Control Systems Engineering 4th Edition Solutions to Skill-Assessment Exercises, Exercises of Control Systems

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Solutions to

Skill-Assessment

Exercises

Control Systems Engineering

4 th^ Edition

Solutions to Skill-Assessment

Exercises

Chapter 2

2.1.

The Laplace transform of t is 1 s^2

using Table 2.1, Item 3. Using Table 2.2, Item 4,

F ( s ) = 1 ( s + 5) 2

Expanding F(s) by partial fractions yields:

F ( s ) = A s

+ B

s + 2

+ C

( s + 3) 2

+ D

( s + 3)

where,

A s s (^) S

→

B = 10

s ( s + 3) (^2) S → − 2

= − 5 C = 10

s ( s + 2) (^) S → − 3

, and

D = ( s + 3) 2 dF ( s ) ds (^) s → − 3

Taking the inverse Laplace transform yields,

f ( t ) = 5 9

− 5 e −^2 t^ + 10 3

te −^3 t^ + 40 9

e −^3 t

2.3.

Taking the Laplace transform of the differential equation assuming zero initial

conditions yields:

s^3 C(s) + 3s 2 C(s) + 7sC(s) + 5C(s) = s 2 R(s) + 4sR(s) + 3R(s)

Collecting terms,

( s^3 + 3 s^2 + 7 s + 5) C ( s ) = ( s^2 + 4 s + 3) R ( s )

Thus,

Chapter 2 3

( s + 1) I 1 ( s ) − sI 2 ( s ) − I 3 ( s ) = V ( s ) − sI 1 ( s ) + (2 s + 1) I 2 ( s ) − I 3 ( s ) = 0 − I 1 ( s ) − I 2 ( s ) + ( s + 2) I 3 ( s ) = 0

Solving the mesh equations for I 2 (s),

I 2 ( s ) =

( s + 1) V ( s ) − 1 − s 0 − 1 − 1 0 ( s + 2) ( s + 1) − s − 1 − s (2 s + 1) − 1 − 1 − 1 ( s + 2)

= ( s

(^2) + 2 s + 1) V ( s ) s ( s^2 + 5 s + 2)

But, V (^) L ( s ) = sI 2 ( s )

Hence,

V (^) L ( s ) = ( s

(^2) + 2 s + 1) V ( s ) ( s^2 + 5 s + 2)

or

V (^) L ( s ) V ( s )

= s

(^2) + 2 s + 1 s^2 + 5 s + 2

Nodal Analysis

Writing the nodal equations,

(^1 s

• 2. V 1 ( s ) − V (^) L ( s ) = V ( s )

− V 1 ( s ) + (^2 s

• 1. V (^) L ( s ) = 1 s

V ( s )

Solving for V (^) L ( s ) ,

V (^) L ( s ) =

(^1

s

• 2. V ( s )

− 1 1 s

V ( s )

(^1 s

− 1 (^2

s

= ( s

(^2) + 2 s + 1) V ( s ) ( s^2 + 5 s + 2)

or

V (^) L ( s ) V ( s )

= s

(^2) + 2 s + 1 s^2 + 5 s + 2

4 Solutions to Skill-Assessment Exercises

Inverting

G ( s ) = − Z^2 ( s ) Z 1 ( s )

= −^100000

(10 5 / s )

= − s Noninverting

G ( s ) = [ Z^1 ( s )^ +^ Z^2 ( s )] Z 1 ( s )

(^10

5 s

(^10

5 s

= s + 1

Writing the equations of motion, ( s^2 + 3 s + 1) X 1 ( s ) − (3 s + 1) X 2 ( s ) = F ( s ) −(3 s + 1) X 1 ( s ) + ( s^2 + 4 s + 1) X 2 ( s ) = 0 Solving for X 2 ( s ) ,

X 2 ( s ) =

( s^2 + 3 s + 1) F ( s ) −(3 s + 1) 0 ( s^2 + 3 s + 1) −(3 s + 1) −(3 s + 1) ( s^2 + 4 s + 1)

= (3 s^ +^ 1) F ( s ) s ( s^3 + 7 s^2 + 5 s + 1)

Hence, X 2 ( s ) F ( s )

= (3 s^ +^ 1) s ( s^3 + 7 s^2 + 5 s + 1) 2.9. Writing the equations of motion, ( s^2 + s + 1) θ 1 ( s ) − ( s + 1) θ 2 ( s ) = T ( s ) −( s + 1) θ 1 ( s ) + (2 s + 2) θ 2 ( s ) = 0 where θ 1 ( s ) is the angular displacement of the inertia. Solving for θ 2 ( s ) ,

θ 2 ( s ) =

( s^2 + s + 1) T ( s ) −( s + 1) 0 ( s^2 + s + 1) −( s + 1) −( s + 1) (2 s + 2)

= ( s^ +^ 1) F ( s ) 2 s^3 + 3 s^2 + 2 s + 1

From which, after simplification,

6 Solutions to Skill-Assessment Exercises

Now find the electrical constants. From the torque-speed equation, set ωm = 0 to find stall torque and set T (^) m = 0 to find no-load speed. Hence, T (^) stall = 200 ω no − load = 25 which, K (^) t Ra

= T^ stall Ea

K (^) b = Ea ω no − load

Substituting all values into the motor transfer function,

θ m ( s ) Ea ( s )

K T

Ra J (^) m s ( s + 1 J (^) m

( Dm + K^ T^ K^ b Ra

s ( s + 15 2

where θ m ( s ) is the angular displacement of the armature.

Now θ L ( s ) = 1 20

θ m ( s ). Thus,

θ L ( s ) Ea ( s )

s ( s + 15 2

Letting θ 1 ( s ) =ω 1 ( s ) / s θ 2 ( s ) =ω 2 ( s ) / s in Eqs. 2.127, we obtain

( J 1 s + D 1 + K s

) ω 1 ( s ) − K s

ω 2 ( s ) = T ( s )

− K s

ω 1 ( s ) + ( J 2 s + D 2 + K s

) ω 2 ( s )

From these equations we can draw both series and parallel analogs by considering these to be mesh or nodal equations, respectively.

Chapter 2 7

Series analog

Parallel analog

2.13.

Writing the nodal equation,

C dv dt

• ir − 2 = i ( t )

But,

C = 1 v = vo +δ v ir = e v^ r^ = e v^ = e v^ o^ +δ v

Substituting these relationships into the differential equation,

d ( vo +δ v ) dt

• e v^ o^ +δ v^ − 2 = i ( t ) (1)

We now linearize e v^.

The general form is

f ( v ) − f ( vo ) ≈ df dv (^) v (^) o

δ v

Substituting the function, f ( v ) = e v^ , with v = vo +δ v yields,

e v^ o^ +δ v^ − e v^ o^ ≈ de^

v dv (^) v (^) o

δ v

Solving for e v^ o^ +δ v^ ,

9

Chapter 3

3.1.

Identifying appropriate variables on the circuit yields

Writing the derivative relations

C 1

dvC 1 dt

= iC 1

L diL dt

= v (^) L

C 2

dvC 2 dt

= iC 2

Using Kirchhoff’s current and voltage laws,

iC 1 = iL + iR = iL + 1 R

( v (^) L − vC 2 ) v (^) L = − vC 1 + vi

iC 2 = iR = 1 R

( v (^) L − vC 2 )

Substituting these relationships into Eqs. (1) and simplifying yields the state

equations as

dvC 1 dt

RC 1

vC 1 + 1 C 1

iL − 1 RC 1

vC 2 + 1 RC 1

vi

diL dt

L

vC 1 + 1 L

vi dvC 2 dt

RC 2

vC 1 − 1 RC 2

vC 2 1 RC 2

vi

where the output equation is

vo = vC 2

Putting the equations in vector-matrix form,

10 Solutions to Skill-Assessment Exercises

x

• =

RC 1

C 1

RC 1

L

RC 2

RC 2

x +

RC 1

L

RC 2

vi ( t )

y = [ 0 0 1 ] x

Writing the equations of motion ( s^2 + s + 1) X 1 ( s ) − sX 2 ( s ) = F ( s ) − sX 1 ( s ) + ( s^2 + s + 1) X 2 ( s ) − X 3 ( s ) = 0 − X 2 ( s ) + ( s^2 + s + 1) X 3 ( s ) = 0 Taking the inverse Laplace transform and simplifying,

x 1

•• = − x 1

• − x 1 + x 2 - + f x 2

•• = x 1

• − x 2 - − x 2 + x 3 x 3

•• = − x 3

• − x 3 + x 2 Defining state variables, z (^) i,

z 1 = x 1 ; z 2 = x 1

• ; z 3 = x 2 ; z 4 = x 2 - ; z 5 = x 3 ; z 6 = x 3 -

Writing the state equations using the definition of the state variables and the inverse transform of the differential equation,

z 1

• = z 2 z 2

= x 1

•• = − x 1

• − x 1 + x 2 - + f = − z 2 − z 1 + z 4 + f z 3

= x 2

= z 4 z 4

= x 2

•• = x 1

• − x 2 - − x 2 + x 3 = z 2 − z 4 − z 3 + z 5 z 5

= x 3

= z 6 z 6

= x 3

•• = − x 3

• − x 3 + x 2 = − z 6 − z 5 + z 3 The output is z 5. Hence, y = z 5. In vector-matrix form,

12 Solutions to Skill-Assessment Exercises

A =

^

^

, B =

, and C = [1.5 0.625].

Evaluating ( s I − A ) yields

( s I − A ) =

s + 4 1. − 4 s

^

Taking the inverse we obtain

( s I − A )−^1 = 1 s^2 + 4 s + 6

s −1. 4 s + 4

^

Substituting all expressions into Eq. (1) yields

G ( s ) = 3 s^ +^5 s^2 + 4 s + 6 3.5. Writing the differential equation we obtain d^2 x dt^2

• 2 x^2 = 10 +δ f ( t ) (1)

Letting x = x (^) o +δ x and substituting into Eq. (1) yields d^2 ( x (^) o +δ x ) dt^2

• 2( x (^) o +δ x ) 2 = 10 +δ f ( t ) (2)

Now, linearize x^2.

( x (^) o +δ x ) 2 − x (^) o^2 = d ( x

dx (^) x (^) o

δ x = 2 x (^) o δ x

from which ( x (^) o +δ x ) 2 = x (^) o^2 + 2 x (^) o δ x (3) Substituting Eq. (3) into Eq. (1) and performing the indicated differentiation gives us the linearized intermediate differential equation, d^2 δ x dt^2

• 4 x (^) o δ x = − 2 x (^) o^2 + 10 +δ f ( t ) (4)

The force of the spring at equilibrium is 10 N. Thus, since F = 2 x^2 , 10 = 2 x (^) o^2 from which x (^) o = 5

Chapter 3 13

Substituting this value of x (^) o into Eq. (4) gives us the final linearized differential

equation.

d^2 δ x dt^2

• 4 5 δ x =δ f ( t )

Selecting the state variables,

x 1 =δ x x 2 =δ x

Writing the state and output equations

x 1

• = x 2 x 2

=δ x

•• = − 4 5 x 1 +δ f ( t ) y = x 1

Converting to vector-matrix form yields the final result as

x

• =

^

x +

δ f ( t )

y = [ 1 0 ] x

Chapter 4 15

a. The second-order approximation is valid, since the dominant poles have a real part of

–2 and the higher-order pole is at –15, i.e. more than five-times further.

b. The second-order approximation is not valid, since the dominant poles have a real part

of –1 and the higher-order pole is at –4, i.e. not more than five-times further.

4.7.

a. Expanding G ( s ) by partial fractions yields G ( s ) = 1 s

s + 20

s + 10

s + 6.

But –0.3023 is not an order of magnitude less than residues of second-order terms (term 2

and 3). Therefore, a second-order approximation is not valid.

b. Expanding G ( s ) by partial fractions yields G ( s ) = 1 s

s + 20

s + 10

s + 6.

But 0.0704 is an order of magnitude less than residues of second-order terms (term 2 and

3). Therefore, a second-order approximation is valid.

4.8.

See Figure 4.31 in the textbook for the Simulink block diagram and the output responses.

4.9.

a. Since s I − A =

s − 2 3 s + 5

^

, ( s I − A )−^1 = 1 s^2 + 5 s + 6

s + 5 2 − 3 s

^

^

. Also,

BU ( s ) =

1 / ( s + 1)

^

^

The state vector is X ( s ) = ( s I − A )−^1 [ x (0) + BU ( s )] = 1 ( s + 1)( s + 2)( s + 3)

2( s^2 + 7 s + 7) s^2 − 4 s − 6

The output is Y ( s ) = [ 1 3 ] X ( s ) = 5 s

(^2) + 2 s − 4 ( s + 1)( s + 2)( s + 3)

s + 1

s + 2

s + 3

Taking the inverse Laplace transform yields y ( t ) = −0.5 e − t^ − 12 e −^2 t^ + 17.5 e −^3 t^.

b. The eigenvalues are given by the roots of s I − A = s^2 + 5 s + 6 , or –2 and –3.

16 Solutions to Skill-Assessment Exercises

a. Since ( s I − A ) =

s − 2 2 s + 5

^

, ( s I − A )−^1 = 1 s^2 + 5 s + 4

s + 5 2 − 2 s

^

. Taking the Laplace

transform of each term, the state transition matrix is given by

Φ( t ) =

e − t^ − 1 3

e −^4 t^^2 3

e − t^ − 2 3

e −^4 t

− 2 3

e − t^ + 2 3

e −^4 t^ − 1 3

e − t^ + 4 3

e −^4 t

b. Since Φ( t −τ) =

e −( t^ −τ^ )^ − 1 3

e −4( t^ −τ^ )^2 3

e −( t^ −τ^ )^ − 2 3

e −4( t^ −τ^ )

− 2 3

e −( t^ −τ^ )^ + 2 3

e −4( t^ −τ^ )^ − 1 3

e −( t^ −τ^ )^ + 4 3

e −4( t^ −τ^ )

and Bu ( τ) =

e −^2 τ

^

^

Φ( t −τ) Bu ( τ) =

e −τ^ e − t^ − 2 3

e^2 τ^ e −^4 t

− 1 3

e −τ^ e − t^ + 4 3

e^2 τ^ e −^4 t

Thus, x ( t )^ = Φ( t ) x (0)^ +^ Φ( t^ −τ) Bu (^ τ) 0

t

∫ d^ τ =

e − t^ − e −^2 t^ − 4 3

e −^4 t

− 5 3

e − t^ + e −^2 t^ + 8 3

e −^4 t

c. y ( t ) = [ 2 1 ] x = 5 e − t^ − e −^2 t

18 Solutions to Skill-Assessment Exercises

N 1 ( s ) N 2 (^ s )^ N 3 ( s ) N 4 ( s )

N 5 ( s ) N 6 ( s )

N 7 ( s )

Draw nodes. R ( s ) (^) N 1 ( s ) N 2 ( s ) N 3 ( s ) N 4 ( s ) (^) C ( s )

N 5 ( s ) N 6 ( s )

N 7 ( s )

Connect nodes and label subsystems.

R ( s ) N 2 ( s ) (^) N 3 ( s ) N 4 ( s ) (^) C ( s )

N 5 ( s ) (^) N 6 ( s )

N 7 ( s )

1

1 s s

− 1

s^ s

(^1 )

− 1

1

1 N 1 ( s ) s

Eliminate unnecessary nodes.

R(s) (^1) s s (^1) s C(s)

(^1) s -s

Forward-path gains are G 1 G 2 G 3 and G 1 G 3.

Chapter 5 19

Loop gains are − G 1 G 2 H 1 , − G 2 H 2 , and − G 3 H 3.

Nontouching loops are [− G 1 G 2 H 1 ][− G 3 H 3 ] = G 1 G 2 G 3 H 1 H 3

and [− G 2 H 2 ][− G 3 H 3 ] = G 2 G 3 H 2 H 3.

Also, ∆ = 1 + G 1 G 2 H 1 + G 2 H 2 + G 3 H 3 + G 1 G 2 G 3 H 1 H 3 + G 2 G 3 H 2 H 3.

Finally, ∆ 1 = 1 and ∆ 2 = 1.

Substituting these values into T ( s ) = C ( s ) R ( s )

T (^) k ∆ k k

yields

T ( s ) = G^1 ( s ) G^3 ( s )[1^ +^ G^2 ( s )] [1 + G 2 ( s ) H 2 ( s ) + G 1 ( s ) G 2 ( s ) H 1 ( s )][1 + G 3 ( s ) H 3 ( s )]

5.5.

The state equations are,

x 1

• = − 2 x 1 + x 2 x 2

= − 3 x 2 + x 3 x 3

= − 3 x 1 − 4 x 2 − 5 x 3 + r y = x 2

Drawing the signal-flow diagram from the state equations yields

1 s

1 s

1 1 s (^1 )

1

-3 -

r x^3 x^2 x^1 y

From G ( s ) = 100( s^ +^ 5) s^2 + 5 s + 6

we draw the signal-flow graph in controller canonical

form and add the feedback.