# Soluções engenharia de sistemas de controle nise quinta ed, Exercícios de Análise de Sistemas de Controlo. Universidade Federal de Uberlândia (UFU)

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Chap 1-Solutions

O N E Introduction

2. Yes - power gain, remote control, parameter conversion; No - Expense, complexity

3. Motor, low pass filter, inertia supported between two bearings

4. Closed-loop systems compensate for disturbances by measuring the response, comparing it to

the input response (the desired output), and then correcting the output response.

5. Under the condition that the feedback element is other than unity

6. Actuating signal

7. Multiple subsystems can time share the controller. Any adjustments to the controller can be

implemented with simply software changes.

8. Stability, transient response, and steady-state error

10. It follows a growing transient response until the steady-state response is no longer visible. The

system will either destroy itself, reach an equilibrium state because of saturation in driving

amplifiers, or hit limit stops.

11. Transient response

12. True

13. Transfer function, state-space, differential equations

14. Transfer function - the Laplace transform of the differential equation

State-space - representation of an nth order differential equation as n simultaneous first-order

differential equations

Differential equation - Modeling a system with its differential equation

SOLUTIONS TO PROBLEMS

1. Five turns yields 50 v. Therefore K = 50 volts

5 x 2π rad = 1.59

2 Chapter 1: Introduction

2.

Thermostat Amplifier and

valves Heater

Temperature difference

Voltage difference

Fuel flow

Actual temperature

Desired temperature

+

-

3.

Desired

roll angle

Input voltage

+

-

Pilot controls

Aileron position control

Error voltage

Aileron position

Aircraft dynamics

Roll rate

Integrate

Roll angle

Gyro Gyro voltage

4.

Speed Error

voltage Desired speed

Input voltage

+

-

transducer Amplifier

Motor and

drive system

Actual speed

Voltage proportional

to actual speed

Dancer position sensor

Dancer dynamics

5.

Desired power

Power Error

voltage

Input voltage

+

-

Transducer Amplifier

Motor and drive

system

Voltage proportional

to actual power

Rod position

Reactor

Actual power

Sensor & transducer

Solutions to Problems 3

6.

Desired student

population +

-

Population error

Desired student

rate

Actual student

rate + -

drop-out rate

Net rate of influx

Integrate

Actual student population

7.

Desired volume +

- Transducer

Volume control circuit

Voltage proportional

to desired volume

Volume error

Voltage representing actual volume Actual

volume

-

+

Transducer -

Speed

Voltage proportional to speed

Effective volume

4 Chapter 1: Introduction

8.

a.

R +V

-V

Differential amplifier

Desired level

+-

Power amplifier

Actuator

Valve

Float

Fluid input

Drain Tank

R +V

-V

b.

Desired level

Amplifiers Actuator and valve

Flow rate in

Integrate

Actual level

Flow rate out

Potentiometer +

-

+

Drain

FloatPotentiometer

-

voltage in

voltage out

Displacement

Solutions to Problems 5

9.

Desired force

Transducer Amplifier Valve Actuator and load

Tire

Actual force+

-

Current Displacement Displacement

10.

Commanded blood pressure

Vaporizer Patient

Actual blood pressure+

-

Isoflurane concentration

11.

+

-

Controller &

motor Grinder

Force Feed rate Integrator

Desired depth Depth

12.

+

-

Coil circuit

Solenoid coil & actuator

Coil current Force Armature

& spool dynamics

Desired position DepthTransducer

Coil voltage

LVDT

13.

a. L di dt

+ Ri = u(t)

6 Chapter 1: Introduction

b. Assume a steady-state solution iss = B. Substituting this into the differential equation yields RB =

1,

from which B = 1 R

. The characteristic equation is LM + R = 0, from which M = - R L

. Thus, the total

solution is i(t) = Ae-(R/L)t + 1 R

. Solving for the arbitrary constants, i(0) = A + 1 R

= 0. Thus, A =

- 1 R

. The final solution is i(t) = 1 R

-- 1 R

e-(R/L)t = 1 R

(1 − e−( R / L) t ) .

c.

14.

a. Writing the loop equation, Ri + L di dt

+ 1 C

idt + vC (0)∫ = v(t)

b. Differentiating and substituting values, d2i dt 2

+ 2 di dt

+ 30i = 0

Writing the characteristic equation and factoring,

M2 + 2 M + 30 = M + 1 + 29 i M + 1 - 29 i .

The general form of the solution and its derivative is

i = e-t cos 29 t A + B sin 29 t e- t

= - A + 29 B e-t cos 29 t - 29 A + B e- t sin 29 tdi dt Using i(0) = 0;

di dt

(0) = vL(0)

L =

1 L

= 2

i 0 A= =0 di dt

(0) = − A + 29B =2

Thus, and A = 0 B = 2 29

.

The solution is

Solutions to Problems 7

i = 2 29

29 e- t sin 29 t

c. i

t

15.a. Assume a particular solution of

Substitute into the differential equation and obtain

Equating like coefficients,

From which, C =

35 53 and D =

10 53 .

The characteristic polynomial is

Thus, the total solution is

Solving for the arbitrary constants, x(0) = A +

35 53 = 0. Therefore, A = -

35 53 . The final solution is

b. Assume a particular solution of

xp = Asin3t + Bcos3t

8 Chapter 1: Introduction

Substitute into the differential equation and obtain

(18A − B)cos(3t) − (A +18B)sin(3t) = 5sin(3t)

Therefore, 18A – B = 0 and –(A + 18B) = 5. Solving for A and B we obtain

xp = (-1/65)sin3t + (-18/65)cos3t

The characteristic polynomial is

M2 + 6 M + 8 = M + 4 M + 2

Thus, the total solution is x = C e- 4 t + D e- 2 t + -

18 65

cos 3 t - 1 65

sin 3 t

Solving for the arbitrary constants, x(0) = C + D − 18 65

= 0 .

Also, the derivative of the solution is

= - 3 65

cos 3 t + 54 65

sin 3 t - 4 C e- 4 t - 2 D e- 2 tdx dt

Solving for the arbitrary constants, x . (0) −

3 65

− 4C − 2D = 0 , or C = − 3

10 and D =

15 26

.

The final solution is x = -

18 65

cos 3 t - 1 65

sin 3 t - 3 10

e- 4 t + 15 26

e- 2 t

c. Assume a particular solution of

xp = A

Substitute into the differential equation and obtain 25A = 10, or A = 2/5.

The characteristic polynomial is

M2 + 8 M + 25 = M + 4 + 3 i M + 4 - 3 i

Thus, the total solution is x =

2 5

+ e- 4 t B sin 3 t + C cos 3 t

Solving for the arbitrary constants, x(0) = C + 2/5 = 0. Therefore, C = -2/5. Also, the derivative of the

solution is

= 3 B -4 C cos 3 t - 4 B + 3 C sin 3 t e- 4 tdx dt

Solutions to Problems 9

Solving for the arbitrary constants, x . (0) = 3B – 4C = 0. Therefore, B = -8/15. The final solution is

x(t) = 2 5

e−4t 8 15

sin(3t) + 2 5

cos(3t )⎛ ⎝ ⎞ ⎠

16.

a. Assume a particular solution of

Substitute into the differential equation and obtain

Equating like coefficients,

From which, C = -

1 5 and D = -

1 10 .

The characteristic polynomial is

Thus, the total solution is

Solving for the arbitrary constants, x(0) = A - 1 5 = 2. Therefore, A =

11 5

. Also, the derivative of the

solution is dx dt

Solving for the arbitrary constants, x . (0) = - A + B - 0.2 = -3. Therefore, B = −

3 5

. The final solution

is

x(t) = − 1 5

cos(2t) − 1

10 sin(2t) + et

11 5

cos(t) − 3 5

sin(t)⎛ ⎝ ⎞ ⎠

b. Assume a particular solution of

xp = Ce-2t + Dt + E

Substitute into the differential equation and obtain

10 Chapter 1: Introduction

Equating like coefficients, C = 5, D = 1, and 2D + E = 0.

From which, C = 5, D = 1, and E = - 2.

The characteristic polynomial is

Thus, the total solution is

Solving for the arbitrary constants, x(0) = A + 5 - 2 = 2 Therefore, A = -1. Also, the derivative of the

solution is dx dt

= (−A + B)et Bte t −10e−2t +1

Solving for the arbitrary constants, x . (0) = B - 8 = 1. Therefore, B = 9. The final solution is

c. Assume a particular solution of

xp = Ct2 + Dt + E

Substitute into the differential equation and obtain

Equating like coefficients, C =

1 4 , D = 0, and 2C + 4E = 0.

From which, C = 1 4 , D = 0, and E = -

1 8 .

The characteristic polynomial is

Thus, the total solution is

Solving for the arbitrary constants, x(0) = A -

1 8 = 1 Therefore, A =

9 8 . Also, the derivative of the

solution is

dx dt

Solving for the arbitrary constants, x . (0) = 2B = 2. Therefore, B = 1. The final solution is

Solutions to Problems 11

17.

+

-

Input transducer

Desired force

Input voltage

Controller Actuator Pantograph dynamics Spring Fup

Spring displacement

Fout

Sensor

T W O Modeling in the Frequency Domain

SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transfer Functions

Finding each transfer function:

Pot: Vi(s) θi(s)

= 10 π

;

Pre-Amp: Vp(s) Vi(s) = K;

Power Amp: Ea(s) Vp(s) =

150 s+150

Motor: Jm = 0.05 + 5( 50250 ) 2

= 0.25

Dm =0.01 + 3( 50250 ) 2

= 0.13

Kt Ra =

1 5

KtKb

Ra = 1 5

Therefore: θm(s) Ea(s) =

Kt RaJm

s(s+ 1

Jm(Dm+ KtKb

Ra )) =

0.8 s(s+1.32)

And: θo(s) Ea(s) =

1 5

θm(s) Ea(s) =

0.16 s(s+1.32)

Transfer Function of a Nonlinear Electrical Network

Writing the differential equation, d(i0 + δi)

dt + 2(i0 +δi)

2 − 5 = v(t) . Linearizing i2 about i0,

(i 0

+δi) 2

- i 0 2

= 2i ⎮ i=i

0

δi = 2i 0

δi.. Thus, (i 0 +δi)

2 = i

0 2

+ 2i 0

δi.

Solutions to Problems 13

Substituting into the differential equation yields, dδi dt + 2i0

2 + 4i0δi - 5 = v(t). But, the

resistor voltage equals the battery voltage at equilibrium when the supply voltage is zero since

the voltage across the inductor is zero at dc. Hence, 2i02 = 5, or i0 = 1.58. Substituting into the linearized

differential equation, dδi dt + 6.32δi = v(t). Converting to a transfer function,

δi(s) V(s) =

1 s+6.32 . Using the

linearized i about i0, and the fact that vr(t) is 5 volts at equilibrium, the linearized vr(t) is vr(t) = 2i2 =

2(i0+δi)2 = 2(i02+2i0δi) = 5+6.32δi. For excursions away from equilibrium, vr(t) - 5 = 6.32δi = δvr(t).

Therefore, multiplying the transfer function by 6.32, yields, δVr(s) V(s) =

6.32 s+6.32 as the transfer function

1. Transfer function

2. Linear time-invariant

3. Laplace

4. G(s) = C(s)/R(s), where c(t) is the output and r(t) is the input.

5. Initial conditions are zero

6. Equations of motion

7. Free body diagram

8. There are direct analogies between the electrical variables and components and the mechanical variables

and components.

9. Mechanical advantage for rotating systems

11. Multiply the transfer function by the gear ratio relating armature position to load position.

12. (1) Recognize the nonlinear component, (2) Write the nonlinear differential equation, (3) Select the

equilibrium solution, (4) Linearize the nonlinear differential equation, (5) Take the Laplace transform of

the linearized differential equation, (6) Find the transfer function.

SOLUTIONS TO PROBLEMS

1.

a. F(s) = estdt 0

∫ = − 1 s

est 0

= 1 s

b. F(s) = testdt 0

∫ = est

s2 (−st −1) 0

∞ = −(st +1)

s2est 0

14 Chapter 2: Modeling in the Frequency Domain

Using L'Hopital's Rule

F(s) t → ∞ = −s

s 3est t →∞ = 0. Therefore, F(s) =

1 s2

.

c. F(s) = sinωtestdt 0

∫ = est

s2 + ω 2 (−ssinωt − ω cosωt)

0

= ω

s2 +ω 2

d. F(s) = cosωtestdt 0

∫ = est

s2 + ω 2 (−scosωt + ω sinωt)

0

= s

s2 +ω 2

2.

a. Using the frequency shift theorem and the Laplace transform of sin ωt, F(s) = ω

(s+a)2+ω2 .

b. Using the frequency shift theorem and the Laplace transform of cos ωt, F(s) = (s+a)

(s+a)2+ω2 .

c. Using the integration theorem, and successively integrating u(t) three times, ⌡⌠dt = t; ⌡⌠tdt = t2 2 ;

⌡ ⌠t2

2dt = t3 6 , the Laplace transform of t

3u(t), F(s) = 6 s4 .

3.a. The Laplace transform of the differential equation, assuming zero initial conditions,

is, (s+7)X(s) = 5s

s2+22 . Solving for X(s) and expanding by partial fractions,

Or,

Taking the inverse Laplace transform, x(t) = - 35 53 e

-7t + ( 35 53 cos 2t +

10 53 sin 2t).

b. The Laplace transform of the differential equation, assuming zero initial conditions, is,

(s2+6s+8)X(s) = 15

s2 + 9 .

Solving for X(s)

X(s) = 15

(s2 + 9)(s2 + 6s + 8)

and expanding by partial fractions,

X(s) = − 3

65

6s + 1 9

9

s2 + 9 −

3 10

1 s + 4

+ 15 26

1 s + 2

Solutions to Problems 15

Taking the inverse Laplace transform,

x(t) = − 18 65

cos(3t) − 1

65 sin(3t) −

3 10

e−4t + 15 26

e−2t

c. The Laplace transform of the differential equation is, assuming zero initial conditions,

(s2+8s+25)x(s) = 10 s

. Solving for X(s)

X s = 10 s s 2 + 8 s + 25

and expanding by partial fractions,

X s = 25 1 s -

2 5

1 s + 4 + 4 9

9

s + 42 + 9 Taking the inverse Laplace transform,

x(t) = 2 5

e−4t 8

15 sin(3t) +

2 5

cos(3t )⎛ ⎝ ⎞ ⎠

4.

a. Taking the Laplace transform with initial conditions, s2X(s)-2s+3+2sX(s)-4+2X(s) = 2

s2+22 .

Solving for X(s),

X(s) = 2s 3 + s2 + 8s + 6

(s 2 + 4)(s2 + 2s + 2) .

Expanding by partial fractions

X(s) = − 1 5

⎛ ⎝

⎞ ⎠

s + 1 4

4

s2 + 4 +

1 5

⎛ ⎝

⎞ ⎠

11(s + 1) − 3 1

1

(s +1)2 +1

Therefore, x(t) = -0.2 cos2t - 0.1 sin2t +e-t (2.2 cost - 0.6 sint).

b. Taking the Laplace transform with initial conditions, s2X(s)-2s-1+2sX(s)-4+X(s) = 5

s+2 + 1 s2 .

Solving for X(s),

Therefore, x(t) = 5e-2t - e-t + 9te-t - 2 + t.

c. Taking the Laplace transform with initial conditions, s2X(s)-s-2+4X(s) = 2 s3 . Solving for X(s),

16 Chapter 2: Modeling in the Frequency Domain

Therefore, x(t) = 9 8 cos2t + sin2t -

1 8 +

1 4 t

2.

5.Program: syms t f=5*t^2*cos(3*t+45); pretty(f) F=laplace(f); F=simple(F); pretty(F) 'b' f=5*t*exp(-2*t)*sin(4*t+60); pretty(f) F=laplace(f); F=simple(F); pretty(F) Computer response:

ans = a 2 5 t cos(3 t + 45) 3 2 s cos(45) - 27 s cos(45) - 9 s sin(45) + 27 sin(45) 10 ----------------------------------------------------- 2 3 (s + 9) ans = b 5 t exp(-2 t) sin(4 t + 60) sin(60) ((s + 2) sin(60) + 4 cos(60)) (s + 2) -5 ------------- + 10 ------------------------------------- 2 2 2 (s + 2) + 16 ((s + 2) + 16)

6.

Program: syms s 'a' G=(s^2+3*s+7)*(s+2)/[(s+3)*(s+4)*(s^2+2*s+100)]; pretty(G) g=ilaplace(G); pretty(g) 'b' G=(s^3+4*s^2+6*s+5)/[(s+8)*(s^2+8*s+3)*(s^2+5*s+7)]; pretty(G) g=ilaplace(G); pretty(g) Computer response:

ans = a 2

Solutions to Problems 17

(s + 3 s + 7) (s + 2) -------------------------------- 2 (s + 3) (s + 4) (s + 2 s + 100) 11 4681 1/2 1/2 - 7/103 exp(-3 t) + -- exp(-4 t) - ----- exp(-t) 11 sin(3 11 t) 54 61182 4807 1/2 + ---- exp(-t) cos(3 11 t) 5562 ans = b 3 2 s + 4 s + 6 s + 5 ------------------------------------- 2 2 (s + 8) (s + 8 s + 3) (s + 5 s + 7) 299 1367 1/2 - --- exp(-8 t) + ---- exp(-4 t) cosh(13 t) 93 417 4895 1/2 1/2 - ---- exp(-4 t) 13 sinh(13 t) 5421 232 1/2 1/2 - ----- exp(- 5/2 t) 3 sin(1/2 3 t) 12927 272 1/2 - ---- exp(- 5/2 t) cos(1/2 3 t) 4309

7.

The Laplace transform of the differential equation, assuming zero initial conditions, is, (s3+3s2+5s+1)Y(s) = (s3+4s2+6s+8)X(s).

Solving for the transfer function, Y(s) X(s)

= s3 + 4s2 + 6s + 8 s3 +3s 2 + 5s +1

.

8.a. Cross multiplying, (s2+2s+7)X(s) = F(s).

Taking the inverse Laplace transform, d2 x dt2

+ 2 dx dt

+ 7x = f(t).

b. Cross multiplying after expanding the denominator, (s2+15s+56)X(s) = 10F(s).

Taking the inverse Laplace transform, d2 x dt2

+ 15 dx dt

+ 56x =10f(t).

c. Cross multiplying, (s3+8s2+9s+15)X(s) = (s+2)F(s).

Taking the inverse Laplace transform, d3 x dt3

+ 8 d2 x dt2

+ 9 dx dt

+ 15x = df (t) dt

+2f(t).

9.

= s5 + 2s 4 + 4s3 + s2 + 3

s6 + 7s5 + 3s4 + 2s3 + s2 + 3 . The transfer function is

C(s) R(s)

18 Chapter 2: Modeling in the Frequency Domain

Cross multiplying, (s6+7s5+3s4+2s3+s2+3)C(s) = (s5+2s4+4s3+s2+3)R(s).

Taking the inverse Laplace transform assuming zero initial conditions, d6c dt6

+ 7 d5c dt5

+ 3 d 4c dt4

+ 2 d3c dt3

+ d2c dt2

+ 3c = d5r dt5

+ 2 d 4r dt4

+ 4 d3r dt3

+ d2r dt2

+ 3r.

10.

= s4 + 2s3 + 5s2 + s +1

s5 + 3s 4 + 2s3 + 4s2 + 5s + 2 . The transfer function is

C(s) R(s)

Cross multiplying, (s5+3s4+2s3+4s2+5s+2)C(s) = (s4+2s3+5s2+s+1)R(s).

Taking the inverse Laplace transform assuming zero initial conditions, d5c dc5

+ 3 d 4c dt4

+ 2 d3c dt3

+ 4 d2c dt2

+ 5 dc dt

+ 2c = d 4r dt4

+ 2 d3r dt3

+ 5 d2r dt2

+ dr dt

+ r.

Substituting r(t) = t3, d5c dc5

+ 3 d 4c dt4

+ 2 d3c dt3

+ 4 d2c dt2

+ 5 dc dt

+ 2c

= 18δ(t) + (36 + 90t + 9t2 + 3t3) u(t).

11. Taking the Laplace transform of the differential equation, s2X(s)-s+1+2sX(s)-2+3x(s)=R(s).

Collecting terms, (s2+2s+3)X(s) = R(s)+s+1.

Solving for X(s), X(s) = R(s) s 2 + 2s +3

+ s +1

s 2 + 2s +3 .

The block diagram is then,

12.

Program: 'Factored' Gzpk=zpk([-15 -26 -72],[0 -55 roots([1 5 30])' roots([1 27 52])'],5) 'Polynomial' Gp=tf(Gzpk)

Computer response: ans = Factored Zero/pole/gain: 5 (s+15) (s+26) (s+72) -------------------------------------------- s (s+55) (s+24.91) (s+2.087) (s^2 + 5s + 30) ans =

Solutions to Problems 19

Polynomial Transfer function: 5 s^3 + 565 s^2 + 16710 s + 140400 -------------------------------------------------------------------- s^6 + 87 s^5 + 1977 s^4 + 1.301e004 s^3 + 6.041e004 s^2 + 8.58e004 s

13. Program: 'Polynomial' Gtf=tf([1 25 20 15 42],[1 13 9 37 35 50]) 'Factored' Gzpk=zpk(Gtf) Computer response: ans = Polynomial Transfer function: s^4 + 25 s^3 + 20 s^2 + 15 s + 42 ----------------------------------------- s^5 + 13 s^4 + 9 s^3 + 37 s^2 + 35 s + 50 ans = Factored Zero/pole/gain: (s+24.2) (s+1.35) (s^2 - 0.5462s + 1.286) ------------------------------------------------------ (s+12.5) (s^2 + 1.463s + 1.493) (s^2 - 0.964s + 2.679)

14. Program: numg=[-10 -60]; deng=[0 -40 -30 (roots([1 7 100]))' (roots([1 6 90]))']; [numg,deng]=zp2tf(numg',deng',1e4); Gtf=tf(numg,deng) G=zpk(Gtf) [r,p,k]=residue(numg,deng) Computer response:

Transfer function: 10000 s^2 + 700000 s + 6e006 ----------------------------------------------------------------------------- s^7 + 83 s^6 + 2342 s^5 + 33070 s^4 + 3.735e005 s^3 + 2.106e006 s^2 + 1.08e007 s Zero/pole/gain: 10000 (s+60) (s+10) ------------------------------------------------ s (s+40) (s+30) (s^2 + 6s + 90) (s^2 + 7s + 100) r = -0.0073 0.0313 2.0431 - 2.0385i 2.0431 + 2.0385i -2.3329 + 2.0690i -2.3329 - 2.0690i 0.5556 p =

20 Chapter 2: Modeling in the Frequency Domain

-40.0000 -30.0000 -3.5000 + 9.3675i -3.5000 - 9.3675i -3.0000 + 9.0000i -3.0000 - 9.0000i 0 k = []

15.

Program: syms s '(a)' Ga=45*[(s^2+37*s+74)*(s^3+28*s^2+32*s+16)]... /[(s+39)*(s+47)*(s^2+2*s+100)*(s^3+27*s^2+18*s+15)]; 'Ga symbolic' pretty(Ga) [numga,denga]=numden(Ga); numga=sym2poly(numga); denga=sym2poly(denga); 'Ga polynimial' Ga=tf(numga,denga) 'Ga factored' Ga=zpk(Ga) '(b)' Ga=56*[(s+14)*(s^3+49*s^2+62*s+53)]... /[(s^2+88*s+33)*(s^2+56*s+77)*(s^3+81*s^2+76*s+65)]; 'Ga symbolic' pretty(Ga) [numga,denga]=numden(Ga); numga=sym2poly(numga); denga=sym2poly(denga); 'Ga polynimial' Ga=tf(numga,denga) 'Ga factored' Ga=zpk(Ga) Computer response:

ans = (a) ans = Ga symbolic 2 3 2 (s + 37 s + 74) (s + 28 s + 32 s + 16) 45 ----------------------------------------------------------- 2 3 2 (s + 39) (s + 47) (s + 2 s + 100) (s + 27 s + 18 s + 15) ans = Ga polynimial Transfer function: 45 s^5 + 2925 s^4 + 51390 s^3 + 147240 s^2 + 133200 s + 53280 --------------------------------------------------------------------------------

Solutions to Problems 21

s^7 + 115 s^6 + 4499 s^5 + 70700 s^4 + 553692 s^3 + 5.201e006 s^2 + 3.483e006 s + 2.75e006 ans = Ga factored Zero/pole/gain: 45 (s+34.88) (s+26.83) (s+2.122) (s^2 + 1.17s + 0.5964) ----------------------------------------------------------------- (s+47) (s+39) (s+26.34) (s^2 + 0.6618s + 0.5695) (s^2 + 2s + 100) ans = (b) ans = Ga symbolic 3 2 (s + 14) (s + 49 s + 62 s + 53) 56 ---------------------------------------------------------- 2 2 3 2 (s + 88 s + 33) (s + 56 s + 77) (s + 81 s + 76 s + 65) ans = Ga polynimial Transfer function: 56 s^4 + 3528 s^3 + 41888 s^2 + 51576 s + 41552 -------------------------------------------------------------------------------- s^7 + 225 s^6 + 16778 s^5 + 427711 s^4 + 1.093e006 s^3 + 1.189e006 s^2 + 753676 s + 165165 ans = Ga factored Zero/pole/gain: 56 (s+47.72) (s+14) (s^2 + 1.276s + 1.111) --------------------------------------------------------------------------- (s+87.62) (s+80.06) (s+54.59) (s+1.411) (s+0.3766) (s^2 + 0.9391s + 0.8119)

16.

a. Writing the node equations, Vo Vi

s +

Vo s

+ Vo = 0. Solve for Vo Vi

= 1

s + 2 .

b. Thevenizing,

22 Chapter 2: Modeling in the Frequency Domain

Using voltage division, Vo (s) = Vi (s)

2

1 s

1 2

+ s + 1 s

. Thus, Vo (s) Vi(s)

= 1

2s2 + s + 2

17. a.

Writing mesh equations

(s+1)I1(s) – I2(s) = Vi(s)

-I1(s) + (s+2)I2(s) = 0

But, I1(s) = (s+2)I2(s). Substituting this in the first equation yields,

(s+1)(s+2)I2(s) – I2(s) = Vi(s) or

I2(s)/Vi(s) = 1/(s2 + 3s + 1)

But, VL(s) = sI2(s). Therefore, VL(s)/Vi(s) = s/(s2 + 3s + 1).

b.

Solutions to Problems 23

i1(t) i2(t)

(2 + 2

s )I1(s) − (1+

1 s )I2(s) = V(s)

−(1+ 1 s )I1(s) + (2 +

1 s

+ 2s )I2 (s) = 0

Solving for I2(s):

I2 (s) =

2(s +1) s

V (s)

s +1 s

0

2(s +1) s

s +1 s

s +1

s 2s2 + 2s +1

s

= V (s)s

4s2 +3s +1

Therefore, VL(s) V(s) = 2s

I2(s) V(s) =

2s2 4s2+3s+1

18.

a.

Writing mesh equations,

(2s + 1)I1(s) – I2(s) = Vi(s)

-I1(s) + (3s + 1 + 2/s)I2(s) = 0

Solving for I2(s),

24 Chapter 2: Modeling in the Frequency Domain

I2 (s) =

2s +1 Vi(s) −1 0

2s + 1 −1

−1 3s2 + s + 2

s

Solving for I2(s)/Vi(s), I2 (s) Vi(s)

= s

6s3 + 5s2 + 4s + 2

But Vo(s) = I2(s)3s. Therefore , G(s) = 3s2/(6s3 + 5s2 +4s + 2).

b. Transforming the network yields,

Writing the loop equations,

(s + s

s2 + 1 )I1(s) −

s s2 +1

I2 (s) − sI3(s) = Vi(s)

s

s2 +1 I1(s) + (

s s2 + 1

+1 + 1 s

)I2 (s) − I3 (s) = 0

sI1(s) − I2 (s) + (2s +1)I3 (s) = 0 Solving for I2(s),

I2 (s) = s(s2 + 2s + 2)

s 4 + 2s 3 + 3s2 + 3s + 2 Vi(s)

But, Vo(s) = I2(s)

s = (s 2 + 2s + 2)

s 4 + 2s3 + 3s2 + 3s + 2 Vi(s) . Therefore,

Vo (s) Vi(s)

= s2 + 2s + 2

s4 + 2s3 + 3s2 + 3s + 2

19.

a. Writing the nodal equations yields,

Solutions to Problems 25

VR(s) −Vi(s)

2s + VR(s)

1 + VR (s) − VC (s)

3s = 0

− 1 3s

VR(s) + 1 2

s + 1 3s

⎛ ⎝

⎞ ⎠ VC (s) = 0

Rewriting and simplifying,

6s + 5

6s VR(s) −

1 3s

VC (s) = 1 2s

Vi(s)

− 1 3s

VR(s) + 3s2 + 2

6s ⎛ ⎝ ⎜ ⎞

VC (s) = 0

Solving for VR(s) and VC(s),

VR(s) =

1 2s

Vi (s) − 1

3s

0 3s2 + 2

6s 6s + 5

6s − 1

3s − 1

3s 3s2 + 2

6s

; VC (s) =

6s + 5 6s

1 2s

Vi(s)

− 1

3s 0

6s + 5 6s

− 1 3s

− 1 3s

3s 2 + 2 6s

Solving for Vo(s)/Vi(s) Vo (s) Vi(s)

= VR(s) − VC (s)

Vi(s) =

3s2

6s3 + 5s2 + 4s + 2

b. Writing the nodal equations yields,

(V1(s) − Vi (s))

s + (s

2 +1) s

V1(s) + (V1(s) − Vo (s)) = 0

(Vo (s) − V1(s)) + sVo(s) + (Vo (s) − Vi (s))

s = 0

Rewriting and simplifying,

(s + 2 s

+1)V1(s) − Vo (s) = 1 s Vi(s)

V1(s) + (s + 1 s

+ 1)Vo (s) = 1 s

Vi (s)

26 Chapter 2: Modeling in the Frequency Domain

Solving for Vo(s)

Vo(s) = (s 2 + 2s + 2)

s 4 + 2s3 + 3s2 + 3s + 2 Vi(s).

Hence, Vo (s) Vi(s)

= (s2 + 2s + 2)

s4 + 2s3 + 3s2 + 3s + 2

20.

a.

Mesh: (2+2s)I1(s) - (1+2s)I2(s) - I3(s) = V(s)

- (1+2s)I1(s) + (7+5s)I2(s) - (2+3s)I3(s) = 0

-I1(s) - (2+3s)I2(s) + (3+3s+ 5 s )I3(s) = 0

Nodal:

V1(s) - V(s) + V1(s) (1+2s) +

(V1(s) - Vo(s)) 2+3s = 0

(Vo(s) - V1(s)) 2+3s +

Vo(s) 4 +

(Vo(s) - V(s)) 5 s

= 0

or 6s 2 +12s + 5 6s2 + 7s + 2

V1(s) - 1

3s + 2 Vo(s) = V(s)

− 1

3s + 2 V1(s) +

1 20

12s2 + 23s + 30 3s + 2

Vo(s) = s 5

V(s)

b. Program: syms s V %Construct symbolic object for frequency

Solutions to Problems 27

%variable 's' and V. 'Mesh Equations' A2=[(2+2*s) V -1 -(1+2*s) 0 -(2+3*s) -1 0 (3+3*s+(5/s))] %Form Ak = A2. A=[(2+2*s) -(1+2*s) -1 -(1+2*s) (7+5*s) -(2+3*s) -1 -(2+3*s) (3+3*s+(5/s))] %Form A. I2=det(A2)/det(A); %Use Cramer's Rule to solve for I2. G1=I2/V; %Form transfer function, G1(s) = I2(s)/V(s). G=4*G1; %Form transfer function, G(s) = V4(s)/V(s). 'G(s) via Mesh Equations' %Display label. pretty(G) %Pretty print G(s) 'Nodal Equations' A2=[(6*s^2+12*s+5)/(6*s^2+7*s+2) V -1/(3*s+2) s*(V/5)] %Form Ak = A2. A=[(6*s^2+12*s+5)/(6*s^2+7*s+2) -1/(3*s+2) -1/(3*s+2) (1/20)*(12*s^2+23*s+30)/(3*s+2)] %Form A. I2=simple(det(A2))/simple(det(A)); %Use Cramer's Rule to solve for I2. G1=I2/V; %Form transfer function, G1(s) = I2(s)/V(s). 'G(s) via Nodal Equations' %Display label. pretty(G) %Pretty print G(s) Computer response: ans = Mesh Equations A2 = [ 2+2*s, V, -1] [ -1-2*s, 0, -2-3*s] [ -1, 0, 3+3*s+5/s] A = [ 2+2*s, -1-2*s, -1] [ -1-2*s, 7+5*s, -2-3*s] [ -1, -2-3*s, 3+3*s+5/s] ans = G(s) via Mesh Equations 2 3 15 s + 12 s + 5 + 6 s 4 -------------------------- 2 3 120 s + 78 s + 65 + 24 s ans = Nodal Equations A2 = [ (6*s^2+12*s+5)/(2+7*s+6*s^2), V] [ -1/(2+3*s), 1/5*s*V] A =

28 Chapter 2: Modeling in the Frequency Domain

[ (6*s^2+12*s+5)/(2+7*s+6*s^2), -1/(2+3*s)] [ -1/(2+3*s), (3/5*s^2+23/20*s+3/2)/(2+3*s)] ans = G(s) via Nodal Equations

2 3 15 s + 12 s + 5 + 6 s 4 --------------------------

3 2 24 s + 78 s + 120 s + 65 21.

a.

Z1(s) = 5x10 5 + 10

6

s

Z2 (s) = 10 5 +

106

s

Therefore,

- Z2 sZ1 s

= - 15 s + 10 s + 2

b. Z1 s 100000 1

1×10−6 s += = 100000 s 10+

s

Z2 s 100000 1 1×10−6 s 1

100000 +

+= = 100000 s 20+ s 10+

Therefore, G s Z2

Z1 −= = s 20+ s

s 10+ 2 −

22. a. Z1 s 200000 1

1×10−6 s +=

Z2 s 100000 1 1×10−6 s

+=

Therefore,

G s Z1 Z2+ Z1

= = 3 2

s 20 3

+

s 5+ .

Solutions to Problems 29

b.

Z1(s) = 2x10 5 +

5x1011

s

5x105 + 10 6

s

Z2 (s) = 5x10 5 +

1011

s

105 + 10 6

s

Therefore, Z1(s) + Z2 (s)

Z1(s) =

7 2

(s + 3.18)(s + 11.68) (s + 7)(s + 10)

23.

Writing the equations of motion, where x2(t) is the displacement of the right member of springr,

(s2+s+1)X1(s) -X2(s) = 0

-X1(s) +X2(s) = F(s)

(s2+s)X1(s) = F(s)

From which, X1(s) F(s)

= 1

s(s +1) .

24. Writing the equations of motion,

(s 2 + s +1)X1 (s) − (s +1)X2 (s) = F(s)

−(s +1)X1(s) + (s 2 + s +1)X2(s) = 0

Solving for X2(s),

X2 (s) =

(s 2 + s + 1) F(s) −(s +1) 0

⎣ ⎢ ⎢

⎦ ⎥ ⎥

(s2 + s +1) −(s +1) −(s +1) (s2 + s +1)

⎣ ⎢ ⎢

⎦ ⎥ ⎥

= (s +1)F(s)

s2(s2 + 2s + 2)

From which,

X2 (s) F(s)

= (s +1)

s2 (s2 + 2s + 2) .

25. Let X1(s) be the displacement of the left member of the spring and X3(s) be the displacement of the mass. Writing the equations of motion

30 Chapter 2: Modeling in the Frequency Domain

2x1(s) − 2x2 (s) = F(s) −2X1(s) + (5s + 2)X2(s) − 5sX3(s) = 0

−5sX2 (s) + (10s 2 + 7s)X3(s) = 0

Solving for X2(s),

X2(s) = ⎪ ⎪ ⎪

5s2+10 -10

-10 5 1 s+10

⎪ ⎪ ⎪

⎪ ⎪ ⎪

5s2+10 F(s)

-10 0

⎪ ⎪ ⎪

= s(s2+50s+2)

10F(s)

Thus, X2 (s) F(s)

= 1 10

(10s + 7) s(5s +1)

26.

s 3 s 2+ +2 X s1 s 1+ X s2− 0=

s 1+ X1 s− s 2 2 s 1+ + X2 s+ F s=

Solving for X1(s); X1

0 s 1+− F s2 2 s 1+ +

s2 3 s 2+ + s 1+− s 1+− s2 2 s 1+ +

= = F s s3 4 s2 4 s 1+ + +

. Thus, X1 F s

1 s3 4 s2 4 s 1+ + +

=

27. Writing the equations of motion,

(s 2 + s +1)X1 (s) − sX2 (s) = 0

sX1(s) + (s 2 + 2s +1)X2 (s) − X3 (s) = F(s)

X2(s) + (s 2 + s +1)X3(s) = 0

Solving for X3(s),

X3 (s) =

(s2 + s + 1) − s 0 −s (s2 + 2s +1) F(s) 0 −1 0

(s2 + s +1) −s 0 −s (s2 + 2s +1) −1 0 −1 (s 2 + s + 1)

= F(s)

s(s3 + 3s2 + 3s + 3)

Solutions to Problems 31

From which, X3 (s) F(s)

= 1

s(s3 + 3s2 + 3s + 3) .

28.a.

(s2 + 2s +1)X1(s) − 2sX2 (s) − X3 (s) = F(s)

−2sX1(s) + (s 2 + 4s)X2(s) − sX3(s) = 0

−X1(s) − sX2 (s) + (s +1)X3(s) = 0 Solving for X2(s),

X 2(s) =

(s2 + 2s +1) F(s) −1 −2s 0 −s −1 0 s + 1

∆ =

−F(s) −2s −s −1 s +1 ∆

or,

X 2(s) F(s)

= 2s + 3

s(s3 + 6s2 + 9s + 3) b.

(4s2 + s+ 4) X1(s) − (s+ 1) X2(s) − 3X 3(s) = 0

−(s+1) X1(s) + (2s 2 + 5s+1) X2(s) − 4sX3(s) = F(s)

−3X1(s) − 4sX2 (s) + (4s+3) X3(s) = 0

Solving for X3(s),

X 3(s) =

(4s2 +s+ 4) −(s+1) 0 −(s+1) (2s2 + 5s+1) F(s)

−3 −4s 0 ∆

= −F(s)

(4s2 + s+ 4) −(s+1) −3 −4s

or

X3(s) F(s) =

16s3 + 4s2 +19s+3 32s5 + 48s4 +114 s3 +18s2

29. Writing the equations of motion,

(s 2 + 2s + 2)X1(s) − X2(s) − sX3 (s) = 0

X1(s) + (s 2 + s +1)X2(s) − sX3 (s) = F(s)

sX1(s) − sX2 (s) + (s 2 + 2s +1) = 0

32 Chapter 2: Modeling in the Frequency Domain

30.a.

Writing the equations of motion,

(s 2 + 9s + 8)θ1 (s) − (2s + 8)θ2(s) = 0

−(2s + 8)θ1(s) + (s 2 + 2s + 11)θ2 (s) = T (s)

b. Defining

θ1(s) = rotation of J1 θ2 (s) = rotation between K1 and D1 θ3(s) = rotation of J3 θ4 (s) = rotation of right - hand side of K2

the equations of motion are

(J1s

2 + K1)θ1(s) − K1θ2 (s) = T (s) −K1θ1(s) + (D1s + K1 )θ2 (s) − D1sθ3(s) = 0

D1sθ2 (s) + (J2s 2 + D1s + K2 )θ3(s) − K2θ4(s) = 0

K2θ3(s) + (D2s + (K2 + K3))θ4 (s) = 0

31.

Writing the equations of motion,

(s 2 + 2s + 1)θ1 (s) − (s + 1)θ2 (s) = T (s) −(s + 1)θ1(s) + (2s + 1)θ2(s) = 0

Solving for θ2 (s)

θ2 (s) =

(s2 + 2s +1) T (s) −(s +1) 0

(s2 + 2s +1) −(s +1) −(s +1) (2s +1)

= T (s)

2s(s +1)

Hence,

θ2 (s) T(s)

= 1

2s(s + 1)

32.

Reflecting impedances to θ3,

Solutions to Problems 33

(Jeqs2+Deqs)θ3(s) = T(s) ( N4 N2 N3N1

)

Thus,

θ3(s) T (s)

=

N4N2 N3 N1

Jeq s 2 + Deqs

where

Jeq = J4+J5+(J2+J3) N4 N3

⎝ ⎜ ⎜

⎠ ⎟ ⎟

2 + J1

N4 N2 N3 N1

⎝ ⎜ ⎜

⎠ ⎟ ⎟

2 , and

Deq = (D4 + D5 ) + (D2 + D3)( N4 N3

)2 + D1( N4N 2 N 3N1

)2

33. Reflecting all impedances to θ2(s),

{[J2+J1(N2N1 )

2 +J3 (N3N4 )

2]s2 + [f2+f1(N2N1 ) 2

+f3(N3N4 ) 2]s + [K(N3N4 )

2]}θ2(s) = T(s)N2N1

Substituting values,

{[1+2(3)2+16(14 ) 2]s2 + [2+1(3)2+32(14 )

2]s + 64(14 ) 2}θ2(s) = T(s)(3)

Thus,

θ2(s) T(s) =

3 20s2+13s+4

34. Reflecting impedances to θ2,

200 + 3 50 5

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

+ 200 5 25

x 50 5

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2⎡

⎣ ⎢ ⎢

⎦ ⎥ ⎥ s 2 + 1000

5 25

x 50 5

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2⎡

⎣ ⎢ ⎢

⎦ ⎥ ⎥ s + 250 + 3

50 5

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2⎡

⎣ ⎢ ⎢

⎦ ⎥ ⎥

= 50 5

⎛ ⎝ ⎜

⎞ ⎠ ⎟ T (s)

Thus,

θ2 (s) T(s)

= 10

1300s2 + 4000s + 550

35. Reflecting impedances and applied torque to respective sides of the spring yields the following equivalent circuit:

34 Chapter 2: Modeling in the Frequency Domain

Writing the equations of motion,

θ2(s) - θ3(s) = 4T(s)

-θ2(s) + (s+1)θ3(s) = 0

Solving for θ3(s),

θ 3

(s ) =

1 4 T (s )

-1 0

1 -1 -1 (s+ 1 )

= 4 T (s )

s

Hence, θ3(s) T(s) =

4 s . But, θ4(s) =

1 5 θ3(s). Thus,

θ4(s) T(s) =

4/5 s .

36. Reflecting impedances and applied torque to respective sides of the viscous damper yields the following

equivalent circuit:

Writing the equations of motion,

(s 2 + s)θ2 (s) − sθ3(s) = 10T (s) −sθ2 (s) + (s +1)θ3 (s) −θ4 (s) = 0 −θ3 (s) + (s +1)θ4 (s) = 0

Solving for θ4 (s) ,

Solutions to Problems 35

θ4 (s) =

s(s +1) −s 10T (s) −s (s +1) 0 0 −1 0

s(s +1) −s 0 −s (s +1) −1 0 −1 (s +1)

= s10T (s)

s(s +1) −s 0 −s (s +1) −1 0 −1 (s +1)

Thus,

θ4 (s) T(s)

= 10

s(s +1)2

But, θ L(s) = 5θ4 (s) . Hence,

θ4 (s) T(s)

= 50

s(s +1)2

37. Reflect all impedances on the right to the viscous damper and reflect all impedances and torques on the

left to the spring and obtain the following equivalent circuit:

Writing the equations of motion,

(J1eqs2+K)θ2(s) -Kθ3(s) = Teq(s)

-Kθ2(s)+(Ds+K)θ3(s) -Dsθ4(s) = 0

-Dsθ3(s) +[J2eqs2 +(D+Deq)s]θ4(s) = 0

where: J1eq = J2+(Ja+J1)(N2N1 ) 2

; J2eq = J3+(JL+J4)(N3N4 ) 2

; Deq = DL(N3N4 ) 2

; θ2(s) = θ1(s)

N1 N2 .

36 Chapter 2: Modeling in the Frequency Domain

38. Reflect impedances to the left of J5 to J5 and obtain the following equivalent circuit:

Writing the equations of motion,

[Jeqs2+(Deq+D)s+(K2+Keq)]θ5(s) -[Ds+K2]θ6(s) = 0

-[K2+Ds]θ5(s) + [J6s2+2Ds+K2]θ6(s) = T(s)

From the first equation, θ6(s) θ5(s)

= Jeqs2+(Deq+D)s+ (K2+Keq)

Ds+K2 . But, θ5(s) θ1(s)

= N1N3 N2N4 . Therefore,

θ6(s) θ1(s)

= N1N3 N2N4 ⎝⎜

⎛ ⎠⎟ ⎞

Jeqs2+(Deq+D)s+ (K2+Keq)

Ds+K2 ,

where Jeq = [J1(N4N2N3N1 ) 2

+ (J2+J3)(N4N3 ) 2

+ (J4+J5)], Keq = K1(N4N3 ) 2

, and

Deq = D[(N4N2N3N1 ) 2

+ (N4N3 ) 2

+ 1].

39. Draw the freebody diagrams,

Solutions to Problems 37

Write the equations of motion from the translational and rotational freebody diagrams,

(Ms2+2fv s+K2)X(s) -fvrsθ(s) = F(s)

-fvrsX(s) +(Js2+fvr2s)θ(s) = 0

Solve for θ(s),

θ(s) =

Ms2+2fvs+K2 F(s)

-fvrs 0

Ms2+2fvs+K2 -fvrs

-fvrs Js 2 +fvr

2s

= fvrF(s)

JMs 3 +(2Jfv+Mfvr

2)s2+(JK 2 +fv

2 r2)s+K

2 fvr

2

From which, θ(s) F(s) =

fvr JMs3+(2Jfv+Mfvr2)s2+(JK2+fv2r2)s+K2fvr2

.

40. Draw a freebody diagram of the translational system and the rotating member connected to the

translational system.

From the freebody diagram of the mass, F(s) = (s2+s+1)X(s). Summing torques on the rotating

member,

(Jeqs2 +Deqs)θ(s) + F(s)2 = Teq(s). Substituting F(s) above, (Jeqs2 +Deqs)θ(s) + (2s2+2s+2)X(s) =

Teq(s). However, θ(s) = X(s)

2 . Substituting and simplifying,

Teq = [(Jeq2 +2)s2 +( Deq

2 +2)s+2]X(s)

38 Chapter 2: Modeling in the Frequency Domain

But, Jeq = 1+1(4)2 = 17, Deq = 1(2)2 +1 = 5, and Teq(s) = 4T(s). Therefore, [ 212 s2 +92 s+2]X(s) =

4T(s). Finally, X(s) T(s) =

8 21

s2 + 9

21 s + 4

21 .

41. Writing the equations of motion,

(J1s2+K1)θ1(s) - K1θ2(s) = T(s) -K1θ1(s) + (J2s2+D3s+K1)θ2(s) +F(s)r -D3sθ3(s) = 0 -D3sθ2(s) + (J2s2+D3s)θ3(s) = 0

where F(s) is the opposing force on J2 due to the translational member and r is the radius of J2. But,

for the translational member,

F(s) = (Ms2+fvs+K2)X(s) = (Ms2+fvs+K2)rθ(s)

Substituting F(s) back into the second equation of motion,

(J1s2+K1)θ1(s) - K1θ2(s) = T(s)

-K1θ1(s) + [(J2 + Mr2)s2+(D3 + fvr2)s+(K1 + K2r2)]θ2(s) -D3sθ3(s) = 0

-D3sθ2(s) + (J2s2+D3s)θ3(s) = 0 Notice that the translational components were reflected as equivalent rotational components by the

square of the radius. Solving for θ2(s), θ2 (s) = K1(J3s

2 + D3s)T(s) ∆

, where ∆ is the

determinant formed from the coefficients of the three equations of motion. Hence,

θ2 (s) T(s)

= K1(J3s

2 + D3s) ∆

Since

X(s) = rθ2 (s), X(s) T (s)

= rK1(J3s

2 + D3s) ∆

42. Kt Ra

= Tstall Ea

= 100 50

= 2 ; Kb = Ea

50 150

= 1 3

Also,

Jm = 2+18( 13 ) 2

= 4; Dm = 2+36( 13 ) 2

= 6.

Thus, θm (s) Ea (s)

= 2 / 4

s(s + 1 4

(6 + 2 3

)) =

1/ 2

s(s + 5 3

)

Solutions to Problems 39

Since θL(s) = 1 3

θm(s),

θL(s) Ea (s)

=

1 6

s(s + 5 3

) .

43. The parameters are:

Kt Ra

= Ts Ea

= 5 5

= 1; Kb = Ea ω

= 5

600 π

2π 1 60

=

1 4

; Jm =16 1 4

⎛ ⎝

⎞ ⎠

2

+ 4 1 2

⎛ ⎝

⎞ ⎠

2

+1 = 3; Dm = 32 1 4

⎛ ⎝

⎞ ⎠

2

= 2

Thus,

θm (s) Ea (s)

=

1 3

s(s + 1 3

(2 + (1)(1 4

))) =

1 3

s(s + 0.75)

Since θ2(s) = 1 4

θm(s),

θ2 (s) Ea (s)

=

1 12

s(s + 0.75) .

44. The following torque-speed curve can be drawn from the data given:

v

T

100

50

500 1000

Therefore, Kt Ra

= Tstall Ea

= 100 10

= 10 ; Kb = Ea

10 1000

= 1

100 . Also, Jm = 5+100( 15 )

2 = 9;

Dm = 1. Thus,

40 Chapter 2: Modeling in the Frequency Domain

θm (s) Ea (s)

=

10 9

s(s + 1 9

(1 + 0.1)) =

10 9

s(s + 0.122) . Since θL(s) =

1 5

θm(s), θL(s) Ea (s)

=

10 45

s(s + 0.122) =

0.222 s(s + 0.122)

.

45. From Eqs. (2.45) and (2.46),

RaIa(s) + Kbsθ(s) = Ea(s) (1)

Also,

Tm(s) = KtIa(s) = (Jms2+Dms)θ(s). Solving for θ(s) and substituting into Eq. (1), and simplifying

yields

Ia (s) Ea (s)

= 1 Ra

(s + Dm Jm

)

s + Ra Dm + KbKt RaJm

(2)

Using Tm(s) = KtIa(s) in Eq. (2),

Tm(s) Ea (s)

= Kt Ra

(s + Dm Jm

)

s + Ra Dm + KbKt RaJm

46. For the rotating load, assuming all inertia and damping has been reflected to the load,

(JeqLs2+DeqLs)θL(s) + F(s)r = Teq(s), where F(s) is the force from the translational system, r=2 is

the radius of the rotational member, JeqL is the equivalent inertia at the load of the rotational load and

the armature, and DeqL is the equivalent damping at the load of the rotational load and the armature.

Since JeqL = 1(2)2 +1 = 5, and DeqL = 1(2)2 +1 = 5, the equation of motion becomes, (5s2+5s)θL(s)

+ F(s)r = Teq(s). For the translational system, (s2+s)X(s) = F(s). Since X(s) = 2θL(s), F(s) =

(s2+s)2θL(s). Substituting F(s) into the rotational equation, (9s2+9s) θL(s) = Teq(s). Thus, the

equivalent inertia at the load is 9, and the equivalent damping at the load is 9. Reflecting these back

to the armature, yields an equivalent inertia of 9 4 and an equivalent damping of

9 4 . Finally,

Kt Ra = 1;

Solutions to Problems 41

Kb = 1. Hence, θm(s) Ea(s) =

4 9

s(s+ 4 9(

9 4+1))

=

4 9

s(s+ 13 9 )

. Since θL(s) = 1 2 θm(s),

θL(s) Ea(s) =

2 9

s(s+ 13 9 )

. But

X(s) = rθL(s) = 2θL(s). therefore, X(s)

E (s)a =

4 9

s(s+ 13 9 )

.

47. The equations of motion in terms of velocity are:

[M1s + ( fv1 + fv 3) + K1 s

+ K2 s

]V1(s) − K2 s

V2(s) − fv 3V3(s) = 0

K2 s

V1(s) + [M2s + ( fv2 + fv4) + K2 s

]V2 (s) − fv4V3 (s) = F(s)

fv3V1 (s) − fv4V2 (s) + [M3s + fV3 + fv4]V3(S) = 0

For the series analogy, treating the equations of motion as mesh equations yields

In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries. For the parallel analogy, treating the equations of motion as nodal equations yields

In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.

42 Chapter 2: Modeling in the Frequency Domain

48. Writing the equations of motion in terms of angular velocity, Ω(s) yields

(J1s + D1 + K1 s

)Ω1(s) − (D1 + K1 s

)Ω2 (s) = T(s)

−(D1 + K1 s

)Ω1(s) + (J2s + D1 + (K1 + K2 )

s )Ω2 (s) = 0

K2 s

Ω2(s) − D2Ω3(s) + (D2 + K2 s

)Ω4 (s) = 0

(J3s + D2 + K3 s

)Ω3(s) − D2Ω4 (s) = 0

For the series analogy, treating the equations of motion as mesh equations yields

In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries. For the parallel analogy, treating the equations of motion as nodal equations yields

In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.

49.

An input r1 yields c1 = 5r1+7. An input r2 yields c2 = 5r2 +7. An input r1 +r2 yields, 5(r1+r2)+7 =

5r1+7+5r2 = c1+c2-7. Therefore, not additive. What about homogeneity? An input of Kr1 yields c =

5Kr1+7 ≠ Kc1. Therefore, not homogeneous. The system is not linear. 50.

a. Let x = δx+0. Therefore,

Solutions to Problems 43

δx . .

+3δx .

+2δx =sin (0+δx)

But, sin (0+δx) = sin 0 + d sinx

dx ⎮ x =0

δx = 0+cosx ⎮ x =0

δx = δx

Therefore, δx . .

+3δx .

+2δx = δx. C ollecting term s, δx . .

+3δx .

+δx = 0 .

b. Let x = δx+π. Therefore, δx ..

+3δx .

+2δx =sin (π+δx)

But, sin (π+δx) = sin π + d sinx

dx ⎮x=π δx = 0+cosx ⎮

x=π δx = −δx

Therefore, δx ..

+3δx .

+2δx = -δx. Collecting terms, δx ..

+3δx .

+3δx = 0 . 51.

If x = 0 + δx,

δx ...

+ 10δx ..

+ 31δx .

+ 30δx = e -(δx)

But e -(δx)

= e-0 + de -x

dx ⎮ x=0

δx = 1 - e-x ⎮ x=0

δx = 1 - δx

Therefore, δx ...

+ 10δx ..

+ 31δx .

+ 30δx =1 - δx, or, δx ...

+ 10δx ..

+ 31δx .

+ 31δx =1. 52.

The given curve can be described as follows:

f(x) = -4 ; -∞<x<-2;

f(x) = 2x; -2<x<2;

f(x) = 4; 2<x<+∞

Thus,

x.. + 15x. +50x = -4

x.. + 15x. + 50x = 2x, or x.. + 15x. +48x = 0

x.. + 15x. + 50x = 4

a.

b.

c. 53.

The relationship between the nonlinear spring’s displacement, xs(t) and its force, fs(t) is

xs(t) = 1 − e fs ( t)

Solving for the force,

fs (t) = − ln(1 − xs (t)) (1)

Writing the differential equation for the system by summing forces,

d2 x(t) dt2

+ dx(t)

dt − ln(1− x(t)) = f (t) (2)

44 Chapter 2: Modeling in the Frequency Domain

Letting x(t) = x0 + δx and f(t) = 1 + δf, linearize ln(1 – x(t)).

ln(1− x) − ln(1 − x0 ) = d ln(1 − x)

dx x =x 0 δx

Solving for ln(1 – x),

ln(1− x) = ln(1 − x0 ) − 1

1 − x x = x0 δx = ln(1− x0) −

1 1− x0

δx (3)

When f = 1, δx = 0. Thus from Eq. (1), 1 = -ln(1 – x0 ). Solving for x0, 1 – x0 = e-1 , or x0 = 0.6321.

Substituting x0 = 0.6321 into Eq. (3),

ln(1- x) = ln(1 – 0.6321) - 1

1- 0.6321 δx = -1 - 2.718δx

Placing this value into Eq. (2) along with x(t) = x0 + δx and f(t) = 1 + δf, yields the linearized

differential equation,

dx dt2

+ dδx dt

+ 1+ 2.718δx = 1 +δf

or

dx dt2

+ dδx dt

+ 2.718δx = δf Taking the Laplace transform and rearranging yields the transfer function,

δx(s) δf (s)

= 1

s2 + s + 2.718

54. First assume there are n plates without the top plate positioned at a displacement of y2(t) where

y2(t) = 0 is the position of the unstretched spring. Assume the system consists of mass M, where M is

the mass of the dispensing system and the n plates, viscous friction, fv, where the viscous friction

originates where the piston meets the sides of the cylinder, and of course the spring with spring

constant, K. Now, draw the freebody diagram shown in Figure (b) where Wn is the total weight of the

n dishes and the piston. If we now consider the current position, y2(0),

Solutions to Problems 45

Restaurant Plate Dispenser

the equilibrium point and define a new displacement, y1(t), which is measured from equilibrium, we

can write the force in the spring as Ky2(t) = Ky2(0) + Ky1(t). Changing variables from y2(t) to y1(t),

we sum forces and get,

M d2y1 dt2

+ fv dy1 dt + Ky1 + Ky2(0) + Wn = 0 (1)

where d2y2 dt2

= d2y1 dt2

and dy2 dt =

dy1 dt . But, Ky2(0) = -Wn , since it is the component of the spring

force that balances the weight at equilibrium when y1 = 0. Thus, the differential equation becomes,

M d2y1 dt2

+ fv dy1 dt + Ky1 = 0 (2)

When the top plate is added, the spring is further compressed by an amount, WD K , where WD is the

weight of the single dish, and K is the spring constant. We can think of this displacement as an initial

condition. Thus, y1(0-) = - WD K and

dy1 dt (0-) =0, and y1(t) = 0 is the equilibrium position of the

spring with n plates rather than the unstretched position. Taking the Laplace transform of equation

(2), using the initial conditions,

46 Chapter 2: Modeling in the Frequency Domain

M(s2Y1(s) + s WD K ) + fv(sY1(s) +

WD K ) + KY1(s) = 0 (3)

or

(Ms2 + fvs + K)Y1(s) = - WD K

(Ms + fv ) (4)

Now define a new position reference, Y(s), which is zero when the spring is compressed with the

initial condition,

Y(s) = Y1(s) + WD Ks (5)

or

Y1(s) = Y(s) - WD Ks (6)

Substituting Y1(s) in Equation (4), we obtain,

(Ms2 + fvs + K)Y(s) = WD

s = F(s) (7)

a differential equation that has an input and zero initial conditions. The schematic is shown in Figure

(c). Forming the transfer function, Y(s) F(s) , we show the final result in Figure (d), where for the

removal of the top plate, F(s) is always a step, F(s) = WD

s .

55. Writing the equations of motion,

(17.2s2 + 160s + 7000)Yf(s) – (130s + 7000)Yh(s) – 0Ycat(s) = Fup(s)

- (130s+7000)Yf(s) + (9.1ss + 130s + 89300)Yh(s) - 82300Ycat(s) = 0

- 0Yf(s) - 82300Yh(s) + 1.6173 x106 Ycat(s) = 0

These equations are in the form AY=F, where det(A) = 2.5314 x 108 (s2 + 15.47 s + 9283) (s2 +

8.119 s + 376.3)

Using Cramer’s rule:

Ycat (s) Fup(s)

= 0.04227(s + 53.85)

(s2 +15.47s + 9283)(s 2 + 8.119s + 376.3)

Yh(s) Fup(s)

= 0.8306(s + 53.85)

(s2 +15.47s + 9283)(s2 + 8.119s + 376.3)

Yh(s) − Ycat (s)

Fup(s) =

0.7883(s + 53.85) (s2 + 15.47s + 9283)(s2 +8.119s + 376.3)

T H R E E Modeling

in the Time Domain

SOLUTIONS TO CASE STUDIES CHALLENGES

Antenna Control: State-Space Representation

For the power amplifier, Ea(s) Vp(s) =

150 s+150 . Taking the inverse Laplace transform, ea

. +150ea =

150vp. Thus, the state equation is

ea •

= −150ea + 150vp

For the motor and load, define the state variables as x1 = θm and x2 = θ . m. Therefore,

x . 1 = x2 (1)

Using the transfer function of the motor, cross multiplying, taking the inverse Laplace transform,

and using the definitions for the state variables,

x . 2 = -

1 Jm (Dm+

KtKa Ra ) x2 +

Kt RaJm ea (2)

Using the gear ratio, the output equation is

y = 0.2x1 (3)

Also, Jm = Ja+5( 1 5 )

2 = 0.05+0.2 = 0.25, Dm = Da+3( 1 5 )

2 = 0.01+0.12 = 0.13, Kt

RaJm = 1

(5)(0.25)

= 0.8, and 1

Jm (Dm+ KtKa

Ra ) = 1.32. Using Eqs. (1), (2), and (3) along with the previous values, the

state and output equations are,

x. = 0 1 0 -1.32

x + 0

0.8 ea ; y = 0.2 0 x

48 Chapter 3: Modeling in the Time Domain

Aquifer:State-SpaceRepresentation

C1 dh1 dt = qi1-qo1+q2-q1+q21 = qi1-0+G2(h2-h1)-G1h1+G21(H1-h1) =

-(G2+G1+G21)h1+G2h2+qi1+G21H1

C2 dh2 dt = qi2-q02+q3-q2+q32 = qi2-qo2+G3(h3-h2)-G2(h2-h1)+0 = G2h1-[G2+G3]h2+G3h3+qi2-qo2

C3 dh3 dt = qi3-qo3+q4-q3+q43 = qi3-qo3+0-G3(h3-h2)+0 = G3h2-G3h3+qi3-qo3

Dividing each equation by Ci and defining the state vector as x = [h1 h2 h3]T

x .

=

−(G1 + G2 + G3 ) C1

G2 C1

0

G2 C2

−(G2 + G3) C2

G3 C2

0 G3 C3

G3 C3

⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥

x +

qi1 + G21H1 C1

qi 2 − qo2 C2

qi3 − qo3 C3

⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥

u(t)

y = 1 0 0 0 1 0 0 0 1

⎢ ⎢

⎥ ⎥ x

where u(t) = unit step function.

1. (1) Can model systems other than linear, constant coefficients; (2) Used for digital simulation

2. Yields qualitative insight

3. That smallest set of variables that completely describe the system

4. The value of the state variables

5. The vector whose components are the state variables

6. The n-dimensional space whose bases are the state variables

7. State equations, an output equation, and an initial state vector (initial conditions)

8. Eight

9. Forms linear combinations of the state variables and the input to form the desired output

10. No variable in the set can be written as a linear sum of the other variables in the set.

11. (1) They must be linearly independent; (2) The number of state variables must agree with the order of

the differential equation describing the system; (3) The degree of difficulty in obtaining the state equations

for a given set of state variables.

12. The variables that are being differentiated in each of the linearly independent energy storage elements

Solutions to Problems 49

13. Yes, depending upon the choice of circuit variables and technique used to write the system equations.

For example, a three -loop problem with three energy storage elements could yield three simultaneous

second-order differential equations which would then be described by six, first-order differential equations.

This exact situation arose when we wrote the differential equations for mechanical systems and then

proceeded to find the state equations.

14. The state variables are successive derivatives.

SOLUTIONS TO PROBLEMS

1. Add the branch currents and node voltages to the network.

Write the differential equation for each energy storage element.

di2 dt

= v1

di4 dt

= v2

dvo dt

= i5

Therefore, the state vector is x = i2 i4 vo

⎢ ⎢ ⎢

⎥ ⎥ ⎥

Now obtain v1, v2, and i5 in terms of the state variables. First find i1 in terms of the state variables.

vi + i1 + i3 + i5 + vo = 0 But i3 = i1 − i2 and i5 = i3 − i4. Thus, −vi + i1 + (i1 − i2 ) + (i3 − i4 ) + vo = 0 Making the substitution for i3 yields −vi + i1 + (i1 − i2 ) + ((i1 − i2 ) − i4 ) + vo = 0

Solving for i 1

50 Chapter 3: Modeling in the Time Domain

i1 = 2 3

i2 + 1 3

i4 − 1 3

vo + 1 3

vi

Thus,

v1 = vi i1 = − 2 3

i2 − 1 3

i4 + 1 3

vo + 2 3

vi

Also,

i3 = i1 − i2 = − 1 3

i2 + 1 3

i4 − 1 3

vo + 1 3

vi

and

i5 = i3 − i4 = − 1 3

i2 − 2 3

i4 − 1 3

vo + 1 3

vi

Finally,

v2 = i5 + vo = − 1 3

i2 − 2 3

i4 + 2 3

vo + 1 3

vi

Using v1, v2, and i5, the state equation is

x

=

− 2 3

− 1 3

1 3

−1 3

− 2 3

2 3

−1 3

− 2 3

−1 3

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

x +

2 3 1 3 1 3

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

vi

y = 0 0 1[ ]x

2. Add branch currents and node voltages to the schematic and obtain,

Write the differential equation for each energy storage element.

Solutions to Problems 51

dv1 dt

= i2

di3 dt

= vL

Therefore the state vector is x = v1 i3

⎣ ⎢ ⎢

⎦ ⎥ ⎥

Now obtain v and i , in terms of the state variables, L 2 vL = v1 − v2 = v1 − iR = v1 − (i3 + 4v1) = −3v1 − i3 i2 = i1 − i3 = (vi v1 ) − i3 = −v1 − i3 + vi Also, the output is

y = i = 4v + iR 1 3 Hence,

x

= −1 −1 −3 −1

⎣ ⎢

⎦ ⎥ x +

1 0

⎣ ⎢

⎦ ⎥ vi

y = 4 1[ ]x

3. Let C1 be the grounded capacitor and C 2 be the other. Now, writing the equations for the energy storage components yields,

diL dt

= vi vC1

dvC1 dt

= i1 − i2

dvC2 dt

= i2 − i3

(1)

Thus the state vector is x = iL vC1 vC2

⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥

. Now, find the three loop currents in terms of the state variables

and the input.

Writing KVL around Loop 2 yields vC = vC1 2 + i2

i2

.Or,

= vC1 − vC

3

2

Writing KVL around the outer loop yields i + i2 = vi Or,

i3 = vi i2 = vi vC1 + vC2

52 Chapter 3: Modeling in the Time Domain

Also, i1 3− i = iL . Hence,

i1 = iL + i3 = iL + vi vC1 + vC2

Substituting the loop currents in equations (1) yields the results in vector-matrix form,

diL dt

dvC1 dt

dvC2 dt

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

= 0 −1 0 1 −2 2 0 2 −2

⎢ ⎢ ⎢

⎥ ⎥ ⎥

iL vC1 vC2

⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥

+ 1 1 −1

⎢ ⎢ ⎢

⎥ ⎥ ⎥ vi

Since vo = i2 = vC −1 vC , the output equation is 2

y = 0 1 1[ ] iL vC1 vC2

⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥

4.

Equations of motion in Laplace:

(s 2 + 3s +1)x1(s) − (s +1)x2 (s) − sx3 (s) = 0

−(s +1)x1(s) + (s 2 + 2s +1)x2 (s) − sx3 (s) = F(s)

sx1(s) − sx2(s) + (s 2 + 3s)x3(s) = 0

Equations of motion in the time domain: d2 x1 dt 2

+ 3 dx1 dt

+ x1 − dx2 dt

x2 − dx3 dt

= 0

d2 x2 dt2

+ 2 dx2 dt

+ x2 − dx1 dt

x1 − dx3 dt

= f (t)

d2 x3 dt2

+ 3 dx3 dt

+ x1 − dx2 dt

dx1 dt

= 0

Define state variables:

z1 = x1 or x1 = z1 (1)

z2 = dx1 dt

or dx1 dt

= z2 (2)

z3 = x2 or x2 = z3 (3)

z4 = dx2 dt

or dx2 dt

= z4 (4)

z5 = x3 or x3 = z5 (5)

z6 = dx3 dt

or dx3 dt

= z5 (6)

Solutions to Problems 53

Substituting Eq. (1) in (2), (3) in (4), and (5) in (6), we obtain, respectively:

dz1 dt

= z2 (7)

dz3 dt

= z4 (8)

dz5 dt

= z6 (9)

Substituting Eqs. (1) through (6) into the equations of motion in the time domain and solving for the

derivatives of the state variables and using Eqs. (7) through (9) yields the state equations:

dz1 dt

= z2

dz2 dt

= −z1 − 3z2 + z3 + z4 + z6

dz3 dt

= z4

dz4 dt

= z1 + z2 − z3 − 2z4 + z6 + f (t)

dz5 dt

= z6

dz6 dt

= z2 + z4 − 3z6

The output is x3 = z5. In vector-matrix form:

z .

=

0 1 0 0 0 0 −1 −3 1 1 0 1 0 0 0 1 0 0 1 1 −1 −2 0 1 0 0 0 0 0 1 0 1 0 1 0 −3

⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥

z +

0 0 0 1 0 0

⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥

f (t)

y = 0 0 0 0 1 0[ ]z

5. Writing the equations of motion,

54 Chapter 3: Modeling in the Time Domain

(s 2 + 2s +1)X1(s) − sX2(s) − (s + 1)X3(s) = 0

sX1(s) + (s 2 + 2s +1)X2 (s) − (s +1)X3 (s) = 0

−(s +1)X1(s) − (s +1)X2 (s) + (s 2 + 2s + 2)X3 (s) = F(s)

Taking the inverse Laplace transform,

x1 ••

+ 2 x1 •

+ x1 − x2 •

x3 •

x3 = 0

x1 •

+ x2 ••

+ 2 x2 •

+ x2 − x3 •

x3 = 0

x1 •

x1 − x2 •

x2 + x3 ••

+ 2 x3 •

+ 2x3 = f (t)

Simplifying,

x1 ••

= −2 x1 •

x1 + x2 •

+ x3 •

+ x3

x2 ••

= x1 •

− 2 x2 •

x2 + x3 •

+ x3

x3 ••

= x1 • + x1 + x2

• + x2 − 2 x3

• − 2x3 + f (t)

Defining the state variables,

z1 = x1; z2 = x1 • ; z3 = x2 ; z4 = x2

• ; z5 = x3; z6 = x3

Writing the state equations using the simplified equations above yields,

z1 •

= x1 •

= z2

z2 •

= x1 ••

= −2z2 − z1 + z4 + z6 + z5

z3 •

= x2 •

= z4

z4 •

= x2 ••

= z2 − 2z4 − z3 + z6 + z5

z5 •

= x3 •

= z6

z6 = •

x3 ••

= z2 + z1 + z4 + z3 − 2z6 − 2z5 + f (t )

Converting to vector-matrix form,

Solutions to Problems 55

z

=

0 1 0 0 0 0 −1 −2 0 1 1 1 0 0 0 1 0 0 0 1 −1 −2 1 1 0 0 0 0 0 1 1 1 1 1 −2 −2

⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥

z +

0 0 0 0 0 1

⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥

f (t)

y = 1 0 0 0 0 0[ ]z

6.

Drawing the equivalent network,

Writing the equations of motion,

(100s2 +100)θ2 −100θ3 = 10T

−100θ2 + (100s 2 + 100s +100)θ3 = 0

Taking the inverse Laplace transform and simplifying,

θ2 ••

+ θ2 −θ3 = 1

10 T

−θ2 +θ3 ••

+ θ3 •

+θ3 = 0

Defining the state variables as

x1 = θ2 , x2 =θ2 •

, x3 = θ3, x4 = θ3 •

Writing the state equations using the equations of motion and the definitions of the state variables

56 Chapter 3: Modeling in the Time Domain

x1 •

= x2

x2 •

= θ2 ••

= −θ2 + θ3 + 1

10 T = −x1 + x3 +

1 10

T

x3 •

= x4

x4 •

= θ3 ••

= θ2 −θ3 −θ3 •

= x1 − x3 − x4 y = 10θ2 =10x1

,

In vector-matrix form,

x

=

0 1 0 0 −1 0 1 0 0 0 0 1 1 0 −1 −1

⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥

x +

0 1

10 0 0

⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥

T

y = 10 0 0 0[ ]x

7. Drawing the equivalent circuit,

Writing the equations of motion,

42θ2(s) − 2θ3 (s) = 20T(s) −2θ2 (s) + (3s + 2)θ3(s) − 3sθ4 (s) = 0 −3sθ3 (s) + 5sθ4 (s) = 0

Taking the inverse Laplace transform,

42θ2 (t) − 2θ3(t) = 20T (t) (1)

−2θ2 (t) + 3θ3 •

(t) + 2θ3 −3θ4 •

(t) = 0 (2)

−3θ3 •

(t) + 5θ4 •

(t) = 0 (3) From (3),

Solutions to Problems 57

θ3 •

(t) = 5 3

θ4 •

(t) and θ3(t) = 5 3

θ4 (t) (4)

assuming zero initial conditions.

From (1)

θ2 (t) = 1

21 θ3 (t) +

10 21

T(t ) = 5

63 θ4 (t) +

10 21

T (t) (5)

Substituting (4) and (5) into (2) yields the state equation (notice there is only one equation),

θ4 •

(t) = − 100 63

θ4( t) + 10 21

T(t)

The output equation is given by,

θL ( t) = 1

10 θ4 (t)

8.

Solving Eqs. (3.44) and (3.45) in the text for the transfer functions X1(s) F(s)

and X2 (s) F(s)

:

X 1 s

0 KF M 2 s2 K+

M 1 s2 D s K+ + KKM 2 s2 K+

= and X 2 s

M 1 s2 D s K+ + 0 KF

M 1 s2 D s K+ + KKM 2 s2 K+

=

Thus, X 1 s

F s K

M 2 M 1 s 4 D M 2 s

3 K M 2 s 2 K M 1 s

2 D K s+ + + + =

and

X 2 s F s

M 1 s 2 D s K+ +

M 2 M 1 s 4 D M 2 s

3 K M 2 s 2 K M 1 s

2 D K s+ + + + =

Multiplying each of the above transfer functions by s to find velocity yields pole/zero cancellation at

the origin and a resulting transfer function that is third order.

9. a. . Using the standard form derived in the textbook,

x

=

0 1 0 0 0 0 1 0 0 0 0 1

−100 −7 −10 −20

⎢ ⎢ ⎢

⎥ ⎥ ⎥

x +

0 0 0 1

⎢ ⎢ ⎢

⎥ ⎥ ⎥

r(t)

c = 100 0 0 0[ ]x

b. Using the standard form derived in the textbook,

58 Chapter 3: Modeling in the Time Domain

x

=

0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1

−30 −1 −6 −9 −8

⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥

x +

0 0 0 0 1

⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥

r(t)

c = 30 0 0 0 0[ ]x

10.Program: 'a' num=100; den=[1 20 10 7 100]; G=tf(num,den) [Acc,Bcc,Ccc,Dcc]=tf2ss(num,den); Af=flipud(Acc); A=fliplr(Af) B=flipud(Bcc) C=fliplr(Ccc) 'b' num=30; den=[1 8 9 6 1 30]; G=tf(num,den) [Acc,Bcc,Ccc,Dcc]=tf2ss(num,den); Af=flipud(Acc); A=fliplr(Af) B=flipud(Bcc) C=fliplr(Ccc) Computer response: ans = a Transfer function: 100 --------------------------------- s^4 + 20 s^3 + 10 s^2 + 7 s + 100 A = 0 1 0 0 0 0 1 0 0 0 0 1 -100 -7 -10 -20 B = 0 0 0 1 C = 100 0 0 0 ans = b

Solutions to Problems 59

Transfer function: 30 ------------------------------------ s^5 + 8 s^4 + 9 s^3 + 6 s^2 + s + 30 A = 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 -30 -1 -6 -9 -8 B = 0 0 0 0 1 C = 30 0 0 0 0

11.

a. Using the standard form derived in the textbook,

x

=

0 1 0 0 0 0 1 0 0 0 0 1

−10 −5 −1 −2

⎢ ⎢ ⎢

⎥ ⎥ ⎥

x +

0 0 0 1

⎢ ⎢ ⎢

⎥ ⎥ ⎥

r(t )

c = 10 5 0 0 0[ ]x

b. Using the standard form derived in the textbook,

x

=

0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 −8 −10 −9

⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥

x +

0 0 0 0 1

⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥

r(t)

c = 3 7 12 2 1[ ]x

12.

Program: 'a' num=[5 10]; den=[1 2 1 5 10] G=tf(num,den) [Acc,Bcc,Ccc,Dcc]=tf2ss(num,den); Af=flipud(Acc); A=fliplr(Af) B=flipud(Bcc)

60 Chapter 3: Modeling in the Time Domain

C=fliplr(Ccc) 'b' num=[1 2 12 7 3]; den=[1 9 10 8 0 0] G=tf(num,den) [Acc,Bcc,Ccc,Dcc]=tf2ss(num,den); Af=flipud(Acc); A=fliplr(Af) B=flipud(Bcc) C=fliplr(Ccc) Computer response: ans = a den = 1 2 1 5 10 Transfer function: 5 s + 10 ---------------------------- s^4 + 2 s^3 + s^2 + 5 s + 10 A = 0 1 0 0 0 0 1 0 0 0 0 1 -10 -5 -1 -2 B = 0 0 0 1 C = 10 5 0 0 ans = b den = 1 9 10 8 0 0 Transfer function: s^4 + 2 s^3 + 12 s^2 + 7 s + 3 ------------------------------ s^5 + 9 s^4 + 10 s^3 + 8 s^2 A = 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 -8 -10 -9

Solutions to Problems 61

B = 0 0 0 0 1 C = 3 7 12 2 1

13.

The transfer function can be represented as a block diagram as follows:

1 s3 + 6s2 + 9s + 4

R(s) X(s) Y(s) s 2 + 3s + 7

Writing the differential equation for the first box:

6 9 4 ( )x x x x r t ••• •• •

+ + + =

Defining the state variables:

1

2

3

x x

x x

x x

••

=

=

=

Thus,

1 2

2 3

3 14 9 6 ( ) 4 9 6 (

x x

x x

2 3 )x x x x r t x x x r t

• • ••

=

=

= − − − + = − − − +

From the second box,

1 23 7 7 3y x x x x x x •• •

3= + + = + + In vector-matrix form:

[ ]

0 1 0 0 0 0 1 0 ( 4 9 6 1

7 3 1

r t

y

• ⎡ ⎤ ⎡ ⎢ ⎥ ⎢= +⎢ ⎥ ⎢ ⎢ ⎥ ⎢− − −⎣ ⎦ ⎣

=

x x

x

) ⎤ ⎥ ⎥ ⎥⎦

14.

62 Chapter 3: Modeling in the Time Domain

a. G(s)=C(sI-A)-1B

A = 0 1 0 0 0 1 -3 -2 -5

; B = 0 0 10

; C = 1 0 0

(sI - A) -1

= 1

s3 + 5s 2 + 2s +3

s2+5s+2 s+5 1

-3 s(s+5) s

-3s -2s-3 s2

Therefore, G(s) = 10

s3+5s2+2s+3 .

b. G(s)=C(sI-A)-1B

A 2 3 8− 0 5 3 3− 5− 4−

= ; B 1 4 6

= ; C 1 3 6, ,=

s I A− 1− 1 s3 3 s2− 27 s− 157+

s2 s− 5− 3 s 52+ 8 s− 49+ 9− s2 2 s 32−+ 3 s 6−

3 s− 15+ 5 s− 1+ s2 7 s− 10+ =

Therefore, G s 49 s 2 349 s− 452+

s3 3 s2− 27 s− 157+ = .

c. G(s)=C(sI-A)-1B

A = 3 −5 2 1 −8 7

−3 −6 2

⎢ ⎢

⎥ ⎥ ; B =

5 −3 2

⎢ ⎢

⎥ ⎥ ; C = 1 −4 3[ ]

(sI A)−1 = 1

s3 + 3s2 + 19s −133

(s2 + 6s + 26) −(5s + 2) (2s −19) (s − 23) (s2 − 5s +12) (7s −19)

−(3s + 30) −(6s − 33) (s2 + 5s −19)

⎢ ⎢

⎥ ⎥

Therefore, G(s) = 23s2 − 48s − 7

s3 + 3s2 +19s −133 .

15.

Program: 'a' A=[0 1 3 0;0 0 1 0;0 0 0 1;-7 -9 -2 -3]; B=[0;5;8;2]; C=[1 3 4 6]; D=0; statespace=ss(A,B,C,D)

Solutions to Problems 63

[num,den]=ss2tf(A,B,C,D); G=tf(num,den) 'b' A=[3 1 0 4 -2;-3 5 -5 2 -1;0 1 -1 2 8;-7 6 -3 -4 0;-6 0 4 -3 1]; B=[2;7;6;5;4]; C=[1 -2 -9 7 6]; D=0; statespace=ss(A,B,C,D) [num,den]=ss2tf(A,B,C,D); G=tf(num,den) Computer response:

ans = a a = x1 x2 x3 x4 x1 0 1 3 0 x2 0 0 1 0 x3 0 0 0 1 x4 -7 -9 -2 -3 b = u1 x1 0 x2 5 x3 8 x4 2 c = x1 x2 x3 x4 y1 1 3 4 6 d = u1 y1 0 Continuous-time model. Transfer function: 59 s^3 - 164 s^2 - 1621 s - 260 ------------------------------- s^4 + 3 s^3 + 2 s^2 + 30 s + 7 ans = b a = x1 x2 x3 x4 x5 x1 3 1 0 4 -2 x2 -3 5 -5 2 -1 x3 0 1 -1 2 8 x4 -7 6 -3 -4 0 x5 -6 0 4 -3 1 b = u1 x1 2

64 Chapter 3: Modeling in the Time Domain

x2 7 x3 6 x4 5 x5 4 c = x1 x2 x3 x4 x5 y1 1 -2 -9 7 6 d = u1 y1 0 Continuous-time model. Transfer function: -7 s^4 - 408 s^3 + 1708 s^2 + 1.458e004 s + 2.766e004 ----------------------------------------------------- s^5 - 4 s^4 - 32 s^3 + 148 s^2 - 1153 s - 4480

16.

Program: syms s 'a' A=[0 1 3 0 0 0 1 0 0 0 0 1 -7 -9 -2 -3]; B=[0;5;8;2]; C=[1 3 4 6]; D=0; I=[1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1]; 'T(s)' T=C*((s*I-A)^-1)*B+D; T=simple(T); pretty(T) 'b' A=[3 1 0 4 -2 -3 5 -5 2 -1 0 1 -1 2 8 -7 6 -3 -4 0 -6 0 4 -3 1]; B=[2;7;6;5;4]; C=[1 -2 -9 7 6]; D=0; I=[1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1]; 'T(s)' T=C*((s*I-A)^-1)*B+D; T=simple(T); pretty(T) Computer response: ans = a

Solutions to Problems 65

ans = T(s)

2 3 -164 s - 1621 s - 260 + 59 s ------------------------------ 4 3 2 s + 3 s + 2 s + 30 s + 7 ans = b ans = T(s) 2 3 4 14582 s + 1708 s - 408 s - 7 s + 27665 ------------------------------------------ 5 4 3 2 s - 4 s - 32 s + 148 s - 1153 s - 4480

17.

Let the input be dθz dt =ωz, x1=θx , x2=θ

. x . Therefore,

x . 1 = x2

x . 2 = -

Kx Jx x1 -

Dx Jx x2 + Jωωz

The output is θx.

In vector-matrix form, θx = x1 . Therefore, y = x1.

x. =

0 1

- K x Jx

- D x Jx

x + 0

Jω ω z

y = 1 0 x

18.

The equivalent cascade transfer function is as shown below.

Ka K

3

s3+ K

2 K

3 s2+

K 1

K 3 s+

K 0

K 3

s+ K

b Ka

d (s) X(s) F (s)

66 Chapter 3: Modeling in the Time Domain

For the first box, x ...

+ K

2 K

3 x..+

K 1

K 3

x. + K

0 K

3 x =

Ka K

3 δ(t).

Selecting the phase variables as the state variables: x

1 =x, x

2 =x. , x

3 =x...

Writing the state and output equations:

x . 1 = x2

x . 2 = x3

x . 3 = -

K0 K3 x1-

K1 K3 x2-

K2 K3 x3+

Ka K3 δ(t)

y = φ(t) = x . +

Kb Ka x =

Kb Ka x1+x2

In vector-matrix form,

x. =

0 1 0 0 0 1

- K

0 K

3 -

K 1

K 3

- K

2 K

3

x+

0 0

Ka K

3

δ(t) ; y = K

b Ka

1 0 x

19.

Since Tm = Jeq dωm

dt + Deqωm, and Tm = Kt ia,

Jeq dωm

dt + Deqωm = Kt ia (1)

Or, dωm

dt = - Deq Jeq ωm +

Kt Jeq ia

But, ωm = N2 N1 ωL.

Substituting in (1) and simplifying yields the first state equation,

dωL dt = -

Deq Jeq ωL +

Kt Jeq

N1 N2 ia

The second state equation is: dθL dt = ωL

Since

ea = Raia+La dia dt +Kbωm = Raia+La

dia dt +Kb

N2 N1 ωL,

the third state equation is found by solving for dia dt . Hence,

dia dt = -

Kb La

N2 N1 ωL -

Ra La ia+

1 La ea

Solutions to Problems 67

Thus the state variables are: x1 = ωL, x2 = θL , and x3 = ia.

Finally, the output is y = θm = N2 N1 θL .

In vector-matrix form,

x. =

- Deq Jeq

0 Kt Jeq

N

1 N

2

1 0 0

- K

b La

N

2 N

1 0 -

Ra La

x +

0 0 1 La

ea ; y = 0 N

2 N

1 0 x

where,

x =

ω L

θ L

ia 20.

Writing the differential equations,

d2x1 dt2

+ dx1 dt + 2x1

2 - dx2 dt = 0

d2x2 dt2

+ dx2 dt -

dx1 dt = f(t)

Defining the state variables to be x1, v1, x2, v2, where v's are velocity,

x . 1 = v1

x . 2 = v2

v . 1 = -v1-2x12+v2

v . 2 = v1-v2+f(t)

Around x1 = 1, x1 = 1+δx1, and x . 1 = δ x

. 1 . Also,

x 1 2 = x

1 2

⎮ x=1

+ dx

1 dt ⎮

x =1 δx

1 = 1+2x

1 ⎮ x=1

δx 1

= 1+2δx 1

Therefore, the state and output equations are,

δx . 1 = v1

x . 2 = v2

v . 1 = -v1-2(1+2δx1)+v2

68 Chapter 3: Modeling in the Time Domain

v . 2 = v1-v2+f(t)

y = x2 In vector-matrix form,

δx 1

.

x. 2

v. 1

v. 2

=

0 0 1 0 0 0 0 1 -4 0 -1 1 0 0 1 -1

δx 1

x 2

v 1

v 2

+

0 0 0 0 -2 0 0 1

1 f(t)

; y = 0 1 0 0

δx 1

x 2

v 1

v 2

where f(t) = 2 + δf(t), since force in nonlinear spring is 2 N and must be balanced by 2 N force on

damper. 21.

Controller:

The transfer function can be represented as a block diagram as follows:

Rc(s) Xc(s) Yc(s)K1

s + b s+ a

Writing the differential equation for the first box,

K1 s + b

and solving for cx

,

1 ( )c c cx bx K r t

= − + From the second box,

1

1

( ) ( ) ( )

c c c c c

c c

y x ax bx K r t ax a b x K r t

= + = − + + = − +

c

Wheels:

The transfer function can be represented as a block diagram as follows:

Rw(s) Xw(s)c s + c

Solutions to Problems 69

Writing the differential equation for the block of the form, c

s + c

and solving for wx

,

( )w w wx cx cr t

= − + The output equation is,

yw = xw Vehicle:

The transfer function can be represented as a block diagram as follows:

Rv(s) Xv(s)1 s

Writing the differential equation for the block, 1 s

and solving for vx

,

( )v vx r t

= The output equation is

yv = xv 22.

A 1.702− 50.72 263.38 0.22 1.418− 31.99−

0 0 14− = ; B

272.06− 0 14

=

For G1(s), C 1 1 0 0, ,= , and

G1(s) = C1(sI-A)-1B

Thus,

G 1 C 1 1 s 3 17.12 s 2 34.935 s 122.43−+ +

s 2 15.418 s 19.852+ + 50.72 s 710.08+ 263.38 s 1249.1−

0.22 s 3.08+ s 2 15.702 s 23.828+ + 31.99 s− 3.4966+

0 0 s 2 3.12 s 8.745−+

B=s

Or

G 1 s 272.06 s 2− 507.3 s− 22888−

s 3 17.12 s 2 34.935 s 122.43−+ + = = 272.06 s

2 1.8647 s 84.128+ +− s 14+ s 1.7834− s 4.9034+

For G2(s), C2 = (0,1,0), and

70 Chapter 3: Modeling in the Time Domain

G2(s) = C2(sI-A)-1B

Thus,

G 2 s C 2 1 s 3 17.12 s 2 34.935 s 122.43−+ +

s 2 15.418 s 19.852+ + 50.72 s 710.08+ 263.38 s 1249.1−

0.22 s 3.08+ s 2 15.702 s 23.828+ + 31.99 s− 3.4966+

0 0 s 2 3.12 s 8.745−+

B=

Or G 2 s

507.71 s− 788.99−

s 3 17.12 s 2 34.935 s 122.43−+ + = = 507.71 s 1.554+−

s 14+ s 1.7834− s 4.9034+

23. Adding displacements to the figure,

xe

z

xsxr

Writing the differential equations for noncontact,

d2 xr dt2

+ 2 dxr dt

+ 2xr xs dxs dt

= u(t)

dxr dt

xr + d2 xs dt2

+ dxs dt

+ xs = 0

Define the state variables as,

x1 = xr ; x2 = xr • ; x3 = xs ; x4 = xs

Writing the state equations, using the differential equations and the definition of the state variables,

we get,

x1 •

= xr

= x2

x2 •

= xr ••

= −2x1 − 2x2 + x3 + x4 + u(t)

x3 •

= xs

= x4

x4 •

= xs ••

= x1 + x2 − x3 − x4

Solutions to Problems 71

Assuming the output to be xs, the output equation is,

y = x3 In vector-matrix form,

x

=

0 1 0 0 −2 −2 1 1 0 0 0 1 1 1 −1 −1

⎢ ⎢ ⎢

⎥ ⎥ ⎥

x +

0 1 0 0

⎢ ⎢ ⎢

⎥ ⎥ ⎥

u(t)

y = 0 0 1 0[ ]x

Writing the differential equations for contact,

d2 xr dt2

+ 2 dxr dt

+ 2xr xs dxs dt

= u(t)

dxr dt

xr + d2 xs dt2

+ dxs dt

+ xs z xe = 0

xs + dz dt

+ z dxe dt

= 0

xs dz dt

+ d2 xe dt 2

+ 2 dxe dt

+ 2xe = 0

Defining the state variables,

x1 = xr ; x2 = xr • ; x3 = xs ; x4 = xs

• ; x5 = z; x6 = z

• ; x7 = xe; x8 = xe

Using the differential equations and the definitions of the state variables, we write the state equations.

x1 •

= x2

x2 •

= −x1 − 2x2 + x3 + x4 +u(t)

x3 •

= x4

x4 •

= x1 + x2 − x3 − x4 + x5 + x7

x5 •

= x6

Differentiating the third differential equation and solving for d2z/dt2 we obtain,

x6 •

= d2z dt 2

= dxs dt

dz dt

+ d2 xe dt2

But, from the fourth differential equation,

d2 xe dt2

= xs + dz dt

− 2 dxe dt

− 2xe = x3 + x6 − 2x8 − 2x7

72 Chapter 3: Modeling in the Time Domain

Substituting this expression back into along with the other definitions and then simplifying yields, x6 •

x6 •

= x4 + x3 − 2x8 − 2x7 Continuing,

x7 •

= x8

x8 •

= x3 + x6 − 2x7 − 2x8

Assuming the output is xs,

y = xs Hence, the solution in vector-matrix form is

x

=

0 1 0 0 0 0 0 0 −1 −2 1 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 −1 −1 1 0 1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 −2 −2 0 0 0 0 0 0 0 1 0 0 1 0 0 1 −2 −2

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

x +

0 1 0 0 0 0 0 0

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

u(t)

y = 0 0 1 0 0 0 0 0[ ]x

24. Writing the equations of motion,

Mf d2y f dt 2

+ ( fvf + fvh ) dyf dt

+ Kh yf fvh dyh dt

Kh yh = fup(t)

fvh dyf dt

Khy f + Mh d 2yh dt2

+ fvh dyh dt

+ (Kh + Ks )yh Ks ycat = 0

Ks yh + (Ks + Kave )ycat = 0

The last equation says that

ycat = Ks

(Ks + Kave ) yh

Defining state variables for the first two equations of motion,

x1 = yh ; x2 = yh

; x3 = yf ; x4 = y f

Solving for the highest derivative terms in the first two equations of motion yields,

Solutions to Problems 73

d2 yf dt2

= − ( fvf + fvh )

M f

dyf dt

Kh M f

yf + f vh Mf

dyh dt

+ Kh Mf

yh + 1

Mf f up(t)

d2 yh dt2

= f vh Mh

dy f dt

+ Kh Mh

yf f vh Mh

dyh dt

− (Kh + Ks ) Mh

yh + Ks Mh

ycat

Writing the state equations,

x1 •

= x2

x2 •

= f vh Mh

x4 + Kh Mh

x3 − f vh Mh

x2 − (Kh + Ks )

Mh x1 +

Ks Mh

Ks (Ks + Kave )

x1

x3 •

= x4

x4 •

= − ( f vf + f vh )

Mf x4 −

Kh Mf

x3 + fvh

M f x2 +

Kh Mf

x1 + 1

Mf fup (t)

The output is yh - ycat. Therefore,

y = yh ycat = yh Ks

(Ks + Kave ) yh =

Kave (Ks + Kave )

x1

Simplifying, rearranging, and putting the state equations in vector-matrix form yields,

x •

=

0 1 0 0 1

Mh Ks

2

(Ks + Kave ) − (Kh + Ks )

⎝ ⎜ ⎞

⎠ ⎟ − fvh

Mh Kh Mh

fvh Mh

0 0 0 1 Kh M f

f vh Mf

Kh Mf

− ( f vf + fvh )

Mf

⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥

x +

0 0 0 1

Mf

⎢ ⎢ ⎢ ⎢

⎥ ⎥ ⎥ ⎥

fup (t)

y = Kave

(Ks + Kave ) 0 0 0

⎣ ⎢ ⎤

⎦ ⎥ x

Substituting numerical values,

x

=

0 1 0 0 −9353 −14.29 769.2 14.29

0 0 0 1 406 7.558 −406 −9.302

⎢ ⎢ ⎢

⎥ ⎥ ⎥

x +

0 0 0

0.0581

⎢ ⎢ ⎢

⎥ ⎥ ⎥

fup(t)

y = 0.9491 0 0 0[ ]x

F O U R Time Response

SOLUTIONS TO CASE STUDIES CHALLENGES

Antenna Control: Open-Loop Response The forward transfer function for angular velocity is,

G(s) = ω0(s) VP(s) =

24 (s+150)(s+1.32)

a. ω0(t) = A + Be-150t + Ce-1.32t

b. G(s) = 24

s2+151.32s+198 . Therefore, 2ζωn =151.32, ωn = 14.07, and ζ = 5.38.

c. ω0(s) = 24

s(s2+151.32s+198) =

Therefore, ω0(t) = 0.12121 + .0010761 e-150t - 0.12229e-1.32t.

d. Using G(s),

ω0 ••

+ 151.32ω0 •

+ 198ω 0 = 24vp (t) Defining,

x1 = ω0

x2 = ω0 •

Thus, the state equations are,

x1 •

= x2

x2 •

= −198x1 −151.32x2 + 24vp(t) y = x1

In vector-matrix form,

x

= 0 1

−198 −151.32 ⎡ ⎣ ⎢

⎤ ⎦ ⎥ x +

0 24

⎡ ⎣ ⎢

⎤ ⎦ ⎥ vp (t); y = 1 0[ ]x

Solutions to Case Studies Challenges 75

e. Program: 'Case Study 1 Challenge (e)' num=24; den=poly([-150 -1.32]); G=tf(num,den) step(G) Computer response: ans = Case Study 1 Challenge (e) Transfer function: 24 ------------------- s^2 + 151.3 s + 198

Ship at Sea: Open-Loop Response

a. Assuming a second-order approximation: ωn2 = 2.25, 2ζωn = 0.5. Therefore ζ = 0.167, ωn = 1.5.

Ts = 4

ζωn = 16; TP =

π ωn 1-ζ2

= 2.12 ;

%OS = e-ζπ / 1 - ζ 2 x 100 = 58.8%; ωnTr = 1.169 therefore, Tr = 0.77.

b. θ s 2.25 s s 2 0.5 s 2.25+ +

= = 1 s

s 0.5+

s 2 0.5 s 2.25+ + −

= 1 s

s 0.25+ 0.25 2.1875

2.1875+

s 0.25+ 2 2.1875+ −

76 Chapter 4: Time Response

= 1 s

s 0.25+ 0.16903 1.479+

s 0.25+ 2 2.1875+ −

Taking the inverse Laplace transform,

θ(t) = 1 - e-0.25t ( cos1.479t +0.16903 sin1.479t)

c. Program: 'Case Study 2 Challenge (C)' '(a)' numg=2.25; deng=[1 0.5 2.25]; G=tf(numg,deng) omegan=sqrt(deng(3)) zeta=deng(2)/(2*omegan) Ts=4/(zeta*omegan) Tp=pi/(omegan*sqrt(1-zeta^2)) pos=exp(-zeta*pi/sqrt(1-zeta^2))*100 t=0:.1:2; [y,t]=step(G,t); Tlow=interp1(y,t,.1); Thi=interp1(y,t,.9); Tr=Thi-Tlow '(b)' numc=2.25*[1 2]; denc=conv(poly([0 -3.57]),[1 2 2.25]); [K,p,k]=residue(numc,denc) '(c)' [y,t]=step(G); plot(t,y) title('Roll Angle Response') xlabel('Time(seconds)') ylabel('Roll Angle(radians)') Computer response: ans = Case Study 2 Challenge (C) ans = (a)

Transfer function: 2.25 ------------------ s^2 + 0.5 s + 2.25 omegan = 1.5000 zeta = 0.1667 Ts = 16

Solutions to Case Studies Challenges 77

Tp = 2.1241 pos = 58.8001 Tr = 0.7801 ans = (b) K = 0.1260 -0.3431 + 0.1058i -0.3431 - 0.1058i 0.5602 p = -3.5700 -1.0000 + 1.1180i -1.0000 - 1.1180i 0 k = [] ans = (c)

78 Chapter 4: Time Response

1.Time constant

2. The time for the step response to reach 67% of its final value

3. The input pole

4. The system poles

5. The radian frequency of a sinusoidal response

6. The time constant of an exponential response

7. Natural frequency is the frequency of the system with all damping removed; the damped frequency of

oscillation is the frequency of oscillation with damping in the system.

8. Their damped frequency of oscillation will be the same.

9. They will all exist under the same exponential decay envelop.

10. They will all have the same percent overshoot and the same shape although differently scaled in time.

11. ζ, ωn, TP, %OS, Ts

12. Only two since a second-order system is completely defined by two component parameters

13. (1) Complex, (2) Real, (3) Multiple real

14. Pole's real part is large compared to the dominant poles, (2) Pole is near a zero

15. If the residue at that pole is much smaller than the residues at other poles

16. No; one must then use the output equation

17. The Laplace transform of the state transition matrix is (sI -A)-1

18. Computer simulation

19. Pole-zero concepts give one an intuitive feel for the problem.

20. State equations, output equations, and initial value for the state-vector

21. Det(sI-A) = 0

SOLUTIONS TO PROBLEMS

1. a. Overdamped Case:

C(s) = 9

s(s2 + 9s + 9)

Expanding into partial fractions,

Taking the inverse Laplace transform,

c(t) = 1 + 0.171 e-7.854t - 1.171 e-1.146t

Solutions to Problems 79

b. Underdamped Case:

K2 and K3 can be found by clearing fractions with K1 replaced by its value. Thus,

9 = (s2 + 3s + 9) + (K2s + K3)s or

9 = s2 + 3s +9 + K2s2 + K3s Hence K2 = -1 and K3 = -3. Thus,

c(t) = 1 - 2 3

e-3t/2 cos( 27 4 t - φ)

= 1 - 1.155 e -1.5t cos (2.598t - φ)

where

φ = arctan ( 3 27

) = 30o

c. Oscillatory Case:

80 Chapter 4: Time Response

The evaluation of the constants in the numerator are found the same way as they were for the

underdamped case. The results are K2 = -1 and K3 = 0. Hence,

Therefore,

c(t) = 1 - cos 3t d. Critically Damped

The constants are then evaluated as

Now, the transform of the response is

c(t) = 1 - 3t e-3t - e-3t 2.

a. C(s) = 5

s(s+5) = 1 s -

1 s+5 . Therefore, c(t) = 1 - e

-5t.

Also, T = 1 5 , Tr =

2.2 a =

2.2 5 = 0.44, Ts =

4 a =

4 5 = 0.8.

b. C(s) = 20

s(s+20) = 1 s -

1 s+20 . Therefore, c(t) = 1 - e

-20t. Also, T = 1

20 ,

Tr = 2.2 a =

2.2 20 = 0.11, Ts =

4 a =

4 20 = 0.2.

Solutions to Problems 81

3.Program: '(a)' num=5; den=[1 5]; Ga=tf(num,den) subplot(1,2,1) step(Ga) title('(a)') '(b)' num=20; den=[1 20]; Gb=tf(num,den) subplot(1,2,2) step(Gb) title('(b)') Computer response: ans = (a) Transfer function: 5 ----- s + 5 ans = (b) Transfer function: 20 ------ s + 20

82 Chapter 4: Time Response

4.

Using voltage division, VC(s) Vi(s) =

1 Cs

(R+ 1

Cs) =

2 (s+2) . Since Vi(s) =

5 s , VC(s) =

10 s(s+2) =

5 s -

5 s+2 .

Therefore vC(t) = 5 - 5e-2t. Also, T = 1 2 , Tr =

2.2 a =

2.2 2 = 1.1, Ts =

4 a =

4 2 = 2.

5.Program: clf num=2; den=[1 2]; G=tf(num,den) step(5*G) Computer response: Transfer function: 2 ----- s + 2

6.

Writing the equation of motion,

(Ms2 + 8s)X(s) = F(s) Thus, the transfer function is,

Solutions to Problems 83

X(s) F(s)

= 1

Ms 2 + 8s

Differentiating to yield the transfer function in terms of velocity,

sX(s) F(s)

= 1

Ms +8 =

1/ M

s + 8 M

Thus, the settling time, Ts, and the rise time, Tr, are given by

Ts = 4

8/ M =

1 2

M; Tr = 2.2

8 / M = 0.275M

7.

Program: Clf M=1 num=1/M; den=[1 8/M]; G=tf(num,den) step(G) pause M=2 num=1/M; den=[1 8/M]; G=tf(num,den) step(G) Computer response: Transfer function: M = 1 Transfer function: 1 ----- s + 8 M = 2 Transfer function: 0.5 ----- s + 4

84 Chapter 4: Time Response

From plot, time constant = 0.125 s.

From plot, time constant = 0.25 s.

8.a. Pole: -2; c(t) = A + Be-2t ; first-order response.

b. Poles: -3, -6; c(t) = A + Be-3t + Ce-6t; overdamped response.

c. Poles: -10, -20; Zero: -7; c(t) = A + Be-10t + Ce-20t; overdamped response.

Solutions to Problems 85

d. Poles: (-3+j3 15 ), (-3-j3 15 ) ; c(t) = A + Be-3t cos (3 15 t + φ); underdamped.

e. Poles: j3, -j3; Zero: -2; c(t) = A + B cos (3t + φ); undamped.

f. Poles: -10, -10; Zero: -5; c(t) = A + Be-10t + Cte-10t; critically damped.

9. Program: p=roots([1 6 4 7 2]) Computer response: p = -5.4917 -0.0955 + 1.0671i -0.0955 - 1.0671i -0.3173

10.

G(s) = C (sI-A)-1 B

A = 8 −4 1

−3 2 0 5 7 −9

⎢ ⎢

⎥ ⎥ ; B =

1 3 7

⎢ ⎢

⎥ ⎥ ; C = 2 8 −3[ ]

(sI A)−1 = 1

s3 − s2 − 91s + 67

(s2 + 7s −18) −(4s + 29) (s − 2) −(3s + 27) (s2 + s − 77) −3

5s − 31 7s − 76 (s 2 − 10s + 4)

⎢ ⎢

⎥ ⎥

Therefore, G(s ) = 5s2 +136s −1777 s3 − s 2 − 91s + 67

.

Factoring the denominator, or using det(sI-A), we find the poles to be 9.683, 0.7347, -9.4179.

11. Program: A=[8 -4 1;-3 2 0;5 7 -9] B=[1;3;7] C=[2 8 -3] D=0 [numg,deng]=ss2tf(A,B,C,D,1); G=tf(numg,deng) poles=roots(deng) Computer response: A = 8 -4 1 -3 2 0 5 7 -9 B = 1

86 Chapter 4: Time Response

3 7 C = 2 8 -3 D = 0 Transfer function: 5 s^2 + 136 s - 1777 --------------------- s^3 - s^2 - 91 s + 67 poles = -9.4179 9.6832 0.7347

12.

Writing the node equation at the capacitor, VC(s) ( 1

R2 + 1 Ls + Cs) +

VC(s) - V(s) R1 = 0.

Hence, VC(s) V(s) =

1 R1

1 R1 +

1 R2 +

1 Ls + Cs

= 10s

s2+20s+500 . The step response is

10 s2+20s+500

.The poles

are at

-10 ± j20. Therefore, vC(t) = Ae-10t cos (20t + φ). 13.

Program: num=[10 0]; den=[1 20 500]; G=tf(num,den) step(G) Computer response:

Transfer function: 10 s ---------------- s^2 + 20 s + 500

Solutions to Problems 87

14.

The equation of motion is: (Ms2+fvs+Ks)X(s) = F(s). Hence, X(s) F(s) =

1 Ms2+fvs+Ks

= 1

s2+s+5 .

The step response is now evaluated: X(s) = 1

s(s2+s+5) =

1/5 s -

1 5 s +

1 5

(s+ 1 2)

2+ 19 4

=

1 5(s+

1 2) +

1 5 19

19 2

(s+ 1 2)

2 + 19 4

.

Taking the inverse Laplace transform, x(t) = 1 5 -

1 5 e

-0.5t ( cos 19 2 t +

1 19

sin 19 2 t )

= 1 5 ⎣

⎡ ⎦ ⎤1 - 2 519 e

-0.5t cos ( 19 2 t - 12.92

o) .

15.

C(s) = ωn2

s(s2+2ζωns+ωn2) =

1 s -

s + 2ζωn s2+2ζωns+ωn2

= 1 s -

s + 2ζωn (s+ζωn)2 + ωn2 - ζ2ωn2

= 1 s -

(s + ζωn) + ζωn (s+ζωn)2 + (ωn 1 - ζ2)2

= 1 s -

(s+ζωn) + ζωn

ωn 1 - ζ2 ωn 1 - ζ2

(s+ζωn)2 + (ωn 1 - ζ2)2

= 1 - e-ζω nt cos ωn 1 - ζ 2 t +

ζ 1 - ζ 2

sin ω n 1 - ζ 2 t

⎝ ⎜

⎠ ⎟ Hence, c(t)

88 Chapter 4: Time Response

= 1 - e-ζωnt 1 + ζ2

1-ζ2 cos (ωn 1 - ζ2 t - φ) = 1 - e-ζωnt

1 1-ζ2

cos (ωn 1 - ζ2 t - φ),

where φ = tan-1 ζ

1 - ζ2

16.

%OS = e-ζπ / 1 - ζ 2 x 100. Dividing by 100 and taking the natural log of both sides,

ln ( %OS 100 ) = -

ζπ 1 - ζ2

. Squaring both sides and solving for ζ2, ζ2 = ln2 (

%OS 100 )

π2 + ln2 (%OS100 ) . Taking the

negative square root, ζ = - ln (

%OS 100 )

π2 + ln2 (%OS100 ) .

17.a.

b.

c.

d.

Solutions to Problems 89

e.

f.

18.a. N/A

b. s2+9s+18, ωn2 = 18, 2ζωn = 9, Therefore ζ = 1.06, ωn = 4.24, overdamped.

c. s2+30s+200, ωn2 = 200, 2ζωn = 30, Therefore ζ = 1.06, ωn = 14.14, overdamped.

d. s2+6s+144, ωn2 = 144, 2ζωn = 6, Therefore ζ = 0.25, ωn = 12, underdamped.

e. s2+9, ωn2 = 9, 2ζωn = 0, Therefore ζ = 0, ωn = 3, undamped.

f. s2+20s+100, ωn2 = 100, 2ζωn = 20, Therefore ζ = 1, ωn = 10, critically damped.

19.

X(s) = 1002

s(s2 +100s+1002) =

1 s -

s+100 (s+50)2+7500

= 1 s -

(s+50) + 50 (s+50)2+7500

= 1 s -

(s+50) + 50 7500

7500

(s+50)2+7500

Therefore, x(t) = 1 - e-50t (cos 7500 t + 50 7500

sin 7500 t)

= 1 - 2 3

e-50t cos (50 3 t - tan-1 1

3 )

20.

a. ωn2 = 16 r/s, 2ζωn = 3. Therefore ζ = 0.375, ωn = 4. Ts = 4

ζωn = 2.667 s; TP =

π ωn 1-ζ2

=

0.8472 s; %OS = e-ζπ / 1 - ζ 2 x 100 = 28.06 %; ωnTr = (1.76ζ3 - 0.417ζ2 + 1.039ζ + 1) = 1.4238;

therefore, Tr = 0.356 s.

90 Chapter 4: Time Response

b. ωn2 = 0.04 r/s, 2ζωn = 0.02. Therefore ζ = 0.05, ωn = 0.2. Ts = 4

ζωn = 400 s; TP =

π ωn 1-ζ2

=

15.73 s; %OS = e-ζπ / 1 - ζ 2 x 100 = 85.45 %; ωnTr = (1.76ζ3 - 0.417ζ2 + 1.039ζ + 1); therefore,

Tr = 5.26 s.

c. ωn2 = 1.05 x 107 r/s, 2ζωn = 1.6 x 103. Therefore ζ = 0.247, ωn = 3240. Ts = 4

ζωn = 0.005 s; TP =

π ωn 1-ζ2

= 0.001 s; %OS = e-ζπ / 1 - ζ 2 x 100 = 44.92 %; ωnTr = (1.76ζ3 - 0.417ζ2 + 1.039ζ +

1); therefore, Tr = 3.88x10-4 s. 21.

Program: '(a)' clf numa=16; dena=[1 3 16]; Ta=tf(numa,dena) omegana=sqrt(dena(3)) zetaa=dena(2)/(2*omegana) Tsa=4/(zetaa*omegana) Tpa=pi/(omegana*sqrt(1-zetaa^2)) Tra=(1.76*zetaa^3 - 0.417*zetaa^2 + 1.039*zetaa + 1)/omegana percenta=exp(-zetaa*pi/sqrt(1-zetaa^2))*100 subplot(221) step(Ta) title('(a)') '(b)' numb=0.04; denb=[1 0.02 0.04]; Tb=tf(numb,denb) omeganb=sqrt(denb(3)) zetab=denb(2)/(2*omeganb) Tsb=4/(zetab*omeganb) Tpb=pi/(omeganb*sqrt(1-zetab^2)) Trb=(1.76*zetab^3 - 0.417*zetab^2 + 1.039*zetab + 1)/omeganb percentb=exp(-zetab*pi/sqrt(1-zetab^2))*100 subplot(222) step(Tb) title('(b)') '(c)' numc=1.05E7; denc=[1 1.6E3 1.05E7]; Tc=tf(numc,denc) omeganc=sqrt(denc(3)) zetac=denc(2)/(2*omeganc) Tsc=4/(zetac*omeganc) Tpc=pi/(omeganc*sqrt(1-zetac^2)) Trc=(1.76*zetac^3 - 0.417*zetac^2 + 1.039*zetac + 1)/omeganc percentc=exp(-zetac*pi/sqrt(1-zetac^2))*100 subplot(223) step(Tc) title('(c)') Computer response: ans = (a)

Solutions to Problems 91

Transfer function: 16 -------------- s^2 + 3 s + 16 omegana = 4 zetaa = 0.3750 Tsa = 2.6667 Tpa = 0.8472 Tra = 0.3559 percenta = 28.0597 ans = (b) Transfer function: 0.04 ------------------- s^2 + 0.02 s + 0.04 omeganb = 0.2000 zetab = 0.0500 Tsb = 400 Tpb =

92 Chapter 4: Time Response

15.7276 Trb = 5.2556 percentb = 85.4468 ans = (c) Transfer function: 1.05e007 ----------------------- s^2 + 1600 s + 1.05e007 omeganc = 3.2404e+003 zetac = 0.2469 Tsc = 0.0050 Tpc = 0.0010 Trc = 3.8810e-004 percentc = 44.9154

Solutions to Problems 93

22.

Program: T1=tf(16,[1 3 16]) T2=tf(0.04,[1 0.02 0.04]) T3=tf(1.05e7,[1 1.6e3 1.05e7]) ltiview Computer response: Transfer function: 16 -------------- s^2 + 3 s + 16 Transfer function: 0.04 ------------------- s^2 + 0.02 s + 0.04 Transfer function: 1.05e007 ----------------------- s^2 + 1600 s + 1.05e007

94 Chapter 4: Time Response

Solutions to Problems 95

23.

a. ζ = - ln (

%OS 100 )

π2 + ln2 (%OS100 ) = 0.56, ωn =

4 ζTs

= 11.92. Therefore, poles = -ζωn ± jωn 1-ζ2

= -6.67 ± j9.88.

b. ζ = - ln (

%OS 100 )

π2 + ln2 (%OS100 ) = 0.591, ωn =

π TP 1-ζ2

= 0.779.

Therefore, poles = -ζωn ± jωn 1-ζ2 = -0.4605 ± j0.6283.

c. ζωn = 4 Ts = 0.571, ωn 1-ζ

2 = π Tp = 1.047. Therefore, poles = -0.571 ± j1.047.

24.

Re =

4 Ts

= 4; ζ = -ln(12.3/100)

π 2 + ln2 (12.3/100) = 0.5549

Re =ζωn = 0.5549ωn = 4; ∴ωn = 7.21

Im = ωn 1 −ζ 2 = 6

G(s) = ωn

2

s2 + 2ζωns + ωn 2 =

51.96 s2 + 8s + 51.96

25. a. Writing the equation of motion yields, (3s2 + 15s + 33)X(s) = F(s)

96 Chapter 4: Time Response

Solving for the transfer function,

X(s) F(s)

= 1/ 3

s2 + 5s +11

b. ωn2 = 11 r/s, 2ζωn = 5. Therefore ζ = 0.754, ωn = 3.32. Ts = 4

ζωn = 1.6 s; TP =

π ωn 1-ζ2

= 1.44

s; %OS = e-ζπ / 1 - ζ 2 x 100 = 2.7 %; ωnTr = (1.76ζ3 - 0.417ζ2 + 1.039ζ + 1); therefore, Tr = 0.69

s.

26. Writing the loop equations,

2s s+ θ 1 s s θ 2 sT s=

s θ 1 ss 1+ θ 2 s+ 0=

Solving for θ2(s),

θ 2 s

s2 s+ T s s− 0

s2 s+ sss 1+

=

( )

= T s s 2 s 1+ +

Forming the transfer function, θ 2 s T s

1 s 2 s 1+ +

=

Thus ωn = 1, 2ζωn = 1. Thus, ζ = 0.5. From Eq. (4.38), %OS = 16.3%. From Eq. (4.42), Ts = 8

seconds. From Eq. (4.34), Tp = 3.63 seconds.

27.

a. 24.542

s(s2 + 4s + 24.542) =

1 s

- s + 4

(s + 2)2 + 20.542 =

1 s

- (s + 2) +

2 4.532

4.532

(s + 2)2 + 20.542 .

Thus c(t) = 1 - e-2t (cos4.532t+0.441 sin 4.532t) = 1-1.09e-2t cos(4.532t -23.80).

b.

Solutions to Problems 97

Therefore, c(t) = 1 - 0.29e-10t - e-2t(0.71 cos 4.532t + 0.954 sin 4.532t)

= 1 - 0.29e-10t - 1.189 cos(4.532t - 53.34o).

c.

Therefore, c(t) = 1 - 1.14e-3t + e-2t (0.14 cos 4.532t - 0.69 sin 4.532t)

= 1 - 1.14e-3t + 0.704 cos(4.532t +78.53o).

28. Since the third pole is more than five times the real part of the dominant pole, s2+1.204s+2.829

determines the transient response. Since 2ζωn = 1.204, and ωn = 2.829 = ωn = 1.682, ζ = 0.358,

%OS = e−ζπ / 1−ζ 2

x100 = 30%, Ts = 4

ζωn = 6.64 sec, Tp =

π ωn 1-ζ2

= 2 sec; ωnTr = 1.4,

therefore, Tr = 0.832.

29.a. Measuring the time constant from the graph, T = 0.0244 seconds.

0

1

2

3

0 0.05 0.1 0.15 0.2 0.25 Time(seconds)

T = 0.0244 seconds

R es

po ns

e

98 Chapter 4: Time Response

Estimating a first-order system, G(s) = K

s+a . But, a = 1/T = 40.984, and K a = 2. Hence, K = 81.967.

Thus,

G(s) = 81.967

s+40.984

b. Measuring the percent overshoot and settling time from the graph: %OS = (13.82-11.03)/11.03 =

25.3%,

0

5

10

15

20

25

0 1 2 3 4 5

R es

po ns

e

Ts = 2.62 seconds

cmax = 13.82

cfinal = 11.03

Time(seconds)

and Ts = 2.62 seconds. Estimating a second-order system, we use Eq. (4.39) to find ζ = 0.4 , and Eq.

(4.42) to find ωn = 3.82. Thus, G(s) = K

s2+2ζωns +ωn2 . Since Cfinal = 11.03,

K ωn2

= 11.03. Hence,

K = 160.95. Substituting all values,

G(s) = 160.95

s2+3.056s+14.59

c. From the graph, %OS = 40%. Using Eq. (4.39), ζ = 0.28. Also from the graph,

Tp = π

ωn 1 −ζ 2

= 4. Substituting ζ = 0.28, we find ωn = 0.818.

Thus,

G(s) = K

s2+2ζωns +ωn2 =

0.669 s2 + 0.458s + 0.669

.

Solutions to Problems 99

30. a.

Since the amplitude of the sinusoids are of the same order of magnitude as the residue of the pole at -

2, pole-zero cancellation cannot be assumed.

b.

Since the amplitude of the sinusoids are of the same order of magnitude as the residue of the pole at -

2, pole-zero cancellation cannot be assumed.

c.

Since the amplitude of the sinusoids are of two orders of magnitude larger than the residue of the pole

at -2, pole-zero cancellation can be assumed. Since 2ζωn = 1, and ωn = 5 = 2.236, ζ = 0.224,

%OS = e−ζπ / 1−ζ 2

x100 = 48.64%, Ts = 4

ζωn = 8 sec, Tp =

π ωn 1-ζ2

= 1.44 sec; ωnTr = 1.23,

therefore, Tr = 0.55.

d.

Since the amplitude of the sinusoids are of two orders of magnitude larger than the residue of the pole

at -2, pole-zero cancellation can be assumed. Since 2ζωn = 5, and ωn = 20 = 4.472, ζ = 0.559,

100 Chapter 4: Time Response

%OS = e−ζπ / 1−ζ 2

x100 = 12.03%, Ts = 4

ζωn = 1.6 sec, Tp =

π ωn 1-ζ2

= 0.847 sec; ωnTr =

1.852, therefore, Tr = 0.414.

31. Program: %Form sC(s) to get transfer function clf num=[1 3]; den=conv([1 3 10],[1 2]); T=tf(num,den) step(T) Computer response: Transfer function: s + 3 ----------------------- s^3 + 5 s^2 + 16 s + 20

%OS = (0.163 - 0.15)

0.15 = 8.67%

32. Only part c can be approximated as a second-order system. From the exponentially decaying cosine the poles are located at s1,2 = −2 ± j9.796 . Thus,

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