chapter 07
TRANSCRIPT
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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 2
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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 3
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 4
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 5
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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 6
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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 7
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 8
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:
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 9
( ),
, ,
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 10
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)
which can be written as:
( ) ( ) ( ) ( ) ( )+ + = + 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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 11
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)
which can be written as:
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.
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 12
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)
which can be written as:
( )
( )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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 13
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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 14
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 15
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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 16
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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 17
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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 18
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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 19
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 20
( )
( )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.
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 21
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.
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 22
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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 23
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 24
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 25
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 26
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 27
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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 28
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 29
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)
which can be written as:
( ) ( ) ( ) ( )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.
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 30
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 31
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 32
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.
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 33
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 34
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.
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 35
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.
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 36
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 37
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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 38
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)
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 39
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:
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 40
>> 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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 41
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 42
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:
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 43
>> 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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 44
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 45
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 46
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 47
Figure P7.2 Output voltages with and without resistor saturation
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 48
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 49
Figure P7.2 Output flow rate as a function of time
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 50
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
Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 7 51
Figure P7.2 Disk rotation angles (in degrees) as functions of time