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Ordinary Differential Equations (ODEs)
1. Notation
The systems we wish to control are modelled by ordinary differential equations
(ODEs) with constant coefficients. In this unit, we review some key concepts from
MATH 375 about ODEs and introduce some new notation. This basic knowledge is
essential for the analysis and design of feedback control systems. In the next unit,
we will build on these concepts to create an intuitive yet rigorous mathematical
framework for this course.
An ODE with constant coefficients has the general form
𝑎𝑎𝑛𝑛 𝑦𝑦 (𝑛𝑛) (𝑡𝑡) + 𝑎𝑎𝑛𝑛−1 𝑦𝑦 (𝑛𝑛−1) (𝑡𝑡) + ⋯ + 𝑎𝑎1 𝑦𝑦 ′ (𝑡𝑡) + 𝑎𝑎0 𝑦𝑦(𝑡𝑡)
= 𝑏𝑏𝑚𝑚 𝑢𝑢(𝑚𝑚) (𝑡𝑡) + 𝑏𝑏𝑚𝑚−1 𝑢𝑢(𝑚𝑚−1) (𝑡𝑡) + ⋯ + 𝑏𝑏1 𝑢𝑢′ (𝑡𝑡) + 𝑏𝑏0 𝑢𝑢(𝑡𝑡),
(1)
for all 𝑡𝑡 ∈ (−∞, ∞), where 𝑦𝑦(𝑡𝑡) is the system output and 𝑢𝑢(𝑡𝑡) is the system input.
These functions of time are called signals and represent changing physical
variables, such as force, position, speed, and voltage. Although these signals are
real valued (i.e. 𝑢𝑢(𝑡𝑡), 𝑦𝑦(𝑡𝑡) ∈ ℝ), we sometimes consider complex-valued signals
𝑢𝑢(𝑡𝑡), 𝑦𝑦(𝑡𝑡) ∈ ℂ, as their real parts give real-valued solutions of (1). The derivative of
𝑦𝑦(𝑡𝑡) is denoted by 𝑦𝑦′(𝑡𝑡), and the nth derivative is denoted 𝑦𝑦 (𝑛𝑛) (𝑡𝑡). The coefficients
𝑎𝑎𝑖𝑖 , 𝑏𝑏𝑖𝑖 are assumed real. The order of the system is given by 𝑛𝑛, and usually, 𝑛𝑛 > 𝑚𝑚.
In this course, ODEs are used to represent 3 types of system: a) plants, which are
physical systems we wish to control; b) controls, which output a dynamic control
signal in response to a measured error (the control ODE is usually implemented in
software); and c) a closed-loop system (CLS) obtained by connecting the control to
the plant. The input to the CLS is usually a reference trajectory that the plant output
is desired to follow, and its output is the tracking error (which is kept small by the
control action).
In this unit, we consider the inputs and outputs to be smooth (i.e., infinitely
differentiable) signals, denoted 𝑢𝑢, 𝑦𝑦 ∈ 𝒞𝒞 ∞ . This notation is read as, “𝑢𝑢 and 𝑦𝑦 are
both elements of the set 𝒞𝒞 ∞ (the set of all smooth signals)”. The set 𝒞𝒞 ∞ is called a
linear space (or vector space), and its elements (smooth signals) are (infinite
dimensional) vectors.
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Note that whereas 𝑦𝑦 ∈ 𝒞𝒞 ∞ is a signal, 𝑦𝑦(𝑡𝑡) ∈ ℝ is a number. The signal 𝑦𝑦′ ∈ 𝒞𝒞 ∞ is
the derivative of 𝑦𝑦, and 𝑦𝑦′(𝑡𝑡) ∈ ℝ is the value of the derivative at time 𝑡𝑡. Thus,
differentiation applies to a signal 𝑦𝑦 ∈ 𝒞𝒞 ∞ , not a number 𝑦𝑦(𝑡𝑡) ∈ ℝ.
Some signals are defined by a formula that maps each 𝑡𝑡 ∈ ℝ into 𝑦𝑦(𝑡𝑡) ∈ ℝ. We can
name these signals by their formulas using the notation 𝑦𝑦 = {𝑦𝑦(𝑡𝑡)}. The curly
braces create a signal 𝑦𝑦 from its formula 𝑦𝑦(𝑡𝑡). For example, 𝑦𝑦 = {5𝑡𝑡 2 } is the signal
defined by 𝑦𝑦(𝑡𝑡) = 5𝑡𝑡 2 for all 𝑡𝑡 ∈ (−∞, ∞). This notation avoids having to assign
variable names (like 𝑦𝑦) to every signal we wish to reference. For example, the
statement “for all 𝑡𝑡 ∈ ℝ, if 𝑦𝑦(𝑡𝑡) = 5𝑡𝑡 2 , then 𝑦𝑦′(𝑡𝑡) = 10𝑡𝑡” can be expressed more
concisely and anonymously as {5𝑡𝑡 2 }′ = {10𝑡𝑡}.
𝑑𝑑
The derivative operator is denoted by the symbol 𝐷𝐷, i.e. 𝐷𝐷 = 𝑑𝑑𝑑𝑑
. Hence, for 𝑦𝑦 ∈ 𝒞𝒞 ∞ ,
𝐷𝐷𝐷𝐷 = 𝑦𝑦 ′ = 𝑦𝑦̇ , where the symbols 𝐷𝐷, prime, and dot all mean differentiation. For
example, 𝐷𝐷{3} = {3}′ = {0}, which says that the derivative (i.e. slope) of the
constant signal {3} is the zero signal. Signals should not be confused with numbers.
For example, the number 3 represents a scalar operator on signals (i.e. scaling by
3) whereas {3} is a signal that has a value of 3 at every time 𝑡𝑡 . Similarly, the signal
𝐷𝐷{3} = {0} is different from the operator 𝐷𝐷3 = 3𝐷𝐷, which multiplies a signal by 3
and then differentiates it (or vice versa, since it does not matter in which order
these 2 operators are applied). An exception to this distinction is that we use the
symbol 0 to mean both the number zero and the signal {0}, as this does not cause
any problems.
Solving (1) is greatly facilitated by operator notation. Since 𝑦𝑦 ′ = 𝐷𝐷𝐷𝐷, 𝑦𝑦 ′′ = 𝐷𝐷2 𝑦𝑦,
and 𝑦𝑦 (𝑛𝑛) = 𝐷𝐷𝑛𝑛 𝑦𝑦, we may write (1) concisely as
𝑝𝑝(𝐷𝐷)𝑦𝑦 = 𝑧𝑧(𝐷𝐷)𝑢𝑢,
(2)
where 𝑝𝑝(𝐷𝐷) = 𝑎𝑎𝑛𝑛 𝐷𝐷𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝐷𝐷(𝑛𝑛−1) + ⋯ + 𝑎𝑎1 𝐷𝐷 + 𝑎𝑎0 and 𝑧𝑧(𝐷𝐷) = 𝑏𝑏𝑚𝑚 𝐷𝐷𝑚𝑚 +
𝑏𝑏𝑚𝑚−1 𝐷𝐷(𝑚𝑚−1) + ⋯ + 𝑏𝑏1 𝐷𝐷 + 𝑏𝑏0 are polynomial differential operators. We may
divide (2) through by 𝑎𝑎𝑛𝑛 to make the leading coefficient of 𝑝𝑝(𝐷𝐷) equal to 1.
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2. Natural Response of a First Order System
If 𝑢𝑢 = 0 in (2), we obtain the homogeneous ODE
𝑝𝑝(𝐷𝐷)𝑦𝑦 = 0.
(3)
The general solution of (3) is called the natural response (or free response or
unforced response) of the system (2) since the input 𝑢𝑢 = 0. The natural response
is important because it defines whether the system is stable and how stable it is,
which is a measure of performance. Fortunately, finding the natural response (i.e.
solving (3)) is easy for any order 𝑛𝑛 of 𝑝𝑝.
If the ODE (3) is first order, it reduces to 𝑦𝑦 ′ (𝑡𝑡) + 𝑎𝑎0 𝑦𝑦(𝑡𝑡) = 0, or if we drop the
subscript on the coefficient 𝑎𝑎0 ,
𝑦𝑦 ′ (𝑡𝑡) + 𝑎𝑎𝑎𝑎(𝑡𝑡) = 0,
(4)
which corresponds to 𝑝𝑝(𝐷𝐷) = 𝐷𝐷 + 𝑎𝑎 in (3). The solution of (4) is 𝑦𝑦(𝑡𝑡) = 𝑐𝑐𝑒𝑒 −𝑎𝑎𝑎𝑎 ,
where 𝑐𝑐 ∈ ℝ is an arbitrary constant. This may be verified by substituting 𝑦𝑦(𝑡𝑡) =
𝑐𝑐𝑒𝑒 −𝑎𝑎𝑎𝑎 into (4) and checking that the left-hand side reduces to zero.
In (3), we say that the polynomial differential operator 𝑝𝑝(𝐷𝐷) annihilates 𝑦𝑦 because
it maps 𝑦𝑦 to zero. Therefore, solving (3) amounts to finding all signals 𝑦𝑦 that are
annihilated by 𝑝𝑝(𝐷𝐷). For the first order case (4) , the proof of this annihilation is
𝑝𝑝(𝐷𝐷)𝑦𝑦 = (𝐷𝐷 + 𝑎𝑎){𝑐𝑐𝑒𝑒 −𝑎𝑎𝑎𝑎 }
= {−𝑎𝑎𝑎𝑎𝑒𝑒 −𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎𝑒𝑒 −𝑎𝑎𝑎𝑎 }
= {0}.
(5)
The general solution 𝑦𝑦(𝑡𝑡) = 𝑐𝑐𝑒𝑒 −𝑎𝑎𝑎𝑎 represents an infinite number of solutions since
𝑐𝑐 is arbitrary. To determine the value of 𝑐𝑐, additional information about the output
is needed, such as the initial value 𝑦𝑦(0) of the output. Setting 𝑡𝑡 = 0 in the
solution 𝑦𝑦(𝑡𝑡) = 𝑐𝑐𝑒𝑒 −𝑎𝑎𝑎𝑎 gives 𝑦𝑦(0) = 𝑐𝑐𝑒𝑒 −𝑎𝑎0 = 𝑐𝑐 ∙ 1 = 𝑐𝑐, and therefore the solution
may be written in terms of the initial condition as 𝑦𝑦(𝑡𝑡) = 𝑦𝑦(0)𝑒𝑒 −𝑎𝑎𝑎𝑎 .
However, in practice, we usually don’t know (or care about) any initial conditions
or the values of arbitrary constants in the natural response. The only important
constant is 𝑎𝑎, which depends not on the input or initial conditions, but only on the
system (4) itself.
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If −𝑎𝑎 < 0, then the solution 𝑦𝑦(𝑡𝑡) = 𝑐𝑐𝑒𝑒 −𝑎𝑎𝑎𝑎 decays to zero and is called stable. If
𝑎𝑎 = 0, then 𝑦𝑦(𝑡𝑡) does not change from its initial value and is called marginally
unstable. If −𝑎𝑎 > 0, then 𝑦𝑦(𝑡𝑡) blows up exponentially and is called unstable. A key
requirement of feedback control is to give a stable closed-loop system.
3. Natural Response of an n-th Order ODE
Let us now generalize the results of the previous section to higher order systems.
The roots (or zeros) of a polynomial function 𝑝𝑝(𝑠𝑠) are the values of 𝑠𝑠 ∈ ℂ for which
𝑝𝑝(𝑠𝑠) = 0.The Fundamental Theorem of Algebra (FTA) states that a polynomial 𝑝𝑝 of
order 𝑛𝑛 has 𝑛𝑛 complex roots (including multiplicities). If the coefficients of the
polynomial are real, then its roots are either real or occur in complex conjugate
pairs.
A consequence of the FTA is that a polynomial differential operator 𝑝𝑝(𝐷𝐷) factors
into a product of real polynomials that are either first order (𝐷𝐷 − 𝑎𝑎), which has a
real root 𝑎𝑎, or second order (𝐷𝐷2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐), which has 2 complex conjugate roots.
If a root of 𝑝𝑝(𝐷𝐷) is repeated, then the corresponding factor will also be repeated,
such as (𝐷𝐷 − 𝑎𝑎)2 or (𝐷𝐷2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐)3 .
Suppose 𝑝𝑝(𝐷𝐷) factors as 𝑝𝑝(𝐷𝐷) = 𝑝𝑝1 (𝐷𝐷)𝑝𝑝2 (𝐷𝐷). If 𝑝𝑝1 (𝐷𝐷)𝑦𝑦1 = 0, then 𝑝𝑝(𝐷𝐷)𝑦𝑦1 =
𝑝𝑝2 (𝐷𝐷)𝑝𝑝1 (𝐷𝐷)𝑦𝑦1 = 𝑝𝑝2 (𝐷𝐷)0 = 0. Similarly, if 𝑝𝑝2 (𝐷𝐷)𝑦𝑦2 = 0, then 𝑝𝑝(𝐷𝐷)𝑦𝑦2 = 0. Also,
since 𝑝𝑝(𝐷𝐷) is a linear operator, we conclude that 𝑝𝑝(𝐷𝐷)(𝑐𝑐1 𝑦𝑦1 + 𝑐𝑐2 𝑦𝑦2 ) =
𝑐𝑐1 𝑝𝑝(𝐷𝐷)𝑦𝑦1 + 𝑐𝑐2 𝑝𝑝(𝐷𝐷)𝑦𝑦2 = 0 for any constants 𝑐𝑐1 , 𝑐𝑐2 . Thus, 𝑦𝑦 = 𝑐𝑐1 𝑦𝑦1 + 𝑐𝑐2 𝑦𝑦2 is a
solution of the homogeneous ODE 𝑝𝑝(𝐷𝐷)𝑦𝑦 = 0.
Applying the above reasoning to all irreducible factors of 𝑝𝑝(𝐷𝐷) gives us the general
solution of 𝑝𝑝(𝐷𝐷)𝑦𝑦 = 0 as the linear combination of all signals that are annihilated
by a factor of 𝑝𝑝(𝐷𝐷). Since these irreducible factors are only 1st or 2nd order
polynomials, possibly repeated, we only need to find out what signals are
annihilated by 1st or 2nd order polynomial differential operators to obtain the
natural response of any system.
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4. Natural Response of an n-th Order ODE with Real Roots
The simplest case is when all roots of 𝑝𝑝 are all real and distinct. If 𝑎𝑎1 is a real root
of 𝑝𝑝, then 𝐷𝐷 − 𝑎𝑎1 is a factor of 𝑝𝑝(𝐷𝐷). Since {𝑐𝑐1 𝑒𝑒 𝑎𝑎1𝑡𝑡 } is annihilated by 𝐷𝐷 − 𝑎𝑎1 , it
solves 𝑝𝑝(𝐷𝐷)𝑦𝑦 = 0. If 𝑎𝑎2 is another real root of 𝑝𝑝, then 𝐷𝐷 − 𝑎𝑎2 is a factor of 𝑝𝑝(𝐷𝐷),
and hence {𝑐𝑐2 𝑒𝑒 𝑎𝑎2𝑡𝑡 } is another solution. Since 𝑝𝑝(𝐷𝐷) is a linear operator, it also
annihilates the linear combination {𝑐𝑐1 𝑒𝑒 𝑎𝑎1𝑡𝑡 + 𝑐𝑐2 𝑒𝑒 𝑎𝑎2𝑡𝑡 }. If 𝑝𝑝 is a second order, then
𝑎𝑎1 and 𝑎𝑎2 are the only roots of 𝑝𝑝(𝐷𝐷), and {𝑐𝑐1 𝑒𝑒 𝑎𝑎1𝑡𝑡 + 𝑐𝑐2 𝑒𝑒 𝑎𝑎2𝑡𝑡 } is the general solution.
Example 1: Find the general solution 𝑦𝑦 of
(𝐷𝐷3 + 5𝐷𝐷2 + 6𝐷𝐷)𝑦𝑦 = 0.
(6)
𝑦𝑦 = {𝑐𝑐1 + 𝑐𝑐2 𝑒𝑒 −2𝑡𝑡 + 𝑐𝑐3 𝑒𝑒 −3𝑡𝑡 }.
(7)
Solution: Since 𝐷𝐷3 + 5𝐷𝐷2 + 6𝐷𝐷 = 𝐷𝐷(𝐷𝐷 + 2)(𝐷𝐷 + 3), and since 𝐷𝐷{𝑒𝑒 0𝑡𝑡 } = 𝐷𝐷{1} =
0 and (𝐷𝐷 + 2){𝑒𝑒 −2𝑡𝑡 } = 0 and (𝐷𝐷 + 3){𝑒𝑒 −3𝑡𝑡 } = 0, the general solution is
Let us now consider the case where 𝑝𝑝(𝐷𝐷) has a repeated real root. The simplest
case is when the repeated root is 𝑎𝑎 = 0. If this root is repeated 𝑛𝑛 times, then 𝑝𝑝(𝐷𝐷)
includes the factor (𝐷𝐷 − 𝑎𝑎)𝑛𝑛 = (𝐷𝐷 − 0)𝑛𝑛 = 𝐷𝐷𝑛𝑛 , which represents 𝑛𝑛-times
differentiation. Recall from calculus that the derivative operator 𝐷𝐷 annihilates
constant signals, 𝐷𝐷2 annihilates linear signals, and more generally, 𝐷𝐷𝑛𝑛 annihilates
any polynomial signal of order 𝑛𝑛 − 1. Hence, 𝐷𝐷𝑛𝑛 �𝑐𝑐𝑛𝑛−1 𝑡𝑡 (𝑛𝑛−1) + ⋯ + 𝑐𝑐1 𝑡𝑡 + 𝑐𝑐0 � = 0,
and the solution of 𝑝𝑝(𝐷𝐷)𝑦𝑦 = 0 includes a general polynomial of order 𝑛𝑛 − 1.
Before tackling the case of a nonzero real root 𝑎𝑎 repeated 𝑛𝑛 times, let us first derive
the exponential shift formula. Consider the derivative of {𝑓𝑓(𝑡𝑡)𝑒𝑒 𝑎𝑎𝑎𝑎 } where 𝑓𝑓 ∈ 𝒞𝒞 ∞
(i.e. 𝑓𝑓 belongs to the space of infinitely differentiable signals). We apply the product
rule to obtain
𝐷𝐷{𝑓𝑓(𝑡𝑡)𝑒𝑒 𝑎𝑎𝑎𝑎 } = {𝑓𝑓 ′ (𝑡𝑡)𝑒𝑒 𝑎𝑎𝑎𝑎 + 𝑓𝑓(𝑡𝑡)𝑎𝑎𝑒𝑒 𝑎𝑎𝑎𝑎 }
Applying (8) repeatedly gives
= {𝑒𝑒 𝑎𝑎𝑎𝑎 ((𝐷𝐷 + 𝑎𝑎)𝑓𝑓)(𝑡𝑡)}.
𝐷𝐷𝑛𝑛 {𝑓𝑓(𝑡𝑡)𝑒𝑒 𝑎𝑎𝑎𝑎 } = {𝑒𝑒 𝑎𝑎𝑎𝑎 ((𝐷𝐷 + 𝑎𝑎)𝑛𝑛 𝑓𝑓)(𝑡𝑡)},
(8)
(9)
which tells us that we can move the exponential function outside of the
differentiation operator if we shift 𝐷𝐷𝑛𝑛 to (𝐷𝐷 + 𝑎𝑎)𝑛𝑛 before applying it to the second
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signal 𝑓𝑓 in the product. Applying (9) to each term of a polynomial differential
operator 𝑝𝑝(𝐷𝐷) gives
𝑝𝑝(𝐷𝐷){𝑓𝑓(𝑡𝑡)𝑒𝑒 𝑎𝑎𝑎𝑎 } = {𝑒𝑒 𝑎𝑎𝑎𝑎 (𝑝𝑝(𝐷𝐷 + 𝑎𝑎)𝑓𝑓)(𝑡𝑡)}.
(10)
If 𝑓𝑓 is a polynomial of order 𝑛𝑛 − 1 and 𝑝𝑝(𝐷𝐷) = (𝐷𝐷 − 𝑎𝑎)𝑛𝑛 , we obtain
(𝐷𝐷 − 𝑎𝑎)𝑛𝑛 {𝑓𝑓(𝑡𝑡)𝑒𝑒 𝑎𝑎𝑎𝑎 } = {𝑒𝑒 𝑎𝑎𝑎𝑎 ((𝐷𝐷 − 𝑎𝑎 + 𝑎𝑎)𝑛𝑛 𝑓𝑓)(𝑡𝑡)}
= {𝑒𝑒 𝑎𝑎𝑎𝑎 (𝐷𝐷𝑛𝑛 𝑓𝑓)(𝑡𝑡)}
= {𝑒𝑒 𝑎𝑎𝑎𝑎 (0)}
= 0,
(11)
since 𝐷𝐷𝑛𝑛 annihilates any polynomial 𝑓𝑓 of order 𝑛𝑛 − 1. Hence, the general solution
of (𝐷𝐷 − 𝑎𝑎)𝑛𝑛 𝑦𝑦 = 0 is
𝑦𝑦 = {𝑓𝑓(𝑡𝑡)𝑒𝑒 𝑎𝑎𝑎𝑎 },
(12)
where 𝑓𝑓 is any real polynomial signal of order 𝑛𝑛 − 1.
Example 2: Find the general solution 𝑦𝑦 of
𝐷𝐷2 (𝐷𝐷 + 2)3 (𝐷𝐷 + 3)𝑦𝑦 = 0.
(13)
𝑦𝑦 = {𝑏𝑏0 + 𝑏𝑏1 𝑡𝑡 + (𝑐𝑐0 + 𝑐𝑐1 𝑡𝑡 + 𝑐𝑐2 𝑡𝑡 2 )𝑒𝑒 −2𝑡𝑡 + 𝑑𝑑𝑒𝑒 −3𝑡𝑡 }.
(14)
Solution: Applying (12) to each factor in this equation gives
5. Natural Response of an ODE with Complex Roots
First consider a second order polynomial with imaginary roots: 𝑝𝑝(𝐷𝐷) = (𝐷𝐷2 + 𝜔𝜔2 ),
where 𝜔𝜔 ∈ ℝ is a real constant. The two roots of 𝑝𝑝 are ±𝑖𝑖𝑖𝑖, where 𝑖𝑖 2 = −1. Since
𝐷𝐷2 {sin 𝜔𝜔𝜔𝜔} = {−𝜔𝜔2 sin 𝜔𝜔𝜔𝜔}, we obtain (𝐷𝐷2 + 𝜔𝜔2 ){𝑐𝑐1 sin 𝜔𝜔𝜔𝜔} = 0. Similarly,
𝐷𝐷2 {cos 𝜔𝜔𝜔𝜔} = {−𝜔𝜔2 cos 𝜔𝜔𝜔𝜔} gives (𝐷𝐷2 + 𝜔𝜔2 ){𝑐𝑐2 cos 𝜔𝜔𝜔𝜔} = 0. Together, these give
the general solution of 𝑝𝑝(𝐷𝐷)𝑦𝑦 = 0 for 𝑝𝑝(𝐷𝐷) = (𝐷𝐷2 + 𝜔𝜔2 ), namely
𝑦𝑦 = {𝑐𝑐1 cos 𝜔𝜔𝜔𝜔 + 𝑐𝑐2 sin 𝜔𝜔𝜔𝜔}.
(15)
In other words, since (𝐷𝐷2 + 𝜔𝜔2 ) annihilates both {sin 𝜔𝜔𝜔𝜔} and {cos 𝜔𝜔𝜔𝜔}, it
annihilates any linear combination of these signals.
Page |7
Now let us consider the more general case of a second order polynomial 𝑝𝑝(𝑠𝑠) with
complex roots (not just imaginary ones). Such a polynomial may be written in the
form 𝑝𝑝(𝑠𝑠) = (𝑠𝑠 + 𝑎𝑎)2 + 𝜔𝜔2 , which has 2 complex conjugate roots at 𝑠𝑠 = −𝑎𝑎 ± 𝑖𝑖𝑖𝑖
(check this by substitution).
Let us now use exponential shift to solve ((𝐷𝐷 + 𝑎𝑎)2 + 𝜔𝜔2 )𝑦𝑦 = 0. Since (𝐷𝐷 + 𝑎𝑎)2 +
𝜔𝜔2 is obtained by shifting 𝐷𝐷 to 𝐷𝐷 + 𝑎𝑎 in the expression 𝐷𝐷2 + 𝜔𝜔2 , we apply
exponential shift (10) to the known solution (15) of (𝐷𝐷2 + 𝜔𝜔2 )𝑦𝑦 = 0 to obtain the
solution 𝑦𝑦 of ((𝐷𝐷 + 𝑎𝑎)2 + 𝜔𝜔2 )𝑦𝑦 = 0, namely
𝑦𝑦 = {𝑒𝑒 −𝑎𝑎𝑎𝑎 (𝑐𝑐1 cos 𝜔𝜔𝜔𝜔 + 𝑐𝑐2 sin 𝜔𝜔𝜔𝜔)}.
(16)
Verify this by substituting this solution into ((𝐷𝐷 + 𝑎𝑎)2 + 𝜔𝜔2 )𝑦𝑦 = 0 and applying
exponential shift.
Example 3: Find the general solution 𝑦𝑦 of
(𝐷𝐷2 + 4𝐷𝐷 + 13)𝑦𝑦 = 0.
(17)
𝑦𝑦 = {𝑒𝑒 −2𝑡𝑡 (𝑐𝑐1 cos 3𝑡𝑡 + 𝑐𝑐2 sin 3𝑡𝑡)}.
(18)
Solution: We first express 𝑝𝑝(𝐷𝐷) = 𝐷𝐷2 + 4𝐷𝐷 + 13 in the form (𝐷𝐷 + 𝑎𝑎)2 + 𝜔𝜔2 .
Setting these equal and comparing the 𝐷𝐷 terms on both sides gives 𝑎𝑎 = 4/2 = 2.
Plugging 𝑎𝑎 = 2 into the second expression and comparing the constant terms gives
𝜔𝜔2 = 9, and so 𝜔𝜔 = 3. Substituting these values into our general solution (16) gives
Finally, let us consider the case when the roots of 𝑝𝑝(𝑠𝑠) are complex and repeated
𝑛𝑛 times. These roots occur as a complex conjugate pair 𝜆𝜆, 𝜆𝜆̅, where 𝜆𝜆 = −𝑎𝑎 + 𝑖𝑖𝑖𝑖.
Hence, 𝑝𝑝(𝐷𝐷) = (𝐷𝐷 − 𝜆𝜆)𝑛𝑛 (𝐷𝐷 − 𝜆𝜆̅)𝑛𝑛 = ((𝐷𝐷 + 𝑎𝑎)2 + 𝜔𝜔2 )𝑛𝑛 . Since the exponential
shift rules (10) and (11) still hold when 𝑎𝑎 is complex, we will follow the same
approach as in (11) to show that 𝑝𝑝(𝐷𝐷) annihilates Re�𝑓𝑓(𝑡𝑡)𝑒𝑒 𝜆𝜆𝜆𝜆 � = {𝑓𝑓(𝑡𝑡) 𝑒𝑒 𝑎𝑎𝑎𝑎 cos 𝜔𝜔𝜔𝜔},
for any real polynomial 𝑓𝑓(𝑡𝑡) of order 𝑛𝑛 − 1. Since 𝑝𝑝(𝐷𝐷) = ((𝐷𝐷 + 𝑎𝑎)2 + 𝜔𝜔2 )𝑛𝑛 is a
real operator, it commutes with the Re operator (which extracts the real part of a
number or signal). This gives
𝑝𝑝(𝐷𝐷){𝑓𝑓(𝑡𝑡) 𝑒𝑒 𝑎𝑎𝑎𝑎 cos 𝜔𝜔𝜔𝜔} = 𝑝𝑝(𝐷𝐷)Re�𝑓𝑓(𝑡𝑡)𝑒𝑒 𝜆𝜆𝜆𝜆 �
= Re(𝑝𝑝(𝐷𝐷)�𝑓𝑓(𝑡𝑡)𝑒𝑒 𝜆𝜆𝜆𝜆 �)
Page |8
= Re((𝐷𝐷 − 𝜆𝜆̅)𝑛𝑛 (𝐷𝐷 − 𝜆𝜆)𝑛𝑛 �𝑓𝑓 (𝑡𝑡)𝑒𝑒 𝜆𝜆𝜆𝜆 �)
𝑛𝑛
= Re(�𝐷𝐷 − 𝜆𝜆̅� �𝑒𝑒 𝜆𝜆𝜆𝜆 ((𝐷𝐷 − 𝜆𝜆 + 𝜆𝜆)𝑛𝑛 𝑓𝑓)(𝑡𝑡)�)
𝑛𝑛
= Re(�𝐷𝐷 − 𝜆𝜆̅� �𝑒𝑒 𝜆𝜆𝜆𝜆 (𝐷𝐷𝑛𝑛 𝑓𝑓)(𝑡𝑡)�)
𝑛𝑛
= Re(�𝐷𝐷 − 𝜆𝜆̅� �𝑒𝑒 𝜆𝜆𝜆𝜆 (0)�)
= 0.
(19)
Similarly, {𝑓𝑓(𝑡𝑡) 𝑒𝑒 𝑎𝑎𝑎𝑎 sin 𝜔𝜔𝜔𝜔} = Im�𝑓𝑓(𝑡𝑡)𝑒𝑒 𝜆𝜆𝜆𝜆 � , and since 𝑝𝑝(𝐷𝐷) is real,
𝑝𝑝(𝐷𝐷)Im�𝑓𝑓(𝑡𝑡)𝑒𝑒 𝜆𝜆𝜆𝜆 � = Im(𝑝𝑝(𝐷𝐷)�𝑓𝑓(𝑡𝑡)𝑒𝑒 𝜆𝜆𝜆𝜆 �). Applying the process of
(19) but taking imaginary parts instead of real parts gives 𝑝𝑝(𝐷𝐷){𝑓𝑓(𝑡𝑡) 𝑒𝑒 𝑎𝑎𝑎𝑎 sin 𝜔𝜔𝜔𝜔} =
0. Hence, 𝑝𝑝(𝐷𝐷) = ((𝐷𝐷 + 𝑎𝑎)2 + 𝜔𝜔2 )𝑛𝑛 annihilates
𝑦𝑦 = {𝑒𝑒 −𝑎𝑎𝑎𝑎 (𝑓𝑓1 (𝑡𝑡) cos 𝜔𝜔𝜔𝜔 + 𝑓𝑓2 (𝑡𝑡) sin 𝜔𝜔𝜔𝜔)},
(20)
where 𝑓𝑓1 (𝑡𝑡) and𝑓𝑓2 (𝑡𝑡) are arbitrary real polynomials of order 𝑛𝑛 − 1. If 𝜔𝜔 = 0, this
solution reduces to that obtained in (12) for a real repeated root −𝑎𝑎, and if the root
is not repeated, then 𝑓𝑓1 (𝑡𝑡) and 𝑓𝑓2 (𝑡𝑡) reduce to real constants.
Example 4: Find the general solution 𝑦𝑦 of
(𝐷𝐷2 + 4𝐷𝐷 + 13)3 𝑦𝑦 = 0.
(21)
Solution: Since (18) gave the solution 𝑦𝑦1 (𝑡𝑡) = {𝑒𝑒 −2𝑡𝑡 (𝑐𝑐1 cos 3𝑡𝑡 + 𝑐𝑐2 sin 3𝑡𝑡)} for the
case of non-repeated roots, we simply replace each coefficient with a general
second order polynomial to obtain the general solution
𝑦𝑦 = {𝑒𝑒 −2𝑡𝑡 ((𝑐𝑐10 + 𝑐𝑐11 𝑡𝑡 + 𝑐𝑐12 𝑡𝑡 2 ) cos 3𝑡𝑡 + (𝑐𝑐20 + 𝑐𝑐21 𝑡𝑡 + 𝑐𝑐22 𝑡𝑡 2 ) sin 3𝑡𝑡)}. (22)
Note that this solution has 6 undetermined coefficients because it is the general
solution to a sixth order ODE. The values of these can be found if 6 initial conditions
are provided.
Page |9
6. Natural Signals
A signal is called natural if it is the natural response of some linear, constant
coefficient ODE (2). Equivalently, every natural signal is annihilated by some
polynomial differential operator 𝑝𝑝(𝐷𝐷). Let 𝑁𝑁 denote the space of all natural signals.
We have shown that a natural signal is a sum of terms that each have the form (20),
which we call a natural term. Each natural term is annihilated by an operator 𝑝𝑝(𝐷𝐷)
that has a root 𝜆𝜆 = −𝑎𝑎 + 𝑖𝑖𝑖𝑖 of multiplicity 𝑛𝑛; i.e., 𝑝𝑝(𝜆𝜆) = 0. Both constants 𝑎𝑎 and
𝜔𝜔 appear in the natural term (20).
The roots of 𝑝𝑝 in (2) are called the poles of the ODE. The roots of 𝑧𝑧 are called the
zeros of the ODE. Hence, the natural response is determined by the poles. Once
these poles are found (e.g. by factoring 𝑝𝑝 manually or using Matlab), we obtain its
natural response by writing a natural term (20) for each pole and adding up all the
terms.
The simplest case of (20) is when 𝜔𝜔 = 0, which gives sin 𝜔𝜔𝜔𝜔 = 0 and cos 𝜔𝜔𝜔𝜔 = 1.
In addition, if the multiplicity 𝑛𝑛 = 1, then then 𝑓𝑓1 (𝑡𝑡) and 𝑓𝑓2 (𝑡𝑡) reduce to real
constants, and hence the natural response reduces to the first order response
𝑦𝑦(𝑡𝑡) = 𝑐𝑐𝑒𝑒 −𝑎𝑎𝑎𝑎 considered in our first example.
Try sketching the first order response 𝑦𝑦(𝑡𝑡) = 𝑐𝑐𝑒𝑒 −𝑎𝑎𝑎𝑎 . Since this solution to the ODE
is valid for all 𝑡𝑡 ∈ (−∞, ∞), the time axis should include negative times. If −𝑎𝑎 > 0,
then 𝑦𝑦(𝑡𝑡) = 𝑐𝑐𝑒𝑒 −𝑎𝑎𝑎𝑎 is small for large negative times and grows exponentially to the
right, called an unstable response. The plot interpolates the point 𝑦𝑦(0) = 𝑐𝑐.
Although we call 𝑐𝑐 the initial condition, it makes more sense to call it the middle
condition since, from the viewpoint of negative times, it occurs in the future.
If −𝑎𝑎 < 0, then 𝑦𝑦(𝑡𝑡) = 𝑐𝑐𝑒𝑒 −𝑎𝑎𝑎𝑎 grows large to the left and decays to the right, called
a stable response. This response is the mirror image about the 𝑦𝑦 axis of the previous
case. If the sign of the constant 𝑐𝑐 is opposite to what you assumed in your first plot,
then the plot is flipped about the time axis.
If 𝑛𝑛 > 1, then the pole −𝑎𝑎 is repeated 𝑛𝑛 times, and 𝑓𝑓1 (𝑡𝑡) and 𝑓𝑓2 (𝑡𝑡) become
polynomials of order 𝑛𝑛 − 1. Still assuming that 𝜔𝜔 = 0, these polynomials have little
effect on the growth rate at large positive or negative times (unless 𝑎𝑎 = 0) since
exponential growth or decay is much faster than polynomial growth. However,
these polynomials do affect the response near the origin, such as the value of 𝑦𝑦(0).
P a g e | 10
Now try plotting the case of 𝑛𝑛 = 1, 𝑎𝑎 = 0, and 𝜔𝜔 ≠ 0. Then, the response (20)
reduces to 𝑦𝑦 = {𝑐𝑐1 cos 𝜔𝜔𝜔𝜔 + 𝑐𝑐2 sin 𝜔𝜔𝜔𝜔}, which can also be written as 𝑦𝑦 =
{𝐴𝐴 cos(𝜔𝜔𝜔𝜔 + 𝜑𝜑)}, where 𝐴𝐴 = �𝑐𝑐1 2 + 𝑐𝑐2 2 is the resultant amplitude of the cosine
wave and 𝜑𝜑 = atan2(𝑐𝑐2 , 𝑐𝑐1 ) its phase shift. Your plot should therefore be that of a
cosine that has a peak value of 𝐴𝐴, but shifted leftward by a time of 𝜑𝜑/𝜔𝜔.
Finally, suppose both 𝑎𝑎 ≠ 0, and 𝜔𝜔 ≠ 0. Then, the response will be the cosine wave
of the last case but scaled by an exponential 𝑒𝑒 −𝑎𝑎𝑎𝑎 that decays to the right if −𝑎𝑎 <
0 and grows to the right if −𝑎𝑎 > 0. Try sketching this. If 𝑛𝑛 > 1, then the response
will be similar, except near the origin where the polynomial factors have an effect.
A simple way of visualizing all these responses is as the real part of a complex valued
response �𝑓𝑓(𝑡𝑡)𝑒𝑒 𝜆𝜆𝜆𝜆 �, where 𝜆𝜆 ∈ ℂ and 𝑓𝑓(𝑡𝑡) ∈ ℂ is a polynomial with complex
coefficients. Even if 𝑓𝑓(𝑡𝑡) is not constant, the behaviour of this response is primarily
due to the factor of 𝑒𝑒 𝜆𝜆𝜆𝜆 = 𝑒𝑒 −𝑎𝑎𝑎𝑎 𝑒𝑒 𝑖𝑖𝜔𝜔𝑡𝑡 , which traces a spiral in the complex plane. The
𝑒𝑒 𝑖𝑖𝜔𝜔𝑡𝑡 factor gives circular motion at an angular speed of 𝜔𝜔. The 𝑒𝑒 −𝑎𝑎𝑎𝑎 factor is the
radius, which spirals into the origin if −𝑎𝑎 < 0 and away from it if −𝑎𝑎 > 0. The real
part of this complex motion is the horizontal component, which is a decaying or
growing cosine wave.
The most important feature of all these natural responses is their decay (stability)
or growth (instability) as time increases. This depends on the number −𝑎𝑎, which is
the real part of the pole. If there are multiple poles, then the real part of the
rightmost pole determines stability since if that pole is stable, then every pole to
the left of it is also stable. If the rightmost pole is unstable, then even if the other
poles are stable, the total response is still unstable since adding signals that decay
to zero to an unstable signal still gives an unstable signal.
7. Solving ODEs for any Natural Input
Now consider the forced ODE (2) with a natural input 𝑢𝑢 ∈ 𝑁𝑁. We will show that the
output 𝑦𝑦 is also natural and its poles are the poles of the ODE combined with those
of 𝑧𝑧(𝐷𝐷)𝑢𝑢. If 𝑦𝑦 = 𝑦𝑦𝑝𝑝 is a particular solution of (2), then so is 𝑦𝑦 = 𝑦𝑦𝑝𝑝 + 𝑦𝑦ℎ , where 𝑦𝑦ℎ
is any solution to the homogeneous ODE (2). This may be verified by substituting
𝑦𝑦 = 𝑦𝑦𝑝𝑝 + 𝑦𝑦ℎ into (2) and setting 𝑝𝑝(𝐷𝐷)𝑦𝑦ℎ = 0. Hence, (2) has infinitely many
solutions.
P a g e | 11
The input term 𝑧𝑧(𝐷𝐷)𝑢𝑢 of (2) with 𝑢𝑢 ∈ 𝑁𝑁 can be shown to be natural as follows. The
derivative 𝐷𝐷𝐷𝐷 of a polynomial signal 𝑓𝑓 is a polynomial, and hence so is 𝐷𝐷2 𝑓𝑓, etc.
and hence so is 𝑧𝑧(𝐷𝐷)𝑓𝑓. Similarly, 𝐷𝐷{𝑒𝑒 𝑎𝑎𝑎𝑎 } = 𝑎𝑎{𝑒𝑒 𝑎𝑎𝑎𝑎 }, 𝐷𝐷2 {𝑒𝑒 𝑎𝑎𝑎𝑎 } = 𝑎𝑎2 {𝑒𝑒 𝑎𝑎𝑎𝑎 }, etc., and
hence 𝑧𝑧(𝐷𝐷){𝑒𝑒 𝑎𝑎𝑎𝑎 } = 𝑧𝑧(𝑎𝑎){𝑒𝑒 𝑎𝑎𝑎𝑎 }, and so 𝑧𝑧(𝐷𝐷) maps exponentials into exponentials.
Finally, consider a complex natural term �𝑓𝑓 (𝑡𝑡)𝑒𝑒 𝜆𝜆𝜆𝜆 �, where 𝜆𝜆 ∈ ℂ and 𝑓𝑓(𝑡𝑡) ∈ ℂ is a
polynomial with complex coefficients. The exponential shift property (10) gives
𝑧𝑧(𝐷𝐷)�𝑓𝑓(𝑡𝑡)𝑒𝑒 𝜆𝜆𝜆𝜆 � = �𝑒𝑒 𝜆𝜆𝜆𝜆 (𝑧𝑧(𝐷𝐷 + 𝜆𝜆)𝑓𝑓)(𝑡𝑡)� = �𝑔𝑔(𝑡𝑡)𝑒𝑒 𝜆𝜆𝜆𝜆 �, where 𝑔𝑔 = 𝑧𝑧(𝐷𝐷 + 𝜆𝜆)𝑓𝑓 is a
polynomial with complex coefficients. Hence, 𝑧𝑧(𝐷𝐷) maps complex natural signals
into complex natural signals.
Since every real natural signal 𝑢𝑢 is the real part of a complex natural signal 𝑢𝑢𝑐𝑐 , and
since 𝑧𝑧(𝐷𝐷) is a real operator, we have 𝑧𝑧(𝐷𝐷)𝑢𝑢 = 𝑧𝑧(𝐷𝐷)Re(𝑢𝑢𝑐𝑐 ) = Re(𝑧𝑧(𝐷𝐷)𝑢𝑢𝑐𝑐 ), which
is a real natural signal. Hence, 𝑧𝑧(𝐷𝐷) maps real natural signals into real natural
signals. Therefore if 𝑢𝑢 in (2) is a real natural signal, then so is 𝑧𝑧(𝐷𝐷)𝑢𝑢 in (2).
To solve (2), we choose 𝑦𝑦 to have the same form as the input 𝑧𝑧(𝐷𝐷)𝑢𝑢 but with
undetermined coefficients. If 𝑝𝑝(𝐷𝐷) has a root in common with the poles of 𝑧𝑧(𝐷𝐷)𝑢𝑢,
then the order of the corresponding polynomial in 𝑦𝑦 must be 𝑛𝑛 + 𝑚𝑚, where 𝑛𝑛 is the
order of the corresponding polynomial in 𝑧𝑧(𝐷𝐷)𝑢𝑢, and 𝑚𝑚 is the multiplicity of the
common root. This accounts for any reductions in the order of any polynomial
factors in 𝑦𝑦 caused by 𝑝𝑝(𝐷𝐷). Examples will demonstrate.
Example 5: Find the general solution 𝑦𝑦 of
if 𝑢𝑢 = {𝑡𝑡𝑒𝑒 𝑡𝑡 + 𝑒𝑒 2𝑡𝑡 }.
(𝐷𝐷 + 2)(𝐷𝐷 + 1)𝑦𝑦 = (𝐷𝐷 + 3)𝑢𝑢,
(23)
Solution: Applying (𝐷𝐷 + 3) to 𝑢𝑢 gives (𝐷𝐷 + 3){𝑡𝑡𝑒𝑒 𝑡𝑡 + 𝑒𝑒 2𝑡𝑡 } = {(4𝑡𝑡 + 1)𝑒𝑒 𝑡𝑡 + 5𝑒𝑒 2𝑡𝑡 }.
This modified input to (23) produces an output of the form 𝑦𝑦 = 𝑦𝑦𝑝𝑝 =
{(𝑐𝑐0 + 𝑐𝑐1 𝑡𝑡)𝑒𝑒 𝑡𝑡 + 𝑏𝑏𝑒𝑒 2𝑡𝑡 }. Here, the polynomials in the output are chosen to have the
same order as those in the input so that the coefficients can be matched. Applying
(𝐷𝐷 + 2)(𝐷𝐷 + 1) to 𝑦𝑦𝑝𝑝 and equating the result to the input {(4𝑡𝑡 + 1)𝑒𝑒 𝑡𝑡 + 5𝑒𝑒 2𝑡𝑡 }
gives
(6𝑐𝑐0 + 5𝑐𝑐1 + 6𝑐𝑐1 𝑡𝑡)𝑒𝑒 𝑡𝑡 + 12𝑏𝑏𝑒𝑒 2𝑡𝑡 = (1 + 4𝑡𝑡)𝑒𝑒 𝑡𝑡 + 5𝑒𝑒 2𝑡𝑡 .
(24)
Solving for the 3 unknown coefficients gives 𝑐𝑐0 = − 7⁄18, 𝑐𝑐1 = 2/3, 𝑏𝑏 = 5/12.
7
5 2𝑡𝑡
𝑒𝑒 . The general unforced
This gives the particular solution 𝑦𝑦𝑝𝑝 = �23𝑡𝑡 − 18
�𝑒𝑒 𝑡𝑡 + 12
P a g e | 12
solution 𝑦𝑦ℎ is obtained by setting 𝑢𝑢 = 0 in (23), which gives 𝑦𝑦ℎ = 𝑏𝑏1 𝑒𝑒 −𝑡𝑡 + 𝑏𝑏2 𝑒𝑒 −2𝑡𝑡 .
Adding this solution to 𝑦𝑦𝑝𝑝 gives the general solution to the forced ODE as
2
7
(25)
5
𝑦𝑦(𝑡𝑡) = �3𝑡𝑡 − 18� 𝑒𝑒 𝑡𝑡 + 12𝑒𝑒 2𝑡𝑡 + 𝑏𝑏1 𝑒𝑒 −𝑡𝑡 + 𝑏𝑏2 𝑒𝑒 −2𝑡𝑡 .
𝐷𝐷+3
2+3
5 2𝑡𝑡
The term 12
𝑒𝑒 could also be found directly as (𝐷𝐷+2)(𝐷𝐷+1) {𝑒𝑒 2𝑡𝑡 } = (2+2)(2+1) {𝑒𝑒 2𝑡𝑡 } =
{𝑒𝑒 2𝑡𝑡 } since 𝐷𝐷{𝑒𝑒 2𝑡𝑡 } = 2{𝑒𝑒 2𝑡𝑡 }. If initial conditions 𝑦𝑦(0) and 𝑦𝑦′(0) are given, then
𝑏𝑏1 and 𝑏𝑏2 can be determined.
5
12
Example 6: Find the general solution 𝑦𝑦 of
if 𝑢𝑢 = {𝑡𝑡𝑒𝑒 −𝑡𝑡 + 𝑒𝑒 −3𝑡𝑡 }.
(𝐷𝐷 + 2)(𝐷𝐷 + 1)𝑦𝑦 = (𝐷𝐷 + 3)𝑢𝑢,
(26)
Solution: Applying (𝐷𝐷 + 3) to 𝑢𝑢 gives (𝐷𝐷 + 3){𝑡𝑡𝑒𝑒 −𝑡𝑡 + 𝑒𝑒 −3𝑡𝑡 } = {(2𝑡𝑡 + 1)𝑒𝑒 −𝑡𝑡 }. This
modified input to (26) has pole 𝜆𝜆 = −1 in common with a root of 𝑝𝑝(𝐷𝐷) =
(𝐷𝐷 + 2)(𝐷𝐷 + 1) of multiplicity 𝑚𝑚 = 1. The corresponding polynomial (2𝑡𝑡 + 1) in
(𝐷𝐷 + 3)𝑢𝑢 has order 𝑛𝑛 = 1. Hence, the order of the corresponding polynomial in 𝑦𝑦
must be 𝑛𝑛 + 𝑚𝑚 = 2. Therefore, we write the solution 𝑦𝑦 = 𝑦𝑦𝑝𝑝 = {(𝑐𝑐1 𝑡𝑡 + 𝑐𝑐2 𝑡𝑡 2 )𝑒𝑒 −𝑡𝑡 }
with coefficients to be determined. We omitted a constant term 𝑐𝑐0 because this
will be annihilated by (𝐷𝐷 + 1) and will be added to the general solution anyway.
Substituting this solution and (𝐷𝐷 + 3)𝑢𝑢 = {(2𝑡𝑡 + 1)𝑒𝑒 −𝑡𝑡 } into (26) gives
(𝐷𝐷 + 2)(𝐷𝐷 + 1){(𝑐𝑐1 𝑡𝑡 + 𝑐𝑐2 𝑡𝑡 2 )𝑒𝑒 −𝑡𝑡 } = {(2𝑡𝑡 + 1)𝑒𝑒 −𝑡𝑡 }.
(27)
Using exponential shift to evaluate the l.h.s. and comparing its terms with the r.h.s.
gives 𝑐𝑐2 = 1 and 𝑐𝑐1 = −1, so 𝑦𝑦 = 𝑦𝑦𝑝𝑝 = {(𝑡𝑡 2 − 𝑡𝑡)𝑒𝑒 −𝑡𝑡 } solves (26). The general
solution of (26) is this solution plus the general solution of the homogeneous ODE
obtained by setting 𝑢𝑢 = 0 in (26). This gives 𝑦𝑦 = {(𝑡𝑡 2 − 𝑡𝑡 + 𝑐𝑐0 )𝑒𝑒 −𝑡𝑡 + 𝑏𝑏𝑒𝑒 −2𝑡𝑡 }. This
general solution has 2 undetermined coefficients 𝑐𝑐0 and 𝑏𝑏 since (26) is a secondorder ODE. These coefficients can be found from two initial conditions, such as
𝑦𝑦(0) = 𝑦𝑦0 and 𝑦𝑦′(0) = 𝑦𝑦1 . Substituting 𝑦𝑦(𝑡𝑡) = (𝑡𝑡 2 − 𝑡𝑡 + 𝑐𝑐0 )𝑒𝑒 −𝑡𝑡 + 𝑏𝑏𝑒𝑒 −2𝑡𝑡 into
these initial conditions gives 𝑦𝑦0 = 𝑐𝑐0 + 𝑏𝑏 and 𝑦𝑦1 = −(1 + 𝑐𝑐0 + 2𝑏𝑏), which may
then be solved for 𝑐𝑐0 and 𝑏𝑏.
This example illustrates two tricky features that can occur when solving ODEs. The
first is that the term 𝑒𝑒 −3𝑡𝑡 in the input 𝑢𝑢 did not appear in the output 𝑦𝑦. This is
P a g e | 13
because it was annihilated by 𝑧𝑧(𝐷𝐷) = 𝐷𝐷 + 3. The modified input of 𝑧𝑧(𝐷𝐷)𝑢𝑢 =
{(2𝑡𝑡 + 1)𝑒𝑒 −𝑡𝑡 } determines the form of 𝑦𝑦. The other tricky feature is that the output
term (𝑡𝑡 2 − 𝑡𝑡 + 𝑐𝑐0 )𝑒𝑒 −𝑡𝑡 contained a higher-order polynomial than that of the input
term (2𝑡𝑡 + 1)𝑒𝑒 −𝑡𝑡 that produced it. This is because one of the poles of the ODE,
namely 𝜆𝜆 = −1 , happened to match a pole of the input 𝑢𝑢. This is called pure
resonance.
The fact that an ODE has an infinite number of solutions 𝑦𝑦 for every given input 𝑢𝑢
is undesirable. We can avoid this by requiring that present values of the output do
not depend on future values of the input. This causality constraint turns the ODE
into an operator, which maps each input into a unique output.
In the next unit, we will develop an algebra for such operators (and more general
operators) where operators behave just like real or complex numbers, providing a
simple mathematical framework for feedback control analysis and design.
8. Frequency Response
To conclude this unit, let us consider the response of an ODE to 2 special natural
inputs that are closely related: exponentials and sinusoids.
Consider the ODE (2) with a complex exponential input 𝑢𝑢 = {𝑐𝑐𝑒𝑒 𝑠𝑠𝑠𝑠 }, where 𝑐𝑐, 𝑠𝑠 ∈ ℂ.
Let us assume the constraint of the previous section: that the real part of 𝑠𝑠 is
greater than the real part of every pole of the ODE. This implies that 𝑝𝑝(𝑠𝑠) ≠ 0. The
unique solution is then 𝑦𝑦 = 𝐺𝐺(𝑠𝑠)𝑢𝑢, where the complex constant 𝐺𝐺(𝑠𝑠) =
𝑧𝑧(𝑠𝑠)/𝑝𝑝(𝑠𝑠). Verify this solution by substitution. If we use 𝑔𝑔 to denote the operator
from input signals to output signals, then the function 𝐺𝐺(𝑠𝑠) is called the
(operational) Laplace transform of 𝑔𝑔. The number 𝐺𝐺(𝑠𝑠) is the eigenvalue of 𝑔𝑔
corresponding to the eigensignal 𝑢𝑢 = {𝑐𝑐𝑒𝑒 𝑠𝑠𝑠𝑠 }.
A special case of this result is when 𝑠𝑠 = 𝑖𝑖𝑖𝑖. Writing 𝑐𝑐 in polar form as 𝑐𝑐 = |𝑐𝑐|𝑒𝑒 𝑖𝑖∠𝑐𝑐 ,
where ∠𝑐𝑐 denotes the angle that 𝑐𝑐 makes with the real axis, we find the real part
of 𝑢𝑢 = {𝑐𝑐𝑒𝑒 𝑖𝑖𝜔𝜔𝑡𝑡 } to be
𝑢𝑢𝑟𝑟 = Re�𝑐𝑐𝑒𝑒 𝑖𝑖𝜔𝜔𝑡𝑡 �
= Re�|𝑐𝑐|𝑒𝑒 𝑖𝑖∠𝑐𝑐 𝑒𝑒 𝑖𝑖𝜔𝜔𝑡𝑡 �
P a g e | 14
= |𝑐𝑐|Re�𝑒𝑒 𝑖𝑖(𝜔𝜔𝑡𝑡+∠𝑐𝑐) �
= |𝑐𝑐|{cos(𝜔𝜔𝑡𝑡 + ∠𝑐𝑐)},
(28)
which we call a general sinusoid. For example, 𝑐𝑐 = −𝑖𝑖 gives 𝑢𝑢𝑟𝑟 = sin(𝜔𝜔𝑡𝑡). We will
now show that a solution of (2) for a general sinusoidal input is also a general
sinusoid at the same frequency but different amplitude and phase.
If 𝑦𝑦 solves (2) when 𝑢𝑢 = {𝑐𝑐𝑒𝑒 𝑖𝑖𝜔𝜔𝑡𝑡 }, then since 𝑝𝑝 and 𝑧𝑧 are real polynomials,
𝑝𝑝(𝐷𝐷)𝑦𝑦 = 𝑧𝑧(𝐷𝐷)𝑢𝑢
Re(𝑝𝑝(𝐷𝐷)𝑦𝑦) = Re(𝑧𝑧(𝐷𝐷)𝑢𝑢)
𝑝𝑝(𝐷𝐷)Re(𝑦𝑦) = 𝑧𝑧(𝐷𝐷)Re(𝑢𝑢)
𝑝𝑝(𝐷𝐷)𝑦𝑦𝑟𝑟 = 𝑧𝑧(𝐷𝐷)𝑢𝑢𝑟𝑟 ,
and 𝑦𝑦𝑟𝑟 = Re(𝑦𝑦) is obtained from the exponential solution 𝑦𝑦 = 𝐺𝐺(𝑖𝑖𝜔𝜔)𝑢𝑢 as
(29)
𝑦𝑦𝑟𝑟 = Re�𝐺𝐺(𝑖𝑖𝜔𝜔)𝑐𝑐𝑒𝑒 𝑖𝑖𝜔𝜔𝑡𝑡 �
= Re�|𝐺𝐺(𝑖𝑖𝜔𝜔)|𝑒𝑒 𝑖𝑖∠𝐺𝐺(𝑖𝑖𝜔𝜔) |𝑐𝑐|𝑒𝑒 𝑖𝑖∠𝑐𝑐 𝑒𝑒 𝑖𝑖𝜔𝜔𝑡𝑡 �
= |𝑐𝑐||𝐺𝐺(𝑖𝑖𝜔𝜔)|Re�𝑒𝑒 𝑖𝑖(𝜔𝜔𝑡𝑡+∠𝑐𝑐+∠𝐺𝐺(𝑖𝑖𝜔𝜔)) �
= |𝑐𝑐||𝐺𝐺(𝑖𝑖𝜔𝜔)|{cos(𝜔𝜔𝑡𝑡 + ∠𝑐𝑐 + ∠𝐺𝐺(𝑖𝑖𝜔𝜔))}.
(30)
Hence, the amplitude of the output 𝑦𝑦𝑟𝑟 equals the amplitude of the input multiplied
by the magnitude of the complex number 𝐺𝐺(𝑖𝑖𝜔𝜔) = 𝑧𝑧(𝑖𝑖𝜔𝜔)/𝑝𝑝(𝑖𝑖𝜔𝜔), and the phase of
the output equals the phase of the input plus the angle of 𝐺𝐺(𝑖𝑖𝜔𝜔).
Example 7: Find the sinusoidal solution 𝑦𝑦 of
if 𝑢𝑢𝑟𝑟 = {sin 5𝑡𝑡}.
(𝐷𝐷 + 2)(𝐷𝐷 + 1)𝑦𝑦𝑟𝑟 = (𝐷𝐷 + 3)𝑢𝑢𝑟𝑟 ,
Solution: Since 𝐺𝐺(𝑖𝑖𝜔𝜔) =
𝑧𝑧(𝑖𝑖𝜔𝜔)
𝑝𝑝(𝑖𝑖𝜔𝜔)
𝑦𝑦𝑟𝑟 = 0.212{sin(5𝑡𝑡 − 1.533)}.
(31)
3+5𝑖𝑖
= (2+5𝑖𝑖)(1+5𝑖𝑖) = 0.212𝑒𝑒 −1.533 , the solution is
The complex-valued function 𝑓𝑓(𝜔𝜔) = 𝐺𝐺(𝑖𝑖𝜔𝜔) of the real variable 𝜔𝜔 is called the
frequency response of the system (2). Its magnitude is called the magnitude
P a g e | 15
response of the system, and its angle is called the phase response. The magnitude
versus frequency is typically plotted on log-log axes, and the phase versus
frequency is plotted on linear-log axes. These plots are called Bode plots and give
a complete picture of how the system responds to sinusoids. The system bandwidth
is defined as the range of input frequencies 𝜔𝜔 for which |𝐺𝐺(𝑖𝑖𝜔𝜔)| ≥ |𝐺𝐺(0)|/√2,
where the output is considered to respond well to the input.