Chapter 04

Lobontiu: System Dynamics for Engineering Students Solutions: Chapter 4 1

SOLUTIONS TO CHAPTER 4 PROBLEMS Problem 4.1

Figure P4.1 sketches an angular potentiometer. It is formed of a circular resistor

defined by a radius r and total length l calculated in terms of the center angle α (not

shown in the figure) as

l r= α (P4.1)

A wiper moves along a conductive circular guide, which is concentric with the resistor.

Figure P4.1 Angular potentiometer

The voltage generated by the source v is, according to Ohm’s law:

l rv Ri i iA A

= = =ρ ρ α (P4.2)

where R is the total resistance of the circular wire, ρ is the resistivity, and A is the wire

cross-sectional area. The rotary-motion wiper encloses a circuit defined by a center angle

θ whose voltage (measured by a meter for instance) is

l rv R i i iA A

= = =θθ θ

ρ ρ θ (P4.3)

Combination of Eqs. (P4.2) and (P4.3) results in:

R vv vR

= =θθ θ

α (P4.4)

i

v

wiper vθ

+

–

conductive guide

meter

voltage source

resistor

r θ


For v and α being constant, Eq. (P4.4) indicates that the angular displacement angle θ is

proportional to the voltage measured by the meter, vθ.


Problem 4.2

If the 5 resistors, each having a resistance R, are combined in series, the equivalent

resistance would be 5R, which is the maximum value of all possible combinations;

similarly, if the resistors are combined in parallel, the equivalent resistance would be R/5,

which is the minimum value of all possible combinations. However, any combination

needs to have both series and parallel connections. It thus appear that the design of Fig.

P4.1(a) generates the maximum equivalent resistance because it comprises three resistors

in series, whereas the combination of Fig. P4.1(b) yields the minimum equivalent

resistance as it contains four resistors in parallel.

Figure P4.1 Resistor connections for: (a) maximum equivalent resistance; (b) minimum equivalent

resistance

The combination of Fig. P4.1(a) results in the following equivalent resistance:

max73

2 2RR R R= + = × (P4.1)

and the arrangement of Fig. P4.1(b) yields a minimum resistance calculated as:

min5

4 4RR R R= + = × (P4.2)

It can be checked that no other combination of 5 identical resistors is able to result in an

equivalent resistance outside the [5R/4, 7R/2] range; therefore, indeed, Eqs. (P4.1) and

(P4.2) give the extreme equivalent resistances. For R = 120 Ω, the extreme values are

Rmax = 420 Ω and Rmin = 150 Ω.

(a) (b)

R R R

R

R

R

R R

R

R


Problem 4.3

To calculate the equivalent resistance between nodes c and d, the resistances of the

triangle a-b-c of Fig. 4.38 are transformed into the ones corresponding to a reversed Y-

connection, as shown in Fig. P4.1(a), which is further simplified to the circuit of Fig.

P4.1(b).

Figure P4.1 Intermediate resistor connections for equivalent resistance between nodes c and d

As a consequence, the sought resistance is calculated as

125 34514

125 345cd

R RR RR R

= ++

(P4.1)

where

125 15 2

345 45 3

R R RR R R

= + = +

(P4.2)

and

1 414

1 4 5

4 545

1 4 5

1 515

1 4 5

R RRR R R

R RRR R R

R RRR R R

= + +

= + +

=

+ +

(P4.3)

Substitution of Eq. (P4.3) in Eq. (P4.2) and then in Eq. (P4.1) yields

( ) ( ) ( )

( ) ( )( )1 2 3 4 3 4 5 4 5 2 3 4 5 4 5

1 2 3 5 2 3 4 5 4 5cd

R R R R R R R R R R R R R R RR

R R R R R R R R R R+ + + + + + + =

+ + + + + + (P4.4)

Numerically, the equivalent resistance is Rcd = 194.3 Ω.

(a) (b)

c R14

R15

R45

R2

R3

d

R125

R345

R14 c d


Figure P4.2 shows the connections of the resistances between nodes a and b. It

follows that:

2 3 5 1 4

1 1 1 1

abR R R R R R= + +

+ + (P4.5)

which yields:

( )( )( ) ( )( )

5 1 4 2 3

1 2 3 5 2 3 4 5 4 5ab

R R R R RR

R R R R R R R R R R+ +

=+ + + + + +

(P4.6)

whose numeric value is Rab = 78 Ω.

Figure P4.2 Intermediate resistor connections for equivalent resistance between nodes a and b

R2

R4 R1

a b

R3

R5


Problem 4.4

The circuit of Fig. P4.1 shows the resistance arrangement after performing the

conversion from triangle in a reversed-Y connection in the original Fig. 4.39.

Figure P4.1 Resistor connections after conversion from triangle to reversed-Y connection

The resistance between nodes a and b is calculated as

( )( )13 4 23 512 6

13 4 23 5ab

R R R RR R R

R R R R+ +

= + ++ + +

(P4.1)

with

1 212

1 2 3

2 323

1 2 3

1 313

1 2 3

R RRR R R

R RRR R R

R RRR R R

= + +

=+ +

=

+ +

(P4.2)

Substituting Eqs. (P4.2) in Eq. (P4.1) yields

( ) ( )

( ) ( ) ( )( )1 3 4 1 2 3 2 3 5 1 2 31 2

61 2 3 1 2 3 3 1 2 4 5 1 2 3

ab

R R R R R R R R R R R RR RR RR R R R R R R R R R R R R R

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

(P4.3)

and its numerical value is Rab = 62.6 Ω.

R13

R5 R23

R4

R12 R6 a b


Problem 4.5

As demonstrated in the website Chapter 4, the following equations need to be solved

to determine the capacitances that form a reversed-Y connection, and which are C14, C15,

C45 in terms of the capacitances of the original Δ connections, and which are C1, C4, C5

(as indicated in Fig. P4.1):

14 151

14 15

14 452

14 45

45 153

45 15

C C fC CC C f

C CC C f

C C

= +

= +

=

+

(P4.1)

where

4 51 1

4 5

1 54 2

1 5

1 45 3

1 4

37.14

43.33

52

C CC fC C

C CC fC CC CC f

C C

+ = = +

+ = = +

+ = =

+

(P4.2)

Figure P4.1 Partial capacitance circuits for the circuit of Fig. 4.40: (a) original Δ connection; (b)

equivalent reversed-Y connection

The solution to Eqs. (P4.1) is:

C1

C5

C4

C15

C14 C45

(a) (b)


1 2 314

1 2 2 3 3 1

1 2 315

1 2 2 3 3 1

1 2 345

1 2 2 3 3 1

2 65

2 86.67

2 130

f f fCf f f f f f

f f fCf f f f f f

f f fCf f f f f f

= = − + +

= = + −

= =

− +

(P4.3)

The capacitance values in Eqs. (P4.2) and (P4.3) are expressed in μF.

Figure P4.2 Simplifying sequence for capacitance circuit

As shown in Fig. P4.2, the capacitance between nodes a and b is:

( )14 125 345

14 125 345ab

C C CC

C C C+

=+ +

(P4.4)

with

15 2 45 3125 345

15 2 45 3

;C C C CC CC C C C

= =+ +

(P4.5)

With the numerical values of the problem, the following capacitance values are obtained:

C125 = 16.25 μF, C345 = 13.45 μF, and Cab = 20.38 μF.

a

C2

C14

C15

C3 C45

b a

C14

C125

C345

b


Problem 4.6

Figure P4.1 shows the electrostatic forces f1 and f2 applied to the mobile plate pair by

the fixed middle plate after a displacement x occurred.

Figure P4.1 MEMS capacitive actuator with transverse displacement

According to Eq. (4.22), the two forces are:

2 2

1 22 2;2 2

2 2

Av Avf fd dx x

= = − +

ε ε (P4.1)

The net force acting on the mobile plate pair is their difference, namely:

( ) ( )

21 2 2 216

2 2xf f f Adv

d x d x= − =

− +ε (P4.2)

Figure P4.2 illustrates the nonlinear variation of the net force f as a function of the mobile

plates displacement x. The displacement was limited to x = 2 μm when the force reached

approximately 5.5 x 10-17 N.

The minimum gap occurs for

2 500d dx− = (P4.3)

condition which yields x = (249/500)d = 2.49 μm. Using this particular value in Eq.

(P4.2) results in a force of f = 1.408 x 10-13 N. It can be seen that an increase of 0.49 μm

in displacement from the threshold value of 2 μm, which was used in the plot, generates

an increase of 2,560 times in the maximum force.

motion

f1

f2

d/2 - x

d/2 + x

x


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 10-6

0

1

2

3

4

5

6x 10

-17

x (m)

f (N

)

Figure P4.2 Net force acting on mobile plates as a function of displacement


Problem 4.7

The transverse and longitudinal voltages that are picked up transversely and

longitudinally by capacitive sensing are given in Eqs. (4.30) and (4.36) as

0 0

;bt l b

v xv x v vg x x

∆∆ = − ×∆ ∆ = − ×

+ ∆ (P4.1)

The ratio of these two voltages is

0 0

0 0 0

1t

l

v x x x xv g g g

∆ + ∆= = + ×∆

∆ (P4.2)

The relative voltage variation ratio of Eq. (P4.2) can be regarded as a function of Δx

provided x0 and g0 are constant. As such, it can be seen that Δvt = Δvl when Δx = g0 − x0.

The y-axis intercept of the voltage ratio plot is x0/g0, as given in Eq. (P4.2) and shown in

Fig. P4.1.

Figure P4.1 Voltage variation ratio as a function of plate displacement

As a consequence, when x0 ≥ g0 (which means x0/g0 ≥ 1) the transverse sensing is more

sensitive than longitudinal sensing because Δvt/ Δvl ≥ 1. When x0 < g0, there are two

situations possible, namely: for Δx ≤ g0 − x0 it follows that Δvt < Δvl and therefore the

transverse procedure is less sensitive than the longitudinal method; on the other hand, for

Δx > g0 − x0, as seen in Fig. P4.1, Δvt > Δvl indicating that transverse sensing is

preferable to longitudinal sensing.

Δvt/Δvl

Δx

x0/g0

g0 - x0

1


Problem 4.8

The transverse and longitudinal voltage variations are expressed in Eqs. (4.30) and

(4.36) as:

0

0

bt t

ll b

l

vv xg

xv vx x

∆ = − ×∆ ∆∆ = − × + ∆

(P4.1)

From the first Eq. (P4.1), the longitudinal displacement is

0t t

b

gx vv

∆ = − ×∆ (P4.2)

and its numerical value is Δxt = 4 μm. The longitudinal displacement is expressed from

the second Eq. (P4.1) as

0 ll

b l

x vxv v

∆∆ = −

+ ∆ (P4.3)

which yields a numerical value of Δxl = 3.18 μm. It can be seen that Δxt ≠ Δxl due to

imprecision in measuring by the longitudinal capacitive unit. Checking Eq. (P4.3)

indicates that the sources of imprecision in Δxl are unreliable determination of either of

x0, vb or Δvl.

Assuming x0 was assessed erroneously, Eq. (P4.3) yields:

'0

b tt

t

v vx xv+ ∆

= − ∆∆

(P4.4)

where the transverse displacement has been used instead of the longitudinal one. The

absolute value of '0x from Eq. (P4.4) is 44 μm.

Another variant is that the bias voltage was not assessed correctly, but from Eq.

(P4.3), the correct bias voltage is

' 0 lb l

t

x vv vx∆

= − −∆∆

(P4.5)

and its absolute value is 1.95 μV. Note that transverse displacement has been used in Eq.

(P4.5) as the precise measurement.

Eventually, in case the longitudinal voltage has not been detected precisely, its

correct equation is found by expressing it from Eq. (P4.3)


'

0

b tl

t

v xvx x

∆∆ = −

+ ∆ (P4.6)

(where, again, transverse displacement has been used instead of longitudinal

displacement) and its absolute value is 0.2 μV.


Problem 4.9

The transverse and longitudinal capacitive forces are given in Eqs. (4.22) and (4.32)

as

( )

2 2 2

200

;22t l

w v wvf fgg x

= =−

ε ε (P4.1)

where w is the side length of the square capacitive plates. Using the problem specific

relationships

010;3

t

l

f gxf= = (P4.2)

in conjunction with Eq. (P4.1) results in

0940

g w= × (P4.3)

For w = 80 μm, Eq. (P4.3) yields g0 = 18 μm.


Problem 4.10

The same equivalent inductance needs to be between any two of the vertices A, B,

and C for both the Δ connection and the reversed-Y connection of Fig. 4.43. Considering

the nodes A and B in the Δ connection, it is seen that LAB is in parallel with the series

connection of LCA and LBC; similarly, in the reversed-Y connection, LAD and LBD are in

series between nodes A and B; taking into consideration Eqs. (4.45) of the series and

parallel connection of impedances, the above-mentioned equality is written as

( )AB CA BCAD BD

AB CA BC

L L LL L

L L L+

= ++ +

(P4.1)

The following similar equations are formulated with respect to the node pairs B-C and

C-A:

( )

( )

BC AB CABD CD

BC AB CA

CA BC ABCD AD

CA BC AB

L L LL L

L L LL L L

L LL L L

+= + + +

+ = + + +

(P4.2)

Equations (P4.1) and (P4.2) form a system of three equations with three unknowns: the

inductances LAD, LBD and LCD. This system is written as:

( )

( )

( )

1 1 00 1 11 0 1

AB CA BC

AB CA BCAD

BC AB CABD

BC AB CACD

CA BC AB

CA BC AB

L L LL L L

LL L L

LL L L

LL L LL L L

+ + + + = + + +

+ +

(P4.3)

and its solution is obtained in MATLAB® by solving the following vector-matrix

equation using

( )

( )

( )

11 1 00 1 11 0 1

AB CA BC

AB CA BCAD

BC AB CABD

BC AB CACD

CA BC AB

CA BC AB

L L LL L L

LL L L

LL L L

LL L LL L L

−

+ + + + = + + +

+ +

(P4.4)

The result is:


AB CAAD

AB BC CA

BC ABBD

AB BC CA

CA BCCD

AB BC CA

L LLL L L

L LLL L L

L LLL L L

= + +

= + +

=

+ +

(P4.5)

With the numerical values of the problem, the impedances of Eqs. (P4.5) are: LAD = 22.4

mH, LBD = 32 mH, and LCD = 28 mH.


Problem 4.11

Any of the electrical circuits shown in Fig. P4.1 contains five internal rectangles,

which are not divided into any other fragments, and therefore, each of these circuits has

5 meshes inscribed within the same rectangular area and therefore, 5 degrees of freedom.

Figure P4.1 Circuit configurations with 5 meshes

A loop is formed of any mesh combination on condition that component meshes are

adjacent (they share at least one branch) but individual meshes also count as loops. As

such, the circuit shown in Fig. P4.1(a) is formed of 15 loops, as follows: 5 small

rectangles (the meshes), 4 rectangles each comprising two adjacent small rectangles, 3

rectangles formed of three adjacent small rectangles, 2 rectangles determined by four

adjacent small rectangles, and 1 rectangle formed of the five small rectangles.

Following a similar reasoning, the circuit of Fig. P4.1(b) contains 5 meshes, 7 areas

each formed of two adjacent rectangles (meshes), 8 areas comprising each three adjacent

rectangles, 4 areas comprising 4 adjacent rectangles, and 1 area with all five meshes –

the total number of loops is therefore 25.

Eventually, the circuit sketched in Fig. P4.1(c) is formed of 5 + 6 + 7 + 5 + 1 = 24

loops. Consider that each independent branch (circuit line between two different meshes)

has its own current passing through it. There are 11 branches in this circuit (you are

encouraged to detect them), and therefore, 11 currents through them. Assuming voltage

sources (for instance) and resistances are known, 6 nodes out of the total number of 7

nodes provide 6 constraint equation between the 11 currents (the last node, the seventh,

duplicates the equation of another node). As a consequence, the number of independent

(a) (b) (c)


parameters defining the electrical system is 11 – 6 = 5. It follows that the electrical

system has 5 DOF, and 5 currents (for instance) can be these degrees of freedom.

Let us design a circuit like the one of Fig. P4.1(c) with 6 resistors, one voltage source

and defined by 5 DOF such as the circuit of Fig. P4.2. If we consider the currents

already indicated in the figure based on Kirchhoff’s node law, it can be seen that the 5

currents: i (in the voltage source branch), i1, i3, i4, and i6 can be the system’s DOF. Note

that the assumption of a current in the source branch is consistent with a real voltage

source having an internal resistance. Ignoring this current (i = 0) would result in only 4

currents being necessary to describe of the system and therefore, the number of DOF

would be 4.

Figure P4.2 Circuit configuration with 5 meshes, resistors and a voltage source

i1 i1 + i3 i3

i4 i6

R1

+ – v

R2

R4 R5 R3

R6

i

i – i1

i – i1 – i4

i – i1 – i4 – i6

i4 + i6 – i3

A

B


Problem 4.12

(a)

The nodes and currents are shown in Fig. P4.1. According to the rule of thumb, this

electrical system has four meshes and therefore it needs 4 DOF. Indeed, the total number

of branch currents is 11; there are also 8 nodes, out of which only 7 provide independent

relationships between branch currents (by means of Kirchhoff’s node law), which means

there are 7 constraints. As a consequence, the number of DOF or the system

configuration is 11 – 7 = 4, as also determined by the rule-of-thumb reasoning.

Figure P4.1 Currents and nodes in a 4-mesh electrical circuit

(b)

The following current equations are written for the nodes A, B and C based on Fig.

P4.1:

1 2

2 3 4

3 5 6

ii i ii i ii i i

= + = + = +

(P4.1)

At node D, the following equation applies:

1 4 6 7oi i i i= + − (P4.2)

Similar equations are found for nodes E, F, and G:

5 8 9

10 7 9

11 1 8

i i ii i ii i i

= + = + = +

(P4.3)

whereas for node H, the current equation applies:

io2

ii

io1

i1

i3 i5 i8

i4 i6 i9 i11

i7 i10

A

B

C E

D F H

G

i2


2 10 11oi i i= + (P4.4)

Combining the last two Eqs. (P4.3) with Eq. (P4.4) yields:

2 7 9 1 8oi i i i i= + + + (P4.5)

Adding now Eqs. (P4.2) and (P4.5) results in

1 2 1 4 6 8 9o oi i i i i i i+ = + + + + (P4.6)

Using the first Eq. (P4.3) and Eqs. (P4.1), transform Eq. (P4.6) to:

1 2 1 4 6 5 1 4 3 1 2o o ii i i i i i i i i i i i+ = + + + = + + = + = (P4.7)

which demonstrates that that the current entering the electrical system is equal to the

sum of currents exiting the system, as expected.


Problem 4.13

The series connection of any number of identical inductors yields the maximum

equivalent inductance, while their parallel connection generates the minimum equivalent

inductance. Conversely, identical capacitors have to be connected in parallel to produce

the maximum equivalent capacitance, and in series to obtain the minimum equivalent

capacitance. It is also known that inductances and capacitances enter the denominator of

the natural frequency. As a consequence, the electrical circuit of Fig. P4.1(a), which

yields the minimum equivalent inductance and capacitance, generates the maximum

natural frequency, whereas the electrical circuit of Fig. P4.1(b), with inductors in series

and capacitors in parallel, results in the minimum natural frequency.

Figure P4.1 Circuits with inductors and capacitors for: (a) maximum natural frequency; (b) minimum

natural frequency

The equivalent inductance and capacitance for the circuit of Fig. P4.1(a) are:

,min

,min

3

2

e

e

LL

CC

= =

(P4.1)

which leads to the natural frequency

,max,min ,min

1 6n

e eL C LC= =ω (P4.2)

The equivalent inductance and capacitance for the circuit of Fig. P4.1(b) are:

,max

,max

32

e

e

L LC C

= =

(P4.3)

which leads to the natural frequency

L

L

C C

L

C

C

L L L

(a) (b)


,min,max ,max

1 16n

e eL C LC= =

×ω (P4.4)

The ratio of the natural frequencies of Eqs. (P4.2) and (P4.4) is

,max

,min

6n

n

=ωω

(P4.5)


Problem 4.14

Figure P4.1 shows the original circuit with charges on the three branches.

Figure P4.1 Two-mesh electrical circuit with charges

The total electrical energy corresponding to the inductors and capacitors is:

( ) ( )2 22 21 1 2 2 1 2

1 1 1 12 2 2 2

E Lq L q q q q qC C

= + − + + − (P4.1)

The electrical system being conservative, the time derivative of the total energy of Eq.

(P4.1) needs to be zero, which yields:

1 1 1 2 1 2 2 1 2 2 1 21 1 1 1 1 0q Lq Lq Lq q q q Lq Lq q q qC C C C C

+ − + − + − + + − + = (P4.2)

Because the charge rates (currents) cannot e zero at all times, Eq. (P4.2) is satisfied

when its two parentheses are simultaneously zero:

1 2 1 2

1 2 1 2

1 12 0

1 2 0

Lq Lq q qC C

Lq Lq q qC C

− + − =− + − + =

(P4.3)

Solution of the sinusoidal type:

( )( )

1 1

2 2

sin

sin

q Q t

q Q t

=

=

ω

ω (P4.4)

is needed in Eq. (P4.3), which yields:

2 21 2

2 21 2

1 12 0

1 2 0

L Q L QC C

L Q L QC C

− + + − = − + − + =

ω ω

ω ω (P4.5)

C L

L

C

q1 q2

q1 – q2


Equations (P4.5) have nontrivial solution when the determinant of the system is zero,

namely:

2 2

2 2

1 120

1 2

L LC C

L LC C

− + −=

− − +

ω ω

ω ω (P4.6)

Equation (P4.6) has the following roots, which are the electrical system’s natural

frequencies:

,1

,2

3 52

3 52

n

n

LC

LC

− = +

=

ω

ω

(P4.7)

The ratio of the two natural frequencies of Eqs. (P4.7) is:

,2

,1

3 5 2.6183 5

n

n

+= =

−

ωω

(P4.8)


Problem 4.15

Figure P4.1 shows the original circuit with charges on the three branches.

Figure P4.1 Two-mesh electrical circuit with charges

The total energy corresponding to the four electrical components is

( )22 2 21 1 2 1 1 2 2

1 2

1 1 1 12 2 2 2

E q q q L q L qC C

= + − + + (P4.1)

The electrical system is conservative and the energy time derivative is zero, which yields

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

1 1 1 1 1 0L q q q q q L q q q qC C C C C

+ + − + − + =

(P4.2)

Equation (P4.2) is valid at all times when

1 1 1 2

1 2 2

2 2 1 22 2

1 1 1 0

1 1 0

L q q qC C C

L q q qC C

+ + − =

− + =

(P4.3)

Sinusoidal solution of the form

( )( )

1 1

2 2

sin

sin

q Q t

q Q t

=

=

ω

ω (P4.4)

is needed, which, substituted in Eq. (P4.3) leads to

21 1 2

1 2 2

21 2 2

2 2

1 1 1 0

1 1 0

L Q QC C C

Q L QC C

− + + − =

− + − + =

ω

ω

(P4.5)

Equations (P4.5) have nonzero Q1 and Q2 solutions for

C1 L2

L1

C2

q1 q2

q1 – q2


21

1 2 2

22

2 2

1 1 1

01 1

LC C C

LC C

− + + −

=− − +

ω

ω (P4.6)

which is the characteristic equation. For the numerical parameters of this problem, its

roots (the natural frequencies) are ωn1 = 115.79 rad/s and ωn2 = 333.16 rad/s.

The second Eq. (P4.5) gives the amplitude ratio

212 2

2

1Q L CQ

= −ω (P4.7)

For ω = ωn1, the ratio of Eq. (P4.7) becomes

1

(1)1 1

(1)2 2

0.5495n

Q QQ Q

=

= =ω ω

(P4.8)

Equation (P4.8) shows that the charges have either the directions chosen in Fig. P4.1 or

directions opposite to the ones shown in Fig. P4.1; the same equation indicates that the

amplitude of q1 is smaller than the amplitude of q2. The eigenvector is of unit-norm for

( ) ( )2 2(1) (1)1 2 1Q Q+ = (P4.9)

Solving Eqs. (P4.8) and (P4.9) for the two amplitudes results in the following

eigenvector corresponding to ω = ωn1

{ }(1) 0.48160.8764

Q =

(P4.10)

For ω = ωn2, the ratio of Eq. (P4.7) becomes

2

(2)1 1

(2)2 2

2.7295n

Q QQ Q

=

= = −ω ω

(P4.11)

which shows that the directions of the two charges in the circuit of Fig. P4.1 are opposite

(either q1 or q2 opposes the direction in the figure) and that the amplitude of q1 is larger

than the amplitude of q2. For a unit-norm eigenvector it is necessary that

( ) ( )2 2(2) (2)1 2 1Q Q+ = (P4.12)

Solving Eqs. (P4.11) and (P4.12) for the unknown amplitudes generates the eigenvector

related to ω = ωn2


{ }(2) 0.9390.344

Q = −

(P4.13)


Problem 4.16

The mathematical model for the electrical system of Fig. 4.46 is:

1 1 1 2

1 2 2

2 2 1 22 2

1 1 1 0

1 1 0

L q q qC C C

L q q qC C

+ + − =

− + =

(P4.1)

which can be written in vector-matrix form as

1 1 2 2 11

2 2 2

2 2

1 1 10 0

0 1 1 0q C C C qL

L q qC C

+ − + = −

(P4.2)

The natural frequencies are determined using MATLAB® as the eigenvalues of the

dynamic matrix

[ ] [ ] [ ]1D L C−= (P4.3)

The command

>> [V,D] = eig(D)

returns the eigenvectors and eigenfrequencies

v =

0.9390 0.4816

-0.3440 0.8764

D =

1.0e+005 *

1.1100 0

0 0.1341

The eigenvectors are the ones that have been obtained in the previous Problem 4.15; the

square roots of the eigenvalues yield the two natural frequencies determined in Problem

4.15.


Problem 4.17

The mathematical model of a free electrical system (with no voltage or current

source) is given in Eq. (4.91), whose standard format is based on the natural frequency

ωn and the damping ratio ξ, as provided in Eq. (4.93). The damping ratio is expressed as

in Eq. (4.94) as

2R C

L=ξ (P4.1)

The electrical system’s period is connected to its damped frequency as

2 2 2

2 2 2 11 2 1 1

dd n n n

Tf f

= = = =− − −

π π πω ω ξ π ξ ξ

(P4.2)

where fn is the system’s natural frequency (measured in Hz). The damping ratio is

expressed from Eq. (P4.2) as:

2 2

11n df T

= −ξ (P4.3)

whose value is ξ = 0.3049. The natural frequency is

12nf LC

=π

(P4.4)

The critical damping is defined by ξ = 1, and therefore, Eq. (P4.1) yields the expression

of the critical resistance

2crLRC

= (P4.5)

Equations (P4.4) and (P4.5) are solved for L and C, which are:

2

1

4

n cr

cr

Cf R

CRL

= =

π (P4.6)

The electrical components’ parameters are C = 5.3 x 10-7 F and L = 0.0053 H. The actual

resistance is determined now from Eq. (P4.1) as

2 LRC

= ξ (P4.7)

and its value is R = 60.98 Ω.


Problem 4.18

When the circuit resistance is not accounted for, the equation describing the circuit is

( )Ldi tv Ldt

= (P4.1)

whereas the equation

( ) ( )LRLR

di tv L Ri tdt

= + (P4.2)

governs the circuit behavior for both inductance and resistance considered. For zero

initial conditions, the solutions to the two equations above are

( )

( )

L

RtL

LR

vi t tLv vi t eR R

−

= = −

(P4.3)

The inductance and resistance are calculated as

2 2 2 2

2 2

( / 2)4 44

( / 2)

N D N D NDLl N D

l N D NDRA d d

= = = × ×

× = = =

πµ πµ µπ

ρ ρ π ρπ

(P4.4)

The numerical values of L and R are L = 0.25 H and R = 170 Ω. Figure P4.1 shows the

plots of the two currents where the current iLR has a steady-state value (when t → ∞) of

v/R = 0.47 V.


0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x 10-3

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Time (sec)

Cur

rent

s (A

)

iLiLR

Figure P4.1 Currents through ideal inductor and real inductor as functions of time


Problem 4.19

The currents and their positive directions in the electrical circuit are shown in Fig.

P4.1. Because the circuit has four meshes, it follows that it has four DOF; therefore, four

currents can be selected as DOF and four independent algebraic equations are needed to

solve for the currents.

Figure P4.1 Electrical circuit with resistors and constant voltage source with currents

Using Kirchhoff’s node law the following current relationships are expressed

4 2 3

5 3 6

7 4 6 2 3 6

8 5 7 3 6 2 3 6 2

i i ii i ii i i i i ii i i i i i i i i

= − = + = − = − − = + = + + − − =

(P4.1)

The currents i1, i2, i3, and i6 are selected to be the DOF of this system. Kirchhoff’s

voltage law is applied to the four meshes of Fig. P4.1, which results in

( )( )

7 1

5 2 4 3 6 2 7 1

1 2 3 2 6 4 3

3 2 3 6 2 6

00

0

R i vR i R i R i R iR i i R i R i

R i i i R i

= + + − = − + − = − − − =

(P4.2)

The first Eq. (P4.2) allows solving for i1 as i1 = v/R7 = 4 A. The remaining Eqs. (P4.2)

are written as

( )

( )( )

5 6 2 4 3

1 2 1 4 3 2 6

3 2 3 3 2 3 6

0

0

R R i R i v

R i R R i R i

R i R i R R i

+ + =

− + + = − − + =

(P4.3)

which can be written in vector-matrix form as

R2 R4

R1

R3

R5

v +

–

R6

R7

i1 i2 i3

i4

i6

i5

i7 i8


( )( )

25 6 4

1 1 4 2 3

3 3 2 3 6

000

iR R R vR R R R iR R R R i

+ − + = − − +

(P4.4)

Equation (P4.4) allows solving for the unknown currents, namely

( )( )

12 5 6 4

3 1 1 4 2

3 3 2 36

000

i R R R vi R R R R

R R R Ri

−+

= − + − − +

(P4.5)

The currents values are i2 = i8 = 1.426 A, i3 = 1.079 A, i6 = 0.189 A, i4 = 0.347 A, i5 =

1.268 A, and i7 = 0.158 A.


Problem 4.20

Figure P4.1 shows the currents in the circuit branches and the mesh positive

directions. Having three meshes, the electrical system has three DOF, which can be the

currents i1, i2, and i3.

Figure P4.1 Electrical circuit with resistors, inductors, and sinusoidal voltage source

The following equations are obtained after expressing Kirchhoff’s second law for each

mesh

[ ]

[ ]

13 2 2

1 1 2 1 3 2 2

2 1 2 3 1 3

( ) ( ) ( )

( ) ( ) ( ) ( ) 0

( ) ( ) ( ) ( ) 0

di tL R i t v tdt

dL i t i t R i t R i tdtdL i t i t i t R i tdt

+ = − + − = − − − =

(P4.1)

The following MATLAB® code

>>[i1,i2,i3]=dsolve('6*Di1+150*i2=20*sin(6*t),5*Di1...

-5*Di2+100*i3-150*i2=0,8*Di1-8*Di2-8*Di3...

-100*i3=0','i1(0)=0,i2(0)=0,i3(0)=0','t');

>>t=0:0.0001:10;

>>curr1=i1;

>>curr2=i2;

>>curr3=i3;

>>plot(t,curr1)

>>xlabel('Time (sec)')

>>ylabel('i_1 (A)')

R1 L3 L1

L2

R2

v

i1

i1 – i2

i3

i2

i1 – i2 – i3


generates the currents plotted in Fig. P4.2. The currents obtained through solving the

differential equations are:

16.795 70.7051

16.795 70.7052

16.795 70.7053

0.175 0.204cos(6 ) 0.088sin(6 ) 0.027 0.001

0.021cos(6 ) 0.084sin(6 ) 0.018 0.003

0.033cos(6 ) 0.072sin(6 ) 0.035 0.002

t t

t t

t t

i t t e ei t t e ei t t e e

− −

− −

− −

= − + + +

= − + + + = − + + −

(P4.2)

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

Time (sec)

i 1 (A

)

0 1 2 3 4 5 6 7 8 9 10-0.1

-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

Time (sec)

i 2 (A

)


0 1 2 3 4 5 6 7 8 9 10-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

Time (sec)

i 3 (A

)

Figure P4.2 Main currents as functions of time


Problem 4.21

For S1 closed and S2 open, a charge

0q Cv= (P4.1)

charges the capacitor. When switches change states, the charge q0 = 50 x 10-6 F x 80 V =

0.004 C flows (and discharges) through the electrical circuit formed of the resistor,

inductor, and capacitor. The equation governing the time response for the second phase

is

1 0Lq Rq qC

+ + = (P4.2)

The following initial conditions are used

(0) 0; (0) 0.004q q= = (P4.3)

in conjunction with the MATLAB® code

>> dsolve('0.6*D2q+120*Dq+1/(50*10^(-6))*q=0','q(0)=0.004','Dq(0)=0')

and the time-domain charge is obtained

( ) ( ) 1000.0026sin 152.75 0.004cos 152.75 tq t t e−= + (P4.4)

By taking the derivative of q(t) of Eq. (P4.4) and using the MATLAB® command

>>i=diff(q,t)

returns the current

( ) ( ) 1000.871sin 152.75 0.0029cos 152.75 ti t t e−= − + (P4.5)

which is plotted in Fig. P4.1.


0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-0.4

-0.35

-0.3

-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

Time (sec)

i (A

)

Figure P4.1 Current as a function of time


Problem 4.22

When the switch is closed, the current passes through the two resistors, and the

mathematical model is simply:

( )1 2R R i v+ = (P4.1)

which results in a current i = i0 = v/ (R1 + R2) = 0.25 A. The voltage across the capacitor

is equal to the voltage on the resistor R2 (the two components being in parallel), and is

equal to

22

1 2C

Rv iR vR R

= = ×+

(P4.2)

with a value of 50 V. The capacitor charge (its initial value) is therefore

0 Cq Cv= (P4.3)

which is q0 = 0.001 C.

When the switch is opened, the voltage source and the resistor R1 are removed from

the circuit; the initial capacitor charge q0 discharges through C and R2 in their circuit.

The mathematical model for that stage is

21 0R q qC

+ = (P4.4)

Equation (P4.4) is solved symbolically by using the MATLAB® code:

>> syms r2 c q0

>> dsolve('r2*Dq+1/c*q=0','q(0)= q0')

and the time-domain charge is obtained

20

tR Cq q e

−

= (P4.5)

The current is the time derivative of the charge of Eq. (P4.5), namely:

20

2

tR Cqi q e

R C

−

= = − (P4.6)

which is plotted in Fig. P4.1.


0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02-0.25

-0.2

-0.15

-0.1

-0.05

0

Time(sec)

i(A)

Figure P4.1 Current through R2 as a function of time


Problem 4.23

Figure P4.1 shows the circuit with the currents and positive mesh directions. Having

three meshes, the electrical system has three DOF.

Figure P4.1 Electrical circuit with currents and positive mesh directions

Using Kirchhoff’s voltage law, the following equations can be written for the three

meshes:

[ ]

[ ]

[ ] [ ]

21 2 2 3 1

1

21 2 3 2

2

31 1 3 2 3 1 1 2 3

2

( )1 ( ) ( ) ( )

( )1 ( ) ( ) ( ) 0

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

di ti t d t L R i t i t vC dt

di ti t i t i t d tLC dt

di tL R i t R i t i t i t i t i t d tdt C

+ − − =

− − − =

+ + − − − − =

∫

∫

∫

(P4.1)

which can also be written in terms of charges as

( )

232 1

1 2 2 221

22

1 2 3 2 22 2 2

23 3 1

1 1 2 2 1 2 322 2 2

( )( ) ( )1 ( )

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

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

dq td q t dq tq t L R R vC dt dt dt

d q tq t q t q t LC C C dt

d q t d qt dq tL R R R q t q t q tdt dt dt C C C

+ − + =

− − − = + + − − + + =

(P4.2)

Equations (P4.2) represent the mathematical model of the studied electrical system.

v

i1

A

B

C

D

R2

R1 C1

i3

i1 i1 − i2

C2

L1

L2 i2

i3 − i1 i1 − i2 − i3

+ –


Problem 4.24

In terms of charges, the mathematical model of the electrical system of Fig. 4.51 is:

( )

232 1

1 2 2 221

22

1 2 3 2 22 2 2

23 3 1

1 1 2 2 1 2 322 2 2

( )( ) ( )1 ( )

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

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

dq td q t dq tq t L R R vC dt dt dt

d q tq t q t q t LC C C dt

d q t d qt dq tL R R R q t q t q tdt dt dt C C C

+ − + =

− − − = + + − − + + =

(P4.1)

The second Eq. (P4.1) is used first to express the second derivative of q2, which is then

substituted into the first Eq. (P4.1); Equations (P4.1) are rewritten as

311 2 3

2 1 2 2 2 2 2 2

22

1 2 322 2 2 2 2 2

23 32 1 1 2

1 2 321 1 1 2 1 2 1 2

( )( ) 1 1 1 1 1 1( ) ( ) ( )

( ) 1 1 1( ) ( ) ( )

( ) ( )( ) 1 1 1( ) ( ) ( )

dq tdq t q t q t q t vdt dt R C C R C R C R

d q t q t q t q td t L C L C L C

d q t d qtR dq t R R q t q t q tdt L dt L dt L C L C L C

= − + + × + × + ×

= × − × − ×

+= × − × + × − × − ×

(P4.2)

In short-form to be used in Simulink®, Eqs. (P4.2) are written as

311 1 2 2 2 3 3

22

1 2 32

23 31

1 2 3 1 3 2 3 32

( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

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

dq tdq t a q t a q t a q t a vdt dt

d q t bq t bq t bq tdt

d q t d qtdq tc c c q t c q t c q tdt dt dt

= − + + + = − −

= − + − −

(P4.3)

where

1 2 32 1 2 2 2 2

2 2

2 1 21 2 3

1 1 1 2

1 1 1 1 12.921; 1.623; 0.0045

1 178.571

1183.333; 475; 297.619

a a aR C C R C R

bL CR R Rc c cL L L C

= + = = = = =

= = +

= = = = = =

(P4.4)

The Simulink® diagram solving Eqs. (P4.3) is shown in Fig. P4.1 and the time variations

of the three currents are plotted in Fig. P4.2.


v

-K-

-K-

-K-

-K-

-K-

-K-

-K-

b1

-K-

-K-

-K-

-K-

-K-

Scope2

Scope1

Scope

1s

Integrator4

1s

Integrator3

1s

Integrator2

1s

Integrator1

1s

Integrator

Add2

Add1

Add

i_1 q_1

i_2

q_2

i_3 q_3

Figure P4.1 Simulink® diagram integrating Eqs. (P4.3)


Figure P4.2 Time variation of currents by Simulink® integration


Problem 4.25

Figure P4.1 shows the currents through the circuit.

Figure P4.1 Electrical circuit with current source and currents

The system has three meshes, and, therefore it possesses three DOF. By the node

analysis method, one of the nodes a, b, c or d can be selected as reference node, and, as a

consequence, the voltages of the three remaining nodes can be the system’s DOF.

Assuming the voltage at node c is vc = 0, the voltages va, vb and vd play the role of DOF.

The currents shown in Figure P4.1 are related as

1 2

1 3 4

5 2 4

i i ii i ii i i

= + = + = +

(P4.1)

Equations (P4.1) are expressed in terms of significant voltages as

1 2

1 3 4

5 2 4

a b a d

a b b b d

d a d b d

v v v viR R

v v v v vR R R

v v v v vR R R

− −= +

− −

= + − −

= +

(P4.2)

Equations (P4.2) can be rearranged in the form

a

c

R4 R1

R2

R5 R3

i

b d i1

i2

i4

i3

i5


1 2 1 2

1 1 3 4 4

2 4 2 4 5

1 1 1 1

1 1 1 1 1 0

1 1 1 1 1 0

a b d

a b d

a b d

v v v iR R R R

v v vR R R R R

v v vR R R R R

+ − − =

− + + + =

+ − + + =

(P4.3)

Equations (P4.3) are written in vector-matrix form as

1 2 1 2

1 1 3 4 4

2 4 2 4 5

1 1 1 1

1 1 1 1 1 00

1 1 1 1 1

a

b

d

R R R Rv iv

R R R R Rv

R R R R R

+ − −

− + + =

− + +

(P4.4)

The unknown voltage vector can be found as

1

1 2 1 2

1 1 3 4 4

2 4 2 4 5

1 1 1 1

1 1 1 1 1 00

1 1 1 1 1

a

b

d

R R R Rv iv

R R R R Rv

R R R R R

−

+ − −

= − + + − + +

(P4.5)

Using the numerical values of the problem and MATLAB®, the following voltages are

obtained: va = 0.1209 V, vb = 0.0643 V, and vd = 0.0835 V. The currents are therefore:

11

22

33

44

55

a b

a d

b

b d

d

v viR

v viR

viRv vi

RviR

−=

−

= = −

=

=

(P4.6)


their numerical values being i1 = 1.1 mA, i2 = 1.9 mA, i3 = 1.6 mA, i4 = 0.48 mA, and i5

= 1.4 mA.


Problem 4.26

Figure P4.1 shows the currents in the branches of the electrical circuit.

Figure P4.1 Electrical network comprising resistors and a current source to be modeled by the node

analysis method

For the circuit of Fig. P4.1, if node B is considered to be the reference node with zero

voltage, it follows the entire line BD is grounded, and therefore, the voltage of node D is

also zero. By using Kirchhoff’s node law, two node equations can be written when the

voltages vA and vC are associated with nodes A and C, namely:

1 2 1 2

2 3 4

2 3 4

0

or0 0

A CA

A C C C

v vvii i i R Ri i i v v v v

R R R

−− = += + = + − − − = +

(P4.1)

Equation (P4.1) can be written as:

1 2 2

2 2 3 4

1 1 1

1 1 1 1 0

A C

A C

v v iR R R

v vR R R R

+ − =

− + + + =

(P4.2)

and, further, in vector-matrix form as

1 2 2

2 2 3 4

1 1 1

1 1 1 1 0A

C

R R R v iv

R R R R

+ − = − + +

(P4.3)

The voltages vA and vC are determined from Eqs. (P4.3) by using matrix inversion and

multiplication with MATLAB® as:

R2

R1 R3

C

D

i3 i1 i2

R4 A

B

i4

i

–


( )1 2 3 2 4 3 4

1 3 1 4 2 3 2 4 3 4

1 3 4

1 3 1 4 2 3 2 4 3 4

A

C

R R R R R R R iv

R R R R R R R R R RR R R iv

R R R R R R R R R R

+ += + + + +

= + + + +

(P4.4)

Once the voltages vA and vC have been obtained, the 4 currents shown in Fig. P4.1 can be

determined as:

1 2 3 41 2 3 4

; ; ;A C C CA v v v vvi i i iR R R R

−= = = = (P4.5)

For the parameters of this problem, the following numerical values are obtained: vA =

0.02 V, vC = 0.003 V, i1 = 0.663 mA, i2 = 0.338 mA, i3 = 0.296 mA, and i4 = 0.043 mA.


Problem 4.27

Figure P4.1 shows the operational amplifier circuit and the current passing through

the two electrical components.

Figure P4.1 Operational amplifier in a negative feedback circuit with input capacitor and feedback

resistor

Because the operational amplifier has infinite input impedance, there is no current

flowing through the op amp and therefore, the current through the capacitor is equal to

the current through the feedback branch. This current is expressed in two ways, for the

input branch and for the feedback branch, namely:

( )0

0

i

o

di C vdt

viR

= − − =

(P4.1)

Equations (P4.1) took into account that the plus and minus input voltages are equal to

zero due to negative feedback and ground connection of the positive input terminal.

These equations result in:

io

dvv RCdt

= − (P4.2)

which indicates the effect of this particular op amp circuit is that of differentiation.

Taking into account that vi = ct (with c = 80), the solution to Eq. (P4.2) is

ov RCc= − (P4.3)

and therefore the output voltage is constant. Figure P4.2 plots the input and output

voltages for the parameter numerical values of this problem.

vi vo

R

C

i

+ –

i


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-600

-500

-400

-300

-200

-100

0

100

200

Time (sec)

Inpu

t an

d ou

tput

vol

tage

s (V

)

vi

vo

Figure P4.2 Input and output voltages in terms of time


Problem 4.28

Figure P4.1 shows the operational amplifier circuit and the currents passing through

its components.

Figure P4.1 Operational amplifier with three components

The three currents highlighted in Figure 4.1 are related as:

1 2i i i= + (P4.1)

These currents are connected to their components and relevant voltages as

[ ]

[ ]

1

2

( ) 0

1( ) ( ) 0

0 ( ) ( )( )

i

i

o o

v tiR

i t v t dtLd v t dv ti t C C

dt dt

− = = − −

= = −

∫ (P4.2)

Equations (P4.2) took into account that the minus and plus input terminals are under zero

voltage since there is negative feedback and the plus terminal is connected to the ground.

Combining Eqs. (P4.1) and (P4.2) yields:

( )1 1( ) ( ) oi i

dv tv t v t d t CR L dt

+ = −∫ (P4.3)

which, after using the numerical value of vi and integration, can be written as

( ) 1( ) 1 40 4,000 ln 1 0.01

1 0.01odv t t kdt RC t LC

= − × − × + ++

(P4.4)

At t = 0, i = 0, and therefore, dvo(t)/dt = 0 – as follows from the third Eq. (P4.2). The

constant of Eq. (P4.4) is found as k1 = 40/(RC). Integrating now Eq. (P4.4) yields:

C

R

L + –

vi vo

i

i1

i2


( ) ( ) ( ) 1 24,000 4,000( ) ln 1 0.01 100 ln 1 0.01 100ov t t t t t k t kRC LC

= − × + − × + + − − + + (P4.5)

The output voltage is zero at t = 0, and therefore the constant k2 is determined from Eq.

(P4.5) as k2 = 400,000/(LC). As a consequence, the output voltage equation is

( ) ( )40 1 100 4,000 1 100 4,000( ) ln 1 0.01 ln 1 0.01ov t t t t tC R L C R L LC = + − + + − × +

(P4.6)

For the numerical values of this problem, the output voltage is plotted in Figure P4.2. It

can be seen that vo increases nonlinearly to very large negative values quite fast.

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-1200

-1000

-800

-600

-400

-200

0

200

Time (sec)

v o (V

)

Figure P4.2 Output voltage as a function of time


Problem 4.29

Figure P4.1 shows the electrical circuit after transforming the Δ (delta) connection

formed of the original resistors R1, R4, and R5 into the reversed-Y connection formed of

R14, R15, and R45.

Figure P4.1 Equivalent electrical circuit with input and output voltages

The new resistances are calculated as

1 414

1 4 5

4 545

1 4 5

1 515

1 4 5

R RRR R R

R RRR R R

R RRR R R

= + +

= + +

=

+ +

(P4.1)

The currents i1 and i2 are determined as functions of i by solving the equation system

( ) ( )1 2

15 2 1 45 3 2

i i iR R i R R i+ =

+ = + (P4.2)

as

45 31

45 3 15 2

15 22

45 3 15 2

R Ri iR R R R

R Ri iR R R R

+ = + + + + = + + +

(P4.3)

The input and output voltages are expressed as:

( )14 15 2 1

2 1 3 2

i

o

v R i R R iv R i R i= + +

= −

(P4.4)

a

b

i

i1

i2

vi

vo c

R14

R15

R45

R2

R3

d


Substituting i1 of the first Eq. (P4.3) into vi of Eq. (P4.4) yields the following current:

( )( )15 2 45 3 114

15 2 45 3

1 1i i

e

i v vR R R R RRR R R R

= × = ×+ +

++ + +

(P4.5)

whereas using the two currents of Eqs. (P4.3) in vo of Eq. (P4.4) results in:

2 45 3 152

15 2 45 3o e

R R R Rv i R iR R R R

−= × = ×

+ + + (P4.6)

Figure P4.2 shows the main operations that are performed in Simulink®, and which

are based on Eqs. (P4.5) and (P4.6), in order to plot the output voltage vo as a function of

vi and the saturation nonlinearity of R1.

v_i

Scope

Product1

Product

2

R_e2 in

1

1/R_e1 in

i

v _o

Figure P4.2 Main Simulink® diagram for solving for and plotting the output voltage

Figures P4.3 and P4.4 indicate the calculations that are necessary to apply in Simulink®

to obtain the equivalent resistances Re1 and Re2, and the plot of the output voltage. Please

note that the saturation limits corresponding to 1/Re1 that enters the first product of Fig.

P4.2 have to be set as vmax/R1 and vmin/R1 respectively; similarly, the saturation limits

corresponding to Re2 that enters the second product of Fig. P4.2 have to be set as vmax

and vmin, respectively, where vmax = 30 V and vmin = −20 V. As the plot of Fig. P4.4

shows it, the saturation on R1 does not alter the sinusoidal shape of the output voltage.


v_i

Scope

Saturation1

Saturation

-K-

R_5

310

R_4 + R_5

310

R_4 + R5

-K-

R_4 * R_5

-K-

R_4 * R5

-K-

R_4

210

R_3

390

R_2 + R_3

180

R_2

210

R_1 (linear)

-K-

R5 210

R3

390

R2 + R3

180

R2

Product9

Product8

Product7

Product6

Product5

Product4

Product3

Product2

Product1

Product

1

uMath

Function5

1

uMath

Function4

1

uMath

Function3

1

uMath

Function2

1

uMath

Function1

1

uMath

Function

Add7

Add6Add5

Add4

Add3

Add2

Add1Add

1/210

1/R_1 (linear)

i

v _o

R_1s (sat)

R_1s + R_4 + R_5 1/(R_1s + R_4 + R_5)

R_1s* R_4

R_14

R_1s * R_5

R_45

R_15

1/(R_2+R_3+R_45+R_15)

(R_3+R_45)*(R_2+R_15)/(R_2+R_3+R_45+R_15)

R_2+R_15

R_3 + R_45

(R_3+R_45)*(R_2+R_15)

R_1sR_1s

1/(R_2+R_3+R_45+R_15)

R_3*R_15

R_2*R_45

R_2*R_45-R_3*R_15)

Figure P4.3 Full Simulink® diagram for the calculation of vo, Re1 of Eq. (P4.5) and Re2 of Eq. (P4.6)


Figure P4.4 Time-domain plot of output voltage vo


Problem 4.30

The following mathematical model describes the dynamics of the electrical system of

Fig. 4.48:

[ ]

[ ]

13 2 2

1 1 2 1 3 2 2

2 1 2 3 1 3

( ) ( ) ( )

( ) ( ) ( ) ( ) 0

( ) ( ) ( ) ( ) 0

di tL R i t v tdt

dL i t i t R i t R i tdtdL i t i t i t R i tdt

+ = − + − = − − − =

(P4.1)

which is written to comply with Simulink® integration capabilities as

1 22

3 3

2 1 2 12 3

1 1

3 1 2 13

2

( ) 1( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

di t R i t v tdt L L

di t di t R Ri t i tdt dt L L

di t di t di t R i tdt dt dt L

= − +

= − +

= − −

(P4.2)

Figure P4.1 shows the Simulink® diagram solving Eqs. (P4.2).

Sine Wave

Scope2

Scope1

Scope

1s

Integrator2

1s

Integrator1

1s

Integrator

-K-

Gain4

20

Gain3

30

Gain2

-K-

Gain1

25

Gain

Add2

Add1

Add

di_1/dt

di_2/dt

di_3/dt

i_1

i_2

i_3

Figure P4.1 Simulink® diagram for calculation and plotting of the currents


Figure P4.2 illustrates the three currents that are plotted as functions of time.

Figure P4.2 Simulink® plots of currents

Chapter 04

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Transcript of Chapter 04