Classical Solution

Whenf (t)≡0, Eq.2.4ais known as the homogeneous (or complementary) equation. We shall first solve the homogeneous equation. Let the solution of the homogeneous equation bey_c(t), that is,

Q(D)y_c(t)=0 or

(Dⁿ+an−1Dⁿ⁻¹+ · · · +a1D+a0)yc(t)=0

We first show that ifyp(t)is the solution of Eq.2.4a, thenyc(t)+yp(t)is also its solution. This follows from the fact that

Q(D)y_c(t)=0 Ify_p(t)is the solution of Eq.2.4a, then

Q(D)y_p(t)=P (D)f (t) Addition of these two equations yields

Q(D)

yc(t)+yp(t)

=P (D)f (t)

Thus,yc(t)+yp(t)satisfies Eq.2.4aand therefore is the general solution of Eq.2.4a. We callyc(t) the complementary solution andyp(t)the particular solution. In system analysis parlance, these components are called the natural response and the forced response, respectively.

Complementary Solution (The Natural Response) The complementary solutionyc(t)is the solution of

Q(D)y_c(t)=0 (2.5a)

or

Dⁿ+an−1Dⁿ⁻¹+ · · · +a1D+a0

yc(t)=0 (2.5b)

A solution to this equation can be found in a systematic and formal way. However, we will take a short cut by using heuristic reasoning. Equation2.5ab shows that a linear combination ofy_c(t)and

itsnsuccessive derivatives is zero, not at some values oft, but for allt. This is possible if and only if yc(t)and all itsnsuccessive derivatives are of the same form. Otherwise their sum can never add to zero for all values oft. We know that only an exponential functione^λt has this property. So let us assume that

y_c(t)=ce^λt is a solution to Eq.2.5ab. Now

Dy_c(t) = dy_c

dt =cλe^λt D²yc(t) = d²y_c

dt² =cλ²e^λt

· · · · Dⁿy_c(t) = dⁿyc

dtⁿ =cλⁿe^λt Substituting these results in Eq.2.5ab, we obtain

c

λⁿ+an−1λⁿ⁻¹+ · · · +a1λ+a0

e^λt =0

For a nontrivial solution of this equation,

λⁿ+a_n−1λⁿ⁻¹+ · · · +a1λ+a0=0 (2.6a) This result means thatce^λt is indeed a solution of Eq.2.5aprovided thatλsatisfies Eq.2.6aa. Note that the polynomial in Eq.2.6aa is identical to the polynomialQ(D)in Eq.2.5ab, withλreplacing D. Therefore, Eq.2.6aa can be expressed as

Q(λ)=0 (2.6b)

WhenQ(λ)is expressed in factorized form, Eq.2.6ab can be represented as

Q(λ)=(λ−λ1)(λ−λ2)· · ·(λ−λ_n)=0 (2.6c) Clearlyλhasnsolutions: λ1,λ2,. . .,λ_n. Consequently, Eq.2.5ahasnpossible solutions: c1e^λ1t, c2e^λ2t,. . .,c_ne^λnt, withc1,c2,. . .,c_nas arbitrary constants. We can readily show that a general solution is given by the sum of thesensolutions,¹so that

y_c(t)=c1e^λ1t+c2e^λ2t+ · · · +c_ne^λnt (2.7)

1To prove this fact, assume thaty1(t),y2(t),. . .,yn(t)are all solutions of Eq.2.5a. Then Q(D)y1(t) = 0

Q(D)y2(t) = 0

· · · · Q(D)yn(t) = 0

Multiplying these equations byc1, c2, . . . , cn, respectively, and adding them together yields Q(D)

c1y1(t)+c2y2(t)+ · · · +cnyn(t)

=0

This result shows thatc1y1(t)+c2y2(t)+ · · · +cnyn(t)is also a solution of the homogeneous Eq.2.5a.

wherec1,c2,. . .,cnare arbitrary constants determined bynconstraints (the auxiliary conditions) on the solution.

The polynomialQ(λ)is known as the characteristic polynomial. The equation

Q(λ)=0 (2.8)

is called the characteristic or auxiliary equation. From Eq.2.6ac, it is clear thatλ1,λ2,. . .,λ_nare the roots of the characteristic equation; consequently, they are called the characteristic roots. The terms characteristic values, eigenvalues, and natural frequencies are also used for characteristic roots.² The exponentialse^λit(i =1,2, . . . , n)in the complementary solution are the characteristic modes (also known as modes or natural modes). There is a characteristic mode for each characteristic root, and the complementary solution is a linear combination of the characteristic modes.

Repeated Roots

The solution of Eq.2.5aas given in Eq.2.7assumes that thencharacteristic rootsλ1,λ2,. . . ,λnare distinct. If there are repeated roots (same root occurring more than once), the form of the solution is modified slightly. By direct substitution we can show that the solution of the equation

(D−λ)²y_c(t)=0 is given by

y_c(t)=(c1+c2t)e^λt

In this case the rootλrepeats twice. Observe that the characteristic modes in this case aree^λt and te^λt. Continuing this pattern, we can show that for the differential equation

(D−λ)^ryc(t)=0 (2.9)

the characteristic modes aree^λt,te^λt,t²e^λt,. . .,t^r−1e^λt, and the solution is y_c(t)= c1+c2t+ · · · +c_rt^r−1

e^λt (2.10)

Consequently, for a characteristic polynomial

Q(λ)=(λ−λ1)^r(λ−λr+1)· · ·(λ−λn)

the characteristic modes aree^λ1t, te^λ1t,. . ., t^r−1e^λt, e^λr+1t, . . .,e^λnt. and the complementary solution is

y_c(t)=(c1+c2t+ · · · +c_rt^r−1)e^λ1t+c_r+1e^λr+1t+ · · · +c_ne^λnt

Particular Solution (The Forced Response): Method of Undetermined Coefficients The particular solutiony_p(t)is the solution of

Q(D)y_p(t)=P (D)f (t) (2.11)

It is a relatively simple task to determineyp(t)when the inputf (t)is such that it yields only a finite number of independent derivatives. Inputs having the forme^ζt ort^r fall into this category. For example,e^ζthas only one independent derivative; the repeated differentiation ofe^ζtyields the same form, that is,e^ζt. Similarly, the repeated differentiation oft^r yields onlyrindependent derivatives.

2The term eigenvalue is German for characteristic value.

The particular solution to such an input can be expressed as a linear combination of the input and its independent derivatives. Consider, for example, the inputf (t)=at²+bt+c. The successive derivatives of this input are2at +b and2a. In this case, the input has only two independent derivatives. Therefore the particular solution can be assumed to be a linear combination off (t)and its two derivatives. The suitable form fory_p(t)in this case is therefore

y_p(t)=β2t²+β1t+β0

The undetermined coefficientsβ0,β1, andβ2are determined by substituting this expression fory_p(t) in Eq.2.11and 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. Table2.1 shows a variety of such inputs and the form of the particular solution corresponding to each input.

We shall demonstrate this procedure with an example.

TABLE 2.1

Inputf (t) Forced Response 1. e^ζt ζ6=λ_i(i=1,2, βe^{ζ t}

· · ·, n)

2. e^ζt ζ=λ_i βte^{ζ t} 3. k (a constant) β (a constant) 4.cos(ωt+θ) βcos(ωt+φ) 5.

t^r+α_r−1t^r−1+ · · · (βrt^r+β_r−1t^r−1+ · · · +α1t+α0

e^{ζ t} +β1t+β0)e^ζt

Note: By definition,yp(t)cannot have any characteristic mode terms. If any termp(t)shown in the right-hand column for the particular solution is also a characteristic mode, the correct form of the forced response must be modified totⁱp(t), whereiis the smallest possible integer that can be used and still can preventtⁱp(t)from having characteristic mode term. For example, when the input ise^ζt, the forced response (right-hand column) has the formβe^ζt. But ife^ζthappens to be a characteristic mode, the correct form of the particular solution isβte^ζt(see Pair 2). Ifte^ζtalso happens to be characteristic mode, the correct form of the particular solution isβt²e^ζt, and so on.

EXAMPLE 2.1:

Solve the differential equation

D²+3D+2

y(t)=Df (t) (2.12)

if the input

f (t)=t²+5t+3 and the initial conditions arey(0⁺)=2andy(˙ 0⁺)=3.

The characteristic polynomial is

λ²+3λ+2=(λ+1)(λ+2)

Therefore the characteristic modes aree^−t ande^−2t. The complementary solution is a linear com-bination of these modes, so that

y_c(t)=c1e^−t+c2e^−2t t≥0

Here the arbitrary constantsc1andc2must be determined from the given initial conditions.

The particular solution to the inputt²+5t+3is found from Table2.1(Pair 5 withζ =0) to be yp(t)=β2t²+β1t+β0

Moreover,yp(t)satisfies Eq.2.11, that is,

D²+3D+2

yp(t)=Df (t) (2.13)

Now

Dy_p(t) = d dt

β2t²+β1t+β0

=2β2t+β1

D²y_p(t) = d² dt²

β2t²+β1t+β0

=2β2

and

Df (t)= d dt

ht²+5t+3

i=2t+5

Substituting these results in Eq.2.13yields

2β2+3(2β2t+β1)+2(β2t²+β1t+β0)=2t+5 or

2β2t²+(2β1+6β2)t+(2β0+3β1+2β2)=2t+5 Equating coefficients of similar powers on both sides of this expression yields

2β2 = 0 2β1+6β2 = 2 2β0+3β1+2β2 = 5

Solving these three equations for their unknowns, we obtainβ0=1,β1=1, andβ2=0. Therefore, yp(t)=t+1 t >0

The total solutiony(t)is the sum of the complementary and particular solutions. Therefore, y(t) = yc(t)+yp(t)

= c1e^−t+c2e^−2t+t+1 t >0 so that

˙

y(t) = −c1e^−t −2c2e^−2t+1

Settingt =0and substituting the given initial conditionsy(0)=2andy(˙ 0)=3in these equations, we have

2 = c1+c2+1 3 = −c1−2c2+1

The solution to these two simultaneous equations isc1=4andc2= −3. Therefore, y(t)=4e^−t −3e^−2t +t+1 t≥0

The Exponential Input e^ζt

The exponential signal is the most important signal in the study of LTI systems. Interestingly, the particular solution for an exponential input signal turns out to be very simple. From Table2.1we see that the particular solution for the inpute^ζthas the formβe^ζt. We now show thatβ =Q(ζ)/P (ζ).³ To determine the constantβ, we substitutey_p(t)=βe^{ζ t}in Eq.2.11, which gives us

Q(D) βe^ζt

=P (D)e^ζt (2.14a)

Now observe that

De^ζt = d dt e^{ζ t}

=ζe^ζt D²e^ζt = d²

dt² e^ζt

=ζ²e^ζt

· · · · D^re^ζt = ζ^re^ζt

Consequently,

Q(D)e^{ζ t}=Q(ζ)e^{ζ t} and P (D)e^ζt =P (ζ )e^{ζ t} Therefore, Eq.2.14aa becomes

βQ(ζ)e^{ζ t}=P (ζ )e^ζt (2.15a)

and

β = P (ζ ) Q(ζ)

Thus, for the inputf (t)=e^ζt, the particular solution is given by

yp(t)=H (ζ)e^ζt t >0 (2.16a)

where

H (ζ)= ^{P (ζ)}_{Q(ζ )} (2.16b)

This is an interesting and significant result. It states that for an exponential inpute^ζtthe particular solutiony_p(t)is the same exponential multiplied byH (ζ)=P (ζ)/Q(ζ). The total solutiony(t) to an exponential inpute^{ζ t}is then given by

y(t)= Xn j=1

cje^λjt+H (ζ)e^ζt

where the arbitrary constantsc1,c2,. . .,cnare determined from auxiliary conditions.

3This is true only ifζis not a characteristic root.

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 f(t) = C

BecauseC = Ce^0t, the constant input is a special case of the exponential inputCe^ζt with ζ =0. The particular solution to this input is then given by

y_p(t) = CH(ζ)e^ζt with ζ =0

= CH(0) (2.17)

The Complex Exponential Input e^jωt Hereζ =jω, and

y_p(t)=H (jω)e^jωt (2.18)

The Sinusoidal Input f(t) =cosω0t

We know that the particular solution for the inpute^±jωt isH (±jω)e^±jωt. Sincecosωt = (e^jωt+e^−jωt)/2, the particular solution tocosωtis

yp(t)= 1 2

hH (jω)e^jωt+H (−jω)e^−jωti

Because the two terms on the right-hand side are conjugates, y_p(t)= Re

hH (jω)e^jωti

But

H (jω)= |H (jω)|e^{j6 H(jω)} so that

y_p(t) = Re

n|H (jω)|e^{j[ωt+6 H(jω)]}o

= |H (jω)|cos

ωt+⁶ H (jω)

(2.19) This result can be generalized for the inputf (t)=cos(ωt+θ). The particular solution in this case is

y_p(t)= |H (jω)|cos

ωt+θ+⁶ H (jω)

(2.20)

EXAMPLE 2.2:

Solve Eq.2.12for the following inputs:

(a) 10e^−3t (b)5 (c) e^−2t (d)10 cos(3t+30^◦).

The initial conditions arey(0⁺)=2,y(˙ 0⁺)=3.

The complementary solution for this case is already found in Example2.1as y_c(t)=c1e^−t+c2e^−2t t≥0

For the exponential inputf (t) =e^ζt, the particular solution, as found in Eq.2.16aisH(ζ)e^ζt, where

H (ζ )= P (ζ )

Q(ζ ) = ζ

ζ²+3ζ +2 (a) For inputf (t)=10e^−3t,ζ = −3, and

y_p(t) = 10H (−3)e^−3t

= 10

−3 (−3)²+3(−3)+2

e^−3t

= −15e^−3t t >0

The total solution (the sum of the complementary and particular solutions) is y(t)=c1e^−t+c2e^−2t−15e^−3t t ≥0 and

˙

y(t)= −c1e^−t−2c2e^−2t+45e^−3t t≥0

The initial conditions arey(0⁺) = 2andy(˙ 0⁺) = 3. Settingt =0 in the above equations and substituting the initial conditions yields

c1+c2−15=2 and −c1−2c2+45=3 Solution of these equations yieldsc1= −8andc2=25. Therefore,

y(t)= −8e^−t+25e^−2t −15e^−3t t ≥0 (b) For inputf (t)=5=5e^0t, ζ =0, and

y_p(t)=5H (0)=0 t >0

The complete solution isy(t) = y_c(t)+y_p(t) = c1e^−t +c2e^−2t. We then substitute the initial conditions to determinec1andc2as explained in Part a.

(c) Hereζ = −2, which is also a characteristic root. Hence (see Pair 2, Table2.1, or the comment at the bottom of the table),

yp(t)=βte^−2t To findβ, we substituteyp(t)in Eq.2.11, giving us

D²+3D+2

yp(t)=Df (t)

or

D²+3D+2

hβte^−2ti

=De^−2t But

Dh βte^−2ti

= β(1−2t)e^−2t D²h

βte^−2ti

= 4β(t−1)e^−2t De^−2t = −2e^−2t Consequently,

β(4t−4+3−6t+2t)e^−2t = −2e^−2t

or

−βe^−2t = −2e^−2t This means thatβ=2, so that

y_p(t)=2te^−2t

The complete solution isy(t)=yc(t)+yp(t)=c1e^−t +c2e^−2t +2te^−2t. We then substitute the initial conditions to determinec1andc2as explained in Part a.

(d) For the inputf (t)=10 cos(3t+30^◦), the particular solution (see Eq.2.20) is y_p(t)=10|H (j3)|cos

3t+30^◦+⁶ H (j3) where

H (j3) = P (j3)

Q(j3)= j3 (j3)²+3(j3)+2

= j3

−7+j9 = 27−j21

130 =0.263e^−j37.9◦ Therefore,

|H (j3)| =0.263, ⁶ H (j3)= −37.9^◦ and

y_p(t) = 10(0.263)cos(3t+30^◦−37.9^◦)

= 2.63 cos(3t−7.9^◦)

The complete solution isy(t)=yc(t)+yp(t)=c1e^−t+c2e^−2t +2.63 cos(3t−7.9^◦). We then substitute the initial conditions to determinec1andc2as explained in Part a.

Dans le document PART I Signals and Systems (Page 43-51)