First-order systems - System Analysis and Control

First-order systems are the simplest type of dynamic system, and they provide a foundation for understanding more complex systems. In this chapter, we will explore the properties of first-order systems, how to analyze their behavior, and how to find their responses to various inputs.

Examples¶

The system order refers to the highest power of the derivative in the system’s differential equation. So a first-order system has only one derivative. There are many such examples across different domains of science and engineering.

Examples of first-order systems

Here are examples of first-order systems from a variety of domains. Click the tabs to see descriptions and equations for each example.

Mass-damper

RC circuit

RL circuit

DC motors

Magnetic relaxation

Optics

Hydraulic

Heat transfer

Radioactive decay

Chemical reactions

Pharmacokinetics

Population growth

Epidemiology

Finance

Mass-damper (with no spring). For example, the linearized cruise-control example we derived in the linearization chapter have the form:

m\dot v(t) + b v(t) = F_a(t)

(1)

Each of the examples above has a single “storage” element that accumulates or depletes a quantity over time (e.g., mass, charge, energy, population). In each case, the rate of change of that quantity depends linearly on the current value of the quantity and possibly on an external input. If there were additional storage elements (e.g., a spring in the mass-damper system, an inductor in the RC circuit, an additional species in the chemical reaction), the system would be of higher order.

Transfer function and canonical form¶

All the examples above have a similar mathematical structure. They can all be modeled as first-order linear ODE relating an output variable $y(t)$ to an input variable $u(t)$ :

a \dot y(t) + b y(t) = c u(t)

(15)

where $a$ , $b$ , and $c$ are constants that depend on the specific system. The system therefore has a transfer function of the form:

G(s) = \frac{Y(s)}{U(s)} = \frac{c}{a s + b} = \frac{\frac{c}{b}}{\frac{a}{b} s + 1}

(16)

Renaming $\tau = \frac{a}{b}$ and $K = \frac{c}{b}$ gives the canonical form of a first-order system:

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

(17)

Returning to the time domain, the corresponding canonical-form ODE is:

\boxed{\tau \dot y(t) + y(t) = K u(t)}

(18)

The two parameters $\tau$ and $K$ have names:

$\tau$ is the time constant of the system, which sets the timescale of the system’s response to changes in the input. It always has units of time (e.g., seconds).
$K$ is the DC gain of the system. Although “DC” stands for “direct current” in electrical engineering, here it refers to the steady-state output. More on this later. The units of $K$ are the units of $y/u$ . For example, if $y$ is a position in meters and $u$ is a force in newtons, then $K$ has units of m/N.

Example: cruise control¶

Consider the cruise control example from the linearization chapter. The linearized equations of motion are:

m \delta \dot v + 2 c v_0\, \delta v = \delta F

(19)

To put this in canonical form, we must have a coefficient of 1 in front of $\delta v$ . Dividing through by $2 c v_0$ gives:

\frac{m}{2 c v_0} \delta \dot v + \delta v = \frac{1}{2 c v_0} \delta F

(20)

Therefore the time constant and DC gain of the cruise control system are:

\boxed{\tau = \frac{m}{2 c v_0}, \quad K = \frac{1}{2 c v_0}}

(21)

Example: ideal DC motor¶

When the load torque $T_L$ is zero, the ideal DC motor model has transfer function:

G(s) = \frac{\Omega(s)}{V_{\textsf{in}}(s)} = \frac{K_m}{JR s + K_m^2}

(22)

where $K_m$ is the motor constant. We need the constant term in the denominator to be 1, so we divide numerator and denominator by $K_m^2$ :

G(s) = \frac{\frac{K_m}{K_m^2}}{\frac{JR}{K_m^2} s + 1} = \frac{\frac{1}{K_m}}{\frac{JR}{K_m^2} s + 1}

(23)

Therefore, the time constant and DC gain of the ideal DC motor are:

\boxed{\tau = \frac{JR}{K_m^2}, \quad K = \frac{1}{K_m}}

(24)

We can check that the units of $\tau$ are correct:

\begin{aligned} [\tau] &= \frac{[J][R]}{[K_m]^2} = \frac{\text{kg}\cdot\text{m}^2 \cdot \Omega}{(\text{N}\cdot\text{m}/\text{A})^2} = \frac{\text{kg}\cdot \text{A}^2 \cdot \Omega }{\text{N}^2} \\ &= \frac{\text{kg}\cdot \text{A}^2 \cdot \Omega }{\text{N} \cdot \text{kg}\cdot \text{m}/\text{s}^2} = \frac{\text{A}^2 \cdot \Omega}{\text{N}\cdot \text{m}/\text{s}} \cdot \text{s} = \text{s} \end{aligned}

(25)

where the last step follows because $\text{A}^2 \cdot \Omega = \text{N}\cdot\text{m}/\text{s} = \text{W}$ are all units of power.

Step response¶

Let’s start with how a first-order system responds to a unit step input. The Laplace transform of a unit step input is $U(s) = \frac{1}{s}$ , so the output in the Laplace domain is:

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

(26)

To find the time-domain response, we can use partial fraction expansion and take the inverse Laplace transform.

\frac{K}{s(\tau s + 1)} = \frac{K/\tau}{s(s+1/\tau)} = \frac{A}{s} + \frac{B}{s + 1/\tau}

(27)

Solving for $A$ and $B$ gives $A = K$ and $B = -K$ . Therefore:

Y(s) = \frac{K}{s} - \frac{K}{s + 1/\tau}

(28)

Taking the inverse Laplace transform gives the step response:

\boxed{y(t) = K \bigl(1 - e^{-t/\tau}\bigr)}

(29)

Since $t$ only appears in the combination $t/\tau$ , we can think of $t$ as being measured in units of $\tau$ . The output is also scaled by $K$ , so we can make a single table that tells us how all first-order systems respond to a unit step input:

Step response of a first-order system to a unit step input

$t$	$y(t)$
$1\,\tau$	$0.6321\,K$
$2\,\tau$	$0.8647\,K$
$3\,\tau$	$0.9502\,K$
$4\,\tau$	$0.9817\,K$
$5\,\tau$	$0.9933\,K$
$\infty$	$1.0000\,K$

We can see that the output starts at 0 and asymptotically approaches $K$ as $t \to \infty$ . The time constant $\tau$ determines how quickly the output approaches its steady-state value $K$ . After one time constant ( $t = \tau$ ), the output has reached about $63\%$ of its final value. After five time constants ( $t = 5\tau$ ), the output is very close to its final value, at about $99\%$ .

We can also look at the slope of the step response at $t=0$ . The slope is given by the derivative of $y(t)$ at $t=0$ :

\dot y(t) = \frac{K}{\tau} e^{-t/\tau} \quad\implies\quad \dot y(0) = \frac{K}{\tau}

(30)

We can use this information to sketch the step response of a first-order system without doing any calculations.

First-order step response. The dashed lines show tangent lines to the curve. The curve always points towards its destination (y=K) at one time constant into the future. — Figure 1:First-order step response. The dashed lines show tangent lines to the curve. The curve always points towards its destination ( $y=K$ ) at one time constant into the future.

Settling time¶

The step response of a first-order system never actually reaches its final value $K$ , but it gets arbitrarily close as time goes on. In practice, we define a settling time as the time it takes for the output to get within a certain percentage of its final value and stay there.

Settling time is an important performance metric for control systems. It tells us how quickly the system responds to changes in the input and reaches its desired output. A smaller settling time means a faster response, which is often desirable in control applications.

Impulse response¶

We can calculate the impulse response in a similar way to the step response. The Laplace transform of a unit impulse input is $U(s) = 1$ , so the output in the Laplace domain is:

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

(31)

Taking the inverse Laplace transform gives the impulse response:

\boxed{y(t) = \frac{K}{\tau} e^{-t/\tau}}

(32)

This is a decaying exponential with the same time constant $\tau$ as the step response, but scaled by $K/\tau$ instead of $K$ . This means the settling time is the same as before. The impulse response starts at $K/\tau$ at $t=0$ and decays to 0 as $t \to \infty$ . Here is a plot of the impulse response:

First-order impulse response. The dashed lines show tangent lines to the curve. The curve always points towards its destination (y=0) at one time constant into the future. — Figure 2:First-order impulse response. The dashed lines show tangent lines to the curve. The curve always points towards its destination ( $y=0$ ) at one time constant into the future.

Physical interpretation¶

When inputs are more complicated than steps or impulses, the response of a first-order system can be more complicated. However, the key feature of a first-order system is that it always tries to “catch up” to the input, but always lags behind. The output will always point towards where the input will be one time constant into the future.

Here is an illustration of this behavior for a more complicated input, and different choices of $\tau$ . We picked $K=1$ for simplicity, but the same behavior would hold for any value of $K$ (just scale the output by $K$ ).

First-order responses (with K=1 and different \tau values) to the same input. The system is always trying to “catch up” to the input, but always lags behind. Smaller \tau values correspond to faster responses. — Figure 3:First-order responses (with $K=1$ and different $\tau$ values) to the same input. The system is always trying to “catch up” to the input, but always lags behind. Smaller $\tau$ values correspond to faster responses.

Another interpretation of a first-order response is that it smooths out the input. The time constant $\tau$ determines how quickly the system can respond to changes in the input. A smaller $\tau$ means a faster response and less smoothing, while a larger $\tau$ means a slower response and more smoothing. A more precise term for this is that first-order systems act as low-pass filters, which we will explore in more detail in later chapters.

Ramp response¶

As a final example, let’s look at the response of a first-order system to a ramp input:

u(t) = t \cdot H(t) = \begin{cases} t & t \geq 0 \\ 0 & t < 0 \end{cases}

(33)

The Laplace transform of a ramp input is $U(s) = \frac{1}{s^2}$ , so the output in the Laplace domain is:

Y(s) = G(s) U(s) = \frac{K}{\tau s + 1} \cdot \frac{1}{s^2} = \frac{K}{s^2(\tau s + 1)}

(34)

The partial fraction decomposition is:

\frac{K}{s^2(\tau s + 1)} = \frac{K/\tau}{s^2 (s + 1/\tau)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s + 1/\tau}

(35)

Solving for $A$ , $B$ , and $C$ gives $A = K$ , $B = K \tau$ , and $C = -K$ . Therefore:

Y(s) = \frac{K}{s} + \frac{K \tau}{s^2} - \frac{K}{s + 1/\tau}

(36)

Taking the inverse Laplace transform gives the ramp response:

\boxed{y(t) = K\tau \left( \frac{t}{\tau} - \bigl(1 - e^{-t/\tau}\bigr) \right)}

(37)

Here is a plot of the ramp response:

First-order ramp response. The system is trying to catch up to the ramp (top dashed line), but always lags behind, eventually settling to a constant offset of K\tau. — Figure 4:First-order ramp response. The system is trying to catch up to the ramp (top dashed line), but always lags behind, eventually settling to a constant offset of $K\tau$ .

Unlike the step and impulse responses, the ramp response does not settle to a constant value, since the input is continuously increasing. Instead, it does its best to catch up but ends up settling to a constant offset from the input ramp. The output lags behind the input by a constant amount $K\tau$ . This lag is known as the steady-state error of the system in response to a ramp input.

What about units?

Step response: Let’s suppose $y(t)$ has units of meters and $u(t)$ has units of volts. A unit step is $u(t) = 1\,\text{V} \cdot H(t)$ since we want the input to have a magnitude of $1\,\text{V}$ for $t \geq 0$ . The step response (29) is therefore multiplied by the extra factor of $1\,\text{V}$ :

y(t) = \underbrace{K\vphantom{\big)}}_{\text{m/V}}\cdot \underbrace{\bigl(1 - e^{-t/\tau}\bigr)}_{\textsf{unitless}} \cdot \underbrace{1\,\text{V}\vphantom{\big)}}_{\text{V}}

(38)

So the step response has units of $\text{m/V} \cdot \text{V} = \text{m}$ , as it should.

Ramp response: For a ramp, the input is $u(t) = t \cdot H(t)$ , which has units of seconds. To give this units of volts, we need to multiply by $1\,\text{V/s}$ . From Eq. (37),

y(t) = \underbrace{K\tau\vphantom{\bigg)}}_{\text{m}\cdot\text{s/V}} \cdot \underbrace{\left( \frac{t}{\tau} - \bigl(1 - e^{-t/\tau}\bigr) \right)}_{\textsf{unitless}} \cdot \underbrace{1\,\text{V/s}\vphantom{\bigg)}}_{\text{V/s}}

(39)

So the ramp response has units of $\text{m}$ , as it should.

Impulse response: For an impulse, the input is $u(t) = \delta(t)$ , which actually has units of $1/\text{s}$ , since $\int_{-\infty}^\infty \delta(t) \, \dd t = 1$ is unitless and $t$ is in seconds. To give this units of volts, we need to multiply by $1\,\text{Vs}$ . From Eq. (32),

y(t) = \underbrace{\frac{K}{\tau}\vphantom{\bigg)}}_{\text{m/V}\cdot\text{s}} \cdot \underbrace{e^{-t/\tau}\vphantom{\bigg)}}_{\textsf{unitless}} \cdot \underbrace{1\,\text{Vs}\vphantom{\bigg)}}_{\text{V}\cdot\text{s}}

(40)

So once again, the output has units of $\text{m}$ , as it should.

Test your knowledge¶

Exercise 1

For the following first-order systems selected from the examples at the start of this section, identify the time constant $\tau$ and DC gain $K$ .

System	ODE
Mass–damper	$m\dot v + b v = F_a$
RC circuit	$RC\,\dot v_c + v_c = v_\textsf{in}$
Ideal DC motor (with $T_L=0$ )	$JR\,\dot\omega + K_m^2\,\omega = K_m\,v_\textsf{in}$
Newton cooling	$\dot T = -k\,(T-T_\infty)$

Solution to Exercise 1 #

We transform each system into canonical form $\tau \dot y + y = K u$ by rescaling appropriatelyand we identify $\tau$ and $K$ :

System	$\tau$	$K$
Mass–damper	$\frac{m}{b}$	$\frac{1}{b}$
RC circuit	$RC$	$1$
Ideal DC motor	$\frac{JR}{K_m^2}$	$\frac{1}{K_m}$
Newton cooling	$\frac{1}{k}$	$T_\infty$

Solution to Exercise 2 #

Since the canonical form is $G(s) = \frac{K}{\tau s + 1}$ , we can identify $K$ and $\tau$ for each motor by comparing to the given transfer functions:

\begin{aligned} G_1(s) &= \frac{20}{0.2 s + 1}, & \tau_1 &= 0.2 & K_1 &= 20 \\ G_2(s) &= \frac{300}{s + 10} = \frac{30}{0.1 s + 1}, & \tau_2 &= 0.1 & K_2 &= 30 \\ G_3(s) &= \frac{80}{0.1 s + 2} = \frac{40}{0.05 s + 1}, & \tau_3 &= 0.05 & K_3 &= 40 \end{aligned}

(42)

The slowest time constant is $\tau_1 = 0.2$ , so this motor will take $4\tau_1 = 0.8$ seconds to settle to its final value. The largest DC gain is $K_3 = 40$ , so this motor will have the largest final value. We can use this information to make the scales on our plot. The result is shown below:

First-order step response comparison. Step responses of the three motors from . Vertical dashed lines indicated the settling time 4\tau for each motor. The slowest motor (blue) takes the longest to settle and hals the smallest final value, while the fastest motor (green) settles the quickest to the largest final value. — Figure 5:First-order step response comparison. Step responses of the three motors from Exercise 2. Vertical dashed lines indicated the settling time $4\tau$ for each motor. The slowest motor (blue) takes the longest to settle and hals the smallest final value, while the fastest motor (green) settles the quickest to the largest final value.