Системы автоматизированного проектирования / Mathcad - Пример1

Вариант 30

Задание1

R1 := 2	R4 := 4	R7	:= 2	R10	:= 10
R2 := 1	R5 := 3	R8	:= 5	R11	:= 10
R3 := 2	R6 := 4	R9	:= 15	R12	:= 4
U := 110

Rэкв := R1 +

1

1

+

1

+

1

+

1

+

1

1

1

R8 +

R9

R11 + R12

R3 +

R5 +

R10

1

+

1

1

+

1

R2

R7

R4

R6

U

I :=

= 33.997

I1 := I

Rэкв

Up := U - I1×R1 = 42.006

I3 :=

Up

=

15.002

Ur34 := I3×

1

= 12.002

Ur34

I4 :=

= 3

R3 +

1

1

+

1

R4

1

+

1

R4

R2

R2

R4

I5× R5 +

1

I2 :=

Ur34

= 12.002

1

1

R2

Up

+

I5 :=

= 9.694

I7

:=

R6

R7

I5× R5 +

1

1

R5 +

R7

1

1

1

+

1

+

R6

R7

R6

R7

I6 :=

R6

I8 :=

Up

= 2.1

I9 := I8

R8 +

R9

I10 :=

Up

= 4.201

R10

I11 :=

Up

= 3

I12 := I11

R11

+ R12

I1 - I3 - I5 - I8 - I10 - I11 = 3.109 ´ 10− 15

Задание 2

E1 := 220	R7	I1 := 0 I6 := 0 I7 := 0
		:= 8
U6 := 40	R8	:= 0.5
R1 := 1	R9	:= 0.5
R6 := 1

Given

E1 = I1×(R1 + R8 + R9) + I6×R6 + U6

0 = R6×I6 + U6 + I7×R7

0 = I1 + I7 - I6

63.846

Find(I1 , I6 , I7) = 52.308-11.538

Задание 3

I1 := 0 I6 := 0	I7 := 0
I11 := 0	I22 := 0
Given
E1 = I11×(R1 + R8 + R9 + R6) + I22×R6 + U6
0 = I11×R6 + U6 + I22×(R6			+ R7)
		63.846
I := Find(I11, I22) =
		-11.538
I1 := I0 = 63.846		I6	:= I0 + I1 = 52.308	I7 := I1 = -11.538

Задание 4

y1 :=		1
	R1 + R8 + R9
y6 :=		1
	R6
y7 :=		1
	R7
fi1 := 0							fi2 := 10
I1 := 0			I6 := 0			I7 :=
							0
Given
I1 + I7 - I6 = 0
I1 = (fi2 + E1)×y1
I6 = (-1×fi2 - U6)×y6
I7 = fi2×y7
					-92.308
Find(fi2 , I1 , I6 , I7)				= 63.846
					52.308
					-11.538

Задание 5

Резонанс напряжений

Найти: f, Rk, Xk, W, Qk, C, Lk, cosfik + Вект диаграмма

U := 38

f := 50

Uc := 88

I := 19

Zcmod :=

Uc

= 4.632 XL := Zcmod = 4.632

I

Rk :=

U

= 2

XC := Zcmod

I

Zk :=

XL + Rk

= 6.632

-1

C :=

= -6.873 ´ 10− 4

L :=

XL

= 0.015

2×π×f ×XC

1×2π×f

cosfik :=

XL

= 0.698

Zk

Z := Rk + j XL - j XC = 2

W := U×-I = -722

Задание 6

U := 100

f := 50

R1

:= 4

XL3 := 7

+ Векторная диаграмма

R3

:= 24

Xc1 := 4

R4

:= 3

Xc4 := 4

R5

:= 20

Z3 := R3 + j ×XL3

Z4 := R4 - j ×Xc4 Z5 := R5

Z1 := R1 - j ×Xc1

Z := Z1 +

1

= 7.178 - 6.269i

1

+

1

+

1

Z5

Z3

Z4

I1 :=

U

= 7.903 + 6.902i

Z

U1 := I1×Z1 = 59.22 - 4.003i

Uпарал :=

1

×I1 = 40.78 + 4.003i

Проверка напряжения:

1

+

1

+

1

U1 + Uпарал = 100

Z3

Z5

Z4

I3 := Uпарал = 1.611 - 0.303i

Z3

I4 :=	Uпарал
	Z4
I5 :=	Uпарал	= 2.039 + 0.2i
	Z5
Проверка тока
S1 :=	U1×I1		− 15
I1 - (I3 + I4 + I5)			= 1.776 ´ 10
			S3 := Uпарал×I3	S4 := Uпарал×I4	S5 := Uпарал×I5

Баланс мощностей

S := S1 + S3 + S4 + S5 = 790.288 + 690.219i

Sсети := U×I1 = 790.288 + 690.219i

Задание 7

								зам B1, B3, B4 замкнуты в
U := 100				f := 50				начальный момент времени
								Переходный процесс при
R1			:= 4	L3 := 0.22				замыкании B1
R3			:= 24	C1 := 200×10− 6				I1 := 0	I1 := 0
R4			:= 3	C4 := 500×10− 6				I3 := 0	I4 := 0
R5			:= 20
Given
I1 - I3 - I4 - I5 = 0
	U	= I1× R1 +			1		+ I3×(R3 + p×L3)
	p				p×C1

	U	= I1× R1 +			1		+ I4×(R4)
	p				p×C1
	U	= I1× R1 +			1		+ I5×R5
	p				p×C1
				6325.0×p + 765000.0
				418.0×p2 + 364850.0×p + 3.825e7
					37500.0
			209.0×p2 + 182425.0×p + 1.9125e7
Find(I1 , I3 , I4 , I5) ®				2750.0×p + 300000.0
			209.0×p2 + 182425.0×p + 1.9125e7
				825.0×p + 90000.0
				418.0×p2 + 364850.0×p + 3.825e7

				25×		27664609	×t
6325.0×p + 765000.0		invlaplace , p	25×	636286007×cosh					- 121319× 27
						418
		®
418.0×p2 + 364850.0×p + 3.825e7		simplify

Задание 8

ORIGIN := 1

r1

:= 1000 r2 := 20

L := 0.01

C := 10− 6

Em := 10

ω := 10000

:= -r2 = -2 ´ 103

1

-1

6

-1

3

a11

a12 :=

= 100

a21 := C = -1

´ 10

a22 :=

= -1 ´ 10

L

r1×C

L

b1 :=

1

= 1 ´ 103

e(t) :=

if (sin(ω×t) ³ 0 , Em, -Em)

r1×C

10

e(t) 0

− 10

0

1×10− 3

2×10− 3

3×10− 3

t

F(t , x) :=

a11×x1 + a12×x2

x :=

0

a21×x

+ a22×x + b1×e(t)

0

1

2

Z := Rkadapt(x, 0 , 0.04 , 2000, F)

n :=

0 .. 20000

	0.06
	0.04
	0.02
Zn , 2	0
	− 0.02
	− 0.04
	− 0.06	1×10− 3	2×10− 3	3×10− 3
	0
			Zn , 1

6
4
2
Zn , 3 0
− 2
− 4
− 6	1×10− 3	2×10− 3	3×10− 3
0
		Zn , 1