HW6, MAE 171. Spring 2005. UCI

Nasser Abbasi

Problem 6-15

Consider system

MATH

The digital controller $D\left( z\right) $ is described by difference equation MATH

part(a)

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 $z=1$ for the case of a discrete system, or the number of poles at $s=0\,$ in the case of the continuous system. So, determine the open loop tf and find its poles.

solution: First find $D\left( z\right) $, From the difference equation, take the Z transform, we obtain

MATH

Hence the open loop transfer function is now MATH

For $T=1$

MATH

Hence we see that there is one pole at $z=1$, hence the

system type is 1

part(b)

Find the steady state response for a unit step input without finding $C\left( z\right) $

Solution method: Apply the final value theorem:

solution:

MATH

Since $R\left( z\right) $ is a unit step, then MATH then we get

MATH

Another way to solve this is to find steady state error, and subtract that from the input. As follows

MATH

But MATH

hence

MATH

Since $R\left( z\right) $ is a unit step, then MATH then we get

MATH

and using value for $L\left( z\right) $ we found above, then we get

MATH

Since steady state error is zero, then steady state response must be equal to input, which is $1$, hence

$c_{ss}=1$
as found above.

part(c)

Find approximate time for system to reach steady state.




Solution method: Use relation MATH, where $r$ is the length of the dominant pole of the closed loop tf. Then set time to reach steady state as $4\tau .$

Solution: Time to reach steady state within $2\%$ final value is $4\tau $ where $\tau $ is the time constant of the closed loop transfer function. Hence need to determine $\tau $.

But

MATH

Where $r$ is the length of the dominant pole and $T$ is the sampling time.

The closed loop transfer function is

MATHZeros of the denominator at $z=-0.212\,,z=0.949$

Then

the dominant pole $r$ is $0.949$

Hence MATH

Hence it takes MATH

to reach steady state (using $2\%$ criterion)

part(d)

Find the unit step response for the system and verify parts (b) and (c)

solution method: Find response by evaluating MATH

solution:

MATHMATH

MATH

MATH

MATH

Hence MATH

HenceMATH

We see that as MATH, $c[n]\rightarrow 1$ which verifies part(b)

To verify part(c), set $n$ to be the closest integer to $74.68$ , which is the time to reach within $2\%$ of final value. Since these are sample numbers, and the sample time $T=1\sec $, then set $n=75$ and find $c[n]$ for $n=75$, we get

$\ $

MATH

We see that at $c[75]$ is within the $2\%$ 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)


Figure

Problem 6-17

Consider system

MATH

The digital controller $D\left( z\right) $ is described by difference equation MATH

part(a)

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 $z=1$ for the case of a discrete system, or the number of poles at $s=0\,$ in the case of the continuous system. So, determine the open loop tf and find its poles.

solution:

From problem 6-15,

MATH

Hence the open loop transfer function is now MATH

For $T=1$

MATH

Hence we see that now there are no pole at $z=1$, hence the

system type is 0

part(b)

Find the steady state response for a unit step input without finding $C\left( z\right) $

Apply the final value theorem:

MATH

Since $R\left( z\right) $ is a unit step, then MATH then we get

MATH

part(c)

Find approximate time for system to reach steady state.

Time to reach steady state within $2\%$ final value is $4\tau $ where $\tau $ is the time constant of the closed loop transfer function. Hence need to determine $\tau $.

But

MATH

Where $r$ is the length of the dominant pole and $T$ is the sampling time.

The closed loop transfer function is

MATH

Zeros of the denominator at $z=0.634\ $

Then

$r$ is $0.634$

Hence MATH

Hence it takes MATH

to reach steady state (using $2\%$ criterion)

part(d)

Find the unit step response for the system and verify parts (b) and (c)

MATH

MATH

MATH

MATH

Hence

MATH

HenceMATH

We see

that as MATH, MATH
which verifies part(b)

To verify part(c), set $n$ to be the closest integer to MATH , which is the time to reach within $2\%$ of final value. Since these are sample numbers, and the sample time $T=1\sec $, then set $n=9$ and find $c[n]$ for $n=9$, we get

$\ $

MATHSince the final value of $c[n]$ is $0.136\,56$, then this error is MATH MATH. This is good enough approximation to final value, so this verifies part(c).

Problem 7-5

Given the following characteristic equations

iii)$z^{2}-0.1z-0.3=0$ vi)$z^{2}-2z+0.99=0$ vii)MATH viii)MATH

In these problems, the equation is expressed in form MATH, $a_{n}>0$

part(a)

Use Jury test to determine stability of each system

iii)$z^{2}-0.1z-0.3=0$

test1: is MATH, i.e. is $1>0.3$, YES, test pass.

test2: is MATH, i.e. is $1^{2}-0.1-0.3>0,$ YES, test pass.

test3: MATH, hence MATH. hence MATHYES, test pass.

Since second order system, no need to construct Jury table.

Hence

system is stable.

vi)$z^{2}-2z+0.99=0$

test1: is MATH, i.e. is $1>0.99$, YES, test pass.

test2: is MATH, i.e. is $1^{3}-2+0.99>0,$ NO, test pass.

Hence

NOT stable system.

vii)MATH

test1: is MATH, i.e. is $1>0.35$, YES, test pass.

test2: is MATH, i.e. is MATH NO, test fails

Hence

NOT stable system.

viii)MATH

test1: is MATH, i.e. is $1>0.45$, YES, test pass.

test2: is MATH, i.e. is MATH is $0.05>0, $YES, test pass.

test3: MATH, hence MATH. hence MATHYES, test pass.

construct jury table

$a_{0}$ $-0.45$ $1.41$ $-1.9$ $1$
$a_{n}$ $1$ $-1.9$ $1.41$ $-0.45$
$m_{0}$ $-0.7975$ $-0.555$

is MATH, is $0.7975>0.555$, YES. test pass.

Hence

system is stable.

part(b)

List the natural response terms for each of the systems.

Solution method: roots of characteristic equation that appear at MATH result in natural response terms of the form MATH

solution:

iii)$z^{2}-0.1z-0.3=0$

zeros are $z=-0.5,z=0.6$

Hence for the first pole, we have $r=0.5,\theta =\pi $, hence this pole contributes a term of the form

MATH

For the first pole, we have $r=0.6,\theta =0$, hence this pole contributes a term of the form

MATH

So if we let $y[n]$ be the natural response, we can write MATH

vi)$z^{2}-2z+0.99=0$

zeros are MATH

Hence for the first pole, we have $r=1.1,\theta =0$, hence this pole contributes a term of the form

MATH

For the first pole, we have $r=0.9,\theta =0$, hence this pole contributes a term of the form

MATH

So if we let $y[n]$ be the natural response, we can write

MATH

vii)MATH

zeros are $z=0.7,z=0.5,z=1.0$

First pole contributes a term of the form

MATH

Second pole contributes a term of the form

MATH

Third pole contributes a term of the form

MATH

So if we let $y[n]$ be the natural response, we can write

MATH

viii)MATH

zeros are MATH

First pole $r=0.878,\theta =0,$ so it contributes a term of the form

MATH

Second pole MATH, MATH $rad.$ So this pole contributes a term of the form

MATH

Third pole contributes the same term as above.

So if we let $y[n]$ be the natural response, we can write

MATH

part(c)

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)$z^{2}-2z+0.99=0$

vii)MATH

For system vi, the natural response term that causes the instability is as shown below

MATH

This terms comes from the pole at $z=1.1$




For system vii, the natural response term that causes the marginal instability is as shown below

MATH

This term comes from the pole at $z=1$

Problem 7-6

Consider the system shown below. Let the digital controller be MATH.

MATH

part(a)

write the closed loop system characteristic equation.

The open loop transfer function is now MATH

For $T=1$

MATH

The closed loop transfer function is

MATH

Hence closed loop characteristic equation is

MATH

part(b)

determine the range of $K$ for which the closed loop system is stable

Here $n=1$, and $a_{n}=1$, MATH

For stability need MATH, hence MATH, hence MATH, hence

MATH

The second test MATH yields MATH

Third test MATH yields

MATH

Hence

$-1<k<2.1646$

Part(c)

Supposed that $K$ is set to the lower limit found in part(b), find the natural response term and illustrate the marginal instability

$\ $

set $K=-1$

Hence the closed loop system characteristic equation is $z-0.368-\ 0.632=0$ or

MATH

Hence we have a pole at $z=1$ which maps to the origin in the S-plane. A pole at $z=1$ means the system is marginally unstable. This pole will contribute a term of the form MATH to the natural solution. Since there is only one term here, we see that the natural response is simply $A\cos \phi $, which will not decay, hence the system will continue to oscillate as it is marginally unstable.

part(d)

Repeat part(c) for the upper limit of $K$

set $K=2.1646$

Hence the closed loop system characteristic equation is MATH or

MATH

Hence we have a pole at $z=-1$. This pole will contribute a term of the form MATH to the natural solution. Since there is only one term here, we see that the natural response is simply $A\cos \phi $, which will not decay, hence the system will continue to oscillate as it is marginally unstable.

part(e)

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 $K$ will result in an unstable system and for values just inside the range, the step response is stable.

This is the output.


Figure

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));