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Linear Systems with Constant Coefficients

Module by: Thanos Antoulas, JP Slavinsky. E-mail the authors

Summary: Linear Systems with constant coefficients.

Analyzing Linear Constant Coefficient Systems

Constant coefficient linear systems describe LTI systems and form the beginnings of the study of state-space systems. In general, an nn-th order system such as this can be described by the following differential equation, where uu and yy represent the input and output variables, respectively:

dny⁢td t n+α n - 1 ⁢dn−1y⁢td t n−1+ … +α1⁢d1y⁢td t 1+α0⁢y⁢t=βm⁢dmu⁢td t m+ β m - 1 ⁢dm−1u⁢td t m−1+ … +β1⁢d1u⁢td t 1+β0⁢u⁢t t n y t α n - 1 t n1 y t … α1 t 1 y t α0 y t βm t m u t β m - 1 t m1 u t … β1 t 1 u t β0 u t
(1)

This is a linear differential equation with real-valued constant coefficients αiαi and βjβj. We can represent this equation with a much more compact notation by thinking of the differential operator as a variable ss, where the degree of the derivative corresponds to the power of ss. We will define q⁢sqs and p⁢s p s as nn-th and mm-th order polynomials in ss.

q⁢s=sn+ α n - 1 ⁢sn−1+…+α1⁢s+α0 q s s n α n - 1 s n1 … α1 s α0
(2)
p⁢s=βm⁢sm+ β m - 1 ⁢sm−1+…+β1⁢s+β0 p s βm s m β m - 1 s m1 … β1 s β0
(3)

If we go ahead and say that q⁢sq s and p⁢sp s will take the differential operator as arguments, we can rewrite Equation 1 as

q⁢ⅆⅆ⁢t⁢y⁢t=p⁢ⅆⅆ⁢t⁢u⁢t q ⅆ ⅆ t y t p ⅆ ⅆ t u t
(4)

Looking at differential equations in terms of polynomials such as q⁢sq s and p⁢sp s right away reminds us the Laplace transform. Using the Laplace transform can often simplify the calculations involved in system analysis.

Example 1

By using Laplace domain tools to examine the RLC circuit seen before, the differential equation describing the behavior of this system is easy to deduce.

Figure 1: The resistor and inductor in series are combined in the impedance Z1⁢s Z1 s, and the resistor and capacitor in parallel form the impedance Z2⁢s Z2 s.
RLC circuit: 2nd order (Impedances Labeled)
RLC circuit: 2nd order (Impedances Labeled) (rlc_imp.png)
Y⁢sU⁢s=Z2⁢sZ1⁢s+Z2⁢s Y s U s Z2 s Z1 s Z2 s
(5)

where

Z1⁢s=1+s2 Z1 s 1 s 2
(6)

and

Z2⁢s=112+s3 Z2 s 1 1 2 s 3
(7)

This yields

Y⁢s=6s2+72⁢s+9⁢U⁢s Y s 6 s 2 7 2 s 9 U s
(8)

By multiplying both sides by the denominator of the fraction and taking the inverse Laplace transform, the final differential equation describing the system is determined:

d2y⁢td t 2+72⁢d1y⁢td t 1+9⁢y⁢t=6⁢u⁢t t 2 y t 72 t 1 y t 9 y t 6 u t
(9)

The results from this example encourage us to apply Laplace techniques to earlier equations in this section. By taking the Laplace transform of Equation 4, we find that the transfer function of this system is simply the ratio of the two polynomials p⁢sp s and q⁢sq s.

Non-zero Initial Conditions

By saying above that the Laplace transform of ddty⁢t t1 y t is s⁢Y⁢s s Y s we are assuming that the system has zero initial conditions . Taking a more general approach, a factor to characterize the initial conditions must be included:

ℒ⁢dy⁢tdt=s⁢Y⁢s−y⁢0- ℒ t y t s Y s y 0-
(10)

Note: 0-0- here means the instant immediately preceding time t=0t0; it represents the time right before we start paying attention to the system. Also, some readers may be worried that by using y⁢0- y 0- in this equation, we are mixing frequency domain and time domain variables here. However, as y⁢0- y 0- is simply a constant, there is no time domain component involved.

Example 2

Let's look at an example to see how these non-zero initial conditions come into play. A first order system is described by the following differential equation:

d1y⁢td t 1+α0⁢y⁢t=u⁢t t 1 y t α0 y t u t
(11)

When we take the Laplace transform of the derivative of y⁢tyt, we must remember to include a term that represents the initial values of the system output.

s⁢Y⁢s−y⁢0-+α0⁢Y⁢s=U⁢s s Y s y 0- α0 Y s U s
(12)

By combining the Y⁢sYs terms we get

(s+α0)⁢Y⁢s−y⁢0-=U⁢s s α0 Y s y 0- U s
(13)

If we say that q⁢s=s+α0 qs s α0 and p⁢s=1 ps 1 , and define r⁢s rs as y⁢0- y 0- , we can rearrange terms to get an expression relating Y⁢s Y s to U⁢s U s that takes the initial conditions into account:

Y⁢s=p⁢sq⁢s⁢U⁢s+r⁢sq⁢s Y s p s q s U s r s q s
(14)

What we have here is the Laplace domain solution to a differential equation describing a dynamical system. There are two terms in this solution: one that relies on the input and one that does not. These parts correspond to the particular and homogeneous solutions, respectively. Taking the inverse Laplace transform, we can write (14) as:

y⁢t=ypart⁢t+yhomo⁢t y t ypart t yhomo t
(15)

Here, ypart⁢t ypart t corresponds to p⁢sq⁢s⁢U⁢s p s q s U s and yhomo⁢t yhomo t corresponds to r⁢sq⁢s r s q s . This makes a lot of sense. The particular solution (forced response) depends on a combination of q⁢s q s , which describes how the system behaves independently, and p⁢s⁢U⁢s p s U s , which describes how the system reacts to the input UU. The homogeneous solution (natural response) depends on a combination of q⁢s q s and r⁢s r s , the latter of which contains information about the initial conditions of the system.

Example 3

Let's say that we know the homogeneous solution, y⁢t y t , to a differential equation describing a system.

y⁢t=et y t t
(16)

Goal: Using this solution, we want to try and figure out the system's q⁢ⅆⅆ⁢t q ⅆ ⅆ t function given zero initial conditions.

Solution:

From above, we know that for a homogeneous solution

y⁢t=r⁢ⅆⅆ⁢tq⁢ⅆⅆ⁢t y t r ⅆ ⅆ t q ⅆ ⅆ t
(17)

We can clear the denominator by moving the q⁢ⅆⅆ⁢t q ⅆ ⅆ t to the left-hand side. And since we have zero initial conditions, r⁢ⅆⅆ⁢t r ⅆ ⅆ t goes to 0:

q⁢ⅆⅆ⁢t⁢y⁢t=0 q ⅆ ⅆ t y t 0
(18)

The solution can quickly be determined by inspection because we know that the derivative of et t is et t . Therefore a solution of q⁢ⅆⅆ⁢t=ⅆⅆ⁢t−1 q ⅆ ⅆ t ⅆ ⅆ t 1 would work. However, a more systematic approach will be necessary for more difficult situations. We will investigate this approach here.

Again, we will do our work in the Laplace domain. By equating the Laplace transform of our homogeneous solution with the ratio of r⁢s r s and q⁢s q s as discussed above, we have:

Y⁢s=1s−1=r⁢sq⁢s Y s 1 s 1 r s q s
(19)

Directly, we can see the solution for q⁢s q s : by simply setting the denominators equal to each other, q⁢s=s−1 q s s 1 . This, of course, is the Laplace transform of the solution of q⁢ⅆⅆ⁢t q ⅆ ⅆ t that we found by inspection above.

Example 4

Now that we have the basics down, we'll look at a more complicated example. We are given

y⁢t=A⁢ea⁢t+B⁢t⁢eb⁢t y t A a t B t b t
(20)

Goal: We would like to find the differential equation whose homogeneous solution is y⁢t y t .

Solution:

Again, we take the Laplace transform of y⁢t y t , and then combine the two resultant fractions into one ratio of polynomials:

Y⁢s=As−a+Bs−b2=A⁢s−b2+B⁢(s−a)(s−a)⁢s−b2=r⁢sq⁢s Y s A s a B s b 2 A s b 2 B s a s a s b 2 r s q s
(21)

Next, we equate the denominators of the last two fractions to find q⁢s q s :

q⁢s=(s−a)⁢s−b2=s3−(2⁢b+a)⁢s2+(b2+2⁢a⁢b)⁢s−a⁢b2 q s s a s b 2 s 3 2 b a s 2 b 2 2 a b s a b 2
(22)

Recalling the start of this module, multiplying q⁢s q s by Y⁢s Y s and taking the inverse Laplace transform will yield the differential equation whose homogeneous solution is y⁢t y t :

d3dt3y⁢t−(2⁢b+a)⁢d2y⁢td t 2+(b2+2⁢a⁢b)⁢d1y⁢td t 1−a⁢b2⁢y⁢t=0 t 3 y t 2 b a t 2 y t b 2 2 a b t 1 y t a b 2 y t 0
(23)

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