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16.323 Principles of Optimal Control

Spring 2008

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 16.323 Lecture 12

Stochastic Optimal Control

• Kwaknernaak and Sivan Chapter 3.6, 5

• Bryson Chapter 14

• Stengel Chapter 5

Spr 2008 16.323 12–1

Stochastic Optimal Control
• Goal: design optimal compensators for systems with incomplete and
noisy measurements
– Consider this first simplified step: assume that we have noisy system
with perfect measurement of the state.

• System dynamics:
ẋ(t) = A(t)x(t) + Bu(t)u(t) + Bw (t)w(t)
– Assume that w(t) is a white Gaussian noise 20 ⇒ N (0, Rww )
– The initial conditions are random variables too, with
E[x(t0)] = 0, and E[x(t0)xT (t0)] = X0
– Assume that a perfect measure of x(t) is available for feedback.

• Given the noise in the system, need to modify our cost functions from
before ⇒ consider the average response of the closed-loop system
1 tf T
� � �
1 T
Js = E x (tf )Ptf x(tf ) + (x (t)Rxx(t)x(t) + uT (t)Ruu(t)u(t))dt
2 2 t0
– Average over all possible realizations of the disturbances.

• Key observation: since w(t) is white, then by definition, the corre­

lation times-scales are very short compared to the system dynamics
– Impossible to predict w(τ ) for τ > t, even with perfect knowledge
of the state for τ ≤ t
– Furthermore, by definition, the system state x(t) encapsulates all
past information about the system
– Then the optimal controller for this case is identical to the deter­
ministic one considered before.

20 16.322 Notes

June 18, 2008

Spr 2008 16.323 12–2
Spectral Factorization
• Had the process noise w(t) had “color” (i.e., not white), then we need
to include a shaping filter that captures the spectral content (i.e.,
temporal correlation) of the noise Φ(s)
– Previous picture: system is y = G(s)w1, with white noise input
w1 y
� G(s) �

– New picture: system is y = G(s)w2, with shaped noise input

w2 y
� G(s) �

• Account for the spectral content using a shaping filter H(s), so that
the picture now is of a system y = G(s)H(s)w1, with a white noise
input
w1 w̃2 y
� H(s) � G(s) �

– Then must design filter H(s) so that the output is a noise w̃2 that
has the frequency content that we need

• How design H(s)? Spectral Factorization – design a stable mini­

mum phase linear transfer function that replicates the desired spectrum
of w2.
– Basis of approach: If e2 = H(s)e1 and e1 is white, then the spec­
trum of e2 is given by
Φe2 (jω) = H(jω)H(−jω)Φe1 (jω)
where Φe1 (jω) = 1 because it is white.

June 18, 2008

Spr 2008 16.323 12–3

• Typically Φw2 (jω) will be given as an expression in ω 2, and we factor

that into two parts, one of which is stable minimum phase, so if
2σ 2α2
Φw2 (jω) = 2
ω√ + α2 √
2σα 2σα
= · = H(jω)H(−jω)
α + jω α − jω
√
2σα
so clearly H(s) = s+α which we write in state space form as
√
ẋH = −αxH + 2ασw1
w2 = xH

• More generally, the shaping filter will be

ẋH = AH xH + BH w1
w2 = CH xH
which we then augment to the plant dynamics, to get:
� ��
A Bw CH
� � � � � � �
ẋ x Bu 0
= + u+ w1
ẋH 0 AH xH 0 B H
� �
� � x

y = Cy 0

xH
where the noise input w1 is a white Gaussian noise.

• Clearly this augmented system has the same form as the original system
that we analyzed - there are just more states to capture the spectral
content of the original shaped noise.

June 18, 2008

Spr 2008 Disturbance Feedforward 16.323 12–4

• Now consider the stochastic LQR problem for this case.

– Modify the state weighting matrix so that
� �
Rxx 0
R̃xx =
0 0
⇒ i.e. no weighting on the filter states – Why is that allowed?
– Then, as before, the stochastic LQR solution for the augmented
system is the same as the deterministic LQR solution (6–9)
� �
� � x
u = − K Kd
xH
– So the full state feedback controller requires access to the state in
the shaping filter, which is fictitious and needs to be estimated

• Interesting result is that the gain K on the system states is com­

pletely independent of the properties of the disturbance
– In fact, if the solution of the steady state Riccati equation in this
case is partitioned as
� �
Pxx PxxH
Paug =
PxHx PxHxH
it is easy to show that
� Pxx can be solved for independently, and
� Is the same as it would be in the deterministic case with the dis­
turbances omitted 21
– Of course the control inputs that are also based on xH will improve
the performance of the system ⇒ disturbance feedforward.

21 K+S pg 262

June 18, 2008

Spr 2008 Performance Analysis 16.323 12–5

• Recall that the specific initial conditions do not effect the LQR con­
troller, but they do impact the cost-to-go from t0
– Consider the stochastic LQR problem, but with w(t) ≡ 0 so that
the only uncertainty is in the initial conditions
– Have already shown that LQR cost can be written in terms of the
solution of the Riccati equation (4–7):
1 T
JLQR = x (t0)P (t0)x(t0)
2 � �
1 T
⇒ Js = E x (t0)P (t0)x(t0)
2
1 � T
�
= E trace[P (t0)x(t0)x (t0)]
2
1
= trace[P (t0)X0]
2
which gives expected cost-to-go with uncertain IC.

• Now return to case with w �= 0 – consider the average performance

of the stochastic LQR controller.

• To do this, recognize that if we apply the LQR control, we have a

system where the cost is based on zT Rzzz = xT Rxxx for the closed-
loop system:
ẋ(t) = (A(t) − Bu(t)K(t))x(t) + Bw (t)w(t)
z(t) = Cz (t)x(t)

• This is of the form of a linear time-varying system driven by white

Gaussian noise – called a Gauss-Markov Random process22.
22 Bryson 11.4

June 18, 2008

 Spr 2008 16.323 12–6

• For a Gauss-Markov system we can predict the mean square value

of the state X(t) = E[x(t)x(t)T ] over time using X(0) = X0 and

Ẋ(t) = [A(t) − Bu(t)K(t)] X(t) + X(t) [A(t) − Bu(t)K(t)]T + Bw Rww BwT

– Matrix differential Lyapunov Equation.

• Can also extract the mean square control values using

E[u(t)u(t)T ] = K(t)X(t)K(t)T

• Now write performance evaluation as:

� � tf �
1 T T T
Js = E x (tf )Ptf x(tf ) + (x (t)Rxx (t)x(t) + u (t)Ruu (t)u(t))dt
2 t0
� � � tf ��
1 T T T
= E trace Ptf x(tf )x (tf ) + (Rxx (t)x(t)x (t) + Ruu (t)u(t)u (t))dt
2 t0
� � tf �
1 T
= trace Ptf X(tf ) + (Rxx (t)X(t) + Ruu (t)K(t)X(t)K(t) )dt
2 t0

• Not too useful in this form, but if P (t) is the solution of the LQR
Riccati equation, then can show that the cost can be written as:
� � tf �
1
Js = trace P (t0)X(t0) + (P (t)Bw Rww BwT )dt
2 t0

– First part, 12 trace {P (t0)X(t0)} is the same cost-to-go from the

uncertain initial condition that we identified on 11–5
– Second part shows that the cost increases as a result of the process
noise acting on the system.

June 18, 2008

Spr 2008 16.323 12–7

• Sketch of Proof: first note that

� tf
d
P (t0)X(t0) − Ptf X(tf ) + (P (t)X(t))dt = 0
t0 dt

1 � �
Js = trace Ptf X(tf ) + P (t0)X(t0) − Ptf X(tf )
2 �� tf �
1
+ trace {Rxx(t)X(t) + Ruu(t)K(t)X(t)K(t)T }dt
2
�� t0tf �
1
+ trace {Ṗ (t)X(t) + P (t)Ẋ(t)}dt
2 t0
−1 T
and (first reduces to standard CARE if K(t) = Ruu Bu P (t))

−Ṗ (t)X(t) = (A − BuK(t))T P (t)X(t) + P (t)(A − BuK(t))X(t)

+RxxX(t) + K(t)T RuuK(t)X(t)

P (t)Ẋ(t) = P (t)(A − BuK(t))X(t) + P (t)X(t)(A − BuK(t))T

+P (t)Bw Rww BwT
• Rearrange terms within the trace and then cancel terms to get final
result.

June 18, 2008

Spr 2008 16.323 12–8
Steady State Values
• Problems exist if we set t0 = 0 and tf → ∞ because performance will
be infinite
– Modify the cost to consider the time-average
1
Ja = lim Js
tf →∞ tf − t0

– No impact on necessary conditions since this is still a fixed end-time

problem.
– But now the initial conditions become irrelevant, and we only need
focus on the integral part of the cost.

• For LTI system with stationary process noise (constant Rww ) and well-
posed time-invariant control problem (steady gain u(t) = −Kssx(t))
mean square value of state settles down to a constant
lim X(t) = Xss
tf →∞

0 = (A − BuKss) Xss + Xss (A − BuKss)T + Bw Rww BwT

– Can show that time-averaged mean square performance is
1 � T
�
Ja = trace [Rxx + KssRuuKss]Xss
2
1
≡ trace[PssBw Rww BwT ]
2

• Main point: this gives a direct path to computing the expected

performance of a closed-loop system
– Process noise enters into computation of Xss

June 18, 2008

Spr 2008 Missile Example 16.323 12–9

• Consider a missile roll attitude control system with ω the roll angular
velocity, δ the aileron deflection, Q the aileron effectiveness, and φ
the roll angle, then
1 Q
ω̇ = − ω + δ + n(t)
δ̇ = u φ̇ = ω
τ τ
where n(t) is a noise input.

• Then this can be written as:

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤
δ̇ 0 0 0 δ 1 0

⎣ ω̇ ⎦ = ⎣ −1/τ Q/τ 0 ⎦ ⎣ ω ⎦ + ⎣ 0 ⎦ u + ⎣ 1 ⎦ n

φ̇ 0 1 0 φ 0 0

• Use τ = 1, Q = 10, Ruu = 1/(π)2 and

(π/12)2 0
⎡ ⎤
0
Rxx = ⎣ 0 0 0 ⎦
0 0 (π/180)2
then solve LQR problem to get feedback gains:
K=lqr(A,B,Rxx,Ruu)
K = [26.9 29.0 180.0]
• Then if n(t) has a spectral density of 1000 (deg/sec2)2· sec 23

• Find RMS response of the system from

X=lyap(A-B*K,Bw*Rww*Bw’)
⎡ ⎤
95 −42 −7
X = ⎣ −42 73 0⎦
−7 0 0.87
�
and that E[φ2] ≈ 0.93deg

23 Process noise input to a derivative of ω, so the units of n(t) must be deg/sec2 , but since E[n(t)n(τ )] = R
ww δ(t − τ ) and
δ(t)dt = 1, then the units of δ(t) are 1/sec and thus the units of Rww are (rad/sec2 )2 · sec=rad2 /sec3
�

June 18, 2008

Spr 2008 Full Control Problem 16.323 12–10

• Goal: design an optimal controller for a system with incomplete

and noisy measurements

• Setup: for the system (possibly time-varying)

ẋ = Ax + Buu + Bw w
z = Cz x
y = Cy x + v
with
– White, Gaussian noises w ∼ N (0, Rww ) and v ∼ N (0, Rvv ), with
Rww > 0 and Rvv > 0
– Initial conditions x(t0), a stochastic vector with E[x(t0)] = x̄0 and
E[(x(t0) − x̄0)(x(t0) − x̄0)T ] = Q0 so that
x(t0) ∼ N (x̄0, Q0)

• Cost:
� � tf �
1 T 1
J =E x (tf )Ptf x(tf ) + (zT (t)Rzzz(t) + uT (t)Ruuu(t))dt
2 2 t0

with Rzz > 0, Ruu > 0, Ptf ≥ 0

• Stochastic Optimal Output Feedback Problem: Find

u(t) = f [y(τ ), t0 ≤ τ ≤ t] t0 ≤ t ≤ tf
that minimizes J

• The solution is the Linear Quadratic Gaussian Controller, which uses

– LQE (10–15) to get optimal state estimates x̂(t) from y(t) using
gain L(t)
– LQR to get the optimal feedback control u(t) = −K(t)x
– Separation principle to implement u(t) = −K(t)x̂(t)

June 18, 2008

Spr 2008 Solution: LQG 16.323 12–11

• Regulator: u(t) = −K(t)x̂(t)

−1 T
K(t) = Ruu Bu P (t)
−1 T
−Ṗ (t) = AT P (t) + P (t)A + CzT RzzCz − P (t)BuRuu Bu P (t)
P (tf ) = Ptf

• Estimator from:
x̂˙(t) = Ax̂ + Buu + L(t)(y(t) − Cy x̂(t))
where x̂(t0) = x̄0 and Q(t0) = Q0
−1
Q̇(t) = AQ(t) + Q(t)AT + Bw Rww BwT − Q(t)CyT Rvv Cy Q(t)
−1
L(t) = Q(t)CyT Rvv

• A compact form of the compensator is:

ẋc = Acxc + Bcy
u = −Ccxc
with xc ≡ x̂ and

Ac = A − BuK(t) − L(t)Cy

Bc = L(t)

Cc = K(t)

• Valid for SISO and MIMO systems. Plant dynamics can also be time-
varying, but suppressed for simplicity.
– Obviously compensator is constant if we use the steady state regu­
lator and estimator gains for an LTI system.

June 18, 2008

Spr 2008 Infinite Horizon LQG 16.323 12–12

• Assuming LTI plant

• As with the stochastic LQR case, use time averaged cost
– To ensure that estimator settles down, must take t0 → −∞ and
tf → ∞, so that for any t, t0 � t � tf
¯ = lim 1
J J
tf →∞ tf − t0
t0 →−∞

– Again, this changes the cost, but not the optimality conditions

• Analysis of J¯ shows that it can be evaluated as

J¯ = E[zT (t)Rzzz(t) + uT (t)Ruuu(t)]
T
= Tr[PssLssRvv Lss + QssCzT RzzCz ]

= Tr[PssBw Rww BwT + QssKss

T
RuuKss]
where Pss and Qss are the steady state solutions of
−1 T
AT Pss + PssA + CzT RzzCz − PssBuRuu Bu Pss = 0
−1
AQss + QssAT + Bw Rww BwT − QssCyT Rvv Cy Qss = 0
with
−1 T −1
Kss = Ruu Bu Pss and Lss = QssCyT Rvv

• Can evaluate the steady state performance from the solution of 2

Riccati equations
– More complicated than stochastic LQR because J¯ must account for
performance degradation associated with estimation error.
– Since in general x̂(t) �= x(t), have two contributions to the cost
� Regulation error x �= 0
� Estimation error x̃ �= 0

June 18, 2008

Spr 2008 Interpretations 16.323 12–13

• Note that
J¯ = Tr[PssLssRvv LTss + QssCzT RzzCz ]
= Tr[PssBw Rww BwT + QssKssT
RuuKss]
both of which contain terms that are functions of the control and
estimation problems.

• To see how both terms contribute, let the regulator get very fast
⇒ Ruu → 0. A full analysis requires that we then determine what
¯ But what is clear is that:
happens to Pss and thus J.
lim J¯ ≥ Tr[QssCzT RzzCz ]
Ruu →0

which is independent of Ruu

– Thus even in the limit of no control penalty, the performance is
lower bounded by term associated with estimation error Qss.
• Similarly, can see that limRvv →0 J¯ ≥ Tr[PssBw Rww BwT ] which is re­
lated to the regulation error and provides a lower bound on the per­
formance with a fast estimator
– Note that this is the average cost for the stochastic LQR problem.

• Both cases illustrate that it is futile to make either the estimator or

regulator much “faster” than the other
– The ultimate performance is limited, and you quickly reach the
“knee in the curve” for which further increases in the authority of
one over the other provide diminishing returns.
– Also suggests that it is not obvious that either one of them should
be faster than the other.

• Rule of Thumb: for given Rzz and Rww , select Ruu and Rvv so that
the performance contributions due to the estimation and regulation
error are comparable.

June 18, 2008

Spr 2008 Separation Theorem 16.323 12–14

• Now consider what happens when the control u = −Kx is changed

to the new control u = −Kx̂ (same K).
– Assume steady state values here, but not needed.
– Previous looks at this would have analyzed the closed-loop stability,
as follows, but we also want to analyze performance.
plant : ẋ = Ax + Buu + Bw w
z = Cz x
y = Cy x + v
compensator : ẋc = Acxc + Bcy
u = −Ccxc

• Which give the closed-loop dynamics

� � � �� � � �� �
ẋ A −BuCc x Bw 0 w
= +
ẋc BcCy Ac xc 0 Bc v
� �
� � x
z = Cz 0
xc
� �
� � x
y = Cy 0 +v
xc

• It is not obvious that this system will even be stable: λi(Acl) < 0?
– To analyze, introduce n = x − xc, and the similarity transform
� � � � � �
I 0 x x
T = = T −1 ⇒ =T
I −I n xc
so that Acl ⇒ T AclT −1 ≡ Acl and when you work through the
math, you get
� �
A − BuK BuK
Acl =
0 A − LCy

June 18, 2008

Spr 2008 16.323 12–15

• Absolutely key points:

1. λi(Acl) ≡ λi(Acl)
2. Acl is block upper triangular, so can find poles by inspection:
det(sI − Acl) = det(sI − (A − BuK)) · det(sI − (A − LCy ))

The closed-loop poles of the system consist of the

union of the regulator and estimator poles

– This shows that we can design any estimator and regulator sepa­
rately with confidence that the combination will stabilize the system.
� Also means that the LQR/LQE problems decouple in terms of
being able to predict the stability of the overall closed-loop system.

• Let Gc(s) be the compensator transfer function (matrix) where

u = −Cc(sI − Ac)−1Bcy = −Gc(s)y
– Reason for this is that when implementing the controller, we often
do not just feedback −y(t), but instead have to include a reference
command r(t)
– Use servo approach and feed back e(t) = r(t) − y(t) instead

r e u y
� � Gc(s) � G(s) �

– So now u = Gce = Gc(r−y), and if r = 0, then have u = Gc(−y)

• Important points:
– Closed-loop system will be stable, but the compensator dynamics
need not be.
– Often very simple and useful to provide classical interpretations of
the compensator dynamics Gc(s).

June 18, 2008

Spr 2008 Performance Optimality 16.323 12–16

• Performance optimality of this strategy is a little harder to establish

– Now saying more than just that the separation principle is a “good”
idea ⇒ are trying to say that it is the “best” possible solution

• Approach:
– Rewrite cost and system in terms of the estimator states and dy­
namics ⇒ recall we have access to these
– Design a stochastic LQR for this revised system ⇒ full state feed­
back on x̂(t)

• Start with the cost (use a similar process for the terminal cost)
E[zT Rzzz] = E[xT Rxxx] {±x̂}
= E[(x − x̂ + x̂)T Rxx(x − x̂ + x̂)] {x̃ = x − x̂}
= E[x̃T Rxxx̃] + 2E[x̃T Rxxx̂] + E[x̂T Rxxx̂]

• Note that x̂(t) is the minimum mean square estimate of x(t) given
y(τ ), u(τ ), t0 ≤ τ ≤ t.
– Key property of that estimate is that x̂ and x̃ are uncorrelated24
E[x̃T Rxxx̂] = trace[E{x̃x̂T }Rxx] = 0

• Also,
E[x̃T Rxxx̃] = E[trace(Rxxx̃x̃T )] = trace(RxxQ)
where Q is the solution of the LQE Riccati equation (11–11)

• So, in summary we have:

E[xT Rxxx] = trace(RxxQ) + E[x̂T Rxxx̂]
24 Gelb, pg 112

June 18, 2008

Spr 2008 16.323 12–17

• Now the main part of the cost function can be rewritten as

� � tf �
1
J = E (zT (t)Rzzz(t) + uT (t)Ruuu(t))dt
2
� �t0tf �
1
= E (x̂T (t)Rxxx̂(t) + uT (t)Ruuu(t))dt
2 t0
1 tf
�
+ (trace(RxxQ))dt
2 t0
– The last term is independent of the control u(t) ⇒ it is only a
function of the estimation error
– Objective now is to choose the control u(t) to minimize the first
term

• But first we need another key fact25: If the optimal estimator is

x̂˙(t) = Ax̂(t) + Buu(t) + L(t)(y(t) − Cy x̂(t))
then by definition, the innovations process
i(t) ≡ y(t) − Cy x̂(t)
is a white Gaussian process, so that i(t) ∼ N (0, Rvv + Cy QCyT )

• Then we can rewrite the estimator as

x̂˙(t) = Ax̂(t) + Buu(t) + L(t)i(t)
which is an LTI system with i(t) acting as the process noise through
a computable L(t).

25 Gelb, pg 317

June 18, 2008

Spr 2008 16.323 12–18

• So combining the above, we must pick u(t) to minimize

� � tf �
1
J =E (x̂T (t)Rxxx̂(t) + uT (t)Ruuu(t))dt +term ind. of u(t)
2 t0
subject to the dynamics
x̂˙(t) = Ax̂(t) + Buu(t) + L(t)i(t)
– Which is a strange looking Stochastic LQR problem
– As we saw before, the solution is independent of the driving process
noise
u(t) = −K(t)x̂(t)
– Where K(t) is found from the LQR with the data A, Bu, Rxx, and
Ruu, and thus will be identical to the original problem.

• Combination of LQE/LQR gives performance optimal result.

June 18, 2008

Spr 2008 Simple Example 16.323 12–19

� �
0 1
� � � �
0 0
ẋ = x+ u+ w
0 0 1 1
� �
1 0
z = x
0 1
� �
y = 1 0 x+v
where in the LQG problem we have
� �
1 0
Rzz = Ruu = 1 Rvv = 1 Rww = 1
0 1
• Solve the SS LQG problem to find that
Tr[PssLssRvv LTss] = 8.0 Tr[QssCzT RzzCz ] = 2.8

Tr[PssBw Rww BwT ] = 1.7 T

Tr[QssKss RuuKss] = 9.1

• Suggests to me that we need to improve the estimation error ⇒ that
Rvv is too large. Repeat with
� �
1 0
Rzz = Ruu = 1 Rvv = 0.1 Rww = 1
0 1

Tr[PssLssRvv LTss] = 4.1 Tr[QssCzT RzzCz ] = 1.0

Tr[PssBw Rww BwT ] = 1.7 T

Tr[QssKss RuuKss] = 3.7

and
� �
1 0
Rzz = Ruu = 1 Rvv = 0.01 Rww = 1
0 1

Tr[PssLssRvv LTss] = 3.0 Tr[QssCzT RzzCz ] = 0.5

Tr[PssBw Rww BwT ] = 1.7 T
Tr[QssKss RuuKss] = 1.7

June 18, 2008

Spr 2008 16.323 12–20

• LQG analysis code

A=[0 1;0 0];%

Bu=[0 1]’;%

Bw=[0 1]’; %

Cy=[1 0];%

Cz=[1 0;0 1];%

Rww=1;%

Rvv=1;%

Rzz=diag([1 1]);%

Ruu=1;%

[K,P]=lqr(A,Bu,Cz*Rzz*Cz’,Ruu);%

[L,Q]=lqr(A’,Cy’,Bw*Rww*Bw’,Rvv);L=L’;%

N1=trace(P*(L*Rvv*L’))%

N2=trace(Q*(Cz’*Rzz*Cz))%

N3=trace(P*(Bw*Rww*Bw’))%

N4=trace(Q*(K’*Ruu*K))%

[N1 N2;N3 N4]

June 18, 2008

Spr 2008 16.323 12–21

Stochastic Simulation
• Consider the linearized longitudinal dynamics of a hypothetical heli­
copter. The model of the helicopter requires four state variables:
– θ(t):fuselage pitch angle (radians)
– q(t):pitch rate (radians/second)
– u(t):horizontal velocity of CG (meters/second)
– x(t):horizontal distance of CG from desired hover (meters)

The control variable is:

– δ (t): tilt angle of rotor thrust vector (radians)

Figure by MIT OpenCourseWare.

Figure 12.1: Helicopter in Hover

• The linearized equation of motion are:
θ̇(t) = q(t)

q̇(t) = −0.415q(t) − 0.011u(t) + 6.27δ(t) − 0.011w(t)

u̇(t) = 9.8θ(t) − 1.43q(t) − .0198u(t) + 9.8δ(t) − 0.0198w(t)

ẋ(t) = u(t)

– w(t) represents a horizontal wind disturbance

– Model w(t) as the output of a first order system driven by zero
mean, continuous time, unit intensity Gaussian white noise ξ(t):
ẇ(t) = −0.2w(t) + 6ξ(t)

June 18, 2008

Spr 2008 16.323 12–22

• First, treat original (non-augmented) plant dynamics.

– Design LQR controller so that an initial hover position error, x(0) =
1 m is reduced to zero (to within 5%) in approximately 4 sec.

Figure 12.2: Results show that Ruu = 5 gives reasonable performance.

• Augment the noise model, and using the same control gains, form the
closed-loop system which includes the wind disturbance w(t) as part
of the state vector.
• Solve necessary Lyapunov equations to determine the (steady-state)
variance of the position hover error, x(t) and rotor angle δ(t).
– Without feedforward:
� �
2
E[x ] = 0.048 E[δ 2] = 0.017
• Then design a LQR for the augmented system and repeat the process.
– With feedforward:
� �
2
E[x ] = 0.0019 E[δ 2] = 0.0168

June 18, 2008

Spr 2008 16.323 12–23

• Now do stochastic simulation of closed-loop system using Δt = 0.1.

– Note the subtly here that the design was for a continuous system,
but the simulation will be discrete
– Are assuming that the integration step is constant.
– Need to create ζ using the randn function, which gives zero mean
unit variance Gaussian noise.
– To scale it correctly
√ for a discrete simulation, multiply the 26output
of randn by 1/ Δt, where Δt is the integration step size.
– Could also just convert the entire system to its discrete time equiv­
alent, and then use a process noise that has a covariance
Qd = Rww /Δt

26 Franklin and Powell, Digital Control of Dynamic Systems

June 18, 2008

Spr 2008 16.323 12–24

Figure 12.3: Stochastic Simulations with and without disturbance feedforward.

June 18, 2008

Spr 2008 16.323 12–25

Helicopter stochastic simulation

1 % 16.323 Spring 2008

2 % Stochastic Simulation of Helicopter LQR

3 % Jon How

4 %

5 clear all, clf, randn(’seed’,sum(100*clock));

6 % linearized dynamics of the system

7 A = [ 0 1 0 0; 0 -0.415 -0.011 0;9.8 -1.43 -0.0198 0;0 0 1 0];

8 Bw = [0 -0.011 -0.0198 0]’;

9 Bu = [0 6.27 9.8 0]’;

10 Cz = [0 0 0 1];

11 Rxx = Cz’*Cz;

12 rho = 5;

13 Rww=1;

14
15 % lqr control

16 [K,S,E]=lqr(A,Bu,Rxx,rho);

17 [K2,S,E]=lqr(A,Bu,Rxx,10*rho);

18 [K3,S,E]=lqr(A,Bu,Rxx,rho/10);

19
20 % initial response with given x0

21 x0 = [0 0 0 1]’;

22 Ts=0.1; % small discrete step to simulate the cts dynamics

23 tf=8;t=0:Ts:tf;

24 [y,x] = initial(A-Bu*K,zeros(4,1),Cz,0,x0,t);

25 [y2,x2] = initial(A-Bu*K2,zeros(4,1),Cz,0,x0,t);

26 [y3,x3] = initial(A-Bu*K3,zeros(4,1),Cz,0,x0,t);

27 subplot(211), plot(t,[y y2 y3],[0 8],.05*[1 1],’:’,[0 8],.05*[-1 -1],’:’,’LineWidth’,2)

28 ylabel(’x’);title(’Initial response of the closed loop system with x(0) = 1’)

29 h = legend([’LQR: \rho = ’,num2str(rho)],[’LQR: \rho = ’,num2str(rho*10)],[’LQR: \rho = ’,num2str(rho/10)]);

30 axes(h)

31 subplot(212), plot(t,[(K*x’)’ (K2*x2’)’ (K3*x3’)’],’LineWidth’,2);grid on

32 xlabel(’Time’), ylabel(’\delta’)

33 print -r300 -dpng heli1.png

34
35 % shaping filter

36 Ah=-0.2;Bh=6;Ch=1;

37 % augment the filter dyanmics

38 Aa = [A Bw*Ch; zeros(1,4) Ah];

39 Bua = [Bu;0];

40 Bwa = [zeros(4,1); Bh];

41 Cza = [Cz 0];

42 Ka = [K 0]; % i.e. no dist FF

43 Acla = Aa-Bua*Ka; % close the loop using NO dist FF

44 Pass = lyap(Acla,Bwa*Rww*Bwa’); % compute SS response to the dist

45 vx = Cza*Pass*Cza’; % state resp

46 vd = Ka*Pass*Ka’; % control resp

47
48 zeta = sqrt(Rww/Ts)*randn(length(t),1); % discrete equivalent noise

49 [y,x] = lsim(Acla,Bwa,Cza,0,zeta,t,[x0;0]); % cts closed-loop sim

50 %

51 % second simulation approach: discrete time

52 %

53 Fa=c2d(ss(Acla,Bwa,Cza,0),Ts); % discretize the closed-loop dynamics

54 [dy,dx] = lsim(Fa,zeta,[],[x0;0]); % stochastic sim in discrete time

55 u = Ka*x’; % find control commands given the state response

56
57 % disturbance FF

58 [KK,SS,EE]=lqr(Aa,Bua,Cza’*Cza,rho); % now K will have dist FF

59 Acl=Aa-Bua*KK;

60 PP=lyap(Acl,Bwa*Rww*Bwa’);

61 vxa = Cza*PP*Cza’;

62 vda = KK*PP*KK’;

63 [ya,xa] = lsim(Acl,Bwa,Cza,0,zeta,t,[x0;0]); % cts sim

64 F=c2d(ss(Acl,Bwa,Cza,0),Ts); % discretize the closed-loop dynamics

65 [dya,dxa] = lsim(F,zeta,[],[x0;0]); % stochastic sim in discrete time

66 ua = KK*xa’; % find control commands given the state response

67

June 18, 2008

Spr 2008 16.323 12–26

68 figure(2);

69 subplot(211)

70 plot(t,y,’LineWidth’,2)

71 hold on;

72 plot(t,dy,’r-.’,’LineWidth’,1.5)

73 plot([0 max(t)],sqrt(vx)*[1 1],’m--’,[0 max(t)],-sqrt(vx)*[1 1],’m--’,’LineWidth’,1.5);

74 hold off

75 xlabel(’Time’);ylabel(’y(t)’);legend(’cts’,’disc’)

76 title(’Stochastic Simulation of Helicopter Response: No FF’)

77 subplot(212)

78 plot(t,u,’LineWidth’,2)

79 xlabel(’Time’);ylabel(’u(t)’);legend(’No FF’)

80 hold on;

81 plot([0 max(t)],sqrt(vd)*[1 1],’m--’,[0 max(t)],-sqrt(vd)*[1 1],’m--’,’LineWidth’,1.5);

82 hold off

83

84 figure(3);

85 subplot(211)

86 plot(t,ya,’LineWidth’,2)

87 hold on;

88 plot(t,dya,’r-.’,’LineWidth’,1.5)

89 plot([0 max(t)],sqrt(vxa)*[1 1],’m--’,[0 max(t)],-sqrt(vxa)*[1 1],’m--’,’LineWidth’,1.5);

90 hold off

91 xlabel(’Time’);ylabel(’y(t)’);legend(’cts’,’disc’)

92 title(’Stochastic Simulation of Helicopter Response: with FF’)

93 subplot(212)

94 plot(t,ua,’LineWidth’,2)

95 xlabel(’Time’);ylabel(’u(t)’);legend(’with FF’)

96 hold on;

97 plot([0 max(t)],sqrt(vda)*[1 1],’m--’,[0 max(t)],-sqrt(vda)*[1 1],’m--’,’LineWidth’,1.5);

98 hold off

99

100 print -f2 -r300 -dpng heli2.png

101 print -f3 -r300 -dpng heli3.png

June 18, 2008

Spr 2008 LQG for Helicopter 16.323 12–27

• Now consider what happens if we reduce the measurable states and

use LQG for the helicopter control/simulation
• Consider full vehicle state measurement (i.e., not the disturbance
state)
Cy = [ I4 0 ]
• Consider only partial vehicle state measurement
� �
0 1 0 0 0
Cy =
0 0 0 1 0
• Set Rvv small.

June 18, 2008

Spr 2008 16.323 12–28

Figure 12.4: LQR with disturbance feedforward compared to LQG

June 18, 2008

Spr 2008 16.323 12–29

Figure 12.5: Second LQR with disturbance feedforward compared to LQG

June 18, 2008

Spr 2008 16.323 12–30

Helicopter LQG

1 % 16.323 Spring 2008

2 % Stochastic Simulation of Helicopter LQR - from Bryson’s Book

3 % Jon How

4 %

5 clear all, clf, randn(’seed’,sum(100*clock));

6 set(0,’DefaultAxesFontName’,’arial’)

7 set(0,’DefaultAxesFontSize’,12)

8 set(0,’DefaultTextFontName’,’arial’)

9 % linearized dynamics of the system state=[theta q dotx x]

10 A = [ 0 1 0 0; 0 -0.415 -0.011 0;9.8 -1.43 -0.0198 0;0 0 1 0];

11 Bw = [0 -0.011 -0.0198 0]’;

12 Bu = [0 6.27 9.8 0]’;

13 Cz = [0 0 0 1];

14 Rxx = Cz’*Cz; Rww=1;

15 rho = 5;

16 % lqr control

17 [K,S,E]=lqr(A,Bu,Rxx,rho);

18
19 % initial response with given x0

20 x0 = [0 0 0 1]’;

21 Ts=0.01; % small discrete step to simulate the cts dynamics

22 tf=20;t=0:Ts:tf;nt=length(t);

23 % Now consider shaped noise with shaping filter

24 Ah=-0.2;Bh=6;Ch=1;

25 % augment the filter dyanmics

26 Aa = [A Bw*Ch; zeros(1,4) Ah];

27 Bua = [Bu;0];

28 Bwa = [zeros(4,1); Bh];

29 Cza = [Cz 0];

30 x0a=[x0;0];

31 %zeta = Rww/sqrt(Ts)*randn(length(t),1); % discrete equivalent noise

32 zeta = sqrt(Rww/Ts)*randn(length(t),1); % discrete equivalent noise

33
34 %%%% Now consider disturbance FF

35 [KK,SS,EE]=lqr(Aa,Bua,Cza’*Cza,rho); % now K will have dist FF

36 Acl=Aa-Bua*KK;

37 PP=lyap(Acl,Bwa*Rww*Bwa’);

38 vxa = Cza*PP*Cza’; %state

39 vda = KK*PP*KK’; %control

40 %

41 [ya,xa] = lsim(Acl,Bwa,Cza,0,zeta,t,x0a); % cts sim

42 F=c2d(ss(Acl,Bwa,Cza,0),Ts); % discretize the closed-loop dynamics

43 [dya,dxa] = lsim(F,zeta,[],x0a); % stochastic sim in discrete time

44 ua = KK*xa’; % find control commands given the state response

45
46 %%%% Now consider Output Feedback Case

47 % Assume that we can only measure the system states

48 % and not the dist one

49 FULL=1;

50 if FULL

51 Cya=eye(4,5); % full veh state

52 else

53 Cy=[0 1 0 0;0 0 0 1]; % only meas some states

54 Cya=[Cy [0;0]];

55 end

56 Ncy=size(Cya,1);Rvv=(1e-2)^2*eye(Ncy);

57 [L,Q,FF]=lqr(Aa’,Cya’,Bwa*Rww*Bwa’,Rvv);L=L’;% LQE calc

58 %closed loop dyn

59 Acl_lqg=[Aa -Bua*KK;L*Cya Aa-Bua*KK-L*Cya];

60 Bcl_lqg=[Bwa zeros(5,Ncy);zeros(5,1) L];

61 Ccl_lqg=[Cza zeros(1,5)];Dcl_lqg=zeros(1,1+Ncy);

62 x0_lqg=[x0a;zeros(5,1)];

63 zeta_lqg=zeta;

64 % now just treat this as a system with more sensor noise acting as more

65 % process noise

66 for ii=1:Ncy

67 zeta_lqg = [zeta_lqg sqrt(Rvv(ii,ii)/Ts)*randn(nt,1)];% discrete equivalent noise

June 18, 2008

Spr 2008 16.323 12–31

68 end

69 [ya_lqg,xa_lqg] = lsim(Acl_lqg,Bcl_lqg,Ccl_lqg,Dcl_lqg,zeta_lqg,t,x0_lqg); % cts sim

70 F_lqg=c2d(ss(Acl_lqg,Bcl_lqg,Ccl_lqg,Dcl_lqg),Ts); % discretize the closed-loop dynamics

71 [dya_lqg,dxa_lqg] = lsim(F_lqg,zeta_lqg,[],x0_lqg); % stochastic sim in discrete time

72 ua_lqg = [zeros(1,5) KK]*xa_lqg’; % find control commands given the state estimate

73

74 %LQG State Perf Prediction

75 X_lqg=lyap(Acl_lqg,Bcl_lqg*[Rww zeros(1,Ncy);zeros(Ncy,1) Rvv]*Bcl_lqg’);

76 vx_lqg=Ccl_lqg*X_lqg*Ccl_lqg’;

77 vu_lqg=[zeros(1,5) KK]*X_lqg*[zeros(1,5) KK]’;

78

79 figure(3);clf

80 subplot(211)

81 plot(t,ya,’LineWidth’,3)

82 hold on;

83 plot(t,dya,’r-.’,’LineWidth’,2)

84 plot([0 max(t)],sqrt(vxa)*[1 1],’m--’,[0 max(t)],-sqrt(vxa)*[1 1],’m--’,’LineWidth’,1);

85 hold off

86 xlabel(’Time’);ylabel(’y(t)’);legend(’cts’,’disc’)

87 title(’Stochastic Simulation of Helicopter Response: with FF’)

88 subplot(212)

89 plot(t,ua,’LineWidth’,2)

90 xlabel(’Time’);ylabel(’u(t)’);legend(’with FF’)

91 hold on;

92 plot([0 max(t)],sqrt(vda)*[1 1],’m--’,[0 max(t)],-sqrt(vda)*[1 1],’m--’,’LineWidth’,1);

93 axis([0 tf -0.2 .6])

94 hold off

95 print -f3 -r300 -dpng heli_lqg_1.png;

96

97 figure(4);clf

98 subplot(211)

99 plot(t,ya_lqg,’LineWidth’,3)

100 hold on;

101 plot(t,dya_lqg,’r-.’,’LineWidth’,2)
102 plot([0 max(t)],sqrt(vx_lqg)*[1 1],’m--’,[0 max(t)],-sqrt(vx_lqg)*[1 1],’m--’,’LineWidth’,1);
103 hold off
104 xlabel(’Time’);ylabel(’y(t)’);legend(’cts’,’disc’)
105 title([’Stochastic Simulation of Helicopter Response: LQG R_{v v} = ’,num2str(Rvv(1,1))])
106 subplot(212)
107 plot(t,ua_lqg,’LineWidth’,2)
108 xlabel(’Time’);ylabel(’u(t)’);%legend(’with FF’)
109 if FULL
110 legend(’Full veh state’)
111 else
112 legend(’Pitch rate, Horiz Pos’)
113 end
114 hold on;
115 plot([0 max(t)],sqrt(vu_lqg)*[1 1],’m--’,[0 max(t)],-sqrt(vu_lqg)*[1 1],’m--’,’LineWidth’,1);
116 axis([0 tf -0.2 .6])
117 hold off
118 if FULL
119 print -f4 -r300 -dpng heli_lqg_2.png;
120 else
121 print -f4 -r300 -dpng heli_lqg_3.png;
122 end

June 18, 2008

Spr 2008 16.323 12–32
Example: 747
• Bryson, page 209 Consider the stabilization of a 747 at 40,000 ft
and Mach number of 0.80. The perturbation dynamics from elevator
angle to pitch angle are given by
θ(s) 1.16(s + 0.0113)(s + 0.295)
= G(s) = 2
δe(s) [s + (0.0676)2][(s + 0.375)2 + (0.882)2]

1. Note that these aircraft dynamics can be stabilized with a simple

lead compensator
δe(s) s + 0.6
= 3.50
θ(s) s + 3.6

2. Can also design an LQG controller for this system by assuming that
Bw = Bu and Cz = Cy , and then tuning Ruu and Rvv to get a
reasonably balanced performance.
– Took Rww = 0.1 and tuned Rvv
Pole−Zero Map

4
lead
LQG

1
Imaginary Axis

−1

−2

−3

−4
−4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0
Real Axis

Figure 12.6: B747: Compensators

June 18, 2008

Spr 2008 16.323 12–33

RL of B747 system with the given Lead Comp

4.5

3.5

3
Imaginary Axis

2.5

1.5

0.5

−4.5 −4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0

Real Axis

RL of B747 system with the LQG Comp

4.5

3.5

3
Imaginary Axis

2.5

1.5

0.5

−4.5 −4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0

Real Axis

Figure 12.7: B747: root locus (Lead on left, LQG on right shown as a function of
the overall compensator gain)

June 18, 2008

Spr 2008 16.323 12–34

3. Compare the Bode plots of the lead compensator and LQG designs
2
10
G
Gclead
Gclqg
0
10

−2
10 −3 −2 −1 0 1
10 10 10 10 10
Freq (rad/sec)

100
G
Gclead
0 Gclqg

−100

−200 −3 −2 −1 0 1
10 10 10 10 10
Freq (rad/sec)
2
10
G
Looplead
Looplqg
0
10

−2
10 −3 −2 −1 0 1
10 10 10 10 10
Freq (rad/sec)

100
G
Looplead
0
Looplqg

−100

−200

−300 −3 −2 −1 0 1
10 10 10 10 10
Freq (rad/sec)

Figure 12.8: B747: Compensators and loop TF

June 18, 2008

Spr 2008 16.323 12–35

4. Consider the closed-loop TF for the system

3
10
G
Gcllead
2
10 Gcllqg

1
10

0
10

−1
10

−2
10

−3
10

−4
10 −3 −2 −1 0 1
10 10 10 10 10
Freq (rad/sec)

Figure 12.9: B747: closed-loop TF

5. Compare impulse response of two closed-loop systems.

1.5
G
Gcllead
1
Gcl
lqg
y(t)

0.5

−0.5
0 1 2 3 4 5 6 7 8 9 10
Time

4
Gclead
Gclqg
2
u(t)

−2
0 1 2 3 4 5 6 7 8 9 10
Time

Figure 12.10: B747: Impulse response

6. So while LQG controllers might appear to be glamorous, they are

actually quite ordinary for SISO systems.
– Where they really shine is that it this simple to design a MIMO
controller.

June 18, 2008

 Spr 2008 16.323 12–36

B747 LQG

1 % 16.323 B747 example

2 % Jon How, MIT, Spring 2007

3 %

4 clear all

5 set(0,’DefaultAxesFontName’,’arial’)

6 set(0,’DefaultAxesFontSize’,12)

7 set(0,’DefaultTextFontName’,’arial’)

8
9 gn=1.16*conv([1 .0113],[1 .295]);

10 gd=conv([1 0 .0676^2],[1 2*.375 .375^2+.882^2]);

11 % lead comp given

12 kn=3.5*[1 .6];kd=[1 3.6];

13
14 f=logspace(-3,1,300);

15 g=freqresp(gn,gd,2*pi*f*sqrt(-1));

16
17 [nc,dc]=cloop(conv(gn,kn),conv(gd,kd)); % CLP with lead

18 gc=freqresp(nc,dc,2*pi*f*sqrt(-1)); % CLP with lead

19 %roots(dc)

20 %loglog(f,abs([g gc]))

21
22 %get state space model

23 [a,b,c,d]=tf2ss(gn,gd);

24 % assume that Bu and Bw are the same

25 % take y=z

26 Rzz=1;Ruu=0.01;Rww=0.1;Rvv=0.01;

27 [k,P,e1] = lqr(a,b,c’*Rzz*c,Ruu);

28 [l,Q,e2] = lqe(a,b,c,Rww,Rvv);

29 [ac,bc,cc,tdc] = reg(a,b,c,d,k,l);

30 [knl,kdl]=ss2tf(ac,bc,cc,tdc);

31 N1=trace(P*(l*Rvv*l’))%

32 N2=trace(Q*(c’*Rzz*c))%

33 N3=trace(P*(b*Rww*b’))%

34 N4=trace(Q*(k’*Ruu*k))%

35 N=[N1 N2 N1+N2;N3 N4 N3+N4]

36
37 [ncl,dcl]=cloop(conv(gn,knl),conv(gd,kdl)); % CLP with lqg

38 gcl=freqresp(ncl,dcl,2*pi*f*sqrt(-1)); % CLP with lqg

39 [[roots(dc);0;0;0] roots(dcl)]

40 figure(2);clf;

41 loglog(f,abs([g gc gcl])) % mag plot of closed loop system

42 setlines(2)

43 legend(’G’,’Gcl_{lead}’,’Gcl_{lqg}’)

44 xlabel(’Freq (rad/sec)’)

45
46 Gclead=freqresp(kn,kd,2*pi*f*sqrt(-1));

47 Gclqg=freqresp(knl,kdl,2*pi*f*sqrt(-1));

48
49 figure(3);clf;

50 subplot(211)

51 loglog(f,abs([g Gclead Gclqg])) % Bode of compesantors

52 setlines(2)

53 legend(’G’,’Gc_{lead}’,’Gc_{lqg}’)

54 xlabel(’Freq (rad/sec)’)

55 axis([1e-3 10 1e-2 1e2])

56 subplot(212)

57 semilogx(f,180/pi*unwrap(phase([g])));hold on

58 semilogx(f,180/pi*unwrap(phase([Gclead])),’g’)

59 semilogx(f,180/pi*unwrap(phase([Gclqg])),’r’)

60 xlabel(’Freq (rad/sec)’)

61 hold off

62 setlines(2)

63 legend(’G’,’Gc_{lead}’,’Gc_{lqg}’)

64
65 figure(6);clf;

66 subplot(211)

67 loglog(f,abs([g g.*Gclead g.*Gclqg])) % Bode of Loop transfer function

June 18, 2008

 Spr 2008 16.323 12–37

68 setlines(2)

69 legend(’G’,’Loop_{lead}’,’Loop_{lqg}’)

70 xlabel(’Freq (rad/sec)’)

71 axis([1e-3 10 1e-2 1e2])

72 subplot(212)

73 semilogx(f,180/pi*unwrap(phase([g])));hold on

74 semilogx(f,180/pi*unwrap(phase([g.*Gclead])),’g’)

75 semilogx(f,180/pi*unwrap(phase([g.*Gclqg])),’r’)

76 xlabel(’Freq (rad/sec)’)

77 hold off

78 setlines(2)

79 legend(’G’,’Loop_{lead}’,’Loop_{lqg}’)

80

81 % RL of 2 closed-loop systems

82 figure(1);clf;rlocus(conv(gn,kn),conv(gd,kd));axis(2*[-2.4 0.1 -0.1 2.4])

83 hold on;plot(roots(dc)+sqrt(-1)*eps,’md’,’MarkerFaceColor’,’m’);hold off

84 title(’RL of B747 system with the given Lead Comp’)

85 figure(4);clf;rlocus(conv(gn,knl),conv(gd,kdl));axis(2*[-2.4 0.1 -0.1 2.4])

86 hold on;plot(roots(dcl)+sqrt(-1)*eps,’md’,’MarkerFaceColor’,’m’);hold off

87 title(’RL of B747 system with the LQG Comp’)

88

89 % time simulations

90 Ts=0.01;

91 [y1,x,t]=impulse(gn,gd,[0:Ts:10]);

92 [y2]=impulse(nc,dc,t);

93 [y3]=impulse(ncl,dcl,t);

94 [ulead]=lsim(kn,kd,y2,t); % noise free sim

95 [ulqg]=lsim(knl,kdl,y3,t); % noise free sim

96

97 figure(5);clf;

98 subplot(211)

99 plot(t,[y1 y2 y3])

100 xlabel(’Time’)
101 ylabel(’y(t)’)
102 setlines(2)
103 legend(’G’,’Gcl_{lead}’,’Gcl_{lqg}’)
104 subplot(212)
105 plot(t,[ulead ulqg])
106 xlabel(’Time’)
107 ylabel(’u(t)’)
108 setlines(2)
109 legend(’Gc_{lead}’,’Gc_{lqg}’)
110
111 figure(7)
112 pzmap(tf(kn,kd),’g’,tf(knl,kdl),’r’)
113 legend(’lead’,’LQG’)
114
115 print -depsc -f1 b747_1.eps;jpdf(’b747_1’)
116 print -depsc -f2 b747_2.eps;jpdf(’b747_2’)
117 print -depsc -f3 b747_3.eps;jpdf(’b747_3’)
118 print -depsc -f4 b747_4.eps;jpdf(’b747_4’)
119 print -depsc -f5 b747_5.eps;jpdf(’b747_5’)
120 print -depsc -f6 b747_6.eps;jpdf(’b747_6’)
121 print -depsc -f7 b747_7.eps;jpdf(’b747_7’)
122

June 18, 2008