(a)

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y_0 = dsolve(‘D2y+4*Dy+3*y=0’,‘y(0)=3’,‘Dy(0)=-7’,‘t’) y_0 = 1/exp(t) + 2/exp(3*t)

For k = 3, the zero-input response is therefore y0(t) = e−^t +2e−3*^t* . (b)

y_0 = dsolve(‘D2y+4*Dy+4*y=0’,‘y(0)=3’,‘Dy(0)=-7’,‘t’) y_0 = 3/exp(2*t) - t/exp(2*t)

For k = 4, the zero-input response is therefore y0(t) = 3e−2*^t* −te−2*^t* . (c)

$$y_0 = dsolve('D2y+4*Dy+40*y=0', 'y(0)=3', 'Dy(0)=-7', 't')$$

\n $y_0 = (3*cos(6*t))/exp(2*t) - sin(6*t)/(6*exp(2*t))$

For ^{k =} 40, the zero-input response is therefore ^y0(t) ^{= 3}e−2^t cos(6t)^{− 1 6} e−2*^tsin(6t*).

DR ILL 2.1 Finding the Zero-Input Response of a First-Order System

Find the zero-input response of an LTIC system described by (D + 5)y(t) = x(t) if the initial condition is y(0) = 5.

ANSWER

y0(t) = 5e−5*^t t* ≥ 0

DR ILL 2.2 Finding the Zero-Input Response of a Second-Order System

Letting y0(0) = 1 and y˙0(0) = 4, solve

(D^{2 +}2D)y0(t) ^{= 0}

ANSWER

y0(t) = 3−2e−2*^t t* ≥ 0

PRACTICAL INITIAL CONDITIONS AND THE MEANING OF 0− AND 0+

In Ex. 2.1 the initial conditions y0(0) and y˙0(0) were supplied. In practical problems, we must derive such conditions from the physical situation. For instance, in an RLC circuit, we may be given the conditions (initial capacitor voltages, initial inductor currents, etc.).

From this information, we need to derive y0(0), y˙0(0), … for the desired variable as demonstrated in the next example.

In much of our discussion, the input is assumed to start at t = 0, unless otherwise mentioned. Hence, t = 0 is the reference point. The conditions immediately before t = 0 (just before the input is applied) are the conditions at t = 0−, and those immediately after t = 0 (just after the input is applied) are the conditions at t = 0⁺ (compare this with the historical time frames BCE and CE). In practice, we are likely to know the initial conditions at t = 0⁻ rather than at t = 0+. The two sets of conditions are generally different, although in some cases they may be identical.

The total response y(t) consists of two components: the zero-input response y0(t) [response due to the initial conditions alone with x(t) = 0] and the zero-state response resulting from the input alone with all initial conditions zero. At t = 0−, the total response y(t) consists solely of the zero-input response y0(t) because the input has not started yet. Hence the initial conditions on y(t) are identical to those of y0(t). Thus, y(0−) = y0(0−), y˙(0−) = ˙y0(0−), and so on. Moreover, y0(t) is the response due to initial conditions alone and does not depend on the input x(t). Hence, application of the input at t = 0 does not affect y0(t). This means the initial conditions on y0(t) at t = 0⁻ and 0⁺ are identical; that is, y0(0−), y˙0(0−), … are identical to y0(0+), y˙0(0+), …, respectively. It is clear that for y0(t), there is no distinction between the initial conditions at t = 0−, 0, and 0+. They are all the same. But this is not the case with the total response y(t), which consists of both the zero-input and zero-state responses. Thus, in general, y(0−) = y(0+), y˙(0−) = ˙y(0+), and so on.

EXAMPLE 2.4 Consideration of Initial Conditions

A voltage x(t) = 10e−3*^t u*(t) is applied at the input of the RLC circuit illustrated in Fig. 2.2a. Find the loop current y(t) for t ≥ 0 if the initial inductor current is zero [y(0−) = 0] and the initial capacitor voltage is 5 volts [vC(0−) = 5].

The differential (loop) equation relating y(t) to x(t) was derived in Eq. (1.29) as

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

The zero-state component of y(t) resulting from the input x(t), assuming that all initial conditions are zero, that is, y(0−) = vC(0−) = 0, will be obtained later in Ex. 2.9. In this example we shall find the zero-input reponse y0(t). For this purpose, we need two initial conditions, y0(0) and y˙0(0). These conditions can be derived from the given initial conditions, y(0−) = 0 and vC(0−) = 5, as follows. Recall that y0(t) is the loop current when the input terminals are shorted so that the input x(t) = 0 (zero-input), as depicted in Fig. 2.2b. We now compute y0(0) and y˙0(0), the values of the loop current and its derivative at t = 0, from the initial values of the inductor current and the capacitor voltage. Remember that the inductor current cannot change instantaneously in the absence of an impulsive voltage. Similarly, the capacitor voltage cannot change instantaneously in the absence of an impulsive current. Therefore, when the input terminals are shorted at t = 0, the inductor current is still zero and the capacitor voltage is still 5 volts. Thus,

y0(0) = 0

Figure 2.1 Circuits for Ex. 2.4.

To determine y˙0(0), we use the loop equation for the circuit in Fig. 2.2b. Because the voltage across the inductor is L(dy0/dt) or y˙0(t), this equation can be written as follows:

$$\dot{y}_0(t) + 3y_0(t) + v_C(t) = 0$$

Setting t = 0, we obtain

$$\dot{y}_0(0) + 3y_0(0) + v_C(0) = 0$$

But y0(0) = 0 and vC(0) = 5. Consequently,

y˙0(0) = −5

Therefore, the desired initial conditions are

$$y_0(0) = 0$$

and $\dot{y}_0(0) = -5$

Thus, the problem reduces to finding y0(t), the zero-input component of y(t) of the system specified by the equation (D² +3D+2)y(t) = Dx(t), when the initial conditions are y0(0) = 0 and y˙0(0) = −5. We have already solved this problem in Ex. 2.1a, where we found

$$y_0(t) = -5e^{-t} + 5e^{-2t} \qquad t \ge 0$$

This is the zero-input component of the loop current y(t).

It is interesting to find the initial conditions at t = 0⁻ and 0⁺ for the total response y(t). Let us compare y(0−) and y˙(0−) with y(0+) and y˙(0+). The two pairs can be compared by writing the loop equation for the circuit in Fig. 2.2a at t = 0⁻ and t = 0+. The only difference between the two situations is that at t = 0−, the input x(t) = 0, whereas at t = 0+, the input x(t) = 10 [because x(t) = 10e−3*^t* ]. Hence, the two loop equations are

$$\dot{y}(0^-) + 3y(0^-) + v_C(0^-) = 0$$ $$\dot{y}(0^+) + 3y(0^+) + v_C(0^+) = 10$$

The loop current y(0+) = y(0−) = 0 because it cannot change instantaneously in the absence of impulsive voltage. The same is true of the capacitor voltage. Hence, vC(0+) = vC(0−) = 5. Substituting these values in the foregoing equations, we obtain y˙(0−) = −5 and y˙(0+) = 5. Thus,

y(0−) = 0, y˙(0−) = −5 and y(0+) = 0, y˙(0+) = 5 (2.10)

DR ILL 2.3 Zero-Input Response of an RC Circuit

In the circuit in Fig. 2.2a, the inductance L = 0 and the initial capacitor voltage vC(0) = 30 volts. Show that the zero-input component of the loop current is given by y0(t) = −10e−2t/³ for t ≥ 0.

INDEPENDENCE OF THE ZERO-INPUT AND ZERO-STATE RESPONSES

In Ex. 2.4 we computed the zero-input component without using the input x(t). The zero-state response can be computed from the knowledge of the input x(t) alone; the initial conditions are assumed to be zero (system in zero state). The two components of the system response (the zero-input and zero-state responses) are independent of each other. The two worlds of zero-input response and zero-state response coexist side by side, neither one knowing or caring what the other is doing. For each component, the other is totally irrelevant.

ROLE OF AUXILIARY CONDITIONS IN SOLUTION OF DIFFERENTIAL EQUATIONS

The solution of a differential equation requires additional pieces of information (the auxiliary conditions). Why? We now show heuristically why a differential equation does not, in general, have a unique solution unless some additional constraints (or conditions) on the solution are known.

Differentiation operation is not invertible unless one piece of information about y(t) is given. To get back y(t) from dy/dt, we must know one piece of information, such as y(0). Thus, differentiation is an irreversible (noninvertible) operation during which certain information is lost. To invert this operation, one piece of information about y(t) must be provided to restore the original y(t). Using a similar argument, we can show that, given d²y/dt2, we can determine y(t) uniquely only if two additional pieces of information (constraints) about y(t) are given. In general, to determine y(t) uniquely from its Nth derivative, we need N additional pieces of information (constraints) about y(t). These constraints are also called auxiliary conditions. When these conditions are given at t = 0, they are called initial conditions.

2.2-1 Some Insights into the Zero-Input Behavior of a System

By definition, the zero-input response is the system response to its internal conditions, assuming that its input is zero. Understanding this phenomenon provides interesting insight into system behavior. If a system is disturbed momentarily from its rest position and if the disturbance is then

162 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS

removed, the system will not come back to rest instantaneously. In general, it will come back to rest over a period of time and only through a special type of motion that is characteristic of the system.† For example, if we press on an automobile fender momentarily and then release it at t = 0, there is no external force on the automobile for t > 0.‡ The auto body will eventually come back to its rest (equilibrium) position, but not through any arbitrary motion. It must do so by using only a form of response that is sustainable by the system on its own without any external source, since the input is zero. Only characteristic modes satisfy this condition. The system uses a proper combination of characteristic modes to come back to the rest position while satisfying appropriate boundary (or initial) conditions.

If the shock absorbers of the automobile are in good condition (high damping coefficient), the characteristic modes will be monotonically decaying exponentials, and the auto body will come to rest rapidly without oscillation. In contrast, for poor shock absorbers (low damping coefficients), the characteristic modes will be exponentially decaying sinusoids, and the body will come to rest through oscillatory motion. When a series RC circuit with an initial charge on the capacitor is shorted, the capacitor will start to discharge exponentially through the resistor. This response of the RC circuit is caused entirely by its internal conditions and is sustained by this system without the aid of any external input. The exponential current waveform is therefore the characteristic mode of the RC circuit.

Mathematically we know that any combination of characteristic modes can be sustained by the system alone without requiring an external input. This fact can be readily verified for the series RL circuit shown in Fig. 2.2. The loop equation for this system is

$$(D+2)y(t) = x(t)$$

It has a single characteristic root λ = −2, and the characteristic mode is e−2*^t* . We now verify that a loop current y(t) = ce−2*^t* can be sustained through this circuit without any input voltage. The input voltage x(t) required to drive a loop current y(t) = ce−2*^t* is given by

$$x(t) = L\frac{dy(t)}{dt} + Ry(t)$$

= $\frac{d}{dt}(ce^{-2t}) + 2ce^{-2t}$
= $-2ce^{-2t} + 2ce^{-2t} = 0$

Figure 2.2 Modes always get a free ride.

^† This assumes that the system will eventually come back to its original rest (or equilibrium) position.

^‡ We ignore the force of gravity, which merely causes a constant displacement of the auto body without affecting the other motion.

Clearly, the loop current y(t) = ce−2*^t* is sustained by the RL circuit on its own, without the necessity of an external input.

THE RESONANCE PHENOMENON

We have seen that any signal consisting of a system’s characteristic mode is sustained by the system on its own; the system offers no obstacle to such signals. Imagine what would happen if we were to drive the system with an external input that is one of its characteristic modes. This would be like pouring gasoline on a fire in a dry forest or hiring a child to eat ice cream. A child would gladly do the job without pay. Think what would happen if he were paid by the amount of ice cream he ate! He would work overtime. He would work day and night, until he became sick. The same thing happens with a system driven by an input of the form of characteristic mode. The system response grows without limit, until it burns out.† We call this behavior the resonance phenomenon. An intelligent discussion of this important phenomenon requires an understanding of the zero-state response; for this reason we postpone this topic until Sec. 2.6-7.

2.3 THE UNIT IMPULSE RESPONSE h(t)

In Ch. 1 we explained how a system response to an input x(t) may be found by breaking this input into narrow rectangular pulses, as illustrated earlier in Fig. 1.27a, and then summing the system response to all the components. The rectangular pulses become impulses in the limit as their widths approach zero. Therefore, the system response is the sum of its responses to various impulse components. This discussion shows that if we know the system response to an impulse input, we can determine the system response to an arbitrary input x(t). We now discuss a method of determining h(t), the unit impulse response of an LTIC system described by the Nth-order differential equation [Eq. (2.1)]

$$\frac{d^N y(t)}{dt^N} + a_1 \frac{d^{N-1} y(t)}{dt^{N-1}} + \dots + a_{N-1} \frac{dy(t)}{dt} + a_N y(t)$$

= $b_{N-M} \frac{d^M x(t)}{dt^M} + b_{N-M+1} \frac{d^{M-1} x(t)}{dt^{M-1}} + \dots + b_{N-1} \frac{dx(t)}{dt} + b_N x(t)$

Recall that noise considerations restrict practical systems to M ≤ N. Under this constraint, the most general case is M = N. Therefore, Eq. (2.1) can be expressed as

$$(DN + a1DN-1 + \dots + aN-1D + aN)y(t) = (b0DN + b1DN-1 + \dots + bN-1D + bN)x(t)$$

(2.11)

Before deriving the general expression for the unit impulse response h(t), it is illuminating to understand qualitatively the nature of h(t). The impulse response h(t) is the system response to an impulse input δ(t) applied at t = 0 with all the initial conditions zero at t = 0−. An impulse input δ(t) is like lightning, which strikes instantaneously and then vanishes. But in its wake, in that single moment, objects that have been struck are rearranged. Similarly, an impulse input δ(t) appears momentarily at t = 0, and then it is gone forever. But in that moment it generates energy storages; that is, it creates nonzero initial conditions instantaneously within the system at

^† In practice, the system in resonance is more likely to go in saturation because of high amplitude levels.

t = 0+. Although the impulse input δ(t) vanishes for t > 0 so that the system has no input after the impulse has been applied, the system will still have a response generated by these newly created initial conditions. The impulse response h(t), therefore, must consist of the system’s characteristic modes for t ≥ 0+. As a result,

h(t) = characteristic mode terms t ≥ 0⁺

This response is valid for t > 0. But what happens at t = 0? At a single moment t = 0, there can at most be an impulse,† so the form of the complete response h(t) is

$$h(t) = A_0 \delta(t) + \text{characteristic mode terms} \qquad t \ge 0 \tag{2.12}$$

because h(t) is the unit impulse response. Setting x(t) = δ(t) and y(t) = h(t) in Eq. (2.11) yields

$$(DN + a1DN-1 + \dots + aN-1D + aN)h(t) = (b0DN + b1DN-1 + \dots + bN-1D + bN)\delta(t)$$

In this equation we substitute h(t) from Eq. (2.12) and compare the coefficients of similar impulsive terms on both sides. The highest order of the derivative of impulse on both sides is N, with its coefficient value as A⁰ on the left-hand side and b⁰ on the right-hand side. The two values must be matched. Therefore, A⁰ = b⁰ and

$$h(t) = b_0 \delta(t) + \text{characteristic modes} \tag{2.13}$$

In Eq. (2.11), if M < N, b⁰ = 0. Hence, the impulse term b0δ(t) exists only if M = N. The unknown coefficients of the N characteristic modes in h(t) in Eq. (2.13) can be determined by using the technique of impulse matching, as explained in the following example.