feedback control - 8.31

new controller output value, U.

		5D
				+
C	+	e	+	U	1	R
			Kp		-------
					5D
		-
					5
					F

int controller(int C, int F){

8.6 PRACTICE PROBLEM SOLUTIONS

1.

x

=

1

---

D----------------------------( B – DM)

F

2.

10

V

10

V

Vo

Vo y

---

---

=

=

m

=

m

-----

-----y F--

2

2

Ns

N

F

– MD

– K

D – K

------

---

d

s

– 0.5kgD – 5 m D – 10m

3.

θ 2

Ks2

=------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

τD4( J1J2) + D3( B1J2 + B2J1) + D2( Ks2( J1 + J2) + B1B2) + D( Ks2( B1 + B2) )----

4.

Vo	D( R1R2C)
-----	= --------------------------------------------------------------------------------------------------

Vi D2( R1LC) + D( L + CR1R2) + ( R1 + R2)

feedback control - 8.32

5.

y( t) = – 40e–2t + 40

6.

Ks	= 10		N
			---
			m
				F
	Kd	= 20		Ns
				------
				m

7.

a)

y

1

--- =

-----------------------------------------

F

D

2

M + DKd

+ Ks

b)

F = Mg

c)

y( t)

= –10.13e–0.25t cos ( 0.968t – 0.253)

+ 9.81

y( t + h)

y( t)

y( t)

d)

=

+ h

0 1

+

0

v( t + h)

v( t)

–1 –0.5

v( t)

9.81

8.

a)A setpoint is a system input that correlates to a desired output for a system with negative feedback control.

b)Feedback in a control system allows an out to be compared to an input. The difference, the error, is used to adjust the control variable and increase the accuracy.

feedback control - 8.33

9.

a)

b)

10.

11.

			1
Vn			---
			A

+

+

Vi

20K

---------------------------------

D2 + D + 20K

A + B

------------------------------------

1 + CB + A + B

Vo

CA

------------

1 + A

12.

a) V+ =		R	= 0.5Vd
	Vd	-------------
		R + R
∑IV-	V- – Vf		V- – Ve	= 0
	= --------------- + ----------------
		R	R
	Ve	= 2V-	– Vf	Vd	+						Ve
	Ve	= 2( 0.5Vd) – Vf					-
	Ve	= Vd – Vf									Vf

feedback control - 8.34

b)

∑

I

=

V-

– 0

V-

– Vs

-------------

+ ---------------- = 0

V-

10

3

10

3

V-

=

0.5Vs

(1)

∑

I

V+

– Ve

V+

– 0

= -----------------

+ -------------- = 0

V+

103

RP

V

1

+

1

=

Ve

+

-------

-----

-------

10

3

R

P

10

3

substitute in (1)

RP

+ 1K

Ve

0.5Vs

-------------------

=

-------

R

3

3

Ve

2RP

Vs

P

( 10

)

10

--------------------

Vs

=

2RP

RP + 103

-----

--------------------

Ve

RP + 103

note: a constant set by the variable resistor Rp

c)d ω

---- dt

ω D +

+ω		Km2		= V				Km		–	Tload
							------------				-----------
		------------
	JRRM					s JRRM					JR
Km2					Km				Tload
------------			= V		------------			–	-----------
				s JRRM					JR
JRRM

ω

JRM

V

Km

–

Tload

=

------------

-----------

J---------------------------------

R

R

M

D + K

2

s JRRM

JR

m

RM

Km

ω

= ---------------------------------

V

------

– T

s RM

load

J

R

R

M

D + K

2

m

feedback control - 8.35

d)

τ

Ks2( θ

1 – θ

2)

∑M = τ

– Ks2( θ 1 – θ 2) = J1θ

1D

2

+

J1

τ – K

s2

( θ

1

– θ ) = J

1

θ

1

D2

2

θ 1( J1D2 + Ks2) +θ 2( –Ks2)

= τ

(1)

Ks2( θ

1 – θ

2)

∑M = Ks2( θ 1 – θ 2) – Bθ 2D = J2θ 2D2

+

J2

Bθ

2D

Ks2( θ 1 – θ 2) – Bθ 2D = J2θ 2D2

θ

( J

D

2

+ K

+ BD) =

θ

( K

)

2

2

s2

1

s2

θ

=

θ

J2D2 + Ks2 + BD

(2)

1

2

-----------------------------------------

substitute (2) into (1)

Ks2

J

2

D2 + BD + K

2

θ

s2

( J D

+ K

) +θ

( –K ) = τ

2

-----------------------------------------

1

s2

2

s2

Ks2

D4( J

1

J ) + D3

( BJ ) + D2( K

s2

( J

1

+ J ) ) + D( BK

)

θ

2

1

2

s2

= τ

2

------------------------------------------------------------------------------------------------------------------------------

Ks2

θ

2

=

Ks2

----

------------------------------------------------------------------------------------------------------------------------------

4

3

2

τ

D

( J1J2) + D

( BJ1)

+ D

( Ks2( J1 + J2) ) + D( BKs2)

ω 2 = Dθ

2

ω

2

=

Ks2

-----

-----------------------------------------------------------------------------------------------------------------------

τ

D

3

( J1J2) + D

2

( BJ1) + D( Ks2( J1 + J2) ) + ( BKs2)

e)

ω	=	D3	( ) + D2	( ) + D( ) + ( )
-----		D------------------------------------------------------------------------------------------4( ) + D3( ) + D2( ) + D( ) + ( )
Vd

feedback control - 8.36

13.

R( t)

=

–0.224e

–t

cos

15

–

1

1

---------t – 0.464

--

+ ----

2

5

2t

S( t)

=

0.224e

–t

cos

15

1

---------t – 0.464

– --

2

5

e( t)	–t		15			1
			---------		+	-----
	= –0.112e cos		2	t – 0.464		10

14.

ω d	= 1.530	rad	K = 1.30
		--------
		s

15.

a)	h( t)	= 2.91e–8.873t		– 22.91e–1.127t + 20
b)	ζ =	1.58	ω n	= 3.162	rad
					--------
					s

16.

a)	Vo	=	1000
	-----		----------------------
	Vi		D + 1000
b)	Vo	=	0.050e–1000t + 0.050 sin ( 105t + 1.561)

17.

int controller(){

}