Chapter 07

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 1

SOLUTIONS TO CHAPTER 7 PROBLEMS Problem 7.1

(a)

Let us consider the input is:

( ) sin(10 )=u t t (P7.1)

and therefore:

( ) 10cos(10 )=u t t (P7.2)

As a consequence, the original equation can be written as:

( ) 0.1 ( ) 75 ( ) 0.1 ( ) ( )+ + = + y t y t y t u t u t (P7.3)

By Laplace transforming Eq. (P7.3) with zero initial conditions results in:

2 2

( ) 0.1 1 10( )( ) 0.1 75 10 750

Y s s sG sU s s s s s

+ += = =

+ + + + (P7.4)

(b)

By noticing that:

( )2 2 22− − −= −t t td te e tedt

(P7.5)

the right-hand side of the differential equation becomes:

( )2 2 2 2 ( ) ( )− − − −− = + = +t t t td de te te te u t u tdt dt

(P7.6)

where the input function is:

2( ) −= tu t te (P7.7)

and therefore the differential equation can be written as:

3 2 1( ) ( ) ( ) ( ) ( )a y t a y t a y t u t u t+ + = + (P7.8)

The Laplace transform with zero initial conditions is applied to Eq. (P7.8), which yields

the transfer function:

3 23 2 1

( ) 1( )( )

Y s sG sU s a s a s a s

+= =

+ + (P7.9)


Problem 7.2

Apply Laplace transforms to the two differential equations with zero initial

conditions, which results in:

( ) ( )( ) ( )

3 21 2 1

21 2 2

2 10 80 ( ) 5 60 ( ) ( )

5 6 0 ( ) 5 6 0 ( ) ( )

s s s Y s s Y s U s

s Y s s s Y s U s

+ + + − + =− + + + + =

(P7.1)

Equations (P7.1) can be written in vector-matrix form as:

( )

( )

3 21 1

22 2

( ) ( )2 10 80 5 60( ) ( )5 60 5 60

Y s U ss s s sY s U ss s s

+ + + − + = − + + +

(P7.2)

Because the transfer function matrix [G(s)] connects the input and output vectors as:

[ ]1 1

2 2

( ) ( )( )

( ) ( )Y s U s

G sY s U s

=

(P7.3)

it follows that:

[ ] ( )( )

13 2

2

2 10 80 5 60( )

5 60 5 60s s s s

G ss s s

− + + + − +

= − + + + (P7.4)

The following MATLAB® code line:

>>pretty(inv([s^3+2*s^2+10*s+80,-5*s-60;-5*s-60,s^2+5*s+60]))

returns the transfer function matrix

[ ]( )

( )

2

5 4 3 2 5 4 3 2

3 2

5 4 3 2 5 4 3 2

5 125 607 80 225 400 1200 7 80 225 400 1200( )

5 12 2 10 807 80 225 400 1200 7 80 225 400 1200

ss ss s s s s s s s s sG s

s s s ss s s s s s s s s s

+ + + + + + + + + + + + + =

+ + + + + + + + + + + + + +

(P7.5)


Problem 7.3

By reducing the inertia and the stiffness to the input shaft (where the toque ma is

applied), the equivalent mechanical system of Fig. P7.1 is obtained.

Figure P7.1 Equivalent rotary mechanical shaft-gear system

The mathematical model of this system is described by the differential equation:

1 1 1( ) ( ) ( ) ( )e a eJ t m t c t k tθ θ θ= − − (P7.1)

where the equivalent mass moment of inertia Je and equivalent stiffness ke are calculated

as shown in Chapter 2 by transferring the inertia and stiffness from the output shaft to

the input one and adding them to the corresponding amounts that reside on the input

shaft, namely:

2

11 2

2

2

11 2

2

e

e

NJ J JN

Nk k kN

= +

= +

(P7.2)

The following transfer function is obtained by Laplace transforming Eq. (P7.1) with zero

initial conditions:

( )2

1 21 2 2 2 2 2 2 2

2 1 1 2 2 2 1 1 2

( ) 1( )( )a e e

s NG sM s J s cs k N J N J s N cs N k N kΘ

= = =+ + + + + +

(P7.3)

The required transfer function relates Θ2(s) to Ma(s) and is determined as

( )1 2 1 1 2

1 2 2 2 2 2 21 2 2 1 1 2 2 2 1 1 2

( ) ( )( ) ( )( ) ( )a

s s N N NG s G sM s s N N J N J s N cs N k N kΘ Θ

= × = × =Θ + + + +

(P7.4)

c

θ1, ma

Je

ke


Problem 7.4

For the electrical system of Fig. P7.1 it is considered that the reference voltages are

vC = vD = 0 and therefore the relevant voltages are vA and vB.

Figure P7.1 Electrical circuit with current source and relevant voltages and currents

The following equations can be written based on KNL (Kirchhoff’s node law) at nodes A

and B:

[ ]

[ ]1

1 2

( ) 1( ) ( ) ( )

( )1 1( ) ( ) ( )

= + − − = +

∫

∫ ∫

AA B

BA B B

v ti t v t v t d tR L

dv tv t v t d t v t d tCL L dt

(P7.1)

Laplace transforming Eqs. (P7.1) with zero initial conditions results in:

1 1

1 1 2

1 1 1( ) ( ) ( )

1 1 1( ) ( ) 0

A B

A B

V s V s I sR L s L s

V s Cs V sL s L s L s

+ − =

− + + + =

(P7.2)

which can be written as:

1

1 1

1 1 2

1 1 1( ) ( )( ) 1 1 1 0

A

B

V s R L s L s I sV s Cs

L s L s L s

− + − = − + +

(P7.3)

The transfer function matrix is the one connecting the Laplace-transformed voltage and

current vectors and is calculated by using the following MATLAB® code:

>> syms s r c l1 l2

>> pretty(inv([1/r+1/l1/s,-1/l1/s;-1/l1/s,1/l1/s+1/l2/s+c*s]))

i L2

A

C R

B

C D

L1

iL1 iL2 iR iC

vA vB

vC = 0 vD = 0


as:

[ ]

( )( ) ( )

( ) ( )

31 2 1 2 2

3 2 3 21 2 2 1 2 1 2 2 1 2

22 1 2 2

3 2 3 21 2 2 1 2 1 2 2 1 2

( )

RL L Cs R L L s RL sL L Cs RL Cs L L s R L L Cs RL Cs L L s R

G sRL s L L s RL s

L L Cs RL Cs L L s R L L Cs RL Cs L L s R

+ + + + + + + + + + = +

+ + + + + + + +

(P7.4)


Problem 7.5

(a)

Because of the negative feedback, the voltages at the negative and positive input

terminals are equal to Vi(s). The current passing through Z2 is identical to the one passing

through Z1, namely:

2 1

( ) ( ) ( ) 0o i iV s V s V sZ Z− −

= (P7.1)

and this equation results in the following transfer function:

1 2 2

1 1

( )( ) 1( )

o

i

V s Z Z ZG sV s Z Z

+= = = + (P7.2)

Equation (P7.2) shows that the output and input voltages have the same the same sign; as

a consequence, the operational amplifier electrical circuit behaves as a non-inverting

amplifier.

(b)

The op amp circuit of Fig. 7.40(b) is of the form of the noninverting amplifier of Fig.

7.40(a) with:

1

1 11

1 11

1

22

2 2

1 1

1

= =

+ +

= +

RC s RZ

R C sRC s

RZR C s

(P7.3)

Substitution of Eq. (P7.3) into Eq. (P7.2) results in the following transfer function:

( )1 2 1 2 1 2

1 2 2 1

( )+ + +

=+

R R C C s R RG s

R R C s R (P7.4)


Problem 7.6

According to Table 7.1, the resistance and capacitance thermometer impedances are:

,

,

( ) ( )( )

( )( )

Θ −Θ = Θ =

bR th

C th

s sZQ s

sZQ s

(P7.1)

where Q(s) is the Laplace transform of the heat flow rate. By eliminating Q(s) between

the two Eqs. (P7.1), the following transfer function is obtained:

,

, ,

( )( )( )

Θ= =Θ +

C th

b R th C th

ZsG ss Z Z

(P7.2)

By taking in consideration that:

,

,1

= =

R th th

C thth

Z R

ZC s

(P7.3)

the transfer function becomes:

( ) 1( )( ) 1

Θ= =Θ +b th th

sG ss R C s

(P7.4)

Let us analyze the impedance-based thermal circuit of Fig. P7.1.

Figure P7.1 Single-stage thermal system

The following s-domain equations can be written for the left and right meshes of this

circuit:

, ,

,

( ) ( ) ( )( ) ( )

Θ = +Θ =

b R th C th

C th

s Z Q s Z Q ss Z Q s

(P7.5)

By taking the side-by-side ratio of the two equations above, the transfer function of Eq.

(P7.2) is obtained, which proves that the thermal circuit of Fig. P7.1 is the correct

graphical model of the thermometer-bath system.

Q(s)

Θ(s) Θb(s)

ZR,th

ZC,th


Problem 7.7

The liquid resistance and capacitance impedances are expressed in terms of pressures

as:

,

,

( ) ( )( )

( )( )

− = =

i oR l

oC l

P s P sZQ s

P sZQ s

(P7.1)

where Q(s) is the Laplace transform of the fluid flow rate. By eliminating Q(s) between

the two Eqs. (P7.1), the following transfer function is obtained:

,

, ,

( )( )( )

= =+C lo

pi R l C l

ZP sG sP s Z Z

(P7.2)

which resembles the transfer function of the thermal system of the previous Problem 7.6.

Because:

,

,1

= =

R l l

C ll

Z R

ZC s

(P7.3)

the pressure-defined transfer function becomes:

( ) 1( )( ) 1

= =+

op

i l l

P sG sP s R C s

(P7.4)

When the head is the output, the following relationship can be written between the

input pressure and an equivalent input head:

( ) ( )=i iP s gH sρ (P7.5)

where ρ is the liquid mass density and g is the gravitational acceleration. The head

definitions of liquid resistance and capacitance impedances are:

,

,

( ) ( )( )

( )( )

− = =

iR l

C l

H s H sZQ s

H sZQ s

(P7.6)

By eliminating Q(s) between the two Eqs. (P7.6) and by using Eq. (P7.5), the following

transfer function is obtained:


( ),

, ,

1( ) C lh

l lR l C l

ZG s

gR C s gg Z Z ρ ρρ= =

++ (P7.7)

which indicates that:

( ) ( )=p hG s gG sρ (P7.8)

Figure P7.1 Single-stage liquid system

The following impedance equations can be written based on the impedance-based liquid

circuit of Fig. P7.1.

, ,

,

( ) ( ) ( )( ) ( )

= + =

i R l C l

o C l

P s Z Q s Z Q sP s Z Q s

(P7.9)

By taking the side-by-side ratio of the two equations above, the transfer function of Eq.

(P7.2) is obtained, which proves that the liquid circuit of Fig. P7.1 is the impedance

model of the tank-pipe system.

Q(s)

Po(s) Pi(s)

ZR,l

ZC,l


Problem 7.8

Figure P7.1 shows the lumped-parameter model of the original MEMS device with a

spring k that represents the spring stiffness and a viscous damper of coefficient c, both

connected to the shuttle mass m.

Figure P7.1 Free mechanical system under nonzero initial conditions

Newton’s second law of motion is applied to the shuttle mass motion as:

[ ] [ ]( ) ( ) ( ) ( ) ( )= − − − − my t c y t u t k y t u t (P7.1)


( ) ( ) ( ) ( ) ( )+ + = + my t cy t ky t cu t ku t (P7.2)

The Laplace transform with zero initial conditions is applied to Eq. (P7.2), which yields

the transfer function:

2

( )( )( )

+= =

+ +Y s cs kG sU s ms cs k

(P7.3)

Let us study the mechanical circuit of Fig. P7.2, which uses complex impedances

and where Zm, Zd and Ze are mass, damping and elastic (spring) impedances; U(s) is

considered an input displacement source.

Figure P7.2 Two-mesh impedance-based mechanical system

y

k c

m

u

U(s)

Zd

Ze

Zm

Y(s)

U(s) – Y(s)


According to Newton’s law of motion in impedance form, the following equation can be

written for the right mesh of the mechanical system of Fig. P7.2:

[ ] [ ]( ) ( ) ( ) ( ) ( ) 0− − − − =m d eZ Y s Z U s Y s Z U s Y s (P7.4)


2

( )( )( )

+ += = =

+ + + +d e

m d e

Z ZY s cs kG sU s Z Z Z ms cs k

(P7.5)

Equation (P7.5) is identical to Eq. (P7.3) and therefore, the impedance model of Fig.

P7.2 is the correct representation of the original MEMS.


Problem 7.9

The impedance electrical system is shown in Fig. P7.1.

Figure P7.1 Two-mesh impedance-based electrical system

Kirchhoff’s voltage law is applied to the two meshes of the impedance circuit of Fig.

P7.1, which results in:

[ ]

[ ]1 1 1 2 1

1 2 2 2 2 2

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

+ − =− − + + =

R C

C R L

Z I s Z I s I s V s

Z I s I s Z I s Z I s V s (P7.1)


( )

( )1 1 2 1

1 2 2 2

( ) ( ) ( )

( ) ( ) ( )

+ − =− + + + =

R C C

C R L C

Z Z I s Z I s V s

Z I s Z Z Z I s V s (P7.2)

Equation (P7.2) is expressed in vector-matrix form as:

1 1 1

2 2 2

( ) ( )( ) ( )

+ − = − + +

R C C

C R L C

Z Z Z I s V sZ Z Z Z I s V s

(P7.3)

The transfer function is therefore:

[ ]1

1

2

( )−+ −

= − + + R C C

C R L C

Z Z ZG s

Z Z Z Z (P7.4)

By taking into account that:

1 1 2 21; ; ;= = = =R R L CZ R Z R Z Ls Z

Cs (P7.5)

and by using the MATLAB® code:

>> syms s c r1 r2 l

>>zr1 = r1; zr2 = r2; zl = l*s; zc = 1/c/s;

>>pretty(inv([zr1+zc,-zc;-zc,zr2+zl+zc]))

V2(s)

ZC

ZR1

I1(s)

I1(s) – I2(s) V1(s)

ZR2

ZL

I2(s)


the following transfer matrix is obtained:

[ ] ( ) ( )

( ) ( )

22

2 21 1 2 1 2 1 1 2 1 2

12 2

1 1 2 1 2 1 1 2 1 2

1 1

( )11

+ + + + + + + + + + = +

+ + + + + + + +

LCs R CsR LCs R R C L s R R R LCs R R C L s R R

G sR Cs

R LCs R R C L s R R R LCs R R C L s R R

(P7.6)


Problem 7.10

Based on the thermal impedance definitions of Table 7.1 and on Fig. 5.23, the

following impedances are expressed in terms of left-room external walls resistance,

right-room external walls resistance, internal wall resistance, left room capacitance and

right room capacitance, respectively:

, ,

1 2 1 2

1 2

1 2

1 2

( ) ( ) ( ) ( ) ( ) ( ); ; ;( ) ( ) ( )

( ) ( );( ) ( ) ( ) ( )

Θ −Θ Θ −Θ Θ −Θ= = =

Θ Θ= =

− +

th o th o th

th th

o oR R R

o o

C Co o

s s s s s sZ Z ZQ s Q s Q s

s sZ ZQ s Q s Q s Q s

(P7.1)

where the time-domain counterparts of the Laplace transforms of the equations above

are shown in Fig. 5.23. The volume flow rates are expressed from the first three Eqs.

(P7.1) and then substituted in the last two Eqs. (P7.1), which yields:

, ,

, ,

1 2

1 2

1 ( ) ( ) ( )

( ) 1 ( ) ( )

+ + Θ − Θ = Θ

− Θ + + + Θ = Θ

th th th th

th o th th th o

th th th th

th th o th th o

C C C Co

R R R R

C C C Co

R R R R

Z Z Z Zs s s

Z Z Z Z

Z Z Z Zs s s

Z Z Z Z

(P7.2)

By using:

, ,

1; ;= = =th o th thR th o R th C

th

Z R Z R ZC s

(P7.3)

in Eqs. (P7.2), they become:

1 2

, ,0

1 2, ,0

1 1 1 1( ) ( ) ( )

1 1 1 1( ) ( ) ( )

+ + Θ − Θ = Θ

− Θ + + + Θ = Θ

th oth o th th th

th oth th o th th

C s s s sR R R R

s C s s sR R R R

(P7.4)

By considering that qi = 0 and applying the Laplace transform to Eqs. (5.153) – which

are the time-domain mathematical model of the two-room system – Eqs. (P7.4) are

obtained.

Consider the impedance thermal system of Fig. P7.1 where Θo(s) represents two

identical temperature sources. Assuming the temperatures Θ1(s) and Θ2(s) are set up at

the two indicated nodes in the same Fig. P7.1 and that the opposite nodes have zero


temperatures, the following Kirchhoff voltage law-type equations (with temperature

being the voltage counterpart and the volume flow rate being the variable corresponding

to electric current) can be written for the three meshes:

( )( ) ( )( )

,

,

1 1

2 1

2 2

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) 0

( ) ( ) ( ) ( )

+ − = Θ + + − − =

+ + = Θ

th o th

th th th

th o th

R o C o o

R C o C o

R o C o o

Z Q s Z Q s Q s s

Z Q s Z Q s Q s Z Q s Q s

Z Q s Z Q s Q s s

(P7.5)

Again, by substituting the Laplace transforms of the flow rates from the first three

Eqs. (P7.1), into Eqs. (P7.5) and by using the impedances of Eqs. (P7.3), the first and

third of Eqs. (P7.5) become Eqs. (P7.4), which form the Laplace-domain mathematical

model of the two-room thermal system.

Figure P7.1 Three-mesh impedance-based thermal system

Equations (P7.4) can be written as:

, ,,

1

, , 2,

1( )( )

( ) ( )1

+ + − ΘΘ = Θ Θ − + +

th o th oth o th

th th o

th o th o oth o th

th th

R RR C s

R R ssR R s s

R C sR R

(P7.6)

Because:

[ ]1

2

( )( )( )

( ) ( )ΘΘ

= Θ Θ

o

o

ssG s

s s (P7.7)

The transfer function matrix [G(s)] is determined by inverting the square matrix of Eq.

(P7.6), which is obtained by means of the MATLAB® code:

>> syms s r r0 c

>> pretty(inv([r0*c*s+r0/r+1,-r0/r;-r0/r,r0*c*s+r0/r+1]))

ZCth

ZRth

Q(s)

Qo1(s) – Q(s)

Qo1(s)

ΘO(s) ZCth ΘO(s)

ZRth,o ZRth,o

Qo2(s) + Q(s)

Qo2(s)


Its expression is:

[ ], , ,

, , ,

( ) ( )( )

( ) ( )

th o th th th o th th o

th o th o th th th o th

R R C s R R RD s D s

G sR R R C s R RD s D s

+ + = + +

(P7.8)

with:

( )2 2 2, , , ,( ) 2 2= + + + +th o th th th o th th o th th o thD s R R C s R C R R s R R (P7.9)


Problem 7.11

The fluid capacitance and resistance impedances are expressed based on Fig. 7.44 as:

1 1, ,

( ) ( ) ( ) ( );( ) ( ) ( )

a aC l R l

i o o

P s P s P s P sZ ZQ s Q s Q s

− −= =

− (P7.1)

The assumption has been used here that the pump only acts as a pressure source

generating no input flow rate of its own. Let us check the liquid system sketched in Fig.

P7.1.

Figure P7.1 Impedance-based liquid system

Equations (P7.1) are retrieved if one writes the equations for the two impedances that

are shown in the liquid circuit of Fig. P7.1 and therefore, the physical system of Fig.

P7.1 models correctly the actual liquid system of Fig 7.44. By expressing Qo(s) of the

second Eq. (P7.1) and substituting it in the first Eq. (7.1) yields:

,1

,

,

( ) ( ) ( )1

C li a

C l

R l

ZP s Q s P sZ

Z

= × ++

(P7.2)

The impedances of interest are:

, ,1 ;C l R l ll

Z Z RC s

= = (P7.3)

and with this particular equations, the output pressure of Eq. (P7.3) becomes:

1( ) ( ) ( )1

li a

l l

RP s Q s P sR C s

= × ++

(P7.4)

which can also be written as:

1

( )( ) 1

( )1al

il l

P sRP sQ sR C s

= +

(P7.5)

ZC,l

ZR,l

Qo(s) Qi(s) – Qo(s)

Qi(s)

Qi(s)

P1(s)

Pa(s)


indicating that the system is a one-input two-output system, the transfer matrix (row

vector in this case) being:

[ ]( ) 11

l

l l

RG sR C s

= +

(P7.6)


Problem 7.12

Based on Fig. 7.43, the following dynamic equations are written for the two rotary

plates which are connected by torsional springs:

1 1 2

2 2 1

( ) 2 ( ) ( ) ( )

( ) 2 ( ) ( ) 0

+ − =

+ − =

tJ t k t k t m t

J t k t k t

θ θ θ

θ θ θ (P7.1)

The Laplace transform with zero initial conditions is applied to Eqs. (P7.1), which

results in:

( )

( )

21 2

21 2

2 ( ) ( ) ( )

( ) 2 ( ) 0

+ Θ − Θ =− Θ + + Θ =

tJs k s k s M s

k s Js k s (P7.2)

Equations (P7.2) can be written as:

2

12

2

( ) ( )2( ) 02

Θ + − = Θ− +

ts M sJs k ksk Js k

(P7.3)

Because the input-output connection for this example is:

[ ]1

2

( ) ( )( )

( ) 0Θ

= Θ ts M s

G ss

(P7.4)

it follows that the transfer function is obtained from Eq. (P7.3) as:

[ ]

2

12 2 4 2 2 2 4 2 2

2 2

2 4 2 2 2 4 2 2

22 4 3 4 3( )

2 24 3 4 3

Js k kJs k k J s Jks k J s Jks kG s

k Js k k Js kJ s Jks k J s Jks k

− + + − + + + + = = − + + + + + +

(P7.5)

Figure P7.1 Two-mesh impedance-based mechanical system

By using the mechanical impedance circuit of Fig. P7.1, the following equations are

formulated based on the impedance-form Newton’s second law of motion:

Ze

Zm

Θ2(s)

Θ1(s) – Θ2(s)

Zm

Ze

Θ1(s)

Mt(s)

Ze


( )

( )1 2

1 2

2 ( ) ( ) ( )

( ) 2 ( ) 0

+ Θ − Θ =− Θ + + Θ =

m e e t

e m e

Z Z s Z s M s

Z s Z Z s (P7.6)

By using:

2;= =m eZ Js Z k (P7.7)

Equations (P7.6) change to Eqs. (P7.2) and this demonstrates the mechanical-impedance

circuit of Fig. P7.1 is a correct model for the MEMS of Fig. 7.45.


Problem 7.13

The following MATLAB® code:

>> f = zpk(0,[-2,-2,-3],2);

>> g = tf(f)

generates the transfer function:

3 2

2( )7 16 12

=+ + +

sG ss s s

(P7.1)

whereas the command line:

>> f = zpk(g)

yields:

( )( )2

2( )3 2

=+ +

sF ss s

(P7.2)

which is the original zero-pole-gain model.


Problem 7.14

The following MATLAB® one-line code:

>>g=[tf(zpk([],[-5,-4],1)),tf(zpk([],[-5,-4],2));…

tf(zpk([-1,-2],[-5,-4],1)),tf(zpk([-2,-2],[-5,-4],3))]

returns the following individual transfer functions: Transfer function from input 1 to output...

1

#1: --------------

s^2 + 9 s + 20

s^2 + 3 s + 2

#2: --------------

s^2 + 9 s + 20

Transfer function from input 2 to output...

2

#1: --------------

s^2 + 9 s + 20

3 s^2 + 12 s + 12

#2: -----------------

s^2 + 9 s + 20

Because the two-input, two-output system is characterized by the following equation:

1 11 12 1

2 21 22 2

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

Y s G s G s U sY s G s G s U s

=

(P7.1)

comparison of this equation to the MATLAB® result shown above indicates that the

assembled transfer function matrix is:

[ ]2 2

2 2

2 2

1 29 20 9 20( )3 2 3 12 12

9 20 9 20

s s s sG ss s s s

s s s s

+ + + +=

+ + + + + + + +

(P7.2)


Problem 7.15

It was shown in Chapter 5 that the differential equation of a pneumatic system is:

( ) ( ) ( )+ =g g o o iR C p t p t p t (P7.1)

which indicates the system is a first-order one. The gain is K = 1 and the time constant is

τ = RgCg. By using the transfer function of the system, G(s) = Po(s)/Pi(s), in conjunction

with Pi(s) = Pi/s2, the expression of Po(s) is obtained, whose inverse Laplace transform

is:

( )t

o ip t P t e−

= − + +

ττ τ (P7.2)

At t1 = 3 seconds, the input and output pressures are:

3

(3) 2

(3) 3

i i

o i

p P

p P e ττ τ−

=

= − + +

(P7.3)

which results in the following connection as specified in the problem:

3

3 2i iP e Pττ τ−

− + + =

(P7.4)

equation which can be rewritten as:

3

1 0e ττ τ−

− + = (P7.5)

The solution to Eq. (P7.5) can be determined using the following MATLAB® code:

>> tau=0:0.00001:2;

>> f=@(tau)1-tau+tau.*exp(-3./tau);

>> y=fzero(f,1)

which returns y =

1.0633

This result indicates the time constant is τ = 1.0633 s. The MATLAB® function fzero

is designed to determine the roots of the function f around a point (such as 1 in this


case); the point belongs to a previously defined vector (interval). The function has to be

defined as a function handle (shown by the @ sign), which enables invoking (calling)

the respective function anywhere in a MATLAB® session.

The problem also asks for the system’s response under an input of pi = 50 atm. The

Laplace transform of the output is in this case

2 2

50 50( )1.0633oP s

s s s sτ= =

+ + (P7.6)

whose time-domain counterpart is

1.0633( ) 50 50t

op t e−

= − (P7.7)

The plot of the output pressure is typical for a first-order system under step input, and is

shown in Fig. P7.1.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

5

10

15

20

25

30

35

40

45

50

p o (at

m)

Figure P7.1 Output pressure as a function of time


Problem 7.16

(a)

The time-domain mathematical model of this first-order thermal system is:

1 1 2( ) ( ) ( )th thR C t t t+ =θ θ θ (P7.1)

where θ1(t) is the indoor temperature, θ2(t) is the (constant) outdoor temperature, Rth is

the walls’ thermal resistance and Cth is the room thermal capacitance. The transfer

function is obtained from applying the Laplace transform with zero initial conditions to

Eq. (P7.1), namely:

1

2

( ) 1( )( ) 1th th

sG ss R C s

Θ= =Θ +

(P7.2)

(b)

Based on Eq. (7.51), the equivalent s-domain forcing function, which takes into

account both the actual forcing (outdoors temperature) and initial condition, is:

22, 2 1 1( ) ( ) (0) (0)e th th th ths s R C R C

sΘ = Θ + = +

θθ θ (P7.3)

As a consequence, the Laplace transform of the output is:

1 21 2, 2

(0)( ) ( ) ( ) th the

th th

R C ss G s sR C s s

+Θ = Θ =

+θ θ

(P7.4)

whose inverse Laplace transform is:

[ ]1 2 2 1( ) (0) th th

tR Ct e

−

= − −θ θ θ θ (P7.5)

The thermal capacitance and resistance are calculated by means of Eqs. (5.127) and

(5.134), respectively as:

2

4 4

th p p

th

C mc w hcl lRkA kwh

= =

= =

ρ (P7.6)

The thermal resistance took into account that the total wall area is formed of four

identical walls, which are connected in parallel since the same thermal flow passes

through each wall. With the numerical values of the problem, the parameters of Eqs.

(P7.6) are: Cth = 4.84 x 105 J/C and Rth = 0.0313 C/W. The time constant of this system

is therefore τ = Rth Cth = 15,125 seconds, which is approximately 4.2 hours. The settling


time, which is approximately four times the time constant, is 60,500 seconds or about

16.8 hours. Figure P7.1 shows the indoor temperature variation as a function of time.

0 1 2 3 4 5 6 7

x 104

20

22

24

26

28

30

32

34

36

38

40

Time (sec)

θ 1 (

deg)

Figure P7.1 Indoor temperature (in degrees Celsius) as a function of time


Problem 7.17

The equivalent impedance of Fig. P7.1 models the actual mechanical chain of Fig.

7.46.

Figure P7.1 Impedance chain for the actual mechanical system

It is obtained as:

1 1 2 2 3 3

1 1 1 1 1( )eq e d e d e dZ s Z Z Z Z Z Z

= + + ++ +

(P7.1)

Equation (P7.1) is based on the particular way of connecting various components of the

mechanical chain either in series or in parallel. At the same time, because the force is the

input and the free end displacement is the output, the following relationship can be

written:

1( ) ( ) ( ) ( )( )eq

Y s G s F s F sZ s

= = (P7.2)

By taking into account that elastic impedances are calculated as Ze = k and damping

impedances as Zd = cs, the following transfer function is obtained from Eqs. (P7.1) and

(P7.2) with the numerical data of this problem:

3 2

3 2

0.12 178.8 48,600 1,800,000( )14.4 10,080 1,440,000s s sG s

s s s+ + +

=+ +

(P7.3)

The plot of Fig. P7.2 shows y(t) (measured in meters) for the sinusoidal force input and

which was calculated by using MATLAB® lsim input function.

Zeq Y(s) F(s)


0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Linear Simulation Results

Time (sec)

Am

plitu

de

Figure P7.1 Free end displacement as a function of time


Problem 7.18

Application of Newton’s second law of motion to the mechanical system of Fig. 7.47

yields the following differential equation:

( ) ( ) ( ) ( )J t c t k t m t= − − + θ θ θ (P7.1)


( ) ( ) ( ) ( )J t c t k t m t+ + = θ θ θ (P7.2)

The transfer function of this system is:

2

( ) 1( )( )sG s

M s Js cs kΘ

= =+ +

(P7.3)

The steady-state output angle is calculated as:

[ ] [ ]0 0

( ) lim ( ) lim ( ) ( )s s

ms s sG s M sk→ →

∞ = Θ = =θ (P7.4)

The stiffness is found as k = m/θ(∞) = 286.48 N-m for the numerical values of the

problem.

According to Eq. (7.51), the equivalent forcing is:

20 4 20( ) ( ) (0 ) 4esU s M s J

s s+

= + = + =ω (P7.5)

The Laplace transform of the angle θ is expressed as:

2

4 20 1 1( ) ( ) ( ) '( )0.1 6 286.48e

ss G s U s G ss s s s

+Θ = = × = ×

+ + (P7.6)

MATLAB® step function can now be used to plot the time response of the system

whose transfer function is G’(s) of the equation above under the action of a step input –

whose Laplace transform is 1/s; the plot is shown in Fig. P7.1.


0 0.05 0.1 0.15 0.2 0.25 0.30

5

10

15

20

25Step Response

Time (sec)

Rot

atio

n an

gle

(deg

)

Figure P7.1 Rotation angle as a function of time


Problem 7.19

The transfer function of the electrical system of Fig. 7.45 is calculated as:

2

21

1

1( ) 1( )( ) 1

o

i

RCs

RV s Z CsG sLV s Z R Ls LCs RC sR

×

+= = − = − = −

+ + + +

(P7.1)

Considering that the input voltage, vi is constant, the initial and final value of the output

voltage are determined by means of the initial- and final-value theorems, respectively as:

2

0 0 02

(0) lim ( ) lim ( ) ( ) lim 01

( ) lim ( ) lim ( ) ( ) lim1

io o is s s

io o i is s s

svv sV s sG s V sLs LCs RC sR

svv sV s sG s V s vLs LCs RC sR

→∞ →∞ →∞

→ → →

= = = = + + + ∞ = = = = + + +

(P7.2)

Figure P7.1 shows the output voltage as a function of time – the plot was obtained by

using MATLAB® step function.

0 50 100 150 200 250 3000

5

10

15

20

25

30Step Response

Time (sec)

Out

put v

olta

ge (

V)

Figure P7.1 Output voltage as a function of time


Problem 7.20

The impedance-based Laplace-domain equation corresponding to this circuit is:

1 ( ) ( )R I s V s

Cs + =

(P7.1)

for any given moment in time when, for a brief instance when the capacitor’s mobile

plate can be assumed to be fixed. For constant input voltage v, the Laplace-transformed

current I(s) can be expressed from Eq. (P7.1) as:

1( ) 11

Cv vI sRCs R s

RC

= = ×+ +

(P7.2)

whose inverse Laplace transform is:

( )t

RCvi t eR

−= (P7.3)

Applying the natural logarithm to Eq. (P7.3) results in:

( )ln Ri t tv RC

= −

(P7.4)

Equation (4.21) shows that the capacitance is expressed in terms of the mobile plate

displacement x as:

0

0

ACg x

=−

ε (P7.5)

By combining Eqs. (P7.4) and (P7.5) results in the following expression of the unknown

x when i1 is used for i(t) and t1 for the generic time t:

0 10

1

lnR A Rix gt v

= +

ε (P7.6)

With the numerical values of this problem, a value of x = 5.95 µm is obtained.


Problem 7.21

The lumped-parameter model of the MEMS is shown in Fig. P7.1 with a spring of

stiffness k, a viscous damper of coefficient c and a point force f acting on the point mass

m.

Figure P7.1 Lumped-parameter model of the MEMS

The shuttle mass motion is described by the equation:

( ) ( ) ( ) ( )+ + = my t cy t ky t f t (P7.1)

Applying the Laplace transform with zero initial conditions to Eq. (P7.1) results in:

2

( ) 1( )( )

= =+ +

Y sG sF s ms cs k

(P7.2)

Equation (7.81) gives the equation of the transfer function which is equivalent to the

original system and which corresponds to a forced response under unit impulse and zero

initial conditions. For this problem, the modified transfer function becomes:

12

(0) (0)'( ) my s cy dG sms cs k

+ +=

+ + (P7.3)

where d1 = 10-6 is the multiplier of the unit impulse.

The stiffness k results from the two beam-spring pairs that are connected in parallel;

each beam-spring is formed of two beams: the long one has a stiffness k1 and the short

one has a stiffness k2 (both are given in Table 3.2 under the fixed-guided boundary

condition); as a consequence, the stiffness is calculated as:

( )3 3 4 4

1 2 1 23 3 3 3

1 2 1 2 1 23 31 2

12 1224 32 2 12 12 64 8

×= = = × =

+ + ++

EI EIk k l l E d Edk EI EIk k l l l l

l l

π π (P7.4)

f

k c

m

y


and its value is k = 0.346 N/m. By using the transfer function of Eq. (P7.3) in

conjunction with MATLAB® impulse input produces the time response of Fig. P7.2.

0 2 4 6 8 10 120

0.5

1

1.5

2

2.5

3

3.5x 10

-6 Impulse Response

Time (sec)

Shu

ttle

mas

s di

spla

cem

ent (

m)

Figure P7.2 Shuttle mass displacement as a function of time

It can be seen that the maximum displacement is approximately 3.1 µm, which is less

than the initial capacitor gap g0 = 8 µm.


Problem 7.22

In Problem 7.3, the transfer function was determined as:

12

( ) 1( )( )a e e

sG sM s J s cs kΘ

= =+ +

(P7.1)

with Je = 0.0048 kg-m2, ke = 365.556 N-m calculated

as2 2

1 11 2 1 2

2 2

;e eN NJ J J k k kN N

= + = +

, and c = 80 N-s-m. By applying the Laplace

transform with zero initial conditions to the differential equation: 1 1 1e e aJ c k m= − − + θ θ θ

representing the mathematical model of the rotary system transferred to the shaft 1,

results in

[ ]1 11 1 12

( ) (0) (0) 1( ) ( ) (0 ) (0 ) ( )a ea e

e e

M s J s cs M s J s c G s sJ s cs k s+ +

Θ = = + + × × ×+ +θ θ

θ θ (P7.2)

Because 1/s is the Laplace transform of the unit step function (the one we want to use

with MATLAB®), it follows that:

11( ) '( )Θ = ×s G ss

(P7.3)

with:

2

1 12

(0) (0)'( ) + +=

+ +e a

e e

J s c s mG sJ s cs k

θ θ (P7.4)

Due to the rigid connection between the two meshing gears it follows that:

21 2

1

(0) (0)=NN

θ θ (P7.5)

with a value of θ1(0) = 36/48 x 3o = 2.25o. The transfer function of Eq. (P7.4) is now

used in MATLAB® in conjunction with the step function, which applies a unit step input

to G’(s) – the result is the plot of Figure P7.1.


0 0.2 0.4 0.6 0.8 1 1.2 1.4

2.26

2.28

2.3

2.32

2.34

2.36

2.38Step Response

Time (sec)

Inpu

t sha

ft ro

tatio

n an

gle

(deg

rees

)

Figure P7.1 Input shaft rotation angle as a function of time


Problem 7.23

The pressure transfer function has been derived in Problem 7.7 as:

( ) 1( )( ) 1

= =+

op

i l l

P sG sP s R C s

(P7.1)

which indicates that the time-domain mathematical model (found by cross-multiplication

in Eq. (P7.1) followed by inverse Laplace transformation of the result) is:

( ) ( ) ( )+ =l l o o iR C p t p t p t (P7.2)

The initial value of the pressure at the tank bottom po(0) is not zero because there is an

initial head h(0); the two parameters are related as:

(0) (0)=op ghρ (P7.3)

The Laplace transform is now applied to Eq. (P7.2) by taking into account the nonzero

initial condition (and by following an approach similar to the one used to solve Example

7.14), which yields:

[ ]( ) (0 ) ( )

( ) ( )( )

+= ×i l l o

o ii

P s R C p G sP s P s

P s (P7.4)

or:

( ) '( ) ( )=o iP s G s P s (P7.5)

where the modified transfer function is:

[ ]( ) (0 ) ( )

'( )( )

+= i l l o

i

P s R C p G sG s

P s (P7.6)

By taking into account that:

2

2 10( )10+

=+i

sP s Ps s

(P7.7)

the transfer function of Eq. (P7.6) becomes:

[ ]( )

2

2

(0) 2 5 (0) 10'( )

2 2 5 1 10+ + +

=+ + +

l l o l l o

l l l l

R C p s R C p P s PG s

PR C s P R C s P (P7.8)

The liquid capacitance and resistance are calculated by means of Eqs. (5.37) and (5.52)

respectively:

2

4

128;4l l

i

d lC Rg d

= =π µρ π

(P7.9)


and their values are: Cl = 7.213 x 10-4 m4s2-kg-1 and Rl = 33,939 kg-m-4-s-1. Figure P7.1

displays the input and output (blue line) pressures as functions of time.

0 10 20 30 40 50 60 70 800

1

2

3

4

5

6x 10

5 Linear Simulation Results

Time (sec)

Inpu

t and

out

put p

ress

ures

(N

/m2 )

Figure P7.1 Input and output pressures as functions of time

The steady-state input and output pressures are calculated as:

0

( )( ) lim ( )

i

o os

p Pp sP s P

→

∞ = ∞ = =

(P7.10)


Problem 7.24

The two Laplace-transformed voltages Va(s) and Vb(s) can be expressed as:

( )( )

( )

31 2 1 2

3 21 2 2 1 2

23 2

1 2 2 1 2

( ) ( )

( ) ( )

A

B

RL L Cs R L L sV s I s

L L Cs RL Cs L L s RRL sV s I s

L L Cs RL Cs L L s R

+ += + + + +

= + + + +

(P7.1)

The Laplace-transformed currents through the four electrical components are:

1 21 2

( ) ( ) ( ) ( )( ) ; ( ) ; ( ) ; ( ) ( )−= = = =A A B B

R L L C BV s V s V s V sI s I s I s I s V s Cs

R L s L s (P7.2)

By combining Eqs. (P7.1) and (P7.2), the latter equation can be written as:

( )( )

( )

( )

( )

31 2 1 2

3 21 2 2 1 2

22

3 21 1 2 2 1 2

23 2

1 2 2 1 2

22

3 21 2 2 1 2

( )( ) 1( )( )

R

L

L

C

iL L Cs i L L sL L Cs RL Cs L L s R

I s iRL Cs iRI s L L Cs RL Cs L L s RI s siRI s L L Cs RL Cs L L s R

iRL CsL L Cs RL Cs L L s R

+ + + + + + + + + + + = ×

+ + + +

+ + + +

(P7.3)

which indicates that each of the time-domain currents can be determined by using

MATLAB® step input. Figure P7.1 plots the currents iR, iL1, iL2 ,and iC (from top to

bottom) as functions of time for the numerical values of this problem. It can be seen that

the nonzero steady-state currents are the ones through L1 and L2 because:

( )1 1 2 2 3 20 0 0

1 2 2 1 2

( ) lim ( ) ( ) lim ( ) limL L L Ls s s

iRi sI s i sI s iL L Cs RL Cs L L s R→ → →

∞ = = ∞ = = =+ + + +

(P7.4)

whereas:

( )( )

( )

31 2 1 2

3 20 01 2 2 1 2

22

3 20 01 2 2 1 2

( ) lim ( ) lim 0

( ) lim ( ) lim 0

R Rs s

C Cs s

iL L Cs i L L si sI s

L L Cs RL Cs L L s R

iRL Csi sI sL L Cs RL Cs L L s R

→ →

→ →

+ +∞ = = = + + + +

∞ = = = + + + +

(P7.5)

The following MATLAB® code has been used to generate Fig. P7.1:


>> num={[i*l1*l2*c,0,i*(l1+l2),0];[i*r*l2*c,0,i*r];[i*r];[i*r*l2*c,0,0]};

>> den=[l1*l2*c,r*l2*c,l1+l2,r];

>> step(tf(num,den))

-0.01

0

0.01

To:

Out

(1)

0

0.01

0.02

To:

Out

(2)

0

0.01

0.02

To:

Out

(3)

0 0.2 0.4 0.6 0.8 1 1.2-0.01

0

0.01

To:

Out

(4)

Step Response

Time (sec)

Am

plitu

de

Figure P7.1 Component currents as functions of time


Problem 7.25

The Laplace transforms of the two input functions are:

1 21 1( ) ; ( )

1= =

+U s U s

s s (P7.1)

The Laplace-transformed outputs are calculated by multiplying the transfer function

matrix components to the inputs of Eq. (P7.1), which results in:

2

1 2 2 4 3 2

2

2 2 2 4 3 2

3 2 1 2 1 3 7 2( ) ;4 25 4 25 1 5 29 251 1 3 1 4 1( )

4 25 4 25 1 5 29 25

s s sY ss s s s s s s s s s

s s sY ss s s s s s s s s s

+ + += × + × = + + + + + + + +

+ + + = × + × = + + + + + + + +

(P7.2)

Consider a unit impulse is applied to the following transfer functions:

1 1 2 2( ) ( ); ( ) ( )= =G s Y s G s Y s (P7.3)

The MATLAB® impulse function is applied to the transfer functions defined above,

and the result is shown in the plots of Figure P7.1.

0 1 2 3 4 5 6 70

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Impulse Response

Time (sec)

Am

plitu

de

y1

y2

Figure P7.1 Outputs y1 and y2 as functions of time


Problem 7.26

The time-domain mathematical model of this mechanical system is derived by means

of Newton’s second law of motion as:

1 1 1 2

2 2 2 1

2 02 0

+ + − = + + − =

my cy ky kymy cy ky ky

(P7.1)

which can be written in the vector-matrix form:

1 1 1

2 2 2

0 0 2 00 0 2 0

− + + = −

y y ym c k km y c y k k y

(P7.2)

Comparison of Eq. (P7.2) to the generic Eq. (7.102), indicates that:

[ ] [ ] [ ] { }2 1 0

0 0 2 0; ; ;

0 0 2 0−

= = = = −

m c k ka a a u

m c k k (P7.3)

As a consequence, the equivalent s-domain forcing vector is calculated based on Eq.

(7.106) for the particular initial conditions and numerical values of this problem as:

{ } { } { } { }0 0 0 0.03 0.14

( ) (0) (0) (0)0 0 0 0.02 0.14e

m m c sU s s y y y

m m c s−

= + + = + (P7.4)

The transfer function matrix is calculated as shown in Eq. (7.105):

[ ]

2

12 4 3 2 4 3 2

2 2

4 3 2 4 3 2

2 80 402 4 164 320 4800 4 164 320 4800( )

2 40 2 804 164 320 4800 4 164 320 4800

s sms cs k k s s s s s s s sG s

k ms cs k s ss s s s s s s s

− + + + + − + + + + + + + + = = − + + + + + + + + + + + +

(P7.5)

The two components of the Laplace-transformed output vector are determined from

{Y(s)} = [G(s)] {Ue(s)} by combining Eqs. (P7.4) and (P7.5), namely:

( )

3 2

1 4 3 2

3 2

2 4 3 2

3 8 292 560( )100 400 16,400 32,000 480,000

2 9 154 280( )

100 400 16,400 32,000 480,000

s s sY ss s s s

s s sY s

s s s s

− + −= + + + +

+ + + = + + + +

(P7.6)

The time-domain responses y1(t) and y2(t) can be plotted by using MATLAB® impulse

input, for which the transfer functions are actually Y1(s) and Y2(s) of the previous

equation. The following code was used to create the two displacements’ variations with

respect to time of Fig. P7.1:


>> num = {[3, - 8, 292, - 560];[2, 18, 308, 560]};

>> den = [100, 400, 16400, 32000, 480000];

>> impulse(tf(num,den))

It can be seen that the steady-state values are both zero, which can be checked by

calculating:

1 10

2 20

( ) lim ( ) 0

( ) lim ( ) 0→

→

∞ = = ∞ = =

s

s

y sY s

y sY s (P7.7)

-0.02

-0.01

0

0.01

0.02

0.03

To:

Out

(1)

0 1 2 3 4 5 6-0.03

-0.02

-0.01

0

0.01

0.02

0.03

To:

Out

(2)

Impulse Response

Time (sec)

Am

plitu

de

Figure P7.1 Output displacements y1 and y2 as functions of time


Problem 7.27

Let us generalize the problem by analyzing the electrical system of Fig. P7.1.

Figure P7.1 Operational amplifier electrical system with generic impedances

The current that passes through Z1 is equal to the current that passes through Z4 and,

similarly, the same current (different from the one just mentioned) passes through Z2 and

Z3. Because the op amp has negative feedback, the positive and negative input terminals

are under identical voltages. It then follows that:

1

1 4

2

2 3

( ) ( )( ) ( )

( ) ( ) ( )

oV s V sV s V sZ Z

V s V s V sZ Z

−− = − =

(P7.1)

By eliminating V between the two Eqs. (P7.1), the following relationship is obtained

connecting the input voltages V1 and V2 to the output voltage Vo:

( )( )

3 1 4 42 1

1 2 3 1

( ) ( ) ( )o

Z Z Z ZV s V s V sZ Z Z Z

+= × − ×

+ (P7.2)

which shows that the output voltage is the difference between fractions of the two input

voltages, and that is why this system is known as a difference amplifier.

For the particular case of our problem:

1 21 1 2 2 3 4 1 2

1; ; ; ; ( ) ; ( ) v vZ R Z R Z Z Ls V s V sCs s s

= = = = = = (P7.3)

With the numerical values of this problem, Eqs. (P7.2) and (P7.3) yield:

2

2

12 5,000300 125o

s sVs s

− − +=

+ (P7.4)

Vo

V1

V2

Z3

Z1

Z2

Z4

+ –

V

V


The output voltage vo(t) can be plotted by using a unit impulse input applied to Vo(s) of

Eq. (P7.4) – this one considered to be the transfer function. The result is plotted in Fig.

P7.2. The steady-state value is calculated as:

2

0 0

12 5000( ) lim lim 40300 125o os s

s sv sVs→ →

− − +∞ = = =

+ (P7.5)

as also seen in Fig. P7.2.

0 5 10 150

5

10

15

20

25

30

35

40Impulse Response

Time (sec)

v o (V

)

Figure P7.2 Output voltage as a function of time


Problem 7.28

The following transfer function has been obtained for the particular electrical circuit

of Fig. 7.40(b):

( )1 2 1 2 1 2

1 2 2 1

( ) 68 185( )( ) 42.5 100

o

i

R R C C s R RV s sG sV s R R C s R s

+ + + += = =

+ + (P7.1)

Figure P7.1 shows the Simulink® diagram allowing to plot the output voltage for the

case where the resistors have saturation nonlinearity, as well as for the case where the

resistors are linear. The saturation essentially does not allow the input voltage (which is

also applied to the two resistors) to exceed the limits of −50 V and 70V; however, only

the upper limit will have an effect as the input voltage has a constant positive value. A

fast assessment of the maximum output voltage can be made by ignoring the capacitors;

in this case, the output voltage is obtained from Eq. (P7.1) as

1 2

1

1.85o i iR Rv v v

R+

= × = (P7.2)

Considering that the nonlinearity is applied to the input voltage, it follows that the

maximum input voltage is 70 V, which results in a maximum output voltage of 70 x 1.85

= 129.5 V. Without the saturation, the maximum output voltage would be 110 x 1.85 =

203.5 V. Both these values are shown in Fig. P7.2 where the saturated and non-saturated

output voltages are plotted, corresponding to the Simulink® diagram of Fig. P7.1.

Figure P7.1 Simulink® diagram for the operational-amplifier circuit


Figure P7.2 Output voltages with and without resistor saturation


Problem 7.29

In Problem 7.11, the following transfer function relationship has been derived

connecting the output voltage P1(s) to the input flow rate Qi(s) and the atmospheric

pressure Pa(s):

1( ) ( ) ( )1

li a

l l

RP s Q s P sR C s

= × ++

(P7.1)

At the same time, the output flow rate was expressed as

1

,

( ) ( )( ) ao

R l

P s P sQ sZ−

= (P7.2)

Combining Eqs. (P7.1) and (P7.2) and taking into account that ZR,l = Rl, results in

1( ) ( )1o i

l l

Q s Q sR C s

= ×+

(P7.3)

which indicates that the transfer function connecting the output flow rate to the input

flow rate is

( ) 1( )( ) 1

o

i l l

Q sG sQ s R C s

= =+

(P7.4)

The liquid resistance and capacitance are calculated as:

4

2

128

4

li

l

lRddC

g

= =

µπ

πρ

(P7.5)

whose values are Rl = 2.5465 x 106 N-s- m-5 and Cl = 8.0143 x 10-5 m5-N-1. The time

constant of this first-order system is τ = 4 Rl Cl = 816 seconds.

Figure P7.1 shows the Simulink® diagram of this liquid system and Fig. P7.2 contains

the time plot of the output flow rate.

Figure P7.1 Simulink® diagram of the pump-tank-pipe liquid system


Figure P7.2 Output flow rate as a function of time


Problem 7.30

In Problem 7.12, the following transfer function matrix has been derived:

[ ]1

2

( ) ( )( )

( ) 0Θ

= Θ ts M s

G ss

(P7.1)

where:

[ ]

2

2 4 2 2 2 4 2 2

2

2 4 2 2 2 4 2 2

24 3 4 3( )

24 3 4 3

Js k kJ s Jks k J s Jks kG s

k Js kJ s Jks k J s Jks k

+ + + + + =

+ + + + +

(P7.2)

Equations (P7.1) and (P7.2) show that:

2

1 2 4 2 2

2 2 4 2 2

2( ) ( )4 3

( ) ( )4 3

t

t

Js ks M sJ s Jks k

ks M sJ s Jks k

+Θ = × + +Θ = × + +

(P7.3)

The plate moment of inertia J and bar stiffness k are calculated as

( )2 2 2

412

32p

w t w tJ

GI Gdkl l

+ = = =

ρ

π (P7.4)

With the numerical parameters of the problem, it is obtained that J = 1.5 x 10-19 kg-m2

and k = 1.26 x 10-13 N-m.

Figure P7.1 illustrates the Simulink® diagram allowing to plot the two time-domain

angles θ1 and θ2, which are shown in Fig. P7.2.

Figure P7.1 Simulink® diagram of the two-disk rotary mechanical system


Figure P7.2 Disk rotation angles (in degrees) as functions of time

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Chapter 07

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