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f

g

1

Fq Z]

Z

Fq

## 3System Description

### The only formal description of the controller available to use was the actual implemented 1200 line Pascal code.

1This work was supported by the Swedish Research Council for Engineering Sciences (TFR) and the Swedish National Board for Industrial and Technical Development (NUTEK), which is gratefully acknowledged.

(2)

### Figure 1: The fighter JAS 39 Gripen.

Landing Gear

Controller Landing Gear

Pilot

Other system units

p[2] a[5]

s[30]

m[5]

## 4 Modeling

f0:::15g

### Timer variables and time conditions in the code have been replaced by binary state variables (flip flops) and corresponding input signals. A time condition becoming true in the original code corre- sponds to the timer input signal triggering the state variable. Once triggered, the state variable will be true until there is an explicit timer reset.

Comment Syntax Domains

B

I

I2!I

I2!I

I2!B

I2!B

I2!B

B !B

B2 !B

B2 !B

:::

I

B

I

B

I

B

### The landing gear controller code is one part of the software loop in the aircraft system. This means that the state of the code is stored until next iteration of the code. If we want to write the system as

x +

=f(xu) y=g(xu)

x

x+

u

y

uy

x

Inputs\Outputs

(3)

u )

x )

) y

) x

+

Inputs )

) Outputs

M(zz+)

z

z+

2

3

### follows the control flow graph of the program. The value of each program expression is determined by the current values of symbols and the actual program expression, i.e. the compilation function is of the form:

:PascalState!State

=fv

1 7!e

1

:::v

n 7!e

n g

vi

ei

?



?

pe = 0

B

B

B

B

B

B

B

@

### y2 := e END;

1

C

C

C

C

C

C

C

A

2Input, state and output variables.

3Boolean expressions are essentially polynomials over the fieldF2.

### with the initial symbol table

=fq7!q c7!c d7!d e7!e 

y17!y1 y27!y2 g

### we will get

 +

= (pe)=fq7!q c7!c d7!d 

e7!e y17!(q ^c )_(:q ^d )

y27!(q ^y2 )_(:q ^e )

### The final Boolean relation is computed from the final symbol table



nal

=fx +

7!f(xu)y7!g(xu)g

M(zz +

)=x +

\$f(xu)^y\$g(xu)

z= xyu]

M(zz+)

## 5Analysis – Verification

M(zz +

)

M(zz +

)^Q

1 (z)^Q

2 (z

+

)=0

M(zz+)

Q1(z)

Q2(z+)

z

z+

(4)

1 0

1

0

0 1

0

0

M(xx +

)^Q1(x)^Q2(x +

)=(x +

1

+1+x1+u1)^x1^x +

1

=1+u

1 :

1+u1 = 0

u1 = 1

0

0

u1 = 1

n

Fp

pn

2n

NP 6=P

n

n

M(zz+)

k

I(z)=0

Rk(z)

R

0

(z):=I(z)

R

k +1

(z):=R

k

(z)_(9~z(R

k

(~z)^M(~zz)))

Rd+1(z)=Rd(z)

d



d

d

n

z = z1:::zn]

2n

n

n

n



Rk(z) = Rk +1(z)





10 000

226

0 1 2

x1

x2

x

1 x

2 State

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### Using this encoding we get the polynomial model

M(xx +

)=((:x1^:x2)^(:x +

1

^x +

2 ))_

((:x

1

^x

2 )^(x

+

1

^:x +

2 ))

0

### as

R

0

(x):=I(x)=:x

1

^:x

2

R

1

(x):=((:x

1 )^(:x

2

))_((:x

1 )^x

2 )

R2(x):=((:x1)^(:x2))_((:x1)^x2)_(x1^(:x2))

R3(x):=((:x1)^(:x2))_((:x1)^x2)_(x1^(:x2))

k = 2

2

M(xx+)

2

Q(z)

Q(z) = 0

S(z)

M(zz+)

2

x1^:x2]:

Verify(M(xx +

)

x1^:x2])=

=9x +

M(xx +

)^(x +

1

^:x +

2 )

=(:x

1 )^x

2 :

1

2

### in one step.

Temporal Algebra Natural Language

Q(z)

Q(z)

Q(z)] Q(z)

### EU

Q1(z)Q2(z)] Q1(z)

Q2(z)

Q(z)] Q(z)

Q(z)] Q(z)

Q(z)] Q(z)

### AU

Q1(z)Q2(z)] Q1(z)

Q2(z)

Q(z)] Q(z)

Q(z)] Q(z)

2

0

f012g



P(z)

P(z)]

P(z)]





A

B

QA(z)

A

QB(z)

(6)

B

QA(z)]!

:QB(z)]:

M(zz+)

R (x)

226 108

^

M(zz +

)=R (x)^M(zz +

)^R (x +

)

M^(zz+)

P(u)

P(u)

M(zz+)

S(z)

M(zz +

)

S(z)

## References

### 1863–1864, 1995.

References

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