HW7, MAE 171. Spring 2005. UCI

Nasser Abbasi

Problem 1

Plot the R.L. (Root Locus)

step 1

Identify the position of the open loop poles and zeros on the Z-plane. Since the open loop tf is then we have a zero at and a pole at

step 2

Identify R.L. on the real line.

Number of poles and zeros must be odd to the right of the line segment for that segment to be on the R.L., hence we see that the R.L. on the real line is as follows.

step 3

Find asymptotes.

Number of asymptotes is $n-m\,\$ where is number of finite open loop poles, and is number of open loop finite zeros.

Hence , so there is one asymptote. The angle the asymptote makes with the real axis is for (this is for .

Since , then we count up to zero only (, so we obtain angle as .

Now find where the center of the asymptote on the real line.

This is given by MATH

step 4 $\qquad$

Find break away and break in points on the real line. Note_1

The characteristic equation is hence , then

Hence $\frac{dK}{dz}=0$ leads to $-z^{2}-2z+2.213=0$ , Solution is

$\allowbreak$ . Since both solutions are on the R.L. then both are valid. The solution

is between 2 poles, hence it is a breakaway point, and the solution

is between 2 zeros (one zeros at $-\infty$ ) hence that is a break-in point. So far we have this plot.

step 5

Find angle of departure of the break away point. (point ).

There are 4 branches at this point. Hence each branch makes $90^{0}$ with the other branch. Since one branch is fixed (from to ), hence R.L. makes $90^{0}$ with the real line. Similarly for angle of arrival at point ). Hence RL is a circle that has on its perimeter the point and the point ) (we are not asked to prove this). Since RL starts at a poles and ends at a zero, then one branch will arrive at the break-in point and moves to the zero located at and the other branch will move to the other zero locate at $z=-\infty$ . Hence RL looks as follows

The center of the RL circle is located at midpoint between and i.e. at point

Now to determine the critical value of gain for stability.

will be zero at a pole and will be $\infty$ at a zero. When RL meets the unit circle the system will become unstable.

We can use Jury tests to find the range of stable K.

The characteristic equation is MATH

Hence we need or or

From requirement that we obtain MATH

From requirement that we obtain

MATH

Which does not provide any new range information for as cancels out.

Since this is a second order system, no need to construct Jury table. We can stop here. Hence the range of for stability is MATH

We can now find where the R.L. cross the unit circle.

Since , when we get MATH Solution is MATH

Hence MATH

But MATH

This is $\omega _{d}$ when the system becomes unstable, ie. at $\zeta =0$ and

$T=0.1\sec$ , what value of gain will yield $\zeta =0.5?$

$\zeta$ is a function of the position of the closed loop discrete system as expressed by the relation . The position of the closed loop poles is a function of (since by definition the closed loop poles move on RL as changes).

The characteristic equation is , then the closed loop poles are the roots of this equation. which are

Taking one of the roots, we obtain MATH

Note that $\zeta =0$ at and $\zeta$ will increase as becomes smaller.

By trial and error decrease and for each solve for in the above equation. Once is found, $r,\theta$ are known. Find and check if the desired value. Continue this process until we reach the required $\zeta =0.5$ value.

This is a simple script which solves this. The answer is MATH

A table is generated showing how $\zeta$ changes with , this is a partial listing showing when $\zeta =0.5$

   k           Damping Ratio

 0.066,      0.4927400303245694

 0.065,      0.49825440726325265

 0.064,      0.5038857217422561

 0.063,      0.5096384660975533

This below is the script with a plot showing how changes with $\zeta$

determine damped natural frequency and the number of samples

With the gain set to yield $\zeta =0.5$ , determine damped natural frequency and the number of samples per cycle of damped sinusoidal oscillation

Since fixes the position of the closed loop pole on the R.L., then we can now find the angle $\theta$ that the pole makes with the real axis in the Z plane.

For , substitute into equation (1) above, we obtain the pole location. MATH Hence MATH

Now we can find $\omega _{d}$

MATH

We can also find $\ \omega _{n}$ (but problem did not ask for it). MATH

Now we find MATH

Since must be an integer (it is the number of samples per cycle) then we must choose or . Which one to choose?

If we choose the lower value, i.e. , this means we choose a larger $\theta$ , i.e. larger gain $K\,$ or smaller $\zeta$ , and the reverse will happen if we select , the pole will be further away from the critical since $\theta$ will be smaller. It is better I think to select the larger value to improve the relative stability by staying away from the edge if the unit circle. So I select

MATH

Problem 2

Specification

$\zeta ,$ the damping ratio of the dominant closed loop poles is
the number of samples per second is 8
, the sampling period is $0.1\sec$

Solution

First, assume there is no compensator $D\left( z\right)$ .

MATH

Note that the uncompensated open loop poles are at

, and its zero at

The uncompensated system characteristic equation is hence the characteristic polynomial is or

Hence uncompensated closed loop poles (roots of this polynomial) are at MATH

therefor

and

It is clear that we need a compensator, since we want $\theta$ to be which is not the case as is.

Now find the desired closed loop location to meet the specifications.

MATH

but MATH

Then from (1) we find

MATH

Hence desired closed loop dominant pole must be located at MATH and at MATH Hence desired closed loop pole location is MATH

Now that we know where the desired closed loop pole must be located, we can add a compensator such that the angle condition is satisfied for this desired pole.

This drawing shows the uncompensated $G\left( z\right)$ poles/zeros and the desired closed loop location.

Now we add the lead compensator such that with its pole and zero we will satisfy the angle condition at $z_{d}$ . Since this is a lead compensator, then its zero will be to the right of its pole.

First let find out the angle deficit before adding the compensator.

With simple geometry we can find the sum of uncompensated angles

, hence pole makes with desired closed loop pole.

,hence zero $\widetilde{z}_{o}$ makes with desired closed loop pole.

Hence MATH

Since the angle condition requires that the sum of angles with the desired closed loop pole be we need to locate the compensator pole and zero to meet this condition.

Put the compensator zero $z_{o}$ on top of the uncompensated open loop pole , and the compensator pole on the real axis such that the angle condition is met. Hence now we have this diagram

The angle $\theta ,\phi$ did not change, hence we can now find angle $\gamma$

MATH

Hence MATH

Therefor we can find the lead compensator pole location MATH

$\$

Therefor the lead compensator now is

MATH

Now we need to find $K_{D}.$ Use magnitude condition on the characteristic equation for the compensated closed loop system.

MATH

Where

Note that when finding the magnitude values we do not consider the canceled poles/zeros). We get

MATH

But

Hence

MATH

Hence MATH

Hence lead compensator transfer function now is MATH

So the compensated open loop T.F. now is

MATH

To verify and to check internal consistency of the computation, find the closed loop pole of the compensated system and verify it is the same as the one we started with. The characteristic polynomial of the compensated system is MATH Hence the close loop poles of the compensated system are the roots of the above polynomial. Solution is: MATH which matches the desired closed loop pole we started with. (see equation (2) above).

This completes the design the lead compensator.

To find the static velocity constant. $K_{v}$

From the open loop for the compensated system, equation (3) above

MATH

The pole at can be approximated to be at , so I can write

MATH

This is a

type one system

MATH

Hence $e_{ss,ramp}=$

Add a lag type compensator to boost $K_{v}$ by a factor of 2

We want MATH

Hence now $e_{ss,ramp}=$

The lag compensator is , let per problem specification. Hence the new open loop now is , hence we want

MATH

But , hence

MATH

Hence MATH

Hence the lag compensator is

Therefor the final compensated open loop now is

MATH

To verify that the closed loop pole still remain at the desired location, the characteristic polynomial is MATH

Solution is MATH

So the desired dominant poles remain close to the same location. Compare with equation (2). Not exactly the same location (see equation 2), but very close.

Error in , the length of desired pole is

MATH

and error in $\theta$ , phase of desired pole

MATH

So shift in position is less than one tenth of a percent.

System should be tested now to verify this slight movement of the closed loop pole does not have adverse effect.

This diagram shows the final poles and zeros of the final compensated system.

Discrete Simulation of unit step response

The step response for the plant only, and the plant+lead and plant+lead+lag is shown below to see the effect of adding the compensators.

The step response of the plant only shows that the response takes longer to reach the desired value, the settling time is longer, as well as the rise time. But adding the compensator does not seem to have an effect of the maximum overshoot $M_{p}$ .

When the lead and lag compensators are added, the response reaches the desired value much faster. The rise time is about 1/2 second compared with about 3.5 seconds with the plant alone. The settling time when compensators are present is about 1 second, while without compensators the settling time is about 6-7 seconds.

When comparing the plant+lead vs plant+lead+lag on the step response, the presence of the lag compensator does not seem to have much of an effect, actually looking at the zoomed area, when using only the lead compensator, the step response is a little better without the presence of the lag compensator.

The effect of the lag compensator is more evident when looking at the steady state error for a ramp response. In that instance, the presence of the lag compensator has improved the steady state error by making it smaller than when using a lead compensator only. Actually the steady state error for tracking a ramp input is half that when using a lead compensator only.

The diagram below shows the impulse response for the plant alone, then plant+lead, then plant+lead+lag for comparison.

Continuous time system simulation

Using simulink we can obtain the continuous time system response to a step input. This is the result.

Problem 3

Given and $T=1\sec$ , find equivalent discrete-time systems using

standard Z transform (impulse invariance)
step invariance
backward difference
forward difference
bilinear z-transform
matched z-transform

Summary of results

This tables summarizes the results

Standard Z transform

step invariance

backward difference

forward difference

bilinear z-transform

matched z-transform

Step response of different emulation methods

To help understand more the emulation methods, a step response using each different emulation method is shown alongside the Laplace transform response.