# 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)

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