控 制 工 程

CH4 解 答

( )( )

2

Figure E4.2 .

A digital audio system is designed to minimize the effect of disturbances and noiseas shown in . As an approximation, we may represent G s K

a Calculate the sensitivity of t

=

( ) ( )( )

2

0

1

.

d

he system due to K

b Calculate the effect of the disturbance noise T s on V .

c What value would you select for K to minimize the effect of the disturbance?

E4.2

( )

( ) ( ) ( )

( ) ( )

( )

2

2

2

2 1 2

2

1 2

1

1

1

1

1

TK

d o

o d

sol :

a The system sensitivity to K isKTS

K T K K

b The transfer function from T s to V s isKV s T sK K

c We would select K , so that the transfer function from

∂= =∂ +

=+

( ) ( ) d oT s to V s is small.

( )

100 4 3 1

1 4

A closed loop system is used to track the sun to obtain max imum power from a photovoltaic array.

The tracking system may be represented by Figure . with H s and G( s )S

where sec onds no min aτ

τ

−

= =+

=

lly, ( a ) Calculate the sensitivity of this system for a small change in .( b ) Calculate the time cons tant of the closed - loop system response.

τ

(a)

1 1 4 +1 = 1001 4 10114 1

4

T T GG

TG

sol :

The system sensitivity to is given byS S S

In this case, we havesS

GH( s ) ss

where . Therefore,

τ τ

τ

=

= =+ ++

+

=

T

4 1 4 1

4 =4 101

G s sSs s

sSs

τ

τ

ττ− −

= =+ +

−+

( ) ( )( )

(b) 100

100 0 99101 41 4 101 11101

4 0 0396 101

c

The closed - loop transfer function is

G s .T sGH s s ss

where the time - cons tant . sec ond.

τ

τ

= = = =+ + ++

= =

E4.3

P4.13

( ) ( ) ( )( ) ( )

( )( )

( )( )

( )

1 2

3 4

2 3 1 4

1 2 3 4

(4.16)

Tk

One form of a closed - loop transfer function isG s KG s

T sG s KG s

a Use Equation to show that

k G G G GS

G kG G kG

b Deter mine the sensitivity of t

+=

+

−=

+ +

( ) Figure P4.13,

he system shown in u sin g the equation

verified in part a .

( ) ( ) ( ) ( ) ( ) ( ) ( )( )

( )( )

1 2 3 4

2 3 1 42 4

1 2 3 4 1 2 3 4

=

d

Tk

sol :

a Let N s G s kG s and T s G s kG s . Then

k G G G GkG kGN k D kSk N k D G kG G kG G kG G kG

= + = +

−∂ ∂⋅ − ⋅ = − =

∂ ∂ + + + +

( ) ( )

( ) ( ) ( )( )

( ) ( )( )( ) ( )( ) ( )( )

21 2

3 4

=1 1

Tk

b The closed - loop transfer function is Then u sin g result from a , we have

k UG s MG H sMG s kUG s G kGT s SkGH s G kG MG s kUG s kGH s

−+ += =

+ + + +

CP4.9

( ) ( )

( ) ( )( )

Figure CP4.8 10 5

100 50

( )

Consider the closed - loop system in ,whose transfer function issG s and H s

s sY s

a Obtain the closed - loop transfer function T s and the unitR s

= =+ +

=

( )

( ) 2

1 0

100 ( ) 100

10

step response;

that is, let R s and assume that N( s ) .s

b Obtain the disturbance response when N s is a sinusoidal input ofs

frequency to rad / s. Assume tω

= =

=+

= ( )

( )

0

( )

hat R s .

c In the steady - state, what is the frequency and peak magnitude of the disturbanceresponse from part b ?

=

( )

( ) ( )( )

( )

( ) ( )( ) ( ) ( )

( )

( )

2

2

2

10 500

1 200 5000

10 500

1 200 5000

sol :

a The closed - loop transfer function is

s sG sT s

GH s s s

G s sY s R s

GH s s s

b The response of the system to the sinusoidal disturbance

Y s

+= =

+ + +

+= =

+ + +

( )( )( ) ( )( )

( )

( )

2 2

10 550100 50

10 51 100 50 501100 50

50 100 200 5000 100

10 0 0095

sGH s ss s

sN s GH s s s ss s

sY ss s s

c sec/ m, peak magnitude .ω

− ⋅− −+ += = =+ + + ++ ⋅

+ +−

= ⋅+ + +

= =