Lecture5-ModellingElectricalSystemsI

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Page : 1 EE406 Control Systems Lecture 5 : Modeling Electrical Systems I

UCSI University

Faculty of Engineering

Kuala Lumpur, Malaysia

Department of Mechatronics

Lecture 5

Modeling Electrical Systems I

Mohd Sulhi bin Azman

Lecturer

Department of Mechatronics

UCSI University

[email protected]

1 August 2011

Contents

• Basic concepts : voltage, current and charge.

• Ohm’s law revisited.

• Resistors, Capacitors, Inductors and their

equivalent systems.

• Modeling electrical circuits.

(2)

Current

• Electrical current, consisting of a flow of

electrons, is defined as the rate of charge flow.

• The formula is given by:

• SI Unit for:

– Charge : Coulomb (C)

– Current : Ampere (A)

– Time : Seconds (s)

dQ

I

dt

=

Voltage

• A quantity that measures the electrical energy.

• Sometimes, it is also called the potential difference.

• Essentially, voltage is the work done or the energy

required to move charges from one point to the other.

• Formula:

• SI Unit:

– Work : Joules (J)

– Voltage : Volt (V)

work

charge

W

V

Q

(3)

Ohm’s Law

• It describes the relationship between the

voltage, current and resistance.

Voltage = Current × Time

V

=

IR

Resistors

• Electrical elements whose primary function is to

oppose (resist) the flow of electric current.

• The symbol used is R and the unit is Ohm (Ω).

(4)

Equivalent System for Resistors

Parallel resistors

Series resistors

1

2

3

n

eq

i

n

R

=

= +

+ + +

∑

⋯

1

2

1

n

i

eq

i

n

R

=

+

+ +

∑

⋯

Capacitor

• Capacitors are generally used as (a) filters or (b) storing

charges.

• The symbol is C and the unit is Farad (F).

• Capacitance is a measure of the quantity of charge that

can be stored for a given voltage across the plates.

Mathematically:

C

Q

C

V

(5)

Capacitor

• The current flowing into the capacitor is given

by:

• The voltage across the capacitor is:

• The energy stored is:

( )

dV

i t

C

dt

=

1

( )

V t

i t dt

C

=

∫

2

1

2

E

=

CV

Equivalent System for Capacitors

Parallel capacitors

Series capacitors

1

2

3

n

eq

i

n

C

=

+

+ + +

∑

⋯

1

2

1

n

i

eq

i

n

C

=

+

+ +

∑

(6)

Inductors

• An inductor or a reactor is a

passive electrical component that

can store energy in a magnetic

field created by the electric

current passing through it.

• An inductor's ability to store

magnetic energy is measured by

its inductance, in units of henries.

• The voltage across the inductor L

is given by:

( )

di t

V

L

dt

=

i t

( )

1

v t dt

( )

L

=

∫

•

The current flowing is

defined by:

Application of Inductors

• Used extensively in analog circuits and digital

signal processing.

• Usually in analog circuit, the inductors are used

together with capacitor for tuning and filtering

application.

• Other applications include, but not limited to:

– Electrical transmission system

(7)

Equivalent System for Inductors

Parallel inductors

Series Inductors

1

2

3

n

eq

i

n

L

=

= +

+ + +

∑

⋯

1

2

1

n

i

eq

i

n

L

=

+

+ +

∑

⋯

Summary

VOLTAGE-CURRENT

CURRENT-VOLTAGE

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Circuit Analysis

• There are many techniques that can be used to

analyze electrical circuits.

• Two famous techniques are discussed in this

lecture:

– Kirchoff’s Current Law

– Kirchoff’s Voltage Law

• Other techniques:

– Voltage divider

– Current divider

Kirchoff’s Current Law (KCL)

• The amount of current flowing into the system

equals the amount of current flowing out. Put

differently, the sum of current equals to zero.

(

)

1

0

in

out

in

out

n

j

i

=

+ −

=

∑

in

(9)

Kirchoff’s Voltage Law (KVL)

• Kirchoff’s Voltage Law, sometimes abbreviated

as KVL, relates on the voltage drop of each

elements in the electrical circuits.

• The formula simply states that the input voltage

is the sum of the voltage at each elements in

electrical circuits.

• It is simply: V

in

=V

out

.

Voltage Divider

• In voltage divider circuits, we usually look at

two impedances that are in series with one

another.

2

1

2

out

in

Z

V

Z

(10)

Current Divider

• In voltage divider circuits, we usually look at

two impedances that are in parallel with one

another.

2

1

2

out

in

R

I

R

=

+

Example 1

(11)

Solution to Example 1

• Start modeling by using KVL:

• Take the forward Laplace transform:

in out

( )

1

( )

R C

v

v t

v

v t

iR

i t dt

C

=

= +

∫

c

1

( )

v t

i t dt

C

=

∫

c

1

( )

1

( )

v t

iR

i t dt

C

v t

i t dt

C





= +



⇒







=







∫

L

1

( )

1

( )

c

V s

R

I s

Cs

V t

I s

Cs





=



+







=

Solution to Example 1

• To find V

c

(s)/V(s), we simply divide:

• Upon simplifying, we obtain:

1

( )

1

( )

c

I s

V t

Cs

V s

R

I s

Cs

=





+









( )

1

( )

1

c

V t

(12)

Example 2

• Find the transfer function for the following

system:

Source : Nise, 2007

• Start modeling by using KVL:

• Take forward Laplace transform & assume zero initial

condition:

(

)

(

)

in out

1 1 1 2 2

1 2 1 2 2

( )

v

v t

i R

i

R

v t

R

R i

i R

=

+ −

=

+

−

Loop 1

(

)

in out 2

2 1 2

2

2 2 1 2

0

v

di

i

i R

L

dt

di

i R

L

i R

dt

=

−

+





=



+



−





Loop 2

(

1 2

)

1 2 2

2

2 2 1 2

( )

0

v t

R

R i

i R

di

i R

L

i R

dt

=

+

−







⇒









=



+



−











L

(

)

(

)

1 2 1 2 2

2 1 2 2

( )

0

R

I s

R I s

V s

R I s

R

Ls I s

+

−

=

−

+

=

V

_out 2

( )

o

di

v t

L

dt

(13)

• Form a matrix:

• To find the expression I

2

(s)/V(s), we use Cramer’s Rule:

(

)

(

)

1 2 2 1

2 2 2

( )

0

R

I s

V s

R

Ls

I

s

+

−



 







=



−

+

 



















(

)

(

)

(

)

(

)(

)

2 1 2

( ) 2

2

1 2 2

2 2

2

2 2

1 2 2 2

( )

0

( )

I s

R

V s

D

R

I s

D

R

Ls

R V s

I s

R

Ls

R

+

−

=

+

−

+

−

=

+

−

• But, earlier:

• And therefore, substituting, gives:

2 2 2

( )

o o o

di

v t

L

dt

V s

LsI s

V s

I s

Ls

=

(

)(

2

)

2

1 2 2 2

( )

o

V s

R Ls

V s

R

Ls

R

−

=

(14)

Example 3

• Find the transfer function I

2

(s)/V(s).

Source : Nise, 2007.

• Write the KVL around each loop:

(

)

1

in

1 2

1 1

1 2

1 1

( )

R L

v

d i

i

v t

i R

L

dt

di

v t

i R

L

dt

=

+

−

=

+





=



+



−





Loop 1

(

)

2

in out

2 1

2 2 2

0

1

0

R C L

v

d i

i

i R

i dt

L

C

dt

=

+ +

−

=

+

∫

+

Loop 2

V

_out

2

1

( )

o

v t

i dt

C

(15)

• Take the forward Laplace transform and assume

zero initial condition.

• We have:

1 2

1 1

2 1

2 2 2

( )

1

0

di

v t

i R

L

dt

di

i R

i dt

L

C

dt









=



+



−



















=

+

−

















∫



L

(

1

)

1 2

1 2 2

( )

1

0

( )

V s

R

Ls I s

LsI s

R

Ls I s

Cs

=

+

−





= −

+



+







• In matrix form:

• Determine I

2

(s)/V(s) by using Cramer’s Rule:

(

1

)

1

2 2

( )

1

( )

0

R

Ls

I s

V s

I s

Ls

R

Ls

Cs



+

−















=











−

+



















(

)

(

)

2 1 ( ) 2 1 2 2

( )

0

( )

1

( )

I s

R

Ls

V s

D

Ls

I s

D

R

Ls

R

Ls

(16)

• And earlier, we note that:

• Substituting gives:

(

)

( )

2

1 2

( )

1

o

LsV s

V s Cs

R

Ls

R

Ls

Cs

=





+



+



−





2

1

( )

o

v t

i dt

C

I s

V s

Cs

I s

V s Cs





=









=

∫

L

• Simplifying gives:

(

)

( )

2

1 2

( )

1

( )

o

V s

LCs

V s

R

Ls

R

Ls

Cs

=





+



+



−

(17)

Method of Impedance

• Some textbooks uses the approach of modeling

via the method of impedance.

• However, I will not be discussing this method –

if you wish to use that method in modeling

electrical systems, you may do so.

• I will stick with my method and the lovely

Cramer’s Rule

☺

State Space Representation

(18)

Example 4

• Consider the following RC circuit. Obtain a state

space matrix.

• Since this circuit only contains one energy-storing

element, it means that there is one state variable that

needs to be considered. The state variable is the voltage

output at the capacitor, vc(t). Therefore:

• Next, we notice that the current flowing in this loop is

the same, therefore:

( )

c

( )

c c

dv t

x

v t

x

v t

dt

=

⇒

ɺ

=

ɺ

( )

R C c C

i

dv t

i

C

Cx

dt

= =

(19)

• Writing the loop equation by KVL gives:

• Re-arranging, gives:

0

( )

in out

s R C

s s

v

v u t

v

v u t

Ri

x

v u t

RCx

x

=

= +

=

+

=

ɺ

+

0

( )

1

( )

s s

s

RCx

x

v u t

x v u t

x

RC

x

v u t

RC

= − +

− +

=

= −

+

ɺ

Solution to Example 4

• And the output that we are interested is the

voltage at the capacitor. Therefore, the output

equation is:

• Therefore, the state equations are:

( )

c

y

=

v t

=

x

0

1

( )

s c

x

v u t

RC

y

v t

x

= −

+

=

(20)

Example 5

• Determine the state space representation for

the following circuit:

Solution to Example 5

• This circuit contain two time-domain variables, namely

the current passing through the inductor and the

voltage output at the capacitor. Therefore, our state

variables are:

• By KCL, we have the following equation:

1 1

2

2 L

L L

C L C

C

di

x

i

x

i

dt

x

v

dv

x

v

dt

=

⇒

=

ɺ

( )

R L C

c c

L

i

u t

v t

dv t

i

C

R

dt

= +

(21)

Solution to Example 5

• Substituting and re-arranging gives:

• Coupling the equation, we have:

2

1 2

2 1 2

( )

1

( )

1

( )

u t

x

C x

R

Cx

x

u t

R

x

u t

C

RC

−

= +

= − −

+

= −

−

+

ɺ

1

2 1 2

1

( )

L

di

x

dt

x

u t

C

RC

=

= −

−

+

ɺ

• But we notice that x

1

is not written in terms of x

1

and x

2

. So, how

do we overcome this problem?

• Earlier, we let x

1

=i

L

, that is the inductor’s current. Now, we also

know that the inductor’s current is given by the following formula:

• Differentiating with respect to time:

.

1

( )

L L

x

i

v t dt

L

= =

∫

1

2

1

( )

1

L

dx

di

x

v t dt

dt

L

x

L

= =

=

ɺ

The derivative of integral

will cancel each other’s out.

(22)

• Therefore, the modified form of state equation is:

• Suppose that my output for this system is the voltage at

the capacitor. Therefore:

1

2

1

2

1

( )

x

L

x

u t

C

RC

=

= −

−

+

ɺ

2

( )

c

L

y

=

v t

=

v t

=

x

Solution to Example 5

• In matrix form, it is:

[ ] [

]

1

2

1

2

0

1

0

( )

1

0 1

0

x

L

x

u t

x

C

RC

x

RC

x

y

x

 





 





=

+

 



−



 













 

 

=

 

+

 

ɺ

(23)

Example 6

• Derive the state-space representation for the

circuit shown below:

• The system can be expressed by the following

equations:

• Rearrange the equation so that the first order

derivative is the subject of the equation. We

must remember that in state-space, the subject

( )

i

c

di t

V t

Ri t

L

V t

dt

dV t

i t

C

dt

=

+

(24)

• Hence:

• In matrix form:

( )

c

i

c

V t

di t

Ri t

dt

L

dV t

i t

dt

C

= −

−

+

=

[

]

( )

1

( )

1

0

( )

0 1

( )

i

c

i t

R L

L

i t

V t

C

V t

i t

y

V t

−













 

=

+













 









 













=





=









ɺ

Next Step

• Textbook reference : Chapter 2.

• Homework 5 has been posted on the course

website. Attempt them. You do not have to

submit Homework 5 as it will not be graded.

(25)

Wise Word

“When you come to the end of your rope, tie a

knot and hang on.”