Digital Signal Processing P2

Lathi California State University, Sacramento 2.1 Differential Equations Classical Solution•Method of Convolution 2.2 Difference Equations Initial Conditions and Iterative Solution •Clas

2 Ordinary Linear Differential

and Difference Equations

B.P Lathi

California State University, Sacramento

2.1 Differential Equations Classical Solution•Method of Convolution 2.2 Difference Equations

Initial Conditions and Iterative Solution •Classical Solution•

Method of Convolution References

2.1 Differential Equations

A function containing variables and their derivatives is called a differential expression, and an equation involving differential expressions is called a differential equation A differential equation is an ordinary differential equation if it contains only one independent variable; it is a partial differential equation

if it contains more than one independent variable We shall deal here only with ordinary differential equations

In the mathematical texts, the independent variable is generallyx, which can be anything such

as time, distance, velocity, pressure, and so on In most of the applications in control systems, the independent variable is time For this reason we shall use here independent variablet for time,

although it can stand for any other variable as well

The following equation

d2y

dt2

4

+ 3dy dt + 5y2(t) = sin t

is an ordinary differential equation of second order because the highest derivative is of the second

order Annth-order differential equation is linear if it is of the form

a n (t) d dt n y n + a n−1 (t) d dt n−1 n−1 y + · · · + a1(t) dy dt

where the coefficientsa i (t) are not functions of y(t) If these coefficients (a i) are constants, the

equation is linear with constant coefficients Many engineering (as well as nonengineering) systems can be modeled by these equations Systems modeled by these equations are known as linear time-invariant (LTI) systems In this chapter we shall deal exclusively with linear differential equations

with constant coefficients Certain other forms of differential equations are dealt with elsewhere in this volume

Role of Auxiliary Conditions in Solution of Differential Equations

We now show that a differential equation does not, in general, have a unique solution unless some additional constraints (or conditions) on the solution are known This fact should not come

as a surprise A function y(t) has a unique derivative dy/dt, but for a given derivative dy/dt

there are infinite possible functionsy(t) If we are given dy/dt, it is impossible to determine y(t)

uniquely unless an additional piece of information abouty(t) is given For example, the solution of

a differential equation

dy

obtained by integrating both sides of the equation is

for any value ofc Equation2.2specifies a function whose slope is 2 for allt Any straight line with

a slope of 2 satisfies this equation Clearly the solution is not unique, but if we place an additional constraint on the solutiony(t), then we specify a unique solution.

For example, suppose we require thaty(0) = 5; then out of all the possible solutions available,

only one function has a slope of 2 and an intercept with the vertical axis at 5 By settingt = 0 in

Equation2.3and substitutingy(0) = 5 in the same equation, we obtain y(0) = 5 = c and

y(t) = 2t + 5

which is the unique solution satisfying both Equation2.2and the constrainty(0) = 5.

In conclusion, differentiation is an irreversible operation during which certain information is lost

To reverse this operation, one piece of information abouty(t) must be provided to restore the original y(t) Using a similar argument, we can show that, givend2y/dt2, we can determiney(t) uniquely only

if two additional pieces of information (constraints) abouty(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.

We discuss here two systematic procedures for solving linear differential equations of the form

in Eq.2.1 The first method is the classical method, which is relatively simple, but restricted to a

certain class of inputs The second method (the convolution method) is general and is applicable

to all types of inputs A third method (Laplace transform) is discussed elsewhere in this volume

Both the methods discussed here are classified as time-domain methods because with these methods

we are able to solve the above equation directly, usingt as the independent variable The method

of Laplace transform (also known as the frequency-domain method), on the other hand, requires

transformation of variablet into a frequency variable s.

In engineering applications, the form of linear differential equation that occurs most commonly

is given by

d n y

dt n + a n−1 d dt n−1 n−1 y + · · · + a1dy

dt + a0y(t)

= b m d dt m m f + b m−1 d dt m−1 m−1 f + · · · + b1df

where all the coefficientsa i andb i are constants Using operational notationD to represent d/dt,

this equation can be expressed as

(D n + a n−1 D n−1 + · · · + a1D + a0)y(t)

= (b m D m + b m−1 D m−1 + · · · + b1D + b0)f (t) (2.4b)

where the polynomialsQ(D) and P (D), respectively, are

Q(D) = D n + a n−1 D n−1 + · · · + a1D + a0

P (D) = b m D m + b m−1 D m−1 + · · · + b1D + b0

Observe that this equation is of the form of Eq.2.1, wherer(t) is in the form of a linear combination

off (t) and its derivatives In this equation, y(t) represents an output variable, and f (t) represents

an input variable of an LTI system Theoretically, the powersm and n in the above equations can

take on any value Practical noise considerations, however, require [1]m ≤ n.

2.1.1 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

(D n + a n−1 D n−1 + · · · + a1D + a0)y c (t) = 0

We first show that ify p (t) is the solution of Eq.2.4a, theny c (t) + y p (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)y c (t) + y p (t)= P (D)f (t)

Thus,y c (t) + y p (t) satisfies Eq.2.4aand therefore is the general solution of Eq.2.4a We cally c (t) the complementary solution and y p (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 solutiony c (t) is the solution of

or

D n + a n−1 D n−1 + · · · + a1D + a0

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

itsn successive derivatives is zero, not at some values of t, but for all t This is possible if and only if

y c (t) and all its n successive derivatives are of the same form Otherwise their sum can never add to

zero for all values oft We know that only an exponential function e 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

D2y c (t) = d2y c

dt2 = cλ2e λt

· · · ·

D n y c (t) = d n y c

dt n = cλ n e λt

Substituting these results in Eq.2.5ab, we obtain

cλ n + a n−1 λ n−1 + · · · + a1λ + a0



e λt = 0 For a nontrivial solution of this equation,

This result means thatce λt is indeed a solution of Eq.2.5aprovided thatλ satisfies Eq.2.6aa Note that the polynomial in Eq.2.6aa is identical to the polynomialQ(D) in Eq.2.5ab, withλ replacing

D Therefore, Eq.2.6aa can be expressed as

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) Clearlyλ has n solutions: λ1,λ2, ., λ n Consequently, Eq.2.5ahasn possible solutions: c1e λ1t,

c2e λ2t, , c n e λ n t, withc1,c2, , c nas arbitrary constants We can readily show that a general solution is given by the sum of thesen solutions,1so that

y c (t) = c1e λ1t + c2e λ2t + · · · + c n e λ n t (2.7)

1 To prove this fact, assume thaty1(t), y2(t), , y n (t) are all solutions of Eq.2.5a Then

Q(D)y1(t) = 0

Q(D)y2(t) = 0

· · · ·

Q(D)y n (t) = 0

Multiplying these equations byc1, c2, , c n, respectively, and adding them together yields

Q(D)c1y1(t) + c2y2(t) + · · · + c n y n (t)= 0 This result shows thatc1y1(t) + c2y2(t) + · · · + c n y n (t) is also a solution of the homogeneous Eq.2.5a

wherec1,c2, , c nare arbitrary constants determined byn constraints (the auxiliary conditions)

on the solution

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

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 λ i t (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 then characteristic roots λ1,λ2, .

,λ nare 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 rootλ repeats twice Observe that the characteristic modes in this case are e λt and

te λt Continuing this pattern, we can show that for the differential equation

the characteristic modes aree λt,te λt,t2e λt, , t r−1 e λt, and the solution is

y c (t) = c1+ c2t + · · · + c r t r−1

Consequently, for a characteristic polynomial

Q(λ) = (λ − λ1) r (λ − λ r+1 ) · · · (λ − λ n )

the characteristic modes aree λ1t, te λ1t, , t r−1 e λt, e λ r+1 t, , e λ n t and the complementary

solution is

y c (t) = (c1+ c2t + · · · + c r t r−1 )e λ1t + c r+1 e λ r+1 t + · · · + c n e λ n t

Particular Solution (The Forced Response): Method of Undetermined Coefficients

The particular solutiony p (t) is the solution of

It is a relatively simple task to determiney p (t) when the input f (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 onlyr independent 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 are 2at + b and 2a 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 ζ 6= 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) β cos (ωt + φ)

5.t r + α r−1 t r−1+ · · · (β r t r + β r−1 t r−1+ · · ·

+ α1t + α0

e ζ t + β1t + β0)e ζt

Note: By definition,y p (t) cannot have any characteristic mode terms If any term p(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 tot i p(t), where i is the smallest possible integer that can

be used and still can preventt i p(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

if the input

f (t) = t2+ 5t + 3

and the initial conditions arey(0+) = 2 and ˙y(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 + 3 is found from Table2.1(Pair 5 withζ = 0) to be

y p (t) = β2t2+ β1t + β0

Moreover,y p (t) satisfies Eq.2.11, that is,

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

h

t2+ 5t + 3i= 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

Solving these three equations for their unknowns, we obtainβ0= 1, β1= 1, and β2= 0 Therefore,

y p (t) = t + 1 t > 0

The total solutiony(t) is the sum of the complementary and particular solutions Therefore,

y(t) = y c (t) + y p (t)

= c1e −t + c2e −2t + t + 1 t > 0

so that

˙y(t) = −c1e −t − 2c2e −2t+ 1

Settingt = 0 and substituting the given initial conditions y(0) = 2 and ˙y(0) = 3 in these equations,

we have

The solution to these two simultaneous equations isc1= 4 and c2= −3 Therefore,

y(t) = 4e −t − 3e −2t + t + 1 t ≥ 0

The Exponential Input eζt

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 substitute y p (t) = βe ζ tin Eq.2.11, which gives us

Q(D)βe ζt

Now observe that

De ζt = dt d e ζ t
= ζe ζt

D2e ζt = d2

dt2 e ζt
= ζ2e ζt

· · · ·

D r e ζt = ζ r e 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

and

β = Q(ζ) P (ζ )

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

where

This is an interesting and significant result It states that for an exponential inpute ζtthe particular

solutiony p (t) is the same exponential multiplied by H (ζ) = P (ζ)/Q(ζ) The total solution y(t)

to an exponential inpute ζ tis then given by

y(t) =

n

X

j=1

c j e λ j t + H (ζ)e ζt

where the arbitrary constantsc1,c2, ., c nare determined from auxiliary conditions

3 This is true only ifζ is 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 = Ce0t, 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

Hereζ = jω, and

We know that the particular solution for the inpute ±jωtisH (±jω)e ±jωt Since cosωt = (e jωt + e −jωt )/2, the particular solution to cos ωt is

y p (t) = 1

2

h

H (jω)e jωt + H (−jω)e −jωti

Because the two terms on the right-hand side are conjugates,

y p (t) = RehH (jω)e jωti But

H (jω) = |H (jω)|e j6 H(jω)

so that

y p (t) = Ren|H (jω)|e j[ωt+6 H(jω)]o

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

is

EXAMPLE 2.2:

Solve Eq.2.12for the following inputs:

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

y p (t) = 10H (−3)e −3t

= 10



−3

(−3)2+ 3(−3) + 2



e −3t

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+) = 2 and ˙y(0+) = 3 Setting t = 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= −8 and c2= 25 Therefore,

y(t) = −8e −t + 25e −2t − 15e −3t t ≥ 0

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

at the bottom of the table),

y p (t) = βte −2t

To findβ, we substitute y p (t) in Eq.2.11, giving us



D2+ 3D + 2y p (t) = Df (t)

D2+ 3D + 2 hβte −2ti

= De −2t

But

Dhβte −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