The effect of zeros - System Analysis and Control

So far, we have analyzed the response and performance of systems based on their poles. For example, we looked at first and second-order systems in canonical forms

G(s) = \frac{K}{\tau s + 1}, \qquad G(s) = \frac{K\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2},

(1)

We saw that the poles determine the stability of the system and transient performance metrics such as rise time, overshoot, and settling time.

We also considered higher-order systems and approximated them by keeping only the dominant poles. In all of these cases, we assumed the system did not contain any zeros (the numerators had no $s$ terms). Now we’ll see what happens when we introduce zeros into the system.

Where do zeros come from?¶

For example, consider a system with two thermal reservoirs.

The temperatures of the reservoirs are $T_1$ and $T_2$ . The reservoirs exchange heat with the environment (constants $k_1$ and $k_2$ ) but are thermally insulated from each other.
A heater injects heat $u$ into the system. A fraction $\alpha_1$ goes to reservoir 1, and a fraction $\alpha_2$ goes to reservoir 2.
A sensor measures the temperature $y$ at some location partway between the two reservoirs. Thus, it measures $\beta_1 T_1 + \beta_2 T_2$ .

The ODE model for this system is:

\begin{aligned} \dot{T}_1 &= -k_1 T_1 + \alpha_1 u \\ \dot{T}_2 &= -k_2 T_2 + \alpha_2 u \\ y &= \beta_1 T_1 + \beta_2 T_2 \end{aligned}

(2)

The transfer function from $u$ to $T_1$ is $\frac{\alpha_1}{s+k_1}$ , and the transfer function from $u$ to $T_2$ is $\frac{\alpha_2}{s+k_2}$ . Therefore, the transfer function from $u$ to $y$ is a linear combination of these two:

\begin{aligned} G(s) &= \frac{\alpha_1\beta_1}{s+k_1} + \frac{\alpha_2\beta_2}{s+k_2} \\ &= \frac{(\alpha_1\beta_1+\alpha_2\beta_2)s + (\alpha_1\beta_1k_2+\alpha_2\beta_2k_1)}{(s+k_1)(s+k_2)} \end{aligned}

(3)

The system has two energy storage elements (the thermal masses of the reservoirs), so it makes sense that the system should have two poles. The pole locations are completely determined by the heat exchange constants $k_1$ and $k_2$ .

The numerator, however, depends on the sensor and actuator parameters $\alpha_i$ and $\beta_i$ . We have some special cases:

If $\alpha_2\beta_2=0$ (either inject nothing into reservoir 2 or do not measure $T_2$ ), the transfer function (3) simplifies to $\frac{\alpha_1\beta_1}{s+k_1}$ . This makes sense; the second reservoir is effectively decoupled from the system.
If $\alpha_1\beta_1=0$ (either inject nothing into reservoir 1 or do not measure $T_1$ ), the transfer function (3) simplifies to $\frac{\alpha_2\beta_2}{s+k_2}$ . This also makes sense; the first reservoir is effectively decoupled from the system.
If $k_1=k_2=k$ , the transfer function simplifies to $\frac{\alpha_1\beta_1+\alpha_2\beta_2}{s+k}$ . The system behaves like a first-order system with a single pole at $s=-k$ and no zeros.
For other combinations, we get a zero whose location depends on the parameters:
$s = -\frac{\alpha_1\beta_1k_2+\alpha_2\beta_2k_1}{\alpha_1\beta_1+\alpha_2\beta_2}$
(4)

Similar behavior occurs in higher-order systems. The poles are determined by the energy storage elements, while the zeros are determined by the way the system is measured and actuated. Next, we will see how the presence of zeros can affect the transient response of a system, even if the poles remain unchanged.

Zeros in second-order systems¶

Consider a second-order system with an added zero:

G(s) = \frac{K\omega_n^2(\tau s + 1)}{s^2 + 2\zeta \omega_n s + \omega_n^2}.

(5)

We added a zero at $s = -\frac{1}{\tau}$ , but the poles and the DC gain remain unchanged. How does this affect the transient response?

Our step response will still be a combination of $e^{-\zeta \omega_n t} \sin(\omega_d t)$ and $e^{-\zeta \omega_n t} \cos(\omega_d t)$ terms, but the coefficients will depend on $\tau$ , which will affect the shape of the response.

Here is a plot showing how the step response changes as we vary $\tau$ . For this example, we fixed $\zeta=0.5$ and $\omega_n=1$ .

Step response of the second-order system for different values of \tau. We fixed \zeta=0.5 and \omega_n=1. The poles remain unchanged, but the zero location changes with \tau, which significantly affects the transient response. — Figure 1:Step response of the second-order system (5) for different values of $\tau$ . We fixed $\zeta=0.5$ and $\omega_n=1$ . The poles remain unchanged, but the zero location changes with $\tau$ , which significantly affects the transient response.

We can observe some interesting effects as we vary $\tau$ :

For $\tau > 0$ (LHP zero), the response becomes faster and more aggressive. The peak time decreases and the overshoot increases. The settling time stays roughly the same.
For $\tau = 0$ (no zero), we get the standard second-order response. This is the baseline case. The overshoot is minimum here.
For $\tau < 0$ (RHP zero), the response initially moves in the wrong direction before eventually settling to the correct steady-state value. This is called inverse response or undershoot. The peak time increases, the overshoot increases, and the settling time stays roughly the same.

The initial slope of the step response is also affected by the zero location. As it turns out,

\dot y(0) = K\omega_n^2 \tau

(6)

so the initial slope is directly proportional to $\tau$ . This explains why the response is initially faster for $\tau > 0$ and initially slower (in the wrong direction) for $\tau < 0$ .

Inverse response in practice

Inverse response is a common phenomenon in many real-world systems.

Consider driving a car in reverse. Let $u$ be the position of the steering wheel and $y$ be the lateral position of the front bumper. If you turn the steering wheel to the right, the front bumper initially moves left before eventually moving right.
Consider an airplane. Let $u$ be the angle of the elevator and $y$ be the vertical position of the center of mass of the plane. If you increase elevator angle (in order to pitch the airplane up and climb), there is an initial downforce due to the elevator angle, which causes the plane to initially drop in altitude before eventually climbing.

In all of these systems, the transfer function from input $u$ to output $y$ contains a RHP zero, which causes the inverse response. This can make the system difficult to control because it behaves like it has a “built-in delay”; it must initially move in the wrong direction before it can move correctly.

Zeros can be moved by changing the way the system is measured or actuated. For example, if we put the elevator at the front of the airplane instead of the back^[1], we would get a LHP zero instead of a RHP zero, so no inverse response.

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 various parameters. This time, try adjusting the value of $\tau$ to see how the zero location affects the transient response.

Zeros and dominant poles¶

We saw in the section on higher-order systems that we can often approximate a higher-order system by keeping only the dominant poles. However, this approximation assumes there are no zeros!We will now see that zeros can disrupt dominant pole approximations. To illustrate, consider a system with two simple poles and one zero:

G(s) = \frac{10}{z} \cdot \frac{(s + z)}{(s + 1)(s + 10)},

(7)

where $z$ is a parameter that we can vary to change the zero location. The poles are fixed at $s=-1$ and $s=-10$ , so the dominant pole approximation would suggest that the transient response should be dominated by the pole at $s=-1$ regardless of the value of $z$ . However, as we will see, this is not always the case.

We included the factor of $\frac{10}{z}$ in front to ensure that $G(s)$ would always have a DC gain of 1 for any choice of $z$ . If we let $z\to\infty$ , we recover the case with no zeros, as the transfer function simplifies to $\frac{10}{(s + 1)(s + 10)}$ . This is also called having “a zero at infinity”.

Now perform a PFE to obtain two first-order terms and place them in canonical form so we can compare their DC gains:

\boxed{ G(s) = \frac{10}{9} \cdot \biggl(\; \underbrace{\frac{1-\frac{1}{z}}{\vphantom{\frac{1}{10}}s + 1}}_{\textsf{slow pole}} - \underbrace{\frac{\frac{1}{10}-\frac{1}{z}}{\frac{1}{10}s + 1}}_{\textsf{fast pole}} \;\biggr)}

(8)

Now let’s look at what happens as we vary $z$ :

If $z\to\infty$ (no zero), the slow pole has a DC gain of $\frac{10}{9}$ and the fast pole has a DC gain of $-\frac{1}{9}$ . As expected, this is the standard dominant pole scenario; the slow pole dominates due to its much larger DC gain.
If $z=10$ (zero at the fast pole), the slow pole has a DC gain of $\frac{10}{9}$ and the fast pole has a DC gain of 0. The fast pole is effectively cancelled by the zero, so the slow pole still dominates. We can see this in the original transfer function (7). When $z=10$ , the numerator has a factor of $(s+10)$ , which cancels the $(s+10)$ in the denominator, leaving us with a single pole at $s=-1$ .
If $z=1$ (zero at the slow pole), the slow pole has a DC gain of 0 and the fast pole has a DC gain of $\frac{10}{9}$ . The slow pole is effectively cancelled by the zero, so the fast pole dominates instead! This is a dramatic change in behavior caused by the zero, even though the poles remain unchanged.
If $z\to 0$ (zero close to the origin), the DC gains of both poles blow up, causing large transient distortions.

We can summarize these observations with the following principles:

Interactive applet¶

The following interactive applet illustrates the effect of zeros on pole dominance. You can move the poles by dragging the “ $\mathsf{x}$ ” markers and move the zero by dragging the “ $\mathsf{o}$ ” marker. The step response will update in real time to show how the transient response changes. The shaded lines emanating from the poles indicate the DC gain of each pole; the longer the line, the larger the DC gain and the more dominant that pole is.

Test your knowledge¶

Solution to Exercise 1 #

We seek a PFE of the form:

G(s) = \frac{A}{s+a} + \frac{B}{s+b}

(10)

Using the cover-up method, we have:

\begin{aligned} A &= \frac{s+z}{s+b}\bigg|_{s=-a} = \frac{z-a}{b-a} \\ B &= \frac{s+z}{s+a}\bigg|_{s=-b} = \frac{z-b}{a-b} \end{aligned}

(11)

Putting the transfer function in canonical form, we get:

\begin{aligned} G(s) &= \frac{z-a}{b-a} \cdot \frac{1}{s+a} + \frac{b-z}{b-a} \cdot \frac{1}{s+b} \\ &= \underbrace{\frac{1}{a}\cdot\frac{z-a}{b-a}}_{\substack{\textsf{DC gain of}\\\textsf{slow pole}}} \cdot \frac{1}{\tfrac{1}{a}s+1} + \underbrace{\frac{1}{b}\cdot \frac{b-z}{b-a}}_{\substack{\textsf{DC gain of}\\\textsf{fast pole}}} \cdot \frac{1}{\tfrac{1}{b}s+1} \end{aligned}

(12)

Setting the DC gains equal to each other, we get:

\frac{1}{a}\cdot\frac{z-a}{b-a} = \frac{1}{b}\cdot \frac{b-z}{b-a} \implies z = \frac{2ab}{a+b}

(13)

Or in other words, the zero must be located at the harmonic mean of the two poles in order for both poles to have equal DC gains. We can rearrange the result to get:

\frac{1}{z} = \frac{1}{2}\biggl(\frac{1}{a} + \frac{1}{b}\biggr)

(14)

So if we think of the poles and zeros in terms of their time constants rather than their actual values, the time constant of the zero must be the average of the time constants of the two poles in order for both poles to have equal DC gains.

Footnotes¶

This is called a canard configuration. It is used in some fighter jets, but it is not common in commercial airplanes due to stability and control issues.
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