CLASSICAL SOLUTION OF DIFFERENTIAL EQUATIONS
2.5 CLASSICAL SOLUTION OF DIFFERENTIAL EQUATIONS
In the classical method we solve differential equation to find the natural and forced components rather than the zero-input and zero-state components of the response. Although this method is relatively simple compared with the method discussed so far, as we shall see, it also has several glaring drawbacks.
As Section 2.4-5 showed, when all the characteristic mode terms of the total system response are lumped together, they form the system's natural response y n (t) (also known as the homogeneous solution or complementary solution). The remaining portion of the response consists entirely of
noncharacteristic mode terms and is called the system's forced response y φ (t) (also known as the particular solution). Equation (2.52b) showed these two components for the loop current in the RLC circuit of Fig. 2.1a .
The total system response is y(t) = y n(t) +y φ (t). Since y(t) must satisfy the system equation [Eq. (2.1)], or But y n (t) is composed entirely of characteristic modes. Therefore so that
The natural response, being a linear combination of the system's characteristic modes, has the same form as that of the zero-input response; only its arbitrary constants are different. These constants are determined from auxiliary conditions, as explained later. We shall now discuss a method of determining the forced response.
2.5-1 Forced Response: The Method of Undetermined Coefficients
It is a relatively simple task to determine y φ(t) , the forced response of an LTIC system, when the input x(t) is such that it yields only a finite number of independent derivatives. Inputs having the form e ζt or t r fall into this category. For example, the repeated differentiation of e ζt yields the same form
as this input; that is, e ζt . Similarly, the repeated differentiation of t r yields only r independent derivatives. The forced response to such an input can be expressed as a linear combination of the input and its independent derivatives. Consider, for example, the input has at 2 + bt + c. The successive derivatives of this input are 2at + b and 2a. In this case, the input has only two independent derivatives. Therefore, the forced response can be
assumed to be a linear combination of x(t) and its two derivatives. The suitable form for y φ(t) in this case is, therefore
The undetermined coefficients β 0 , β 1 and β 2 are determined by substituting this expression for y φ (t)in Eq. (2.53)
and then equating coefficients of similar terms on both sides of the resulting expression. Although this method can be used only for inputs with a finite number of derivatives, this class of inputs includes a wide variety of the most
commonly encountered signals in practice. Table 2.2 shows a variety of such inputs and the form of the forced response corresponding to each input. We shall demonstrate this procedure with an example.
Table 2.2: Forced Response
Open table as spreadsheet
No. Input x(t)
Forced Response
1 e ζt ζ?? i (i = 1,2,...,N)
3 k (a constant)
β (a constant)
4 cos ( ωt + θ)
βcos (ωt + θ)
5 (t r +a r-1 t r-1 + ... + α 1 t+ α 0 )e ζt )
( β r t r + β r-1 t r-1 + ... + β 1 t+ β 0 )e ζt )
Note: By definition, y φ (t) cannot have any characteristic mode terms. If any term appearing in the right-hand column for the forced response is also a characteristic mode of the system, the correct form of the forced response must be modified to t i y φ (t), where i is the smallest possible integer that can be used and still can prevent t i y φ (t) from having a characteristic mode term. For example, when the input is e ζt , the forced
response (right-hand column) has the form βe ζt . But if e ζt happens to be a characteristic mode of the system, the correct form of the forced
response is βte ζt (see pair 2). If te ζt also happens to be a characteristic mode of the system, the correct form of the forced response is βt 2 e ζt ,
and so on.
EXAMPLE 2.10
Solve the differential equation if the input and the initial conditions are y(0 + ) = 2 and y(0 + ) = 3.
The characteristic polynomial of the system is Therefore, the characteristic modes are e −t and e -2t . The natural response is then a linear combination of these modes, so that
Here the arbitrary constants K 1 and K 2 must be determined from the system's initial conditions. The forced response to the input t 2 + 5t + 3, according to Table 2.2 (pair 5 with ζ = 0), is
Moreover, y φ (t) satisfies the system equation [ Eq. (2.53) ]; that is,
Now Now
Substituting these results in Eq. (2.54) yields or Equating coefficients of similar powers on both sides of this expression yields
Solution of these three simultaneous equations yields β 0 = 1, β 1 = 1, and β 2 = 0. Therefore
The total system response y(t) is the sum of the natural and forced solutions. Therefore
so that Setting t = 0 and substituting y(0) = 2 y(0) = 3 in these equations, we have
The solution of these two simultaneous equations is K 1 = 4 and K 2 = −3. Therefore
COMMENTS ON INTIAL CONDITIONS
In the classical method, the initial conditions are required at t = 0 + . The reason is that because at t = 0 − , only the zero-input component exists, and the initial conditions at t = 0 − can be applied to the zero-input component only. In the classical method, the zero-input and zero-state components cannot be separated Consequently, the initial conditions must be applied to the total response, which begins at t = 0 + .
EXERCISE E2.15
An LTIC system is specified by the equation
The input is x(t) = 6t 2 . Find the following: a. The forced response y φ (t) b. The total response y(t) if the initial conditions are y(0 + ) = 25/18 and y(0 + )= −2/3
Answers
a. b.
THE EXPONENTIAL INPUT e ζ(t) The exponential signal is the most important signal in the study of LTIC systems. Interestingly, the forced response for an exponential input signal
turns out to be very simple. From Table 2.2 we see that the forced response for the input e ζt has the form βe ζt . We now show that β = Q(ζ)/P(ζ). [ † ]
To determine the constant β, we substitute y φt in the system equation [ Eq. (2.53) ] to obtain Now observe that
Consequently Therefore, Eq. (2.53) becomes and
Thus, for the input x(t) = e ζt u(t), the forced response is given by
where
This is an interesting and significant result. It states that for an exponential input e ζt the forced response y ζ (t) is the same exponential multiplied by H( ζ) = P(ζ)/Q(ζ). The total system response y(t) to an exponential input e ζt is then given by [ ‡ ]
where the arbitrary constants K 1 ,K 2 ,..., K N are determined from auxiliary conditions. The form of Eq. (2.57) assumes N distinct roots. If the roots are not distinct, proper form of modes should be used.
Recall that the exponential signal includes a large variety of signals, such as a constant ( ζ = 0), a sinusoid (ζ = ± jω), and an exponentially growing or decaying sinusoid ( ζ = σ ± jω). Let us consider the forced response for some of these cases.
THE CONSTANT INPUT x(t) = C
Because C = C e 0t , the constant input is a special case of the exponential input Ce ζt with ζ = 0. The forced response to this input is then given by
THE EXPONENTIAL INPUT e j ωt Here ζ = jω and
THE SINUSOIDA INPUT x(t) = cos ωt We know that the forced response for the input e ±jwt is H(±jw)e ±jwt . Since cos ω t = (e j ωt +e −jwt )/2, the forced response to cos ωt is
Because the two terms on the right-hand side are conjugates, But so that
This result can be generalized for the input x(t) = cos ( ωt + θ). The forced response in this case is
EXAMPLE 2.11
Solve the differential equation if the initial conditions are y(0 + ) = 2 and
and the input is
a. 10e −3t b. 5 −2t c. e d. 10 cos(3t + 30°)
According to Example 2.10 , the natural response for this case is For this case
a. For input x(t) = 10e −3t ζ = −3, and
The total solution (the sum of the forced and the natural response) is and The initial conditions are y(0 + ) = 2 and
. Setting t = 0 in the foregoing equations and then substituting the initial conditions yields
Solution of these equations yields K 1 = −8 and K 2 = 25. Therefore
b. For input x (t) = 5 = 5e 0t , ζ = 0, and The complete solution is K 1 e −t +K 2 e −2t + 2te − t. Using the initial conditions, we determine K 1 and K 2 as in part (a).
c. Here ζ = −2, which is also a characteristic root of the system. Hence (see pair 2, Table 2.2 , or the note at the bottom of the table). To find β, we substitute yφ,(t) in the system equation to obtain or But
Consequently or Therefore, β = 2 so that
The complete solution is K e −t +K 2 e −2u . Using the initial conditions, we determine K 1 and K 2 as in part (a).
d. For the input x(t) = 10cos (3 t + 30°), the forced response [see Eq. (2.61) ] is where
Therefore
and The complete solution is K 1 e −1 +K 2 e −2t + 2.63cos (3t − 7.9°). We then use the initial conditions to determine K 1 and K 2 as in part (a).
EXAMPLE 2.12
Use the classical method to find the loop current y(t) in the RLC circuit of Example 2.2 ( Fig. 2.1) if the input voltage x(t) = 10 −3t e and the initial
conditions are y(0 − ) = 0 and υ c (0 − ) = 5.
The zero-input and zero-state responses for this problem are found in Examples 2.2 and 2.6 , respectively. The natural and forced responses appear in Eq. (2.52b) . Here we shall solve this problem by the classical method, which requires the initial conditions at t = 0 + . These conditions, already found in Eq. (2.15) , are
The loop equation for this system [see Example 2.2 or Eq. (1.55) ] is
The characteristic polynomial is λ 2 +3 λ + 2 = (λ + 1)(λ + 2) Therefore, the natural response is
The forced response, already found in part (a) of Example 2.11 , is The total response is Differentiation of this equation yields Setting t = 0 + and substituting y(0 + ) = 0, y(0 + ) = 5 in these equations yields
Therefore which agrees with the solution found earlier in Eq. (2.52b) .
COMPUTER EXAMPLE C2.4
Solve the differential equation
using the input x(t) = 5t + 3 and initial conditions y 0 (0) = 2 and y 0 (0) = 3.
>>y = dsolve('D2y+3*Dy+2*y=5*t+3', y(0)=2', 'Dy(0)=3', 't'); >>disp (['y(t) = (',char(y), ')u(t) ' ]); y(t) = (-9/4+5/2*t+9*exp(-t)-19/4*exp(-2*t))u(t)
Therefore,
ASSESSMENT OF THE CLASSICAL METHOD
The development in this section shows that the classical method is relatively simple compared with the method of finding the response as a sum of the zero-input and zero-state components. Unfortunately, the classical method has a serious drawback because it yields the total response, which cannot be separated into components arising from the internal conditions and the external input. In the study of systems it is important to be able to express the system response to an input x(t) as an explicit function of x(t). This is not possible in the classical method. Moreover, the classical The development in this section shows that the classical method is relatively simple compared with the method of finding the response as a sum of the zero-input and zero-state components. Unfortunately, the classical method has a serious drawback because it yields the total response, which cannot be separated into components arising from the internal conditions and the external input. In the study of systems it is important to be able to express the system response to an input x(t) as an explicit function of x(t). This is not possible in the classical method. Moreover, the classical
If we must solve a particular linear differential equation or find a response of a particular LTIC system, the classical method may be the best. In the theoretical study of linear systems, however, the classical method is not so valuable.
Caution We have shown in Eq. (2.52a) that the total response of an LTI system can be expressed as a sum of the zero-input and the zero-
state components. In Eq. (2.52b) , we showed that the same response can also be expressed as a sum of the natural and the forced components. We have also seen that generally the zero-input response is not the same as the natural response (although both are made of natural modes). Similarly, the zero-state response is not the same as the forced response. Unfortunately, such erroneous claims are often encountered in the literature.
[ † ] This result is valid only if ζ is not a characteristic root of the system. [ ‡ ] Observe the closeness of Eqs. (2.57) and (2.47) . Why is there a difference between the two equations? Equation (2.47) is the response to an
exponential that started at −∞, while Eq. (2.57) is the response to an exponential that starts at t = 0. As t → ∞, Eq. (2.57) approaches Eq. (2.47) . In Eq. (2.47) , the term y n (t), which starts at t = −∞, has already decayed at t = 0, and hence, is missing.