1.8 SYSTEM [MODEL: INPUT–OUTPUT](#page-8-0) DESCRIPTION

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1.8 SYSTEM MODEL: INPUT–OUTPUT DESCRIPTION

A system description in terms of the measurements at the input and output terminals is called the input–output description. As mentioned earlier, systems theory encompasses a variety of systems, such as electrical, mechanical, hydraulic, acoustic, electromechanical, and chemical, as well as social, political, economic, and biological. The first step in analyzing any system is the construction of a system model, which is a mathematical expression or a rule that satisfactorily approximates the dynamical behavior of the system. In this chapter we shall consider only continuous-time systems. Modeling of discrete-time systems is discussed in Ch. 3.

1.8-1 Electrical Systems

To construct a system model, we must study the relationships between different variables in the system. In electrical systems, for example, we must determine a satisfactory model for the voltage-current relationship of each element, such as Ohm’s law for a resistor. In addition, we must determine the various constraints on voltages and currents when several electrical elements are interconnected. These are the laws of interconnection—the well-known Kirchhoff laws for voltage and current (KVL and KCL). From all these equations, we eliminate unwanted variables to obtain equation(s) relating the desired output variable(s) to the input(s). The following examples demonstrate the procedure of deriving input–output relationships for some LTI electrical systems.

112 CHAPTER 1 SIGNALS AND SYSTEMS

By using the voltage-current laws of each element (inductor, resistor, and capacitor), we can express this equation as

$$\frac{dy(t)}{dt} + 3y(t) + 2\int_{-\infty}^{t} y(\tau) d\tau = x(t)$$

\n(1.27)

Differentiating both sides of this equation, we obtain

$$\frac{d^2y(t)}{dt^2} + 3\frac{dy(t)}{dt} + 2y(t) = \frac{dx(t)}{dt}$$

\n(1.28)

This differential equation is the input–output relationship between the output y(t) and the input x(t).

It proves convenient to use a compact notation D for the differential operator d/dt. This notation can be repeatedly applied. Thus,

$$\frac{dy(t)}{dt} \equiv Dy(t), \qquad \frac{d^2y(t)}{dt^2} \equiv D^2y(t), \qquad \dots, \qquad \frac{d^Ny(t)}{dt^N} \equiv D^Ny(t)$$

With this notation, Eq. (1.28) can be expressed as

$$(D2 + 3D + 2)y(t) = Dx(t)$$

\n(1.29)

The differential operator is the inverse of the integral operator, so we can use the operator 1/D to represent integration.†

$$\int_{-\infty}^{t} y(\tau) d\tau \equiv \frac{1}{D} y(t)$$ $$\frac{d}{dt} \left[ \int_{-\infty}^{t} y(\tau) d\tau \right] = y(t)$$

^† Use of operator 1/D for integration generates some subtle mathematical difficulties because the operators D and 1/D do not commute. For instance, we know that D(1/D) = 1 because

However, (1/D)D is not necessarily unity. Use of Cramer’s rule in solving simultaneous integro-differential equations will always result in cancellation of operators 1/D and D. This procedure may yield erroneous results when the factor D occurs in the numerator as well as in the denominator. This happens, for instance, in circuits with all-inductor loops or all-capacitor cut sets. To eliminate this problem, avoid the integral operation in system equations so that the resulting equations are differential rather than integro-differential. In electrical circuits, this can be done by using charge (instead of current) variables in loops containing capacitors and choosing current variables for loops without capacitors. In the literature this problem of commutativity of D and 1/D is largely ignored. As mentioned earlier, such a procedure gives erroneous results only in special systems, such as the circuits with all-inductor loops or all-capacitor cut sets. Fortunately such systems constitute a very small fraction of the systems we deal with. For further discussion of this topic and a correct method of handling problems involving integrals, see [4].

Consequently, Eq. (1.27) can be expressed as

$$\left(D+3+\frac{2}{D}\right)y(t) = x(t)$$

Multiplying both sides by D to differentiate the expression, we obtain

$$(D^2 + 3D + 2)y(t) = Dx(t)$$

which is identical to Eq. (1.29).

Recall that Eq. (1.29) is not an algebraic equation, and D²+3D+2 is not an algebraic term that multiplies y(t); it is an operator that operates on y(t). It means that we must perform the following operations on y(t): take the second derivative of y(t) and add to it 3 times the first derivative of y(t) and 2 times y(t). Clearly, a polynomial in D multiplied by y(t) represents a certain differential operation on y(t).

EXAMPLE 1.17 Input–Output Equation of a Series RC Circuit

Using operator notation, find the equation relating input to output for the series RC circuit of Fig. 1.35 if the input is the voltage x(t) and output is

• (a) the loop current i(t)
• (b) the capacitor voltage y(t)

(a) The loop equation for the circuit is

$$R i(t) + \frac{1}{C} \int_{-\infty}^{t} i(\tau) d\tau = x(t)$$

or

$$15i(t) + 5 \int_{-\infty}^{t} i(\tau) d\tau = x(t)$$

With operator notation, this equation can be expressed as

$$15 i(t) + \frac{5}{D} i(t) = x(t)$$

\n(1.30)

(b) Multiplying both sides of Eq. (1.30) by D (i.e., differentiating the equation), we obtain

(15D+5)i(t) = Dx(t)

Using the fact that ⁱ(t) ^{= C} dy(t) dt ^{= 1 5}Dy(t), simple substitution yields

$$(3D+1)y(t) = x(t)$$

\n(1.31)

DR ILL 1.17 Input–Output Equation of a Series RLC Circuit with Inductor Voltage as Output

If the inductor voltage vL(t) is taken as the output, show that the RLC circuit in Fig. 1.34 has an input–output equation of (D² +3D+2)vL(t) = D2x(t).

DR ILL 1.18 Input–Output Equation of a Series RC Circuit with Capacitor Voltage as Output

If the capacitor voltage vC(t) is taken as the output, show that the RLC circuit in Fig. 1.34 has an input–output equation of (D² +3D+2)vC(t) = 2x(t).

1.8-2 Mechanical Systems

Planar motion can be resolved into translational (rectilinear) motion and rotational (torsional) motion. Translational motion will be considered first. We shall restrict ourselves to motions in one dimension.

TRANSLATIONAL SYSTEMS

The basic elements used in modeling translational systems are ideal masses, linear springs, and dashpots providing viscous damping. The laws of various mechanical elements are now discussed.

For a mass M (Fig. 1.36a), a force x(t) causes a motion y(t) and acceleration y¨(t). From Newton’s law of motion,

$$x(t) = M\ddot{y}(t) = M\frac{d^2y(t)}{dt^2} = MD^2y(t)$$

The force x(t) required to stretch (or compress) a linear spring (Fig. 1.36b) by an amount y(t) is given by

$$x(t) = Ky(t)$$

where K is the stiffness of the spring.

Figure 1.36 Some elements in translational mechanical systems.

For a linear dashpot (Fig. 1.36c), which operates by virtue of viscous friction, the force moving the dashpot is proportional to the relative velocity y˙(t) of one surface with respect to the other. Thus

$$x(t) = B\dot{y}(t) = B\frac{dy(t)}{dt} = BDy(t)$$

where B is the damping coefficient of the dashpot or the viscous friction.

EXAMPLE 1.18 Input–Output Equation for a Translational Mechanical System

Find the input–output relationship for the translational mechanical system shown in Fig. 1.37a or its equivalent in Fig. 1.37b. The input is the force x(t), and the output is the mass position y(t).

116 CHAPTER 1 SIGNALS AND SYSTEMS

In mechanical systems it is helpful to draw a free-body diagram of each junction, which is a point at which two or more elements are connected. In Fig. 1.37, the point representing the mass is a junction. The displacement of the mass is denoted by y(t). The spring is also stretched by the amount y(t), and therefore it exerts a force −Ky(t) on the mass. The dashpot exerts a force −By˙(t) on the mass, as shown in the free-body diagram (Fig. 1.37c). By Newton’s second law, the net force must be My¨(t). Therefore,

$$M\ddot{y}(t) = -B\dot{y}(t) - Ky(t) + x(t)$$

or

$$(MD2 + BD + K)y(t) = x(t)$$

ROTATIONAL SYSTEMS

In rotational systems, the motion of a body may be defined as its motion about a certain axis. The variables used to describe rotational motion are torque (in place of force), angular position (in place of linear position), angular velocity (in place of linear velocity), and angular acceleration (in place of linear acceleration). The system elements are rotational mass or moment of inertia (in place of mass) and torsional springs and torsional dashpots (in place of linear springs and dashpots). The terminal equations for these elements are analogous to the corresponding equations for translational elements. If J is the moment of inertia (or rotational mass) of a rotating body about a certain axis, then the external torque required for this motion is equal to J (rotational mass) times the angular acceleration. If θ (t) is the angular position of the body, θ (¨ t) is its angular acceleration, and

torque =

$$J\ddot{\theta}(t) = J\frac{d^2\theta(t)}{dt^2} = JD^2\theta(t)$$

Similarly, if K is the stiffness of a torsional spring (per unit angular twist), and θ is the angular displacement of one terminal of the spring with respect to the other, then

torque =

$$K\theta(t)$$

Finally, the torque due to viscous damping of a torsional dashpot with damping coefficient B is

torque =

$$B\dot{\theta}(t)$$

= $BD\theta(t)$

EXAMPLE 1.19 Input–Output Equation for Aircraft Roll Angle

The attitude of an aircraft can be controlled by three sets of surfaces (shown shaded in Fig. 1.38): elevators, rudder, and ailerons. By manipulating these surfaces, one can set the aircraft on a desired flight path. The roll angle ϕ(t) can be controlled by deflecting in the opposite direction the two aileron surfaces as shown in Fig. 1.38. Assuming only rolling motion, find the equation relating the roll angle ϕ(t) to the input (deflection) θ (t).

Figure 1.38 Attitude control of an airplane.

J

The aileron surfaces generate a torque about the roll axis proportional to the aileron deflection angle θ (t). Let this torque be cθ (t), where c is the constant of proportionality. Air friction dissipates the torque Bϕ(˙ t). The torque available for rolling motion is then cθ (t) − Bϕ(˙ t). If J is the moment of inertia of the plane about the x axis (roll axis), then

net torque =

$$J\ddot{\varphi}(t) = c\theta(t) - B\dot{\varphi}(t)$$

and

$$J\frac{d^2\varphi(t)}{dt^2} + B\frac{d\varphi(t)}{dt} = c\theta(t) \qquad \text{or} \qquad (JD^2 + BD)\varphi(t) = c\theta(t)$$

This is the desired equation relating the output (roll angle ϕ(t)) to the input (aileron angle θ (t)).

The roll velocity ω(t) is ϕ(˙ t). If the desired output is the roll velocity ω(t) rather than the roll angle ϕ(t), then the input–output equation would be

$$\frac{d\omega(t)}{dt} + B\omega(t) = c\theta(t) \qquad \text{or} \qquad (JD + B)\omega(t) = c\theta(t)$$

DR ILL 1.19 Input–Output Equation of a Rotational Mechanical System

Torque T (t) is applied to the rotational mechanical system shown in Fig. 1.39a. The torsional spring stiffness is K; the rotational mass (the cylinder’s moment of inertia about the shaft) is J; the viscous damping coefficient between the cylinder and the ground is B. Find the equation relating the output angle θ (t) to the input torque T (t). [Hint: A free-body diagram is shown in Fig. 1.39b.]

1.8-3 Electromechanical Systems

A wide variety of electromechanical systems is used to convert electrical signals into mechanical motion (mechanical energy) and vice versa. Here we consider a rather simple example of an armature-controlled dc motor driven by a current source x(t), as shown in Fig. 1.40a. The torque T (t) generated in the motor is proportional to the armature current x(t). Therefore,

$$\mathcal{T}(t) = K_T x(t)$$

where KT is a constant of the motor. This torque drives a mechanical load whose free-body diagram is shown in Fig. 1.40b. The viscous damping (with coefficient B) dissipates a torque Bθ (˙ t). If J is the moment of inertia of the load (including the rotor of the motor), then the net torque T (t)−Bθ (˙ t) must be equal to Jθ (¨ t):

$$J\ddot{\theta}(t) = \mathcal{T}(t) - B\dot{\theta}(t)$$

Thus,

$$(JD2 + BD)\theta(t) = \mathcal{T}(t) = K_T x(t)$$

which in conventional form can be expressed as

$$J\frac{d^2\theta(t)}{dt^2} + B\frac{d\theta(t)}{dt} = K_T x(t)$$

\n(1.32)

Figure 1.40 Armature-controlled dc motor.