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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:

dnytd t n+α n - 1 dn1ytd t n1++α1d1ytd t 1+α0yt=βmdmutd t m+ β m - 1 dm1utd t m1++β1d1utd t 1+β0ut 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 qsqs and ps p s as nn-th and mm-th order polynomials in ss.

qs=sn+ α n - 1 sn1++α1s+α0 q s s n α n - 1 s n1 α1 s α0
(2)
ps=βmsm+ β m - 1 sm1++β1s+β0 p s βm s m β m - 1 s m1 β1 s β0
(3)

If we go ahead and say that qsq s and psp s will take the differential operator as arguments, we can rewrite Equation 1 as

qtyt=ptut q t y t p t u t
(4)

Looking at differential equations in terms of polynomials such as qsq s and psp 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 Z1s Z1 s, and the resistor and capacitor in parallel form the impedance Z2s Z2 s.
RLC circuit: 2nd order (Impedances Labeled)
RLC circuit: 2nd order (Impedances Labeled) (rlc_imp.png)
YsUs=Z2sZ1s+Z2s Y s U s Z2 s Z1 s Z2 s
(5)

where

Z1s=1+s2 Z1 s 1 s 2
(6)

and

Z2s=112+s3 Z2 s 1 1 2 s 3
(7)

This yields

Ys=6s2+72s+9Us 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:

d2ytd t 2+72d1ytd t 1+9yt=6ut 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 psp s and qsq s.

Non-zero Initial Conditions

By saying above that the Laplace transform of ddtyt t1 y t is sYs 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:

dytdt=sYsy0- 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 y0- y 0- in this equation, we are mixing frequency domain and time domain variables here. However, as y0- 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:

d1ytd t 1+α0yt=ut t 1 y t α0 y t u t
(11)

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

sYsy0-+α0Ys=Us s Y s y 0- α0 Y s U s
(12)

By combining the YsYs terms we get

(s+α0)Ysy0-=Us s α0 Y s y 0- U s
(13)

If we say that qs=s+α0 qs s α0 and ps=1 ps 1 , and define rs rs as y0- y 0- , we can rearrange terms to get an expression relating Ys Y s to Us U s that takes the initial conditions into account:

Ys=psqsUs+rsqs 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:

yt=ypartt+yhomot y t ypart t yhomo t
(15)

Here, ypartt ypart t corresponds to psqsUs p s q s U s and yhomot yhomo t corresponds to rsqs r s q s . This makes a lot of sense. The particular solution (forced response) depends on a combination of qs q s , which describes how the system behaves independently, and psUs p s U s , which describes how the system reacts to the input UU. The homogeneous solution (natural response) depends on a combination of qs q s and rs 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, yt y t , to a differential equation describing a system.

yt=et y t t
(16)

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

Solution:

From above, we know that for a homogeneous solution

yt=rtqt y t r t q t
(17)

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

qtyt=0 q 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 qt=t1 q t 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 rs r s and qs q s as discussed above, we have:

Ys=1s1=rsqs Y s 1 s 1 r s q s
(19)

Directly, we can see the solution for qs q s : by simply setting the denominators equal to each other, qs=s1 q s s 1 . This, of course, is the Laplace transform of the solution of qt q 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

yt=Aeat+Btebt y t A a t B t b t
(20)

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

Solution:

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

Ys=Asa+Bsb2=Asb2+B(sa)(sa)sb2=rsqs 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 qs q s :

qs=(sa)sb2=s3(2b+a)s2+(b2+2ab)sab2 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 qs q s by Ys Y s and taking the inverse Laplace transform will yield the differential equation whose homogeneous solution is yt y t :

d3dt3yt(2b+a)d2ytd t 2+(b2+2ab)d1ytd t 1ab2yt=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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