identyczne komentarze identyczne jak w zadaniu 12

a

3

−

2i

−

8

−

2i

−

i

−

2i

−













:=

z

1

a

1 0

,

a

0 0

,

1

a

a

1 1

,

















⋅

:=

z

0.412

0.147i

+

0.118

−

0.029i

+

0.294

0.176i

+

0.059

0.235i

+













=

zw

z

0 0

,

z

0 1

,

z

1 0

,

⋅

z

1 1

,

zl

+

−

=

solve zl

,

2 zw

⋅

8 i

⋅

zw

⋅

1

−

4 i

⋅

−

+

(

)

34

−

zw

⋅

14

+

5 i

⋅

+

(

)

→

zl

z

1 1

,

z

0 1

,

z

1 0

,

⋅

z

0 0

,

zw

+

−

=

solve zl

,

1

4 i

⋅

+

2 zw

⋅

+

8 i

⋅

zw

⋅

+

(

)

14

5 i

⋅

+

34 zw

⋅

+

(

)

→

2 zw

⋅

8 i

⋅

zw

⋅

1

−

4 i

⋅

−

+

(

)

−

34 zw

⋅

14

−

5 i

⋅

−

(

)

1

4 i

⋅

+

2 zw

⋅

+

8 i

⋅

zw

⋅

+

(

)

14

5 i

⋅

+

34 zw

⋅

+

(

)

=

solve zw

,

1

34

238

85 i

⋅

+

(

)

1

2

⋅

1

−

34

238

85 i

⋅

+

(

)

1

2

⋅

































→

zw

1

34

238

85i

+

⋅

:=

zw

0.461

0.08i

+

=

zl

1

4 i

⋅

+

2 zw

⋅

+

8 i

⋅

zw

⋅

+

(

)

14

5 i

⋅

+

34 zw

⋅

+

(

)

:=

zl

0.105

0.237i

+

=

wzor na P

L

niechce mi sie pisac jak go wyprowadzalem powinien byc ok

Kpmax

Re

1

2

1

zl

zw

+













2

⋅

zl

⋅













1

8 Re zw

(

)

⋅

:=

Kpmax

0.46

=