any input is therefore the sum of the responses of these simpler terms. Since we have already analyzed first and second-order systems in detail, we can in principle analyze the transient response of any system!

That being said, the transient response of higher-order systems can be quite complicated, and high-order systems often behave like lower-order systems in practice. For example, we saw that a DC motor has a second-order transfer function, but its step response looks very much like that of a first-order system. We will now study this phenomenon in more detail.

Example: third-order system¶

Consider the following third-order system and its PFE:

\begin{aligned} G(s) &= \frac{891000}{(s+1)(s+10)(s+100)} \\ &= \frac{1000}{s+1} - \frac{1100}{s+10} + \frac{100}{s+100} \\ &= \frac{1000}{s+1} + \frac{-110}{0.1s+1} + \frac{1}{0.01s+1} \end{aligned}

(17)

We placed each term in canonical form to make it easier to analyze. The step response of this system is the sum of the step responses of each term. Just like with the DC motor, the first pole is both the slowest and has the largest DC gain, so the first term dominates the transient response. The second and third terms are much faster and have much smaller DC gains, so they only contribute small “bumps” to the transient response at early times.

We say that the pole at $s = -1$ is a dominant pole of the system, because it dominates the transient response. The poles at $s = -10$ and $s = -100$ are non-dominant poles because they do not significantly affect the transient response.

We can approximate the response of $G(s)$ by just keeping the dominant pole and ignoring the non-dominant poles. However, we have to do this carefully. If we just keep the first term, we get $\frac{1000}{s+1}$ , which has a DC gain of 1000. However, the original system has a DC gain of $G(0) = 891$ , so this approximation is not very good. Instead, we can keep the first term but reduce its gain to match the DC gain of the original system:

\hat G(s) = \frac{891}{s+1}

(18)

This approximation has the same dominant pole as the original system and the same DC gain, so it captures the transient response much better. In fact, the step response of $\hat G(s)$ is almost indistinguishable from that of $G(s)$ .

The plot above shows the step response of the original system G(s), the approximation \hat G(s) that keeps only the dominant pole but corrects the DC gain, and the approximation that keeps only the dominant pole without correcting the DC gain. We can see that the approximation with corrected DC gain is much better than the one without. — Figure 1:The plot above shows the step response of the original system $G(s)$ , the approximation $\hat G(s)$ that keeps only the dominant pole but corrects the DC gain, and the approximation that keeps only the dominant pole without correcting the DC gain. We can see that the approximation with corrected DC gain is much better than the one without.

General procedure¶

Rule of thumb for dominant poles

A simple pole (or a pair of complex conjugate poles) is said to be dominant if its real part is at least 5 times closer to the imaginary axis than the real parts of all other poles. Some important caveats:

The closer to the imaginary axis a pole is, the more dominant it becomes. So if a pole is 10 times closer to the imaginary axis than all other poles, it is even more dominant than if it were only 5 times closer.
This rule only applies to stable systems, so all poles must be in the left-half plane for the rule to apply. If there are poles in the RHP, then the system is unstable, and the notion of pole dominance does not make sense.
This rule assumes the systems contains only poles (no zeros). Zeros can significantly affect the transient response; we will discuss how zeros affect pole dominance in the next section.

Given a transfer function $G(s)$ , we can find a reduced-order model $\hat G(s)$ that approximates the transient response of $G(s)$ by keeping only the dominant poles. Here are the steps.

Procedure for finding a reduced-order model

Find the poles of $G(s)$ and identify the dominant poles using the rule of thumb above. If there are no dominant poles, then the transient response is complicated and cannot be approximated by a lower-order model. In this case, just keep all poles and analyze the transient response using superposition.
If there are dominant poles, then keep only the dominant poles and ignore the non-dominant poles. This gives us a reduced-order model $\hat G(s)$ .
The reduced-order model $\hat G(s)$ will have a different DC gain than the original system. To fix this, multiply $\hat G(s)$ by a constant gain such that the DC gains match. By the FVT, the DC gain of $G(s)$ is $G(0)$ and the DC gain of $\hat G(s)$ is $\hat G(0)$ , so we can multiply $\hat G(s)$ by $\frac{G(0)}{\hat G(0)}$ to get a new reduced-order model that has the same DC gain as the original system.

Example: DC motor revisited¶

Recall the DC motor case study from earlier. We found that the DC motor had a transfer function

G(s) = \frac{133333}{(s + 21.8)(s + 312)}

(19)

The poles are at $s = -21.8$ and $s = -312$ . The pole at $s = -21.8$ is about 14 times closer to the imaginary axis than the pole at $s = -312$ , so it is a dominant pole. Therefore, we can approximate the transient response of the DC motor by keeping only the dominant pole. Our reduced-order model is of the form

\hat G(s) = \frac{K}{s + 21.8}

(20)

To find the gain $K$ , we can use the FVT to match the DC gain of the original system:

\begin{aligned} \hat G(0) = G(0) &\implies \frac{K}{21.8} = \frac{133333}{21.8 \cdot 312} \\ &\implies K = \frac{133333}{312} \approx 427.3 \end{aligned}

(21)

We can now plot the original system and its reduced-order model to see how well the approximation works.

Figure 2:Step response of the original DC motor system and its reduced-order model that keeps only the dominant pole. The reduced-order model captures the transient response of the original system very well, even though it is a much simpler first-order system.

Example: dominant complex poles¶

Let’s consider the following system:

G(s) = \frac{17a}{(s+a)(s^2 + 2s + 17)}

(22)

The poles are at $s = -a$ and $s = -1 \pm 4j$ . We can conclude the following:

If $a < \frac{1}{5}$ , then the pole at $s = -a$ is dominant, and the transient response looks like that of a first-order system. Specifically, the response should look like that of the system
$G_1(s) = \frac{a}{s+a}.$
(23)
If $a > 5$ , then the poles at $s = -1 \pm 4j$ are dominant, and the transient response looks like that of an underdamped second-order system. Specifically, the response should look like that of the system
$G_2(s) = \frac{17a}{s^2 + 2s + 17}$
(24)
If $\frac{1}{5} < a < 5$ , then there are no dominant poles, and the transient response is complicated and cannot be approximated by a lower-order model.

Let’s verify that this holds true by plotting the step responses of $G(s)$ , $G_1(s)$ , and $G_2(s)$ for different values of $a$ .

The plot above shows the step response of G(s) from Eq. together with the dominant pole approximation of the first-order pole G_1(s) and the second-order pole G_2(s) for different values of a. When a is small, the response of G(s) looks like that of G_1(s), and when a is large, the response of G(s) looks like that of G_2(s). When a is in between, the response of G(s) does not look like either G_1(s) or G_2(s), which confirms our analysis. — Figure 3:The plot above shows the step response of $G(s)$ from Eq. (22) together with the dominant pole approximation of the first-order pole $G_1(s)$ and the second-order pole $G_2(s)$ for different values of $a$ . When $a$ is small, the response of $G(s)$ looks like that of $G_1(s)$ , and when $a$ is large, the response of $G(s)$ looks like that of $G_2(s)$ . When $a$ is in between, the response of $G(s)$ does not look like either $G_1(s)$ or $G_2(s)$ , which confirms our analysis.

Test your knowledge¶

Solution to Exercise 1 #

The poles are at $s = -1$ , $s = -10$ , and $s = -11$ . The pole at $s = -1$ is 10 times closer to the imaginary axis than the pole at $s = -10$ , so it is a dominant pole. The original system has DC gain $G(0) = \frac{220}{1 \cdot 10 \cdot 11} = 2$ . Therefore, the reduced-order model is
$\boxed{\hat G(s) = \frac{2}{s+1}}$
(25)
The poles are at $s = -1 \pm j$ , $s = -0.05 \pm j$ , and $s = -5$ . The poles at $s = -0.05 \pm j$ are about 20 times closer to the imaginary axis than the poles at $s = -1 \pm j$ , so they are dominant poles. The original system has DC gain $G(0) = \frac{1}{2 \cdot 1 \cdot 5} = 0.1$ . Therefore, the reduced-order model is
$\boxed{\hat G(s) = \frac{0.1}{s^2 + 0.1s + 1}}$
(26)
The poles are at $s = -1$ , $s = -2$ , and $s = -3 \pm 10j$ . The poles are all about equally close to the imaginary axis, so there are no dominant poles. Therefore, we cannot find a reduced-order model that approximates the transient response of the system.