Lecture 6: First- and Second-Order Systems

The review lectures gave us transfer functions and a way to invert them. We now start the analysis part of the course. The question is this: given a transfer function, what does the response look like, and how can we tell without computing it in full? The answer is built around poles and zeros. This lecture defines them, studies the two simplest building blocks, first- and second-order systems, and takes a short detour into the number $e$ that governs all of their responses.

Forced and natural responses

The total response of a system splits into two parts:

The forced response is driven by the input. The natural response comes from the initial conditions (the system’s stored energy) and describes how the system behaves on its own. We saw both in Lecture 5: in $y(t) = e^{-at}y_0 + \frac ba(1 - e^{-at})$, the first term is natural and the second is forced.

To guess quickly whether a response is stable, unstable or oscillating, we look at the poles and zeros. When there is no time delay, the poles alone determine stability.

Poles and zeros

Let $G(s) = N(s)/D(s)$.

• Poles are the values of $s$ that make the transfer function infinite, that is, the roots of $D(s) = 0$. They are marked “$\times$” in the complex $s$-plane ($s = \sigma + j\omega$).
• Zeros are the values of $s$ that make the transfer function zero, that is, the roots of $N(s) = 0$. They are marked “$\circ$”.

For example, if $G(s) = N(s)/\big((s+p_1)(s+p_2)\cdots(s+p_n)\big)$, the poles are $-p_1, -p_2, \ldots, -p_n$, and $G(s)\to\infty$ as $s\to-p_i$.

A few additional facts:

• The system is asymptotically stable if and only if all poles have negative real parts, that is, all poles lie in the open left half-plane (LHP). Lecture 10 develops stability in detail.
• Zeros do not affect stability, but they change the amplitude and speed of the response (Lecture 9).
• Common roots of $N$ and $D$ cancel. The cancellation must be taken into account when listing the poles and zeros that actually shape the input-output response.

Example 1. Find the poles and zeros of

Before cancellation, the zeros are $-1$, $-3$, $-3$ and the poles are $-1$ and $1\pm j$. The factor $(s+1)$ cancels, which leaves a double zero at $-3$ and a complex-conjugate pair of poles at $1\pm j$. Those poles have positive real parts, so the system is unstable.

Example 2. Find the poles and zeros of the system described by

where $x(t)$ is the output and $u(t)$ is the input, with zero initial conditions. The transfer function describes only how the input $u$ reaches the output. Taking Laplace transforms,

Note that the constant term $1$ transforms to $1/s$, not $1$. It acts as a constant input and adds its own term to the response, but it does not change $G(s)$. The transfer function $G(s) = 2s/(s(s+2))$ has poles at $0$ and $-2$ and a zero at $0$. The zero cancels the pole at the origin, leaving $G(s) = 2/(s+2)$. (If the equation was meant to read $\ddot x + 2\dot x + x = 2\dot u$, then $G(s) = 2s/(s+1)^2$, with a double pole at $-1$ and a zero at $0$.)

First-order systems

A first-order differential equation gives a first-order system, and its response is an exponential decay or growth.

Demo: pushing a block against friction. A block of mass $m$ slides on a surface with viscous friction force $f_dv$. With the applied force $f$ as the input and the velocity $v$ as the output, Newton’s law gives

The single pole is at $s = -f_d/m$.

• A sharp hit. An impulse $f = 5\delta(t)$ gives $F(s) = 5$ and $v(t) = \mathcal{L}^{-1}\left\{\frac{5/m}{s + f_d/m}\right\} = \frac5m e^{-(f_d/m)t}$: the block jumps to speed $5/m$ and coasts to a stop.
• A continuous push. A step $f = 2mg\,1(t)$ gives $V(s) = \frac{1/m}{s+f_d/m}\cdot\frac{2mg}{s} = \frac{K_1}{s} + \frac{K_2}{s+f_d/m}$ with $K_1 = \lim_{s\to0}sV(s) = 2mg/f_d$ and $K_2 = -2mg/f_d$, so $v(t) = \frac{2mg}{f_d}\left(1 - e^{-(f_d/m)t}\right)$: the block accelerates until friction balances the push.

The generalized first-order system. Every first-order system has the form

With zero initial conditions and a unit step input $U(s) = 1/s$,

The term $-\frac ba e^{-at}$ is the transient, which comes from the pole; $\frac ba$ is the steady-state value, the DC gain of Lecture 5.

• The response decays to steady state (stable) when the pole $-a$ satisfies $\mathrm{Re}(\text{pole}) \lt 0$.
• It grows without bound (unstable) when $\mathrm{Re}(\text{pole}) \gt 0$.

The location of the pole dictates stability, and the value of the pole dictates how fast the response decays or grows.

For a stable first-order system, three performance measures follow directly from $1 - e^{-at}$:

• Time constant $\tau = 1/a$: the time for the step response to reach 63% of the final value.
• Rise time $T_r = 2.2/a$: the time to go from 10% to 90% of the final value ($\ln 9 = 2.197$).
• Settling time $T_s = 4/a$: the time to reach and stay within 2% of the final value.

The dashed line in the figure is the initial slope: the response would reach its final value in one time constant if it kept its initial rate.

About $e$

Why do “63%” and “4 time constants” appear? Both come from the number $e = 2.71828\ldots$, which is worth a short detour.

Exponential growth in nature. A population with unlimited resources grows at a rate proportional to its size,

This is a first-order system with a pole at $s = r \gt 0$: an unstable one. In reality, resources are limited. The logistic growth model slows the growth as $N$ approaches a carrying capacity $K$:

At first the two models agree; then the logistic curve bends over and levels off at $K$.

The logistic S curve. The logistic solution can be written as

where $K$ is the final value, $r$ the logistic growth rate, and $t_0$ the midpoint. Each parameter changes the curve in a distinct way:

The same function, normalized, is the sigmoid $\sigma(x) = 1/(1+e^{-x})$, widely used in deep learning. It transforms an input variable into a value between 0 and 1 that can be read as the probability that a dependent variable is 1 rather than 0.

What is $e$? The Taylor expansion

with $x = 1$ gives

The partial sums converge very quickly:

import math
print(math.e)
for ii in range(10):
    print(sum(1 / math.factorial(k) for k in range(ii)))

Solution concepts of $e^{at}x(0)$. The natural response of $\dot x = ax$ is $e^{at}x(0)$. When $a\lt 0$, define the time constant $\tau \triangleq 1/|a|$. Four numbers are worth memorizing:

After $3\tau$ the transient has decayed to about 5% of its initial size and has approximately converged; after $4\tau$ it is within 2%. This is where $T_s = 4/a$ comes from, and the 63% of the time constant is $1 - e^{-1}$. For the step input, when $a\lt 0$ and $u(t) = 1(t)$, the solution of $\dot x = ax + bu$ with $x(0)=0$ is $x(t) = \frac{b}{|a|}\left(1 - e^{at}\right)$.

Pole locations and the response

The first-order case generalizes. Each pole $\lambda$ contributes a term $e^{\lambda t}$ to the natural response. For a real pole the term is a pure exponential. For a complex pole $\lambda = \sigma + j\omega$, Euler’s formula gives

Complex poles of a real system come in conjugate pairs, and the imaginary parts combine into a real oscillation $e^{\sigma t}\cos(\omega t + \phi)$. The real part $\sigma$ sets the envelope and the imaginary part $\omega$ sets the oscillation frequency. This gives a map from pole location to response shape:

• Real pole in the LHP ($\lambda\in\mathbb{R}$, $\lambda\lt 0$): exponential decay.
• Pole at the origin: a constant.
• Real pole in the RHP: exponential growth.
• Complex pair in the LHP: decaying oscillation.
• Pair on the imaginary axis ($\lambda = \pm j\omega$): sustained, bounded oscillation.
• Complex pair in the RHP: growing oscillation.

Moving left speeds up the decay; moving up or down raises the oscillation frequency.

Second-order systems

A second-order differential equation gives a second-order system. Its response has an exponential decay or growth and possibly an oscillation. The mass-spring-damper of Lecture 3 is the standard example:

Transforming with zero initial conditions,

If the velocity is the output instead, multiplying by $s$ (differentiation in time) gives $\frac{sX(s)}{F(s)} = \frac{s/m}{s^2 + \frac bms + \frac km}$. The denominator, and so the poles, is the same; only a zero at the origin is added.

The generalized second-order system. We write

where $k$ is the DC gain and

• $\omega_n$ is the natural frequency, the frequency of oscillation without damping;
• $\zeta$ is the damping ratio, $\zeta = \dfrac{\text{exponential decay frequency}}{\text{natural frequency}} = \dfrac{\zeta\omega_n}{\omega_n}$;
• $\omega_d = \omega_n\sqrt{1-\zeta^2}$ is the damped frequency of oscillation, defined for $0\lt \zeta\lt 1$.

For the mass-spring-damper, $\omega_n = \sqrt{k/m}$ and $\zeta = b/(2\sqrt{km})$.

Poles. The roots of $s^2 + 2\zeta\omega_ns + \omega_n^2$ are

Four cases. For a unit step input,

and the damping ratio determines which of four responses we get.

1. Overdamped, $\zeta\gt 1$. Two distinct real poles $p_{1,2} = -\zeta\omega_n\pm\omega_n\sqrt{\zeta^2-1}$. Natural response $c_1e^{p_1t} + c_2e^{p_2t}$: a sluggish approach with no overshoot, dominated by the slower pole.
2. Underdamped, $0\lt \zeta\lt 1$. A complex-conjugate pair $p_{1,2} = -\zeta\omega_n\pm j\omega_d$. Natural response $c_1e^{-\zeta\omega_nt}\cos(\omega_dt + \phi)$: an oscillation at the damped frequency $\omega_d$, with phase shift $\phi$, inside a decaying envelope.
3. Undamped, $\zeta=0$. Imaginary poles $p_{1,2} = \pm j\omega_n$. Natural response $c_1\cos(\omega_nt + \phi)$: an oscillation that never decays.
4. Critically damped, $\zeta=1$. A repeated real pole $p_{1,2} = -\omega_n$. Natural response $c_1te^{-\omega_nt} + c_2e^{-\omega_nt}$: the fastest response without overshoot.

As $\zeta$ decreases from 2 to 0, the poles move from the real axis (overdamped), meet at $-\omega_n$ (critically damped), split onto a circle of radius $\omega_n$ (underdamped) and reach the imaginary axis (undamped). The step responses go from sluggish to oscillatory accordingly.

Summary and outlook

• Total response = forced response + natural response. Poles ($\times$) are roots of the denominator, zeros ($\circ$) roots of the numerator; poles decide stability.
• First-order systems $b/(s+a)$: $y = \frac ba(1-e^{-at})$, with $\tau = 1/a$, $T_r = 2.2/a$, $T_s = 4/a$.
• $e^{-1}, e^{-2}, e^{-3}, e^{-4}\approx 37\%, 14\%, 5\%, 2\%$. Exponentials also describe population growth; the logistic curve and the sigmoid are their bounded cousins.
• Each pole contributes $e^{\lambda t}$: the real part sets decay or growth, the imaginary part sets oscillation.
• Second-order systems $k\omega_n^2/(s^2+2\zeta\omega_ns+\omega_n^2)$ are overdamped, critically damped, underdamped or undamped depending on $\zeta$.

The underdamped case is the most important in practice, because most well-tuned control loops behave this way. The next lecture derives its step response exactly and turns it into design specifications: peak time, overshoot and settling time.

References

• N. S. Nise, Control Systems Engineering, 6th ed.: §4.1 Introduction (p. 162); §4.2 Poles, Zeros, and System Response (p. 162); §4.3 First-Order Systems (p. 166); §4.4 Second-Order Systems: Introduction (p. 168); §4.5 The General Second-Order System (p. 173).
• X. Chen and M. Tomizuka, Introduction to Modern Controls, with Illustrations in MATLAB and Python: §6.1.1 The Solution to $\dot x = ax + bu$ (p. 103); §6.1.2 The Irrational Number $e$ (p. 105); §8.2.1 Method of Eigenvalue Locations (p. 144).

Adapted from the ME 545 slides “Time Response, Poles, Zeros, Higher Order Systems” (T. Chu, UW ME 545, 2024) and the additional notes “e and time constant” (X. Chen); state-space material from those slides is left to ME 547.