Mathcad zadanie 3

background image

b

0.501

0.499i

+

4.976

10

3

×

0.015i

+

50.049

50i

+

0.501

0.499i

+

=

b

1

z

0 1

,

z

1 1

,

1

z

z

0 0

,

:=

a

0.501

0.499i

+

4.976

10

3

×

0.015i

+

50.049

50i

+

0.501

0.499i

+

=

a

1

z

1 0

,

z

0 0

,

1

z

z

1 1

,

:=

z

19.992

40.14i

19.992

60.152i

19.992

60.152i

19.992

40.14i

=

z

Z

1

Z

2

Z

4

+

(

)

Z

1

Z

2

+

Z

4

+

Z

3

+

Z

1

Z

2

Z

1

Z

2

+

Z

4

+

Z

3

+

Z

1

Z

2

Z

1

Z

2

+

Z

4

+

Z

3

+

Z

2

Z

1

Z

4

+

(

)

Z

1

Z

2

+

Z

4

+

Z

3

+

:=

Z

2

99.903i

=

Z

4

100

=

Z

1

99.903i

=

Z

3

100.097i

=

Z

4

R

:=

Z

3

1

j ω

C

:=

Z

2

Z

1

:=

Z

1

j ω

L

:=

ω

9.99

10

4

×

=

F

C

0.1 10

6

:=

ω

2πf

:=

H

L

1 10

3

:=

Hz

f

15.9

10

3

×

:=

R

100

:=

zadanie 3

j

1

:=


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