Transient Response from State Space Representation

Example: Find State Transition Matrix of a 2^nd Order System

Find Φ(s) and φ(t) if

Solution:

The inverse of a 2×2 matrix is given here.

To find φ(t) we must take the inverse Laplace Transform of every term in the matrix

We now must perform a partial fraction expansion of each term, and solve

Solution via MatLab
MatLab can be used to find the zero input response of a state space system:

 

Example: Zero Input Response from State Space (2x2)

Find the response for the system defined by:

with

and

Solution:
The zero input problem was solved previously

with the state transition matrix given by

For the given A matrix, Φ(s) and φ(t) were calculated previously (above)

So

and

Solution via Matlab
This problem can also be solve with MatLab

A=[0 1; -2 -3];             %Define Matrices
B=[0; 1;];  C=[1 -1];  D=0;
mySys=ss(A,B,C,D);          %Define State Space system
q0=[1; 2;];                 %Define initial conditions
initial(mySys,q0);          %Plot zero input solution

Example: Evaluation of the Matrix Exponential

Write a closed form expression for e^At if .

Solution:
Since

and we know (from above) that for the A matrix specified that

then

 

It is perhaps surprising that the series form of the matrix exponential

yields such a compact closed form solution, but this makes it possible to evaluate e^At (and φ(t)) precisely and efficiently.

Example: Zero State Solution from State Space (2x2)

Find the zero input solution (q_zi(t) and y_zi(t)) for the system defined by:

with

 

and

Solution:
First we need to find the transfer function from the state space representation

We found Φ(s); earlier.

so

(we can check this with MatLab)

>> mySys=ss([0 1; -2 -3], [0; 1], [1 -1], 0);	% Define system in state space
>> [n,d]=tfdata(mySys,'v')   		% Get numerator and denominator
n =
         0   -1.0000    1.0000
d =
     1     3     2

We also know that

so

The partial fraction expansion can be done by hand or with Matlab.  The Matlab solution is shown.

>> [r,p,k]=residue([1 -1],[1 3 2 0 0])	% Perform partial fraction expansion
r =
    0.7500
   -2.0000
   -1.2500
    0.5000
p =
    -2
    -1
     0
     0
k =  []

Example: Complete Response from State Space (2x2)

Find the response for the system defined by:

with

and

Solution:
The zero input problem was solved previously

 

The zero state problem was also solved previously

 

The complete response is simply the sum of the two

 

Solution via Matlab
A numerical solution can be found with Matlab

t=linspace(0,10);           %Define time vector
A=[0 1; -2 -3];             %Define Matrices
B=[0; 1;];  C=[1 -1];  D=0;
mySys=ss(A,B,C,D);          %Define State Space system

u=t;                        %Define input   
q0=[1; 2;];                 %Define initial conditions

yzi=initial(mySys,q0,t);    %Find zero input response
yzs=lsim(mySys,u,t);        %Find zero state response
yc=yzi+yzs;                 %Find complete response

Note that the complete response converges to the zero state response at long times as the zero input response decays to zero.

Example: Another transient response of a state space system

The system shown is a simplified model of a part of a suspension system of a wheel on a car or motorcyle.  The mass, m, represents the weight of the vehicle supported by the wheel, and the spring and dashpot represent the suspension system.  For our purposes let m=500 kg, k=3000 N/m, b=2500 N-m/s.

  

Find the output if the system starts at rest (the velocity is zero) but x_out(0^-)=0.05 and x_in(t)=0.1·γ(t).

 

Solution:

We must first develop a state space model.  Techniques for doing so are discussed elsewhere.  We will start from the system transfer function (derived on the previous page):

We can transform this to Observable Canonic form as

From the output equation we see that

 

From the top row of the state variable equation we see that:

or

 

 

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Zero State Solution:
We start by finding the state transition matrix.  This could be done by hand, we'll use Matlab's symbolic toolbox:

>> syms s
>> A=[-5 1; -6 0];
>> Phi=inv(s*eye(2)-A)
Phi = 
[  s/(s^2 + 5*s + 6),       1/(s^2 + 5*s + 6)]
[ -6/(s^2 + 5*s + 6), (s + 5)/(s^2 + 5*s + 6)]

The zero input solution is

 

with

 

We again turn to Matlab to find Y_zi(s)

>> C=[1 0]; 
>> q0=[0.05; 0.25]; 
>> Yzi=C*Phi*q0 
Yzi =
 s/(20*(s^2 + 5*s + 6)) + 1/(4*(s^2 + 5*s + 6))

>> pretty(simple(Yzi)) 
        s + 5 
  ----------------- 
       2 
  20 (s  + 5 s + 6)

At this point we could perform a partial fraction expansion, but we will let Matlab do the work

>> [r,p,k]=residue([1 5],20*[1 5 6])
r =
   -0.1000
    0.1500
p =
   -3.0000
   -2.0000
k =     []

So we get

As expected this agrees with the solution obtained using the transfer function (done on previous page).

 

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Zero State Solution:
We start by finding the transfer function, which was derived at the start of the problem. 

(Note, if we didn't already know the transfer function we could always use the relationship H(s)=CΦ(s)B+D.)

Rather than solving this again, we refer to the solution on the previous page.

 

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Complete Solution:

The complete response is simply the sum of the zero input and zero stat response.