mech321 dynamics of engineering system - …buyung/systemdynamiclecture.pdf1 mech321 dynamics of...

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MECH321 Dynamics of Engineering System

Week 4 (Chapter 6) 1. Basic electric circuit theories

2. Mathematical Modeling of Passive Circuits

3. Complex Impedance Approach

4. Mechanical – Electrical analogy

5. Modelling of Active Circuits: Operational – Amplifier Circuits

R

e(t)

+

-

- - -

Two important variables in electrical circuit analysis: • Voltage, e : represents the potential energy in the circuit • Current, i : represents rate of charge q with respect of time. It can be positive or negative, depending upon the direction in which the electrons flow.

i

Direction of current

Direction of flow of electron

Unit : voltage

(V) = J / coulomb

Unit : ampere

(A) = coulomb / s

dt

dqi

Basic Electric Circuit Theories

2

• Resistance: unit ohm () = volt / ampere

• Capacitance: unit farad (F) = coulomb / volt

• Inductance: unit Henry (H) = V.s / A

Rie

i

eR

R

R

Ce

qC

t

CC eidtC

te0

01

dtdi

eL L

dt

diLte L

L )(

Ohm’s law

Basic passive elements

Using Ohm’s law the following rules governing electrical circuits can be derived:

• Series circuits

• Parallel circuits

21

321

RRR

eeee

total

total

21

21

111

RRR

iii

total

total

Ohm’s law

3

Using the series and parallel circuit theories, combined resistance of series/parallel resistors can be calculated:

Ohm’s law

1st Kirchoff’s Law “Current law (node law)”:

the sum of all currents entering (+) and leaving

(-) a node is zero

i1

i2 i3

i4

i5

054321 iiiii

1st Kirchoff’s law

4

Two-loop circuit

0: 112321 RiRiEEloopLeft

0: 23322 RiRiEloopRight

0 riRiE 02121 ririERiE

One-loop

circuit

2nd Kirchoff’s Law “Voltage law (loop law)”: the sum of

the voltages around any loop is zero (a rise in voltage

(+); a drop in voltage (-)

2nd Kirchoff’s law

Mathematical Modeling of Passive Circuits:

LR circuit

LR circuit

ERidt

diL

tL/ReR

Eti

L/RssR

E

RLss

EsI

s

EsIRLs

s

EsRIissIL

1

11

0

From Kirchoff’ voltage law around a loop

Taking Laplace transform of both sides

i(0) = 0

Step input

sGsE

sE

i

o Transfer Function:

5

RC circuit

0

1

1

eidtC

eidtC

Ri i

sEsIsC

sEsIsC

sRI i

0

11

11

1

1)( 0

RCssE

sEsG

i


Loop 1

Loop 2

Taking Laplace transform of both equations

If eo is the output and ei is the input, the transfer function of the system is


RC circuit

LRC circuit

0

1

1

eidtC

eidtC

Ridt

diL i

sEsIsC

sEsIsC

sRIsLsI i

0

11

11

1

1)(

2

0

RCsLCssE

sEsG

i


Taking Laplace transform of both equations

If eo is the output and ei is the input, the transfer function of the system is


LRC circuit

6

Cascaded elements

02

2

2

2

12

1

22

21

1

11

1

011

1

edtiC

dtiC

dtiiC

iR

edtiiC

iR i

sEsIsC

sIsC

sIsIsC

sIR

sEsIsIsC

sIR i

02

2

2

2

12

1

22

21

1

11

1

011

1


Taking Laplace transform of the equations

After eliminating I1(s) and I2(s), the transfer function of the system is obtained.

1

1

11

1)(

212211

2

2211

212211

0

sCRCRCRsCRCR

sCRsCRsCRsE

sEsG

i

Cascaded elements

Example 1 (B-6-9) Obtain the transfer function Eo(s)/Ei(s) of the circuit below.

(solution will be done in the lecture)

Tutorial

7

Complex Impedance Approach

Using this approach transfer functions of simple circuits are obtained directly as the Laplace-transformed equations.

Consider again LRC circuit:

sEsIsC

sEsIsC

sRIsLsI i

0

11

11

Complex impendance of: an inductor

ZL = Ls

A resistor

ZR = R

A capacitor

sCZC

1

)(

)()(

sI

sEsZ Complex Impendance:

Using complex impendances, the Laplace-transformed Equations are:

sEsIZ

sEsIZsIZsIZ

C

iCRL

0

These equations will give the transfer function

1

1)(

2

0

RCsLCsZZZ

Z

sE

sEsG

CRL

C

i

Complex Impedance Approach

8

Example 2 (B-6-12)

Obtain the transfer function Eo(s)/Ei(s) of the circuit below using the

complex-impedance method.


Classroom exercise

Force-voltage analogy

qdt

dqi

edtiC

Ridt

diL

1

eqC

qRqL 1

Mechanical-electrical analogies:

9

Force-voltage analogy

pkxxbxm

eqC

qRqL 1

s

CRL

sCRL

idt

deC

R

eedt

Lso

dt

deCi

R

eiedt

Li

iiii

1

,,1

Force-current analogy


10

Force-current analogy


Pkxxbxm

sidt

deC

R

eedt

L

1

Pdvkbvvm )(

A-6-16 (p.304): Find an electric analogy

0

0

12222

212111111

xxkxb

xxkxkxbxm

01

011

12

2

22

21

2

1

1

111

1

dtiiC

iR

dtiiC

dtiC

iRdt

diL

11

A-6-17: Find a mechanical analogy

01

011

12

2

1222

2

21221

2

1

1

111

1

dtiiC

iiRdt

diL

iiRdtiiC

dtiC

iRdt

diL

0

0

12221222

212212111111

xxbxxkxm

xxbxxkxkxbxm

Modelling of Active Circuits: Operational Amplifier (op-amp) Circuits

21120 eeKeeKe

e 1 e o

e 2

e e o

e +

-

Differential amplifier

Very large

gain 105

Input voltages

difference

Op-amp has a large input impendance, so it draws a negligible current.

12

Op-Amp Amplifier: proportional amplifier

'

0''

0

2

0

1

Kee

R

ee

R

eei

1

2

12

120

/11

/

R

R

K

RR

RR

e

e

i

Applying Kirchoff’s node law

i1 i3

i2 i1 = i2 + i3

But i3 0,

Because K is very large

0

Op-amp has a large input impendance, so it draws a negligible current.

Zf(s) = 1 / Cs

Zi(s) = R

sCRsZ

sZ

sV

sVsGfunctiontransferThe

i

f

i

1

)(

)(

)(

)()(, 0

In the time domain

t

i dtvRC

tv0

0

1)(

Op-Amp Amplifier: integrator

13

Zi(s) = 1/Cs

Zf(s) = R

sCRsZ

sZ

sV

sVsGfunctiontransferThe

i

f

i

)(

)(

)(

)()(, 0

In the time domain,

dt

tdvCRtv i )(

)(0

Op-Amp Amplifier: differentiator

Example 3 (B-6-25) Obtain the transfer function Eo(s)/Ei(s) of the following operational-amplifier shown below.


Classroom exercise

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