First-order case studies - System Analysis and Control

Before we dive into the case studies, let’s review first-order systems. We saw that the canonical transfer function of a first-order system is given by

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

(1)

Any stable first-order system can be put in this form. The DC gain $K$ tells us the steady-state output in response to a unit step input, and the time constant $\tau$ tells us how quickly the system responds to changes in the input. Specifically, the $2\%$ settling time is $4\tau$ .

Pole-zero plots¶

Recall the definition of poles and zeros. The poles of a system are the values of $s$ that make the denominator of the transfer function zero, and the zeros are the values of $s$ that make the numerator zero. Since poles and zeros are complex numbers in general, we can plot them on the complex plane. The poles are typically denoted by “x” and the zeros by “o”. For example,

Transfer function and associated pole-zero plot. The poles are located at s=-4 and s=-2 \pm 3j, and the zero is located at s=-3. — Figure 1:Transfer function and associated pole-zero plot. The poles are located at $s=-4$ and $s=-2 \pm 3j$ , and the zero is located at $s=-3$ .

Remember from our definition of stability that a system is stable if all of its poles have negative real part. On the diagram above, this is the region to the left of the imaginary axis, also called the left-half plane (LHP). A system is unstable if any of its poles are in the right-half plane (RHP). The zeros of a system do not affect stability; they can be in the LHP, RHP, or on the imaginary axis.

For a first-order system, there is only one pole and no zeros. The pole is located at $s = -\frac{1}{\tau}$ , which is always in the LHP since $\tau > 0$ , and therefore stable. The distance of the pole from the imaginary axis (i.e., the real part of the pole) determines how quickly the system responds to changes in the input. The closer the pole is to the imaginary axis, the slower the response; the farther away it is, the faster the response.

Figure 2:Possible pole locations on a pole-zero plot. The farther left in the left-half plane (LHP), the faster the pole. The closer to the imaginary axis, the slower the pole. Poles in the right-half plane (RHP) are unstable.

As we progress through the course, we will see how the location of poles and zeros on the complex plane affects the performance of more complex systems.

RC circuit¶

Consider the RC circuit illustrated below.

The ODE relating the input voltage to the capacitor voltage is given by:

RC \dot{v}_C(t) + v_C(t) = v_\textsf{in}(t)

(2)

This is a first-order ODE with $\tau = RC$ and $K=1$ . Initially, all voltages are zero.

Charging the capacitor¶

At time $t=0$ , we connect a 1 V battery battery, which acts like a step input $v_\textsf{in}(t) = H(t)$ . Using the calculation we did in the previous section, we can find the voltage across the capacitor as a function of time, which is Eq. (29):

\boxed{v_C(t) = 1 - e^{-t/(RC)}}

(3)

The capacitor voltage will rise from 0 V to 1 V, with a time constant of $\tau = RC$ . After a time of $4\tau$ , the capacitor voltage will be within $2\%$ of its final value of 1 V.

Discharging the capacitor¶

With the capacitor charged up to $1$ V, let’s see what happens if we step the battery down to $0$ V. This is equivalent to replacing the battery by a short circuit, and we obtain the figure below.

RC circuit from with the battery stepped down (v_\textsf{in}=0). This will discharge the capacitor through the resistor. — Figure 4:RC circuit from Figure 3 with the battery stepped down ( $v_\textsf{in}=0$ ). This will discharge the capacitor through the resistor.

Let’s solve the problem using the ODE. The ODE is the same as before, but now the input voltage is zero and we have an initial condition:

RC \dot{v}_C(t) + v_C(t) = 0,\qquad v_C(0) = 1

(4)

Taking Laplace transforms of both sides, we obtain

RC \left( s V_C(s) - v_C(0) \right) + V_C(s) = 0

(5)

Solving for $V_C(s)$ and substituting the initial condition, we get

V_C(s) = \frac{RC}{RCs + 1} v_C(0) = \frac{1}{s + 1/(RC)}

(6)

Taking the inverse Laplace transform, we find

\boxed{v_C(t) = e^{-t/(RC)}}

(7)

Comparing this equation to the charging case (3), we now have a decaying exponential instead of a rising exponential, but the time constant is still $\tau = RC$ . After $4$ ms, the capacitor voltage will be within $2\%$ of its final value of $0$ V, about $0.02$ V in this case.

DC motor¶

Let’s consider a more complex example: a small hobby DC motor. Here are the parameters:

Table 1:DC motor parameters

Parameter	Name	Value
$R$	Resistance of motor windings	$0.5$ Ω
$L$	Inductance of motor windings	$1.5$ mH
$K$	Motor constant	$0.05$ Nm/A
$J$	Moment of inertia of rotor	$2.5\cdot 10^{-4}$ kg m $^2$
$b$	Viscous friction coefficient	$10^{-4}$ Nms/rad

Step response¶

We previously derived the ODE for the DC motor, which is given by

J L\, \ddot\omega + (J R + b L)\, \dot\omega + (b R + K^2)\, \omega = K v_\textsf{in} - L\, \dot T_L - R\, T_L

(8)

Let’s assume zero load torque $T_L = 0$ and zero initial conditions. Taking Laplace transforms, we can find the transfer function from input voltage to angular velocity:

G(s) = \frac{\Omega(s)}{V_\textsf{in}(s)} = \frac{K}{J L s^2 + (J R + b L) s + (b R + K^2)}

(9)

Now let’s substitute in the numerical values for the parameters:

G(s) = \frac{0.05}{3.75\cdot 10^{-7} s^2 + 1.2515\cdot 10^{-4} s + 2.55\cdot 10^{-3}}

(10)

Let’s divide the numerator and denominator by $3.75\cdot 10^{-7}$ to clean things up a bit:

G(s) = \frac{133333}{s^2 + 334 s + 6800}

(11)

This is not a first-order system, but a quick calculation of the discriminant of the denominator: $\Delta = 334^2 - 4\cdot 6800 = 111556$ reveals that it is positive, meaning that there are actually two real poles. We can calculate them using the quadratic formula:

s = \frac{-334 \pm \sqrt{111556}}{2} = \begin{cases}-21.8 \\ -312\end{cases}

(12)

Therefore we can factor our transfer function, and find a partial fraction expansion using the cover-up method. The result is

\begin{aligned} G(s) &= \frac{133333}{(s + 21.8)(s + 312)} \\ &= \frac{459.5}{s + 21.8} - \frac{459.5}{s + 312} \\ &= \underbrace{\frac{21.1}{0.046 s + 1}}_{G_1(s)} + \underbrace{\frac{-1.47}{0.0032 s + 1}}_{G_2(s)} \end{aligned}

(13)

In the last step, we put each term of the PFE in canonical form. Now we can apply superposition: Because $G(s) = G_1(s) + G_2(s)$ , the step response of the motor $G$ is the sum of the step responses of $G_1$ and $G_2$ .

The first term $G_1(s)$ has a time constant of $46$ ms and a DC gain of $21.1$ .
The second term $G_2(s)$ has a time constant of $3.2$ ms and a DC gain of $-1.47$ .

The first term has a dominant effect due to its larger DC gain. It also has a much slower time constant, meaning that it will be the main contributor to the step response. Here is what the step response looks like:

DC motor step response decomposition. The motor step response is the sum of the step responses of G_1 and G_2. The first term G_1 dominates the response due to its larger DC gain and slower time constant. — Figure 5:DC motor step response decomposition. The motor step response is the sum of the step responses of $G_1$ and $G_2$ . The first term $G_1$ dominates the response due to its larger DC gain and slower time constant.

We can see in Figure 5 that the main effect of $G_2$ on the total response is to cause a small initial dip in the response, and eventually a smaller steady-state value.

Ideal motor approximation¶

In the ideal motor approximation, we have $L=0$ and $b=0$ , which means that the transfer function simplifies to

G_\textsf{ideal}(s) = \frac{K}{J R s + K^2} = \frac{20}{0.05 s + 1}

(14)

So a single first-order term with a time constant of 50 ms and a DC gain of 20. This is a good approximation to the actual motor, since the time constant of $G_1$ is close to 50 ms and its DC gain is close to 20.

Pole-zero plot¶

The pole-zero plot of the motor is shown below.

We saw that the bulk of the step response was due to the slow pole at $s=-21.8$ . The fast pole at $s=-312$ has a much smaller effect on the step response, since it has a much smaller DC gain and a much faster time constant.

It turns out that this phenomenon of dominant poles is quite common in practice. Later, we will see how to use this idea to simplify the analysis of complex systems by approximating them simpler systems.

Test your knowledge¶

Exercise 1

Consider the RC circuit shown below, which includes battery (voltage $v_s$ ), a push-button switch, and a light-emitting diode (LED). For simplicity, we will assume there is zero voltage drop across the LED ^[1] and that the brightness of the light emitted by the LED is proportional to the current flow through the LED, labeled $i$ on the circuit.

Figure 7:Fader circuit for an LED.

When the button is pressed (and held) down, the circuit is closed. Current flows from the battery into the capacitor and through the LED. Once the capacitor is fully charged, all the current flows through the LED. This has the effect that the LED fades on.

When the button is released, the circuit is open. At this point, the capacitor discharges its stored current into the LED. Once the capacitor is drained, there is no longer any current flowing and the LED is dark. This has the effect that the LED fades off.

How long will it take (in terms of $R_1$ , $R_2$ , and $C$ ) for the LED to fade on? To fade off?

Solution to Exercise 1 #

Button pressed down. In this case, the current is split: $I$ goes through the LED and $I_C$ into the capacitor. We can write down constitutive equations for each element using KCL for node $A$ :

\begin{aligned} V_s - V_A &= R_1 (I + I_C) \\ V_A &= R_2 I \\ V_A &= \frac{1}{Cs} I_C \end{aligned}

(15)

We care about the current $I$ , so let’s eliminate $I_C$ and $V_A$ . Substituting the last equation into the first to eliminate $I_C$ , we get

\begin{aligned} V_s - V_A &= R_1 \left(I + Cs V_A\right) \\ V_A &= R_2 I \end{aligned}

(16)

Now substitute the second equation into the first to eliminate $V_A$ and simplify:

\begin{aligned} V_s - R_2 I &= R_1 \left(I + Cs R_2 I\right) \\ \implies \quad V_s &= I \left(R_1 + R_2 + R_1 R_2 Cs\right) \end{aligned}

(17)

As a differential equation, this is:

R_1 R_2 C \,\frac{\dd i(t)}{\dd t} + (R_1 + R_2) i(t) = v_s(t)

(18)

This is a first-order ODE with time constant $\tau = \frac{R_1 R_2 C}{R_1 + R_2}$ and DC gain $K = \frac{1}{R_1 + R_2}$ . Therefore, the settling time for the LED to fade on is

\boxed{4\tau_\textsf{on} = \frac{4R_1 R_2 C}{R_1 + R_2}}

(19)

Button released. In this case, the current $I$ through the LED is the same as the current $I_C$ discharging the capacitor. The circuit becomes a simple RC circuit identical to the one we analyzed in Figure 4, except that the resistance is now $R_2$ instead of $R$ . The ODE for this case is

R_2 C \,\dot v_C(t) + v_C(t) = 0

(20)

This has time constant $R_2 C$ . However, we care about the current, not the capacitor voltage. It turns out that the current also has the same time constant! To see why, we can solve for the current, which is $i(t) = \frac{v_C(t)}{R_2}$ . Substituting this into the ODE, we get the same ODE:

R_2 C \,\frac{\dd i(t)}{\dd t} + i(t) = 0

(21)

Therefore, the settling time for the LED to fade off is

\boxed{4\tau_\textsf{off} = 4 R_2 C}

(22)

Footnotes¶

In reality, LEDs typically have a voltage drop of around 1.7-2.0 Volts over a wide range of current flows.
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