HW6, MAE 171. Spring 2005. UCI
Nasser Abbasi
Consider system
The digital controller is described by difference equation
Find system type
Solution method: The system type is always taken relative to the open loop transfer function. The type of the system is the number of poles at for the case of a discrete system, or the number of poles at in the case of the continuous system. So, determine the open loop tf and find its poles.
solution: First find , From the difference equation, take the Z transform, we obtain
Hence the open loop transfer function is now
For
Hence we see that there is one pole at , hence the
Find the steady state response for a unit step input without finding
Solution method: Apply the final value theorem:
solution:
Since is a unit step, then then we get
Another way to solve this is to find steady state error, and subtract that from the input. As follows
But
hence
Since is a unit step, then then we get
and using value for we found above, then we get
Since steady state error is zero, then steady state response must be equal to input, which is , hence
Find approximate time for system to reach steady state.
Solution method: Use relation , where is the length of the dominant pole of the closed loop tf. Then set time to reach steady state as
Solution: Time to reach steady state within final value is where is the time constant of the closed loop transfer function. Hence need to determine .
But
Where is the length of the dominant pole and is the sampling time.
The closed loop transfer function is
Zeros of the denominator at
Then
Hence
Hence it takes
to reach steady state (using criterion)
Find the unit step response for the system and verify parts (b) and (c)
solution method: Find response by evaluating
solution:
Hence
Hence
We see that as , which verifies part(b)
To verify part(c), set to be the closest integer to , which is the time to reach within of final value. Since these are sample numbers, and the sample time , then set and find for , we get
We see that at is within the range (one percent on each side of the final value).
This verifies part(c)
This is a MATLAB verification of response to unit step.
clear all; close all;
z=tf('z',1);
sys=(0.63212*z-0.56891)/(z^2-0.73576*z-0.20103)
step(sys)
Consider system
The digital controller is described by difference equation
Find system type
Solution method: The system type is always taken relative to the open loop transfer function. The type of the system is the number of poles at for the case of a discrete system, or the number of poles at in the case of the continuous system. So, determine the open loop tf and find its poles.
solution:
From problem 6-15,
Hence the open loop transfer function is now
For
Hence we see that now there are no pole at , hence the
Find the steady state response for a unit step input without finding
Apply the final value theorem:
Since is a unit step, then then we get
Find approximate time for system to reach steady state.
Time to reach steady state within final value is where is the time constant of the closed loop transfer function. Hence need to determine .
But
Where is the length of the dominant pole and is the sampling time.
The closed loop transfer function is
Zeros of the denominator at
Then
Hence
Hence it takes
to reach steady state (using criterion)
Find the unit step response for the system and verify parts (b) and (c)
Hence
Hence
We see
To verify part(c), set to be the closest integer to , which is the time to reach within of final value. Since these are sample numbers, and the sample time , then set and find for , we get
Since the final value of is , then this error is . This is good enough approximation to final value, so this verifies part(c).
Given the following characteristic equations
iii) vi) vii) viii)
In these problems, the equation is expressed in form ,
Use Jury test to determine stability of each system
iii)
test1: is , i.e. is , YES, test pass.
test2: is , i.e. is YES, test pass.
test3: , hence . hence YES, test pass.
Since second order system, no need to construct Jury table.
Hence
vi)
test1: is , i.e. is , YES, test pass.
test2: is , i.e. is NO, test pass.
Hence
vii)
test1: is , i.e. is , YES, test pass.
test2: is , i.e. is NO, test fails
Hence
viii)
test1: is , i.e. is , YES, test pass.
test2: is , i.e. is is YES, test pass.
test3: , hence . hence YES, test pass.
construct jury table
is , is , YES. test pass.
Hence
List the natural response terms for each of the systems.
Solution method: roots of characteristic equation that appear at result in natural response terms of the form
solution:
iii)
zeros are
Hence for the first pole, we have , hence this pole contributes a term of the form
For the first pole, we have , hence this pole contributes a term of the form
So if we let be the natural response, we can write
vi)
zeros are
Hence for the first pole, we have , hence this pole contributes a term of the form
For the first pole, we have , hence this pole contributes a term of the form
So if we let be the natural response, we can write
vii)
zeros are
First pole contributes a term of the form
Second pole contributes a term of the form
Third pole contributes a term of the form
So if we let be the natural response, we can write
viii)
zeros are
First pole so it contributes a term of the form
Second pole , So this pole contributes a term of the form
Third pole contributes the same term as above.
So if we let be the natural response, we can write
For those systems in part(a) that are found to be either unstable or marginally stable, list the natural response terms in part(b) that yield these results.
From part(a), these are the systems found to be unstable
vi)
vii)
For system vi, the natural response term that causes the instability is as shown below
This terms comes from the pole at
For system vii, the natural response term that causes the
marginal instability is as shown below
This term comes from the pole at
Consider the system shown below. Let the digital controller be .
write the closed loop system characteristic equation.
The open loop transfer function is now
For
The closed loop transfer function is
Hence closed loop characteristic equation is
determine the range of for which the closed loop system is stable
Here , and ,
For stability need , hence , hence , hence
The second test yields
Third test yields
Hence
Supposed that is set to the lower limit found in part(b), find the natural response term and illustrate the marginal instability
set
Hence the closed loop system characteristic equation is or
Hence we have a pole at which maps to the origin in the S-plane. A pole at means the system is marginally unstable. This pole will contribute a term of the form to the natural solution. Since there is only one term here, we see that the natural response is simply , which will not decay, hence the system will continue to oscillate as it is marginally unstable.
Repeat part(c) for the upper limit of
set
Hence the closed loop system characteristic equation is or
Hence we have a pole at . This pole will contribute a term of the form to the natural solution. Since there is only one term here, we see that the natural response is simply , which will not decay, hence the system will continue to oscillate as it is marginally unstable.
Verify the result of this problem by digital computation.
Using Matlab it is shown that the step response for values just outside the range of will result in an unstable system and for values just inside the range, the step response is stable.
This is the output.
MATLAB code:
function nma_test_prob_7_6
close all; clear all;
n=1;
subplot(2,2,n);
do(2.3);
n=n+1;
subplot(2,2,n);
do(2.1);
n=n+1;
subplot(2,2,n);
do(-0.9);
n=n+1;
subplot(2,2,n);
do(-1.1);
function do(k)
z=tf('z');
sys=0.632/(z-0.368+0.632*k);
step(sys);
title(sprintf('k=%2.1f',k));