25

CHAPTER 4 Longitudinal Motion

Read section 4.2.

Longitudinal Equations

0

u

(4.51)

PURE PITCHING MOTION (Section 4.3 on p.139 of Nelson)

EXAMPLE PROBLEM 1 It is desired to develop a test rig for estimating the pitching lift derivative coefficient,

a

L

C

, for horizontal tail designs. The beginning of this development is shown below.

D

D

+

+

+

D

D

D

D

+

+

+

-

=

D

D

D

D

T

w

w

q

w

w

w

w

u

w

u

w

u

w

u

T

T

T

T

Z

M

M

Z

X

Z

M

M

Z

X

q

w

u

M

u

M

Z

M

M

Z

M

M

u

Z

Z

g

X

X

q

w

u

d

d

q

q

d

d

d

d

d

d

d

d

0

0

0

1

0

0

0

0

0

0

0

&

&

&

&

&

&

&

&

&

Figure 4.9 of Nelson Rod-plate assembly constrained to pure pitching motion.

(a) Development of the equation of angular motion:

We begin with:

q

q

M

M

I

M

y

cg

0

0

+

=

=

a

a

q

&

&

.(E1.1)

Note #1: Since the cg of the test rig is constrained, the angle of attack,

a

, and the pitch angle,

q

, are one and the same. Hence,

q

a

=

and

q

&

=

q

.

Note #2: Let

y

I

M

M

1

0

=

D

a

a

and

y

q

I

q

M

M

1

0

=

D

.

From these notes, (E1.1) can be written as:

0

=

-

-

q

q

q

a

M

M

q

&

&

&

.(E1.2)

(b) Development of the relation between

)

,

(

q

M

M

a

and

a

L

C

:

y

L

L

I

Sl

u

C

M

S

u

C

l

M

/

2

1

2

1

)

(

)

(

2

0

2

0

-

=

-

=

r

r

a

a

a

a

a

.(E1.3a)

Recall that

0

/

tan

u

ql

@

a

, so that for small

a

, we have

0

/

u

ql

@

a

. Hence,

y

L

q

L

I

S

u

u

l

C

M

S

u

u

ql

C

l

q

M

/

2

1

2

1

)

(

2

0

0

2

0

0

-

=

-

=

r

r

a

a

.(E1.3b)

From (E1.3a-b) we obtain:

a

M

u

l

M

q

=

0

.(E1.3c)

Substituting (E1.3c) into (E1.2) gives:

0

0

=

+

+

q

q

q

a

a

M

M

u

l

&

&

&

.(E1.4)

(c) Relation between M and the transient response associated with (E1.4):

[Subtitled: How many different way can we estimate

a

L

C

?]

We know that the transient response associated with (4) will be decaying and oscillatory. Hence, we can write the left side of (E1.4) as:

q

w

q

w

V

q

q

q

q

a

a

2

0

2

n

n

M

M

u

l

+

+

=

+

+

D

&

&

&

&

&

&

.(E1.5a)

From (E1.5a), we have:

a

w

M

n

=

.(E1.5b)

We also have

n

n

u

l

M

u

l

w

V

w

a

2

2

0

0

=

=

. This gives:

0

0

2

/

2

/

u

M

u

l

n

a

w

V

=

=

.(E1.5c)

Also, the system time constant is:

a

w

V

t

M

u

n

0

2

1

=

=

.

Now, the transient response associated with (E1.5a) has the form:

)

sin(

)

(

1

1

j

w

q

q

w

V

+

=

-

t

e

t

d

t

n

(E1.6a)

where the pair of constants

)

,

(

1

1

j

q

will depend on the type of specified initial conditions . The initial conditions are not as important as the nature of the response (E1.6a). If we define

n

Vw

t

/

1

D

=

, then (E1.6a) becomes:

)

sin(

)

(

1

/

1

j

w

q

q

t

+

=

-

t

e

t

d

t

.(E1.6b)

The parameter

t

is called the time constant associated with (E1.5a). This gives rise to one method for estimating

a

L

C

from the transient response measurement.

Method 1. Ignore the oscillations and use only the decay envelope

t

q

/

1

t

e

-

: At a time

t

=

t

, this envelope will equal

1

1

1

37

.

0

q

q

@

-

e

. And so, to estimate

t

=

t

, we sketch an exponential envelope on the decaying oscillatory response, and find the time at which this envelope is ~37% of its peak value. Having this estimate, call it

t

)

, we then note that

=

=

y

L

I

l

S

u

C

M

u

l

4

2

1

2

0

0

r

t

a

a

.

And so, our estimate of the magnitude of

a

L

C

, call it

a

L

C

)

is:

t

r

a

)

)

1

4

1

2

0

-

=

y

L

I

l

S

u

C

.

Method 2. Ignore the envelope and use only the oscillation frequency: The oscillation frequency,

d

w

, of the decaying response (E1.6) is:

2

1

V

w

w

-

=

n

d

. From (E1.5b) and (E1.5c) this becomes:

2

0

)

2

/(

1

u

M

M

d

a

a

w

-

=

.

And so, for an estimate of

d

w

, call it

d

w

)

, this equation provides an estimate of

a

M

, call it

a

M

)

. That estimate and (E1.3a) results in the desired estimate of

a

L

C

, call it

a

L

C

)

.

'

1

=

b

'

167

.

0

=

c

Method 3. Use both decay and oscillation frequency information: This method is best described using a plot of (E1.6):

Figure 1. Initial condition response for

p

w

2

=

n

and

01

.

0

=

V

.

The plot on the right in figure 1 shows that for

sec

10

0

=

t

,

o

t

4

.

5

)

(

0

@

q

, and for

sec

20

1

=

t

,

o

t

3

)

(

1

@

q

. The duration,

0

1

t

t

-

between these times is 20 sec., but it is also 10 cycles (or periods), where one period is

2

1

/

2

/

2

V

w

p

w

p

-

=

=

n

d

T

. Hence,

2

0

1

1

/

20

V

w

p

-

=

-

n

t

t

.

Now, from (6a), we have

0

0

0

)

(

t

n

e

t

w

V

q

q

-

@

and

1

1

1

)

(

t

n

e

t

w

V

q

q

-

@

. And so

2

0

1

1

0

1

/

20

)

(

1

0

)

(

)

(

V

V

p

w

V

w

V

w

V

q

q

-

-

-

-

-

=

=

=

e

e

e

e

t

t

t

t

t

t

n

n

n

.

Define

2

1

/

V

V

d

-

=

D

. Then we have

0094

.

0

20

/

)

3

/

4

.

5

ln(

20

/

)

(

)

(

ln

1

0

=

=

=

p

p

q

q

d

t

t

. From the definition of

d

, we have

2

1

/

d

d

V

-

=

. And so, our estimate

0094

.

0

=

d

)

gives

0094

.

0

@

V

)

. This estimate is close to the true value

01

.

0

=

V

, and it would have been closer, had we not used visually-based estimates of

)

(

0

t

q

and

)

(

1

t

q

.

Finally, we can use (E1.3a), (E1.5b) and (E1.5c) to obtain:

2

8

V

r

a

)

)

=

Sl

I

C

y

L

.

Notice that in this method it is not necessary to measure

0

u

, since the above estimate does not involve it. This can offer a significant advantage over other methods, both in terms of accuracy and equipment.

(d) Use of the setup in Figure 4.9 to validate the experimental design: Now that we know how to use the transient response, (6) to estimate the lift coefficient derivative for a given tail design, it is necessary to validate the experimental setup illustrated in Figure 4.9. In that design, we are assuming that the cross-bar bearing friction is negligible. We will also assume that the tube that supports the tail is completely rigid, and that its aerodynamic influence is negligible. Finally, we will assume that we have perfect measurements of the geometric and mass quantities described in that figure. By using a flat plate with known lift properties, we can compute the theoretical value for

a

L

C

. If the experimentally measured transient response matches our theoretical prediction reasonably well, then we can assume that we have a valid setup. We will now proceed to compute the theoretical value for

a

L

C

from the given numerical information.

The moment of inertia, Iy: From the parallel axis theorem, we have:

2

'

l

m

I

I

p

y

y

+

=

.The moment of inertia of the flat plate about its cg, and the moment of the plate about the setup cg are:

2

5

2

3

'

10

16

.

2

)

/

)(

12

/

1

(

)

12

/

1

(

ft

slug

c

g

W

btc

I

p

y

-

=

=

=

-

r

&

2

3

2

2

10

3

.

9

)

/

(

ft

slug

l

g

W

l

m

p

p

-

=

=

-

Hence,

2

3

)

(

10

32

.

9

ft

slug

I

plate

y

-

=

-

. We also have

2

3

)

(

10

60

.

4

ft

slug

I

Ballast

y

-

=

-

. Hence, the moment of inertia of the entire system is:

2

2

10

4

.

1

ft

slug

I

y

-

@

-

.

Numerical values for M and Mq:

For an infinite flat plate, we have

rad

C

L

/

2

)

(

p

a

=

. For a finite plate with aspect ratio, AR, the lift coefficient derivative is:

)]

/(

1

/[

)

(

)

(

AR

C

C

C

L

L

L

p

a

a

a

+

=

. Since

6

=

AR

, we have

rad

C

L

/

7

.

4

=

a

Using these highlighted results and equations (3), we obtain:

2

/

1

.

36

s

M

-

=

a

and

2

/

38

.

1

s

M

q

-

=

Hence, (2) becomes:

0

1

.

36

38

.

1

=

+

+

q

q

q

&

&

&

. It follows the theoretical values for

n

w

and

V

are:

sec

/

0

.

6

rad

n

=

w

and

038

.

0

=

V

. If we give this plate an initial angular displacement,

0

1

10

=

q

, then we should expect the experimentally measured transient response to be similar to the plot below.

00.511.522.533.544.55

-8

-6

-4

-2

0

2

4

6

8

10

time (sec)

th(t)

Plot of Initial Condition Response

Figure 2. Expected description of the experimentally measured transient response of the flat plat to an initial angular displacement of 10o.

Example Problem 4.2 (Nelson p. 151) Given the differential equations:

d

d

2

&

1

10

5

.

0

1

2

2

2

1

1

=

+

-

-

=

-

+

x

x

x

x

x

x

&

&

(a) Rewrite these in state space form

h

B

Ax

x

+

=

&

.

Solution:

d

d

2

&

1

10

5

.

0

1

2

2

2

1

1

+

-

=

-

+

-

=

x

x

x

x

x

x

&

&

, or,

d

-

+

-

-

=

2

1

1

1

10

5

.

0

2

1

2

1

x

x

x

x

&

&

.

(b) Find the free response eigenvalues.

Solution: The eigenvalues of

-

-

=

1

1

10

5

.

0

A

are the values of s that solve determinant equation

0

|

|

=

-

A

sI

. Specifically:

)

(

5

.

9

5

.

0

10

)

1

)(

5

.

(

1

1

10

5

.

|

|

2

s

p

s

s

s

s

s

s

A

sI

D

=

+

-

=

+

-

+

=

-

-

+

=

-

The roots of the characteristic polynomial p(s) are:

07

.

3

25

.

0

2

,

1

i

s

=

.

(c) What do these eigenvalues tell us about the response of the system?

Answer: The real parts of the eigenvalues are positive, and so the system is dynamically unstable. Specifically, the response to an initial disturbance would grow exponentially. This envelop is of the form

t

e

25

.

0

. It is common to compute the time it takes for this amplitude envelope to double. We will now compute this time.

Let

1

25

.

0

1

)

(

t

Ce

t

A

=

, and

2

25

.

0

2

)

(

t

Ce

t

A

=

. Then

)

(

25

.

0

1

2

1

2

)

(

/

)

(

t

t

e

t

A

t

A

-

=

. For amplitude doubling, we have :

2

)

(

25

.

0

1

2

=

-

t

t

e

, or:

=

=

-

)

2

ln(

4

1

2

t

t

2.77 seconds.

(d) Use of Laplace Transforms to obtain the eigenvalues of A:

For students who do not feel confident about matrix algebra, but who have background in the use of the Laplace transforms, here as an alternative solution method:

1

)

(

10

)

(

5

.

0

)

(

]

1

10

5

.

0

[

2

1

1

2

1

1

-

+

-

=

-

+

-

=

s

X

s

X

s

sX

x

x

x

d

&

l

(E2.1a)

2

)

(

)

(

)

(

]

2

[

1

2

2

1

2

2

+

-

=

+

-

=

s

X

s

X

s

sX

x

x

x

d

&

l

(E2.1b)

Equations (E2.1) include two unknowns; namely

)

(

1

s

X

and

)

(

2

s

X

. And so, either unknown can be solved for using purely algebraic operations. We will now solve these equations for each of these unknowns. These two resulting expressions will have something in common, as will be seen.

Solution of (E2.1) for X1(s):

From

2

)

(

)

(

)

(

1

2

2

+

-

=

s

X

s

X

s

sX

, we have

1

2

)

(

)

(

1

2

-

+

-

=

s

s

X

s

X

. Substituting this into (1a) gives:

1

1

2

)

(

10

)

(

5

.

0

)

(

1

1

1

-

-

+

-

+

-

=

s

s

X

s

X

s

sX

)

1

(

]

2

)

(

[

10

)

(

)

1

(

5

.

0

)

(

)

1

(

1

1

1

-

-

+

-

+

-

-

=

-

s

s

X

s

X

s

s

X

s

s

. Gathering terms gives:

s

s

X

s

s

-

=

+

+

-

21

)

(

)

5

.

9

5

.

0

(

1

2

.(E2.2a)

The corresponding differential equation is:

d

d

&

&

&

&

-

=

+

-

21

5

.

9

5

.

0

1

1

1

x

x

x

.(E2.2b)

Solution of (E2.1) for X2(s):

From

2

)

(

)

(

)

(

1

2

2

+

-

=

s

X

s

X

s

sX

, we have

2

)

(

)

1

(

)

(

2

1

+

-

=

s

X

s

s

X

. Substituting this into (E2.1a) gives:

1

)

(

10

]

2

)

(

)

1

)[(

5

.

0

(

2

2

-

=

+

-

+

s

X

s

X

s

s

. Gathering terms gives:

2

2

)

(

)

5

.

9

5

.

0

(

2

2

+

=

+

-

s

s

X

s

s

.(E2.3a)

The corresponding differential equation is:

d

d

+

=

+

-

&

&

&

&

2

5

.

9

5

.

0

2

2

2

x

x

x

.(E2.3b)

Remark 1. Notice that the left sides of equations (E2.2) and (E2.3) are identical. In particular, the 2-D state space system characteristic polynomial,

5

.

9

5

.

0

)

(

2

+

-

=

s

s

s

p

is common to both solutions. Hence, the system eigenvalues can be obtained from either solution.

Remark 2. This method of solving for the system eigenvalues is comparable in terms of mathematical manipulations to the matrix method associated with parts (a) and (b). However, recall that the longitudinal state space system includes four equations in four unknowns. In that situation, to solve for any chosen unknown would require much more work than what was required in this simple setting of two equations in two unknowns.

Remark 3. In this simple example, the Laplace transform method has the advantage of not only leading to the system eigenvalues, but to the solution for each chosen unknown. Specifically:

5

.

9

5

.

0

21

)

(

2

1

+

+

-

+

-

=

s

s

s

s

X

and

5

.

9

5

.

0

2

2

)

(

2

2

+

-

+

=

s

s

s

s

X

These are solutions in the Laplace (i.e. s domain). To obtain the corresponding solutions in the time (i.e. t domain), we can use a table of Laplace transform pairs.

[ e.g. http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms ]

At this point in the course, we will not address this table. Instead, we will simply use Matlab to obtain plots of

)

(

1

t

x

and

)

(

2

t

x

. The plots are:

02468101214161820

-200

-150

-100

-50

0

50

100

150

200

Impulse Response

Time (sec)

Amplitude

Figure 1. Plots of

)

(

1

t

x

(BLUE) and

)

(

2

t

x

(GREEN).

The code is:

%PROGRAM NAME: x12solutions.m

% This code computes plots of x1(t) and x2(t) via Laplace

p = [1 -0.5 9.5];

n1 = [-1 21];

n2 = [2 2];

sys1 = tf(n1,p);

sys2 = tf(n2,p);

impulse(sys1)

pause

hold on

impulse(sys2)

axis([0 20 -200 200])

grid

Discussion of the plots: Both responses explode. This is because the roots of the characteristic polynomial

5

.

9

5

.

0

)

(

2

+

-

=

s

s

s

p

, which are

07

.

3

25

.

0

2

,

1

i

s

=

, have a positive real part (i.e. they are in the right half of the s plane). They both oscillate at the same frequency as they explode. However, it is clear that

)

(

1

t

x

is exploding faster than

)

(

2

t

x

. This is because the right side of (2b) includes the term

)

(

21

21

t

d

d

=

, where

)

(

t

d

is called the unit impulse. In more mathematical terms, it is also called the Dirac delta function.

Approximation of the Phugoid and Short-Period Longitudinal Modes (Nelson p.152)

From p.7 of these notes, we have:

D

D

+

+

+

D

D

D

D

+

+

+

-

=

D

D

D

D

T

w

w

q

w

w

w

w

u

w

u

w

u

w

u

T

T

T

T

Z

M

M

Z

X

Z

M

M

Z

X

q

w

u

M

u

M

Z

M

M

Z

M

M

u

Z

Z

g

X

X

q

w

u

d

d

q

q

d

d

d

d

d

d

d

d

0

0

0

1

0

0

0

0

0

0

0

&

&

&

&

&

&

&

&

&

(1)

This dynamical system is 4-D, and so the characteristic polynomial will have four roots. These roots will occur as two pairs of complex-conjugate roots. Recall that a 2-D dynamical system with a pair of complex-conjugate roots will has a response that is oscillatory. It is characterized by a damping ratio,

V

, and an undamped natural frequency,

n

w

. And so, the above 4-D system will include

)

,

(

1

1

n

w

V

and

)

,

(

2

2

n

w

V

. One of the natural frequencies, say,

1

n

w

will be a very low frequency (i.e. a very long period). This corresponds to what is known as the Phugoid, or long-period longitudinal mode. The other frequency,

1

2

n

n

w

w

>>

, is a high frequency mode, that is simply referred to as the short period longitudinal mode.

Ultimately, we will address the above 4-D dynamical system. That will necessitate matrix algebra. In this section we will make some simplifying assumptions that will allow us to address two separate 2-D systems; one related to the phugoid mode, and the other related to the short period mode. Not only will this alleviate the need for matrix algebra (i.e. using Laplace transforms), but it will highlight the variables that are most significant in relation to each mode.

The phugoid mode: The phugoid mode as a long period mode. Typically, the period is on the order of many seconds (or even minutes). As noted by Nelson: It represents a gradual interchange of potential and kinetic energy about the equilibrium attitude and airspeed. Another description is:

The phugoid has a nearly constant angle of attack but varying pitch, caused by a repeated exchange of airspeed and altitude. It can be excited by an elevator singlet (a short, sharp deflection followed by a return to the centered position) resulting in a pitch increase with no change in trim from the cruise condition. As speed decays, the nose will drop below the horizon. Speed will increase, and the nose will climb above the horizon. Periods can vary from under 30 seconds for light aircraft to minutes for larger aircraft. Microlight aircraft typically show a phugoid period of 1525 seconds, and it has been suggested that birds and model airplanes show convergence between the phugoid and short period modes. A classical model for the phugoid period can be simplified to about (0.85 speed in knots) seconds, but this only really works for larger aircraft. Phugoids are often demonstrated to student pilots as an example of the speed stability of the aircraft and the importance of proper trimming. When it occurs, it is considered a nuisance, and in lighter airplanes (typically showing a shorter period) it can be a cause of pilot-induced oscillation. The phugoid, for moderate amplitude,[1] occurs at an effectively constant angle of attack, although in practice the angle of attack actually varies by a few tenths of a degree. This means that the stalling angle of attack is never exceeded, and it is possible (in the

-

c

b

)

QUESTION 1: Under what condition(s) will both roots be negative?

Answer: _______________________________________________________________________

QUESTION 2: In relation to (6), why is QUESTION 1 important?

Answer: _______________________________________________________________________

QUESTION 3: Assume that

0

,

>

c

b

. Give the expression for the root that is closest to zero.

Answer: ___________________________________________________________________

Question 4: What is the significance of the root given in response to QUESTION 3, in relation to (6)?

Answer: ___________________________________________________________________

Case 2: complex conjugate roots (i.e.

0

4

2

p-/p

pc/p

pb/p

)/ppQUESTION 1: Under what condition(s) will both roots have negative real parts?/ppAnswer: _______________________________________________________________________/ppQUESTION 2: Is it possible that we can have div class="embedded" id="_1380921587"/p0/p

p/p

pc/p

for this case? Explain./ppAnswer: _______________________________________________________________________/ppQUESTION 3: Assume that div class="embedded" id="_1380921204"/p0/p

p,/p

p/p

pc/p

pb/p

. Explain the implications of this case in relation to (6)./ppAnswer: ________________________________________________________________________/ppQUESTION 4: For the case div class="embedded" id="_1380921204"/p0/p

p,/p

p>

c

b

, write

c

bs

s

s

p

+

+

=

2

)

(

as

2

2

2

)

(

n

n

s

s

s

p

w

w

V

+

+

=

. Give the expressions for b and c in terms of

V

and

.

n

w

Answer: ______________________________________________________________________

Conclusions in Relation to (6):

QUESTION 1: Under what conditions will (6) be a stable system?

Answer: ______________________________________________________________________

QUESTION 2: Assuming (6) is a stable system, under what condition(s) will it exhibit an oscillatory response to an initial condition?

Answer: _________________________________________________________________________

QUESTION 3: Assuming that (6) is stable, and exhibits oscillatory behavior, give expressions for

V

and

n

w

in terms of the various parameters in (6).

Answer: _________________________________________________________________________

PROBLEM Investigate how

V

and

n

w

relate to the plane nominal speed,

0

u

.

Aircraft Lateral Dynamics

1. Pure Rolling Motion: Wind tunnel tests are often used to estimate roll moment parameters. Consider a plane that is mounted in a wind tunnel in a way that it is constrained to pure roll motion about the x-axis. Newtons second law gives:

p

p

L

L

I

M

p

a

a

x

x

a

D

+

D

=

D

=

=

=

0

0

d

d

f

d

&

&

.(1a)

Here,

a

d

D

is a small deflection of the ailerons, and

p

D

is a small perturbation of the roll rate. Since the plane cg is constrained, we have:

f

&

D

=

D

p

.(1b)

Equations (1) result in:

a

a

p

x

a

L

p

p

L

p

I

d

d

d

D

=

D

-

D

=

=

0

0

&

.(2)

Before we continue our development in relation to (2), it is worth spending a little time to discuss the properties of a first order constant coefficient differential equation from a systems perspective.

ac

The structure of a first order dynamical system:

From (2), we see that the dynamics of pure roll correspond to a first order dynamical system. One standard form for such a system is:

)

(

)

(

)

(

t

f

g

t

y

t

y

s

=

+

&

t

. (3a)

The transfer function associated with (3a) is obtained by taking its Laplace transform (under zero initial conditions!). Doing this gives:

)

(

)

(

)

(

s

F

g

s

Y

s

sY

s

=

+

t

.(3b)

The system transfer function is then:

1

)

(

)

(

)

(

+

=

=

D

s

g

s

F

s

Y

s

G

s

t

.(3c)

It follows that, for a given input,

)

(

t

f

, we can find the response,

)

(

t

y

, via a table of Laplace

transform pairs, in relation to:

)

(

1

)

(

)

(

)

(

s

F

s

g

s

F

s

G

s

Y

s

+

=

=

t

.(3d)

Use of a Table of Laplace Transforms:

The system unit impulse response: From the table below, entry #1 shows that the Laplace transform of the unit impulse,

)

(

)

(

t

t

f

d

=

is

1

)

(

=

s

F

. For this input, (3d) becomes:

t

t

t

/

1

/

1

)

(

)

(

+

=

+

=

=

s

g

s

g

s

G

s

Y

s

s

.

Again, using entry #4, we obtain:

t

t

/

)

(

)

(

t

s

e

g

t

g

t

y

-

=

=

.(3e)

The system unit-step response: From the table below, entry #2 shows that the Laplace transform of the unit step,

)

(

)

(

t

u

t

f

=

is

s

s

F

/

1

)

(

=

. For this input, (3d) becomes:

)

/

1

(

/

/

)

(

)

(

t

t

+

=

=

s

s

g

s

s

G

s

Y

s

.

Again, using entry #7, we obtain:

)

1

(

)

(

/

t

t

s

e

g

t

y

-

-

=

.(3f)

Remark 1. Both the unit impulse and the unit step response include the term

t

/

t

e

-

. Notice that this term is a consequence of the fact that the system characteristic polynomial,

1

)

(

+

=

s

s

p

t

, has a single root:

t

/

1

1

-

=

s

. And so,

t

/

1

t

t

s

e

e

-

=

. This term will decay to zero as

t

if

0

>

t

, or equivalently, if the root,

t

/

1

1

-

=

s

, is in the Left Half (of the complex) Plane (LHP). Moreover, for

t

4

=

t

it will have the value

02

.

0

4

/

4

@

=

-

-

e

e

t

t

. The quantity

t

4

is commonly termed the

t

4

or 2% settling time.

Remark 2. From (3f), the asymptotic response to the unit step input is:

s

t

s

t

t

ss

g

e

g

t

y

y

=

-

=

=

-

D

)

1

(

lim

)

(

lim

/

t

.(3g)

Since the input has amplitude 1.0 and the response has steady state amplitude gs, the ratio of these amplitudes is simply gs. For this reason, this ratio is called the system static gain.

Summary: A first order system of the type described by (3a) is characterized by two parameters: the system time constant,

t

, and the system static gain,

s

g

. Table of Laplace and Z Transforms

(Please email me if you find an error) Using this table for Z Transforms with Discrete IndicesShortened 2-page pdf of Laplace Transforms and Properties

Entry#

Laplace Domain

Time Domain

Z Domain(t=kT)

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

We are now in a position to gain some insight into (2) without much effort. Re-writing it in the standard form (3a) gives:

a

a

x

p

L

L

p

p

p

L

I

d

d

D

-

=

D

+

D

-

-

-

1

1

&

.(4a)

The system time constant and static gain are:

1

-

-

=

p

L

I

x

t

and

1

-

-

=

p

L

L

g

a

s

d

.(4b)

Some observations-

(O1): Stability requires that

0

>

t

. In other words, we must have

0

|

/

0

p/p

p/p

p=/p

pD/p

pp/p

pp/p

pL/p

./pp(O2): The div class="embedded" id="_1381731758"/pt/p

p4/p

response time is proportional to div class="embedded" id="_1381731786"/px/p

pI/p

, and inversely proportional to div class="embedded" id="_1381736850"/p|/p

p//p

p|/p

pp/p

pL/p

p/p

p/p

./pp(O3): Since the input, div class="embedded" id="_1381731927"/p)/p

p(/p

pt/p

pa/p

pd/p

pD/p

has units of radians, and the output, div class="embedded" id="_1381731992"/p)/p

p(/p

pt/p

pp/p

pD/p

, has units of radians/sec., the static gain, div class="embedded" id="_1381732083"/ps/p

pg/p

, has units div class="embedded" id="_1381732115"/p/p

p/p

p/p

p/p

p/p

p/p

prad/p

prad/p

psec/p

p//p

. Some people would write this simply as div class="embedded" id="_1381732180"/p/p

p/p

p/p

p/p

p/p

p/p

psec/p

p1/p

. But this misses the point that div class="embedded" id="_1381732083"/ps/p

pg/p

is a ratio. /pp(O4): The static gain, div class="embedded" id="_1381732083"/ps/p

pg/p

, is directly proportional to div class="embedded" id="_1381732355"/pa/p

pL/p

pd/p

p/p

p/p

p//p

, and inversely proportional to div class="embedded" id="_1381732409"/pp/p

pL/p

p/p

p/p

p//p

/ppNotation: In Nelsons book, he defines div class="embedded" id="_1381737436"/px/p

pp/p

pI/p

pp/p

pL/p

pL/p

p//p

p)/p

p//p

p(/p

p/p

p/p

p=/p

pD/p

and div class="embedded" id="_1381737521"/px/p

pa/p

pI/p

pL/p

pL/p

pa/p

p//p

p)/p

p//p

p(/p

pd/p

pd/p

p/p

p/p

p=/p

pD/p

. /ppWith this notation, (4) becomes:/pp div class="embedded" id="_1381739405"/pa/p

pp/p

pp/p

pL/p

pL/p

pp/p

pp/p

pL/p

pa/p

pd/p

pd/p

pD/p

p-/p

p=/p

pD/p

p+/p

pD/p

p-/p

p)/p

p//p

p(/p

p)/p

p//p

p1/p

p(/p

p&/p

.(5a)/ppThe system time constant and static gain are:/pp div class="embedded" id="_1381737720"/pp/p

pL/p

p//p

p1/p

p-/p

p=/p

pt/p

and div class="embedded" id="_1381739375"/pp/p

ps/p

pL/p

pL/p

pg/p

pa/p

p//p

pd/p

p-/p

p=/p

.(5b)/ppEXAMPLE PROBLEM 5.1 (Nelson p.183)/ppCalculate the roll response of an F104A aircraft to a 5o step change in aileron deflection. Assume the plane is flying at sea level with a velocity of div class="embedded" id="_1381738315"/ps/p

pm/p

pu/p

p//p

p87/p

p0/p

p=/p

./ppdiv class="embedded" id="_1381738347"/p2/p

p2/p

p1/p

p1/p

p4676/p

p;/p

p7/p

p./p

p6/p

p;/p

p18/p

p;/p

p039/p

p./p

p0/p

p;/p

p285/p

p./p

p0/p

pm/p

pkg/p

pI/p

pm/p

pb/p

pm/p

pS/p

prad/p

pC/p

prad/p

pC/p

px/p

pl/p

pl/p

pa/p

pp/p

p-/p

p=/p

p=/p

p=/p

p=/p

p-/p

p=/p

p-/p

p-/p

pd/p

/ppSolution: The F104A has the following aerodynamic and geometric characteristics:/pp ./ppHence, div class="embedded" id="_1381738455"/ps/p

pu/p

pb/p

pm/p

pN/p

pu/p

pQ/p

p039/p

p./p

p0/p

p2/p

p//p

p;/p

p//p

p4636/p

p5/p

p./p

p0/p

p0/p

p2/p

p2/p

p0/p

p=/p

p=/p

p=/p

pD/p

pr/p

. And so:/pp div class="embedded" id="_1381738628"/p2/p

p1/p

p0/p

p66/p

p./p

p4/p

p//p

p;/p

p3/p

p./p

p1/p

p//p

p)/p

p2/p

p//p

p(/p

p-/p

p-/p

p=/p

p=/p

p-/p

p=/p

p=/p

ps/p

pI/p

pQSb/p

pC/p

pL/p

ps/p

pI/p

pQSb/p

pu/p

pb/p

pC/p

pL/p

px/p

pl/p

px/p

pl/p

pp/p

pa/p

pa/p

pp/p

pd/p

pd/p

./ppFrom (5b) above, we obtain:/pp div class="embedded" id="_1381738950"/ps/p

p77/p

p./p

p0/p

p=/p

pt/p

and div class="embedded" id="_1382300378"/pr/p

ps/p

pr/p

pg/p

ps/p

p//p

p)/p

p//p

p(/p

p5846/p

p./p

p3/p

p-/p

p=/p

./ppFor the aileron input div class="embedded" id="_1381739528"/pr/p

po/p

pa/p

p0873/p

p./p

p0/p

p)/p

p180/p

p//p

p(/p

p5/p

p5/p

p=/p

p=/p

p=/p

pD/p

pp/p

pd/p

, the roll response will achieve a steady state roll rate of div class="embedded" id="_1381739601"/ps/p

pr/p

pp/p

pss/p

p//p

p3129/p

p./p

p0/p

p=/p

pD/p

after approximately div class="embedded" id="_1381739667"/ps/p

p08/p

p./p

p3/p

p4/p

p=/p

pt/p

. From (3f), the mathematical form of this step response is: div class="embedded" id="_1411815674"/p)/p

p1/p

p)(/p

p5/p

p(/p

p5846/p

p./p

p3/p

p)/p

p(/p

p77/p

p./p

p0/p

p//p

pt/p

po/p

pe/p

pt/p

py/p

p-/p

p-/p

p-/p

p=/p

. This response is plotted below./ppp00.511.522.533.5/p

p0/p

p0.5/p

p1/p

p1.5/p

p2/p

p2.5/p

p3/p

p3.5/p

p4/p

pTime (s)/p

pdelp(t)/p

/ppFigure 1. Plot of the F104A roll rate response to a 5o step aileron input./pp2. Pure Yawing Motion: (Nelson p.188)/ppWind tunnel tests are often used to estimate yaw moment parameters. Consider a plane that is mounted in a wind tunnel in a way that it is constrained to pure yaw motion about the z-axis. Newtons second law gives:/pp div class="embedded" id="_1381741189"/pr/p

pr/p

pz/p

pz/p

pN/p

pr/p

pr/p

pN/p

pN/p

pN/p

pI/p

pM/p

pd/p

pd/p

pb/p

pb/p

pb/p

pb/p

py/p

pD/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p+/p

pD/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p+/p

pD/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p+/p

pD/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p=/p

pD/p

p=/p

p/p

p&/p

p&/p

p&/p

p&/p

.(6a)/ppIn general, div class="embedded" id="_1381741437"/pr/p

pd/p

pD/p

is a small deflection of the rudder, div class="embedded" id="_1381741588"/pr/p

pD/p

is a small perturbation of the yaw rate, div class="embedded" id="_1381741621"/pb/p

pD/p

is a small perturbation of the sideslip angle, and div class="embedded" id="_1381741650"/pb/p

p&/p

pD/p

is a small perturbation of the sideslip rate. Since the plane cg is constrained, we have:/pp div class="embedded" id="_1381741727"/pb/p

py/p

pD/p

p-/p

p=/p

pD/p

and div class="embedded" id="_1381741870"/pr/p

pD/p

p=/p

pD/p

p-/p

p=/p

pD/p

pb/p

py/p

p&/p

p&/p

(6b)/ppEquations (6) result in:/pp div class="embedded" id="_1381741997"/pr/p

pr/p

pr/p

pN/p

pN/p

pN/p

pN/p

pd/p

py/p

py/p

py/p

pd/p

pb/p

pb/p

pD/p

p=/p

pD/p

p+/p

pD/p

p-/p

p+/p

pD/p

p&/p

p&/p

p&/p

p&/p

p)/p

p(/p

(7)/ppwhere we have defined div class="embedded" id="_1381742227"/pz/p

pI/p

pN/p

pN/p

p//p

p*)/p

p//p

p(/p

p*/p

p/p

p/p

p=/p

pD/p

. Since it is known that the response to a rudder input, div class="embedded" id="_1381742369"/pr/p

pd/p

pD/p

, will be oscillatory and decaying in nature, (7) represents a second order underdamped dynamical system. The standard form for such a system is:/pp div class="embedded" id="_1381742582"/pr/p

pn/p

ps/p

pn/p

pn/p

pg/p

pd/p

pw/p

py/p

pw/p

py/p

pVw/p

py/p

pD/p

p=/p

pD/p

p+/p

pD/p

p+/p

pD/p

p)/p

p//p

p(/p

p2/p

p2/p

p2/p

p&/p

p&/p

p&/p

.(8a)/ppEquating (7) and (8a) we immediately have the following:/pp div class="embedded" id="_1381777433"/pb/p

pw/p

pN/p

pn/p

p=/p

, div class="embedded" id="_1476094508"/pb/p

pb/p

pV/p

pN/p

pN/p

pN/p

pr/p

p2/p

p-/p

p=/p

p&/p

, div class="embedded" id="_1476094478"/p)/p

p/(/p

p2/p

p)/p

p(/p

p1/p

pr/p

pn/p

pN/p

pN/p

p-/p

p=/p

p=/p

p-/p

pb/p

pVw/p

pt/p

p&/p

and div class="embedded" id="_1381919281"/p1/p

p)/p

p//p

p(/p

p-/p

p=/p

pr/p

pN/p

pN/p

pg/p

ps/p

pd/p

pb/p

.(8b)/ppThe damped natural frequency is:/pp div class="embedded" id="_1476094523"/p2/p

p2/p

p2/p

p)/p

p(/p

p4/p

p2/p

p1/p

p2/p

p1/p

p1/p

pr/p

pr/p

pn/p

pd/p

pN/p

pN/p

pN/p

pN/p

pN/p

pN/p

pN/p

p-/p

p-/p

p=/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p/p

p-/p

p-/p

p=/p

p-/p

p=/p

pb/p

pb/p

pb/p

pb/p

pb/p

pV/p

pw/p

pw/p

p&/p

p&/p

.(8c)/ppHence, the damped natural oscillation period is:/pp div class="embedded" id="_1476094535"/p2/p

p)/p

p(/p

p4/p

p//p

p//p

p2/p

pr/p

pd/p

pd/p

pN/p

pN/p

pN/p

pT/p

p-/p

p-/p

p=/p

p=/p

pb/p

pb/p

pp/p

pw/p

pp/p

p&/p

.(8d)./ppRemark 1. Nelson notes that the term div class="embedded" id="_1381778795"/pb/p

p&/p

pN/p

is usually negligible. In this case, we have:/ppdiv class="embedded" id="_1381777433"/pb/p

pw/p

pN/p

pn/p

p=/p

; div class="embedded" id="_1381779072"/pb/p

pV/p

pN/p

pN/p

pr/p

p2/p

p=/p

; div class="embedded" id="_1381779087"/pr/p

pN/p

p//p

p2/p

p=/p

pt/p

;div class="embedded" id="_1381778938"/pr/p

pN/p

pN/p

pg/p

ps/p

pd/p

pb/p

p=/p

;div class="embedded" id="_1381779187"/p2/p

p4/p

p2/p

p1/p

pr/p

pd/p

pN/p

pN/p

p-/p

p=/p

pb/p

pw/p

;div class="embedded" id="_1381779298"/p2/p

p4/p

p//p

pr/p

pd/p

pN/p

pN/p

pT/p

p-/p

p=/p

pb/p

pp/p

./ppRemark 2. On p.190, Nelson give plots of div class="embedded" id="_1381780110"/pTRIM/p

pt/p

pb/p

pb/p

pD/p

pD/p

p//p

p)/p

p(/p

versus t for various values of div class="embedded" id="_1381780142"/pV/p

. This ratio converges to 1.0 for every div class="embedded" id="_1381780142"/pV/p

. The problem is: the term div class="embedded" id="_1381780289"/pTRIM/p

pb/p

pD/p

is never defined. To figure out what it is, we need only look at the system static gain: div class="embedded" id="_1381778938"/pr/p

pN/p

pN/p

pg/p

ps/p

pd/p

pb/p

p=/p

. Since div class="embedded" id="_1381780692"/pr/p

ps/p

pTRIM/p

pss/p

pt/p

pg/p

pt/p

pd/p

pb/p

pb/p

pb/p

pD/p

p=/p

pD/p

p=/p

pD/p

p=/p

pD/p

pD/p

p/p

p/p

p)/p

p(/p

plim/p

, we have determined the expression for div class="embedded" id="_1381780289"/pTRIM/p

pb/p

pD/p

: div class="embedded" id="_1381780664"/pr/p

ps/p

pTRIM/p

pg/p

pd/p

pb/p

pD/p

p=/p

pD/p

./ppEXAMPLE PROBLEM 5.2 Suppose an airplane is constrained to a pure yawing motion. Use the data for the general aviation airplane in Appendix B, determine the following quantities:/pp(a) The yaw moment equation written in state space form./pp(b) The characteristic equation and eigenvalues for the system./pp(c) The damping ratio, div class="embedded" id="_1381780142"/pV/p

, and undamped natural frequency, div class="embedded" id="_1381781058"/pn/p

pw/p

./pp(d) The response of the plane to a 5o rudder input. Assume initial conditions are: div class="embedded" id="_1381781156"/p./p

p0/p

p)/p

p0/p

p(/p

p)/p

p0/p

p(/p

p=/p

pD/p

p=/p

pD/p

pr/p

pb/p

/ppSolution: For sea level flight condition, the weathercock stability coefficient, the yaw damping coefficient, and the rudder control power coefficient have, respectively, the following values:/pp div class="embedded" id="_1381781318"/pr/p

pC/p

pr/p

pC/p

pr/p

pC/p

pr/p

pr/p

pn/p

pn/p

pn/p

p//p

p072/p

p./p

p0/p

p;/p

p//p

p125/p

p./p

p0/p

p;/p

p//p

p071/p

p./p

p0/p

p-/p

p=/p

p-/p

p=/p

p=/p

pd/p

pb/p

./pp[The derivative div class="embedded" id="_1381781600"/pb/p

p&/p

pn/p

pC/p

is not included in the appendix, and will be assumed to be zero.]/ppFor div class="embedded" id="_1381781607"/ps/p

pft/p

pu/p

p//p

p176/p

p0/p

p=/p

, the dynamic pressure at sea level is: div class="embedded" id="_1381781681"/p2/p

p2/p

p0/p

p//p

p8/p

p./p

p36/p

p5/p

p./p

p0/p

pft/p

plb/p

pu/p

pQ/p

p=/p

p=/p

pD/p

pr/p

./ppThe plane geometry parameters include: div class="embedded" id="_1381781776"/p2/p

p2/p

p3530/p

p;/p

p4/p

p./p

p33/p

p;/p

p184/p

pft/p

pslug/p

pI/p

pft/p

pb/p

pft/p

pS/p

pz/p

p-/p

p=/p

p=/p

p=/p

./ppThese values result in the following dimensional derivative values:/ppdiv class="embedded" id="_1381781966"/p2/p

p//p

p55/p

p./p

p4/p

p)/p

p//p

p(/p

ps/p

pI/p

pQSb/p

pC/p

pN/p

pz/p

pn/p

p=/p

p=/p

pb/p

pb/p

; div class="embedded" id="_1381782042"/ps/p

pI/p

pQSb/p

pu/p

pb/p

pC/p

pN/p

pz/p

pn/p

pr/p

pr/p

p//p

p76/p

p./p

p0/p

p]/p

p//p

p)/p

p2/p

p//p

p[(/p

p0/p

p-/p

p=/p

p=/p

; div class="embedded" id="_1381782177"/p2/p

p//p

p6/p

p./p

p4/p

p)/p

p//p

p(/p

ps/p

pI/p

pQSb/p

pC/p

pN/p

pz/p

pn/p

pr/p

pr/p

p-/p

p=/p

p=/p

pd/p

pd/p

/ppSubstituting these into (7) gives:/pp div class="embedded" id="_1381782357"/pr/p

pd/p

py/p

py/p

py/p

pD/p

p-/p

p=/p

pD/p

p+/p

pD/p

p+/p

pD/p

p6/p

p./p

p4/p

p55/p

p./p

p4/p

p76/p

p./p

p0/p

p&/p

p&/p

p&/p

. (1)/ppWe are now in a position to solve parts (a-d)./pp(a) The yaw moment equation written in state space form./ppSolution: Recall that the state space form is: div class="embedded" id="_1381782620"/pBu/p

pAx/p

px/p

p+/p

p=/p

p&/p

. In relation to (1) above, this will be a 2-D system. Hence, div class="embedded" id="_1381782753"/ptr/p

pt/p

px/p

pt/p

px/p

pt/p

px/p

p)]/p

p(/p

p)/p

p(/p

p[/p

p)/p

p(/p

p2/p

p1/p

p=/p

We know that one state, say, div class="embedded" id="_1381782856"/p)/p

p(/p

p1/p

pt/p

px/p

, will be div class="embedded" id="_1381782890"/p)/p

p(/p

pt/p

py/p

pD/p

. We also know that, since the plane is constrained to pure yawing, div class="embedded" id="_1381782983"/p)/p

p(/p

p)/p

p(/p

pt/p

pr/p

pt/p

pD/p

p=/p

pD/p

py/p

p&/p

. And so, let div class="embedded" id="_1381783048"/p)/p

p(/p

p)/p

p(/p

p2/p

pt/p

pr/p

pt/p

px/p

pD/p

p=/p

. From (1) we have: div class="embedded" id="_1381783104"/pr/p

pr/p

pr/p

pd/p

py/p

pD/p

p-/p

p=/p

pD/p

p+/p

pD/p

p+/p

pD/p

p6/p

p./p

p4/p

p55/p

p./p

p4/p

p76/p

p./p

p0/p

p&/p

. And so, we arrive at:/pp div class="embedded" id="_1381783178"/pr/p

pr/p

pr/p

pd/p

py/p

py/p

pD/p

p/p

p/p

p/p

p/p

p/p

p/p

p-/p

p+/p

p/p

p/p

p/p

p/p

p/p

p/p

pD/p

pD/p

p/p

p/p

p/p

p/p

p/p

p/p

p-/p

p-/p

p=/p

p/p

p/p

p/p

p/p

p/p

p/p

pD/p

pD/p

p0/p

p6/p

p./p

p4/p

p0/p

p1/p

p55/p

p./p

p4/p

p76/p

p./p

p0/p

p&/p

p&/p

./pp(b) The characteristic equation and eigenvalues for the system./ppSolution: The characteristic polynomial is simply: div class="embedded" id="_1381783632"/p55/p

p./p

p4/p

p76/p

p./p

p0/p

p)/p

p(/p

p2/p

p+/p

p+/p

p=/p

ps/p

ps/p

ps/p

pp/p

. The eigenvalues of div class="embedded" id="_1381783425"/p/p

p/p

p/p

p/p

p/p

p/p

p-/p

p-/p

p=/p

p0/p

p1/p

p55/p

p./p

p4/p

p76/p

p./p

p0/p

pA/p

are obtained via Matlab:/pppassembly/p

ptube/p

pplate/p

pof/p

pcg/p

/pp> A=[-.76 -4.55 ; 1 0];

> eig(A)

ans =

-0.3800 + 2.0990i

-0.3800 - 2.0990i

They can also be obtained via:

)

(

Ballast

weight

counter

> p=[1 .76 4.55];

> roots(p)

ans =

-0.3800 + 2.0990i

-0.3800 - 2.0990i

(c) The damping ratio,

V

, and undamped natural frequency,

n

w

.

Solution: These are easily computed:

178

.

0

;

/

13

.

2

=

=

V

w

s

r

n

(d) The response of the plane to a 5o rudder input. Assume initial conditions are:

.

0

)

0

(

)

0

(

=

D

=

D

r

b

Solution: A homework problem?

The Spiral mode Approximation (Nelson p.195)

This mode is characterized by changes in the bank angle,

f

, and the heading angle,

y

. An approximation to this mode can be obtained from the above 4-state system by neglecting the side force equation and

f

D

. Consequently, he obtains the following:

0

=

D

+

D

r

L

L

r

b

b

and

r

N

N

r

r

D

+

D

=

D

b

b

&

.

These result in the following first order system description:

0

=

D

-

+

D

r

L

N

L

N

L

r

r

r

b

b

b

&

The characteristic polynomial is:

-

+

=

b

b

b

L

N

L

N

L

s

s

p

r

r

)

(

.

The system root is:

-

-

=

b

b

b

L

N

L

N

L

s

r

r

1

.

The time constant is:

)

/(

r

r

N

L

N

L

L

b

b

b

t

-

=

.

The Roll Approximation: (Nelson p.198)

This motion is approximated by pure roll. Hence, from p.4.32 of these notes (neglecting inputs), we have:

0

=

D

-

D

p

L

p

p

&

.

The characteristic polynomial is:

p

L

s

s

p

-

=

)

(

.

The system root is:

p

L

s

=

1

.

The time constant is:

p

L

/

1

-

=

t

.

Dutch Roll Approximation (Nelson p.198)

As noted by Nelson, the Dutch roll is characterized primarily by sideslipping and yaw. (i.e. the rolling moment equation can be ignored). As such, the 4-D lateral system becomes the 2-state system:

D

D

+

D

D

-

-

=

D

D

r

a

r

r

r

a

r

N

N

u

Y

r

N

N

u

Y

u

Y

r

d

d

b

b

d

d

d

b

b

0

0

0

0

1

&

&

.

Since it is known that this mode includes oscillations, this is a second order underdamped system.

Example: Conduct the following investigation of the Dutch roll dynamics.

(a) Compute the system characteristic polynomial directly from

|

|

)

(

A

sI

s

p

-

=

.

Solution:

)

(

1

)

(

1

|

1

|

0

0

0

0

0

0

s

p

u

Y

N

N

s

u

Y

s

N

s

N

u

Y

u

Y

s

N

N

u

Y

u

Y

sI

r

r

r

r

r

r

=

-

+

-

-

=

-

-

-

-

=

-

-

-

b

b

b

b

b

b

or

0

0

0

0

2

)

(

u

N

Y

N

u

N

Y

s

u

N

u

Y

s

s

p

r

r

r

b

b

b

b

-

+

+

+

-

=

(b) Assume that the dynamics include natural oscillations (i.e. the system is stable, but underdamped). Compute expressions for the parameters associated with these dynamics.

Solution: Equating the coefficients of

)

(

s

p

with

2

2

2

)

(

n

n

s

s

s

p

w

Vw

+

+

=

gives:

0

0

u

N

Y

N

u

N

Y

r

r

n

b

b

b

w

-

+

=

and

r

n

N

u

Y

u

0

0

1

2

)

(

+

-

=

=

-

b

Vw

t

.

Since

2

2

2

-

=

t

w

V

n

, we have

2

2

2

/

n

w

t

V

-

=

, or

2

2

)

(

-

=

n

tw

V

. Hence, the damping ratio is:

1

0

0

0

0

2

-

-

+

+

-

=

u

N

Y

N

u

N

Y

N

u

Y

u

r

r

r

b

b

b

b

V

.

We could continue to derive expressions for

d

w

and for

d

d

T

w

p

/

2

=

, but we will withhold those for now.

(c) The expressions for the lateral directional derivatives related to the above are:

).

2

/

(

);

2

/

(

);

/

(

);

/

(

0

2

0

u

I

QSb

C

N

mu

QSb

C

Y

I

QSb

C

N

m

QS

C

Y

z

n

r

y

r

z

n

y

r

r

=

=

=

=

b

b

b

b

Recall that

2

0

5

.

0

u

Q

r

D

=

. Investigate the dependence of the above dynamic parameters as a function of the plane velocity,

0

u

. [To be assigned as a homework problem.]

(d) Recall from EXAMPLE PROBLEM 5.3:

2

/

55

.

4

)

/

(

s

I

QSb

C

N

z

n

=

=

b

b

and

s

I

QSb

u

b

C

N

z

n

r

r

/

76

.

0

]

/

)

2

/

[(

0

-

=

=

.

From Table B1. we have:

.

564

.

-

=

b

y

C

Since there is no entry for

r

y

C

, we will assume

.

0

=

r

y

C

Hence, we have:

.

0

&

72

.

45

)

/

(

=

-

=

=

r

y

Y

m

QS

C

Y

b

b

The characteristic polynomial is then:

71

.

4

102

.

1

)

(

2

+

+

=

s

s

s

p

. This yields

s

r

n

/

17

.

2

=

w

and

254

.

0

=

V

. From these figures, give a description of the transient response of

b

D

.

Solution: The time constant is

.

sec

8

.

1

@

t

The damped natural frequency is

s

r

d

/

1

.

2

@

w

, and so the damped natural period is

.

sec

3

@

d

T

Hence, the transient response will decay in about 7 seconds, during which time there will be only about 2.5 periods of oscillation.

(e) Verify your answer in (d) by plotting the impulse response corresponding to a transfer function

)

(

/

)

(

2

s

p

s

H

n

w

=

.

Solution: The Matlab commands needed to produce the plot are:

> p=[1 1.102 4.71];

> sys=tf(4.71,p)

Transfer function:

4.71

--------------------

s^2 + 1.102 s + 4.71

> impulse(sys)

> grid

024681012

-1

-0.5

0

0.5

1

1.5

2

Impulse Response

Time (sec)

Amplitude

(f) Explain why the above analysis did not provide any information about the system static gain.

Answer: In class. Since no defined input, no output, no static gain.

lb

W

lb

W

B

p

6

.

0

3

=

=

The Full State Lateral Equations:

The lateral rigid body equations are found in Table 3.2 on p.108 and on p.193.

r

r

p

v

r

Y

g

r

Y

u

p

Y

v

Y

dt

d

d

f

q

d

D

=

D

-

D

-

+

D

-

D

-

)

cos

(

)

(

0

0

r

a

r

x

xz

p

v

r

a

L

L

r

L

dt

d

I

I

p

L

dt

d

v

L

d

d

d

d

D

+

D

=

D

+

-

D

-

+

D

-

(Table 3.2)

r

a

r

p

z

xz

v

r

a

N

N

r

N

dt

d

p

N

dt

d

I

I

v

N

d

d

d

d

D

+

D

=

D

-

+

D

+

-

D

-

+

+

+

+

+

D

D

D

D

=

D

D

D

D

0

0

0

*

*

*

*

*

*

*

*

r

r

r

r

r

a

a

a

a

L

I

I

N

N

I

I

L

Y

L

I

I

N

N

I

I

L

r

p

v

A

r

p

v

z

xz

x

xz

z

xz

x

xz

lat

d

d

d

d

d

d

d

d

d

f

f

&

&

&

&

EMBED Equation.3

D

D

r

a

d

d

where

+

+

+

+

+

+

-

=

0

0

1

0

0

0

cos

*

*

*

*

*

*

*

*

*

*

*

*

0

0

r

z

xz

r

p

z

xz

p

v

z

xz

v

r

x

xz

r

p

x

xz

p

v

x

xz

v

r

p

v

lat

L

I

I

N

L

I

I

N

L

I

I

N

N

I

I

L

N

I

I

L

N

I

I

L

g

u

Y

Y

Y

A

q

.

The starred derivatives are defined as the un-starred derivatives divided by

)

/(

1

2

z

x

xz

I

I

I

-

.

If the product of inertia

0

=

xz

I

, these equations reduce to:

+

D

D

D

D

=

D

D

D

D

0

0

0

r

r

r

a

a

N

L

Y

N

L

r

p

v

A

r

p

v

lat

d

d

d

d

d

f

f

&

&

&

&

EMBED Equation.3

D

D

r

a

d

d

where

-

=

0

0

1

0

0

0

cos

0

0

r

p

v

r

p

v

r

p

v

lat

N

N

N

L

L

L

g

u

Y

Y

Y

A

q

The author notes that it is sometimes convenient to use the sideslip angle

b

D

instead of the side velocity

v

D

. Since

0

0

1

tan

u

v

u

v

D

@

D

@

D

-

b

, the above equations become (for the product of inertia

0

=

xz

I

):

+

D

D

D

D

=

D

D

D

D

0

/

0

0

0

r

r

r

a

a

N

L

u

Y

N

L

r

p

A

r

p

lat

d

d

d

d

d

f

b

f

b

&

&

&

&

EMBED Equation.3

D

D

r

a

d

d

where

-

=

0

0

1

0

0

0

/

cos

1

)

/

(

/

/

0

0

0

0

0

r

p

v

r

p

v

r

p

v

lat

N

N

N

L

L

L

u

g

u

Y

u

Y

u

Y

A

q

EXAMPLE PROBLEM 5.3

Consider once again, the general aviation aircraft (Table B.1on p.400 of Nelson).

(a) Find the eigenvalues of the 4-D system.

Solution: The state transition matrix is:

-

-

-

-

-

-

=

-

=

0

0

1

0

0

76

.

35

.

49

.

4

0

19

.

2

40

.

8

02

.

16

18

.

1

0

25

.

0

0

1

0

0

0

/

cos

1

)

/

(

/

/

0

0

0

0

0

r

p

v

r

p

v

r

p

v

lat

N

N

N

L

L

L

u

g

u

Y

u

Y

u

Y

A

q

.

Using the Matlab command eig(A) gives:

s1= -8.4322 ; s2= -0.4845 + 2.3329i ; s3=-0.4845 - 2.3329i ; s4=-0.0088

The spiral mode approximation is:

s

L

N

L

N

L

s

r

r

/

144

.

0

-

=

-

-

=

b

b

b

.

The roll mode approximation is:

s

L

s

p

/

44

.

8

-

=

.

The Dutch roll approximation is obtained from

0

0

0

0

2

)

(

u

N

Y

N

u

N

Y

s

u

N

u

Y

s

s

p

r

r

r

b

b

b

b

-

+

+

+

-

=

using the quadratic formula, and is:

i

s

11

.

2

51

.

0

3

,

2

-

=

.

Questions:

(Q1): Which modes correspond to which eigenvalues?

(Q2): Which approximation is the worst, and why?

(Q3): How good is the Dutch roll approximation re: the various dynamic parameters?

The Role of the Eigenvectors of A

1. Introduction

The eigenvalues of A determine the type of response of the system

Bu

Ax

x

+

=

&

to both an input

)

(

t

u

and to an initial condition

0

)

0

(

x

x

D

-

=

. A real eigenvalue relates to an exponential type of response with no natural oscillations, whereas a conjugate pair of eigenvalues relates to an exponential type of response, but with natural oscillations at the damped natural frequency. The question that is addressed in this section is:

In what way(s) do the eigenvectors contribute to the response of the system

0

)

0

(

;

x

x

Ax

x

=

=

-

&

.

Notice that this question restricts attention to the initial condition response. As will be seen, even with this restriction, the answer to this question is mathematically nontrivial. These noted were compiled because I feel that, while both Etkin and Nelson give some insight as to how the eigenvectors influence the transient response, it is vague. These notes are an attempt to reduce that vagueness. They were compiled from the book Linear Systems by Thomas Kailath.

2. A Laplace Transform Approach

The Laplace transform of the n-D system

0

)

0

(

;

x

x

Ax

x

=

=

-

&

(1)

gives the Laplace domain solution

0

1

)

(

)

(

x

A

sI

s

X

-

-

=

.(2)

In these notes we will assume that the eigenvalues of A, call them

n

k

k

s

1

}

{

=

are distinct (i.e. there are no repeated eigenvalues). Using the method of partial fraction expansions (a method that will be covered in AerE331), (2) can be written as:

0

1

1

)

(

)

(

x

R

s

s

s

X

n

k

k

k

=

-

-

=

.(3)

The

n

n

matrix

k

R

is called the residue (matrix) associated with the eigenvalue

.

k

s

The eigenvalue,

k

s

, of the

n

n

system matrix

A

is a scalar that satisfies the equation

k

k

k

p

s

Ap

=

.(4)

The column vector,

k

p

, is called the eigenvector that corresponds to the eigenvalue

k

s

. To be more precise, it is the right eigenvector that corresponds to the eigenvalue

k

s

. Similarly, we have

tr

k

k

tr

k

q

s

A

q

=

.(5)

Here, the row vector

tr

k

q

is called the left eigenvector that corresponds to the eigenvalue

k

s

. [Note the the superscript tr refers to the transpose. Hence,

k

q

is an

1

n

column vector, and its transpose,

tr

k

q

, is a

n

1

row vector. We now present the first of a number of facts that will be given. The interested reader may refer to the book by Kailath for elaborations and/or proofs.

Fact 1: (i)

k

tr

k

k

R

q

p

=

; (ii)

1

=

k

tr

k

p

q

.

From (i), (3) becomes:

=

-

D

=

-

=

-

-

=

-

=

-

=

n

k

k

k

k

k

n

k

tr

k

k

n

k

tr

k

k

k

p

s

s

p

x

q

s

s

x

q

p

s

s

s

X

1

1

0

1

1

0

1

1

)

(

)

(

)

(

)

(

)

(

)

(

a

(6)

where we have defined

0

x

q

tr

k

k

D

=

a

. The inverse Laplace transform of (6) is then:

=

=

n

k

k

t

s

k

p

e

t

x

k

1

)

(

a

where

0

x

q

tr

k

k

D

=

a

.(7)

Definition 1. The kth mode of (1) is

k

t

s

k

k

p

e

t

x

k

a

=

)

(

)

(

.

It follows that (7) is composed of the sum of the n modes

n

k

k

t

x

1

)

(

)}

(

{

=

. From (7), we arrive at one answer to the question in the Introduction of these notes:

(A1): The temporal shape of the kth mode of (7) is controlled solely by the eigenvalue,

k

s

. All of the components of the

1

n

response vector

)

(

t

x

have the same general temporal shape.

%PROGRAM NAME: ex43.m

% Nelson Example 4.3 on p. 155

% In the eqn Xdot = Ax, x = [ u w q th] on p.149

% Phugoid mode: xph = [u th]

%Short Period mode: xsp = [w q];

%==================SPECIFICATIONS=====================================

% Specify initial condition:

x0 = [10 0 0 0]';

%Specify time increment and total observation time:

%NOTE: Phugoid is ~100 sec. and Short period is ~1 sec.

dt = .001; tmax = 100; tvec = 0:dt:tmax; lt = length(tvec);

% Specify A-matrix

A=[-.045 .036 0 -32.2;-.37 -2.02 176 0;.002 -.04 -2.95 0;0 0 1 0];

% ===== 4-D SYSTEM COMPUTATIONS USING EQN.7 on p.48 OF Ch4 NOTES =====

% Compute eiganvalues, and right (col.) and left (row) eigenvectors:

s = eig(A);

display('4-D system eigenvalues:')

disp(s)

pause

[P,D]=eig(A);

Q = P^-1;

% Compute {a(k)}k=1:4 where a(k)=alpha(k) = q(k)tr*x0:

a = Q*x0;

% Compute 4-D initial condition response

x = zeros(4,lt);

for j = 1:lt

for k=1:4

x(:,j) =x(:,j)+ a(k)*exp(s(k)*(j-1)*dt)*P(:,k);

end

end

% Plot real and imaginary parts to verify correctness (i.e imag. parts =0)

for m = 1:4

figure(m)

ur = real(x(m,:)); ui = imag(x(m,:));

plot(tvec,ur,tvec,ui)

xlabel('Time (sec)')

title('Figure # = Initial Condition Response #')

pause

end

%========== 2-D SYSTEM APPROXIMATION COMPUTATIONS =======================

%Compute Phugoid mode u(t) Response (i.e. state #1 of 4-state response

Xu=-.045; Zu=-.369; u0=176; g=32.2; uic=x0(1);

sys = tf(uic*[1 -Xu],[1 -Xu -g*Zu/u0]);

sph2 = roots([1 -Xu -g*Zu/u0]);

display('2-D phugoid eigenvalues:')

disp(sph2)

pause

figure(1)

hold

impulse(sys,tvec,'r')

title('u(t) phugoid response from 4-D Model (BLUE) & 2-D approx.(RED)')

pause

% Compute Short Period mode Response

Zw=-2.02;Mq=-2.05; Mw=-.05; Mwdot=-.0051;wic=x0(2);

a11=Zw; a12=u0; a21=Mw+Mwdot*Zw; a22=Mq+Mwdot*u0;

den = [1 -(a11+a22) (a11*a22-a12*a21)];

num = [1 -(a11+a22)];

sys1 = tf(num,den);

ssp2 = roots(den);

display('2-D short period eigenvalues:')

disp(ssp2)

pause

figure(2)

hold

impulse(sys1,tvec,'r')

title('w(t) short period response from 4-D Model(BLUE) & 2-D approx.(RED)')

pause

%=========================================================================

%=========================================================================

% THE FOLLOWING CODE WAS DEVELOPED TO VERIFY EQN.(7) OF CH4 NOTES VIA

% THE MATLAB FUNCTION: 'ss'= state space

%-------------------------------------------------------------------------

% Compute TF matrix: G(j,k) = output(j)/input(k)

C=eye(4);

sys4 = ss(A,[],C,[]);

g = initial(sys4,x0,tvec);

figure(1)

plot(tvec,g(:,1),'k--','LineWidth',2)

title('u(t) phugoid response:eqn7 (BLUE),2-D approx.(RED),Matlab (BLACK')

Homework 7 PROBLEM 1: Use the above code to investigate the coupling between the four initial condition response variables,

tr

t

t

q

t

w

t

u

t

x

)]

(

)

(

)

(

)

(

[

)

(

q

D

D

D

D

=

for initial conditions having only one nonzero entry. Recall that the phugoid mode uses the approximate 2-D variable

tr

ph

t

t

u

t

x

)]

(

)

(

[

)

(

)

(

q

D

D

=

, while the short period mode uses the approximate 2-D variable

tr

sp

t

q

t

w

t

x

)]

(

)

(

[

)

(

)

(

D

D

=

.

Solution:

To begin this homework, we will investigate the initial condition:

tr

x

]

0

0

0

10

[

0

=

. This plane velocity disturbance will primarily excite the phugoid mode. Since we know that this mode has a time constant

.

sec

23

@

t

, we will specify the observation time to be 100 sec. The initial condition and observation time items in the Matlab code are highlighted in yellow. Running this code results in the following figures:

0102030405060708090100

-8

-6

-4

-2

0

2

4

6

8

10

u(t) phugoid response:eqn7 (BLUE),2-D approx.(RED),Matlab (BLACK

Time (sec)

Amplitude

Figure 1. Plots of

)

(

t

u

D

associated with the 4-D system (BLUE & BLACK) and with the 2-D phugoid approximation (RED), for initial condition

tr

x

]

0

0

0

10

[

0

=

.

Discussion: From the code, we have:

4-D system eigenvalues:

-2.4905 + 2.6110i

0

u

-2.4905 - 2.6110i

-0.0170 + 0.2152i

-0.0170 - 0.2152i

2-D phugoid eigenvalues:

-0.0225 + 0.2589i

-0.0225 - 0.2589i

The most obvious difference relates to the oscillation period

.

sec

/

2

)

(

d

ph

T

w

p

=

For the 4-D system this is

.

sec

29

)

(

4

@

ph

D

T

For the 2-D approximation, it is

.

sec

24

)

(

2

@

ph

D

T

This represents an error of -17%.

0102030405060708090100

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

w(t) short period response from 4-D Model(BLUE) & 2-D approx.(RED)

Time (sec)

Amplitude

Figure 2. Plots of

)

(

t

w

D

associated with the 4-D system (BLUE) and with the 2-D short period approximation (RED), for initial condition

tr

x

]

0

0

0

10

[

0

=

.

Discussion: Since the initial condition is zero for C and

)

(

t

q

D

, the 2-D approximation of the short period response will be zero. Figure 2 shows that, not only is the 4D response not zero, it contains a short period response and a long period response. The latter has the same temporal structure as the phugoid response. This plot illustrates a cross-coupling between these two modes.

Recall that the relation between

)

(

t

w

D

and

)

(

t

a

D

is approximately

0

/

)

(

)

(

u

t

w

t

D

=

D

a

. From this relation, we can get an idea of how this cross-coupling influences the angle of attack. Specifically, for

sec

/

5

.

0

ft

w

=

D

and

sec

/

176

0

ft

u

=

, we have

o

16

.

0

@

D

a

. Hence, we can conclude that this cross-coupling will probably not be noticeable.

Remark: There are other figures included in the code. Also, while the code validates equation (7) on p.48 of these notes for

)

(

t

u

D

[i.e. the black and blue lines match exactly in Figure 1 above], we have not validated it for the other three state variables, nor for other initial conditions. Hence, there are many things that one could continue to pursue in this investigation.

Optional Extra Credit (80pts) : Repeat the investigation of PROBLEM 1 in relation to EXAMPLE PROBLEM 5.3 of Nelson (see p.44 of these notes). Due >

|

|

|

|

. For convenience, assume that we have scaled

k

p

so that

1

|

|

1

2

2

=

=

=

D

n

k

mk

k

p

p

. Then

tr

i

tr

nk

k

k

k

mk

e

p

p

p

p

]

0

0

0

0

[

]

[

2

1

L

L

L

q

@

=

, (13a)

and the n requirements in (12) reduce to the single requirement

b

q

-

=

mk

. It follows that for

mk

i

e

c

q

-

=

0

, the initial condition

tr

m

k

p

c

x

]

0

0

1

0

0

0

[

0

0

L

L

@

=

will excite only the mth mode. If we assume that (13a) is an exact equality, then the condition

1

=

k

tr

k

p

q

requires that

]

0

0

0

0

[

]

[

2

1

L

L

L

mk

i

nk

k

k

tr

k

e

q

q

q

q

q

-

@

=

.(13b)

And so

mk

i

tr

k

k

e

x