(α−120 o )=3sin ( (α+90 o )−cos (α+30 o )=−coswork6000/A00 elec eng fundamentals revisited... · MEBS 6000 2010 Utilities services M.Sc.(Eng) in Building Services Engineering

MEBS 6000 2010 Utilities services M.Sc.(Eng) in Building Services Engineering

Department of Electrical & Electronic Engineering University of Hong Kong

K.F. Chan (Mr.) Page 1 of 26 July 2010 ALL RIGHTS RESERVED

“It is not enough that you should understand about applied science in order that your

work may increase man’s blessings. Concern for man himself and his fate must

always form the chief interest of all technical endeavours…… in order that the

creations of our mind shall be a blessing and not a curse to mankind. Never forget

this in the midst of your diagrams and equations.”

Albert Einstein

MATHEMATICS REVISITED

1) ROOTS OF 2nd ORDER POLYNOMIAL

For a quadratic equation:

02 =++ cbxax

The solution can be written as:

a

acbbx

2

42 −±−=

2)

Trigonometrical function

( ) ( )tt 2cos2

1

2

1cos2 +=

( ) ( )tt 2cos2

1

2

1sin2 −=

sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

( ) ( )oo 30sin3120sinsin +=−− ααα

( ) ( ) ( )ooo 30cos30cos90cos −−=+−+ ααα




3) Complex number manipulation

For an expression involving the complex number j,

jba +1

−−

+=

jba

jba

jba

1

( )22 jba

jba

−−=

22 ba

jba

+−=

where 1−=j

4) EULER IDENTITY

jxjx

jxjx

eex

ej

ej

x

−

−

+=

−=

2

1

2

1cos

2

1

2

1sin

From the above we can see that

xjxe

xjxejx

jx

sincos

sincos

−=+=

−

5) INTEGRATION

)ln(1

)(bxa

bbxa

dx +=+∫




6) DIFFERENTIATION

1) ²

11

xxdx

d −=

2) 2)(

1)

1(

bxab

bxadx

d

+−=

+

3) 32 )(

12

)(

1

bxab

bxadx

d

+−=

+

4) dx

duv

dx

dvu

dx

duv +=

5) ²

)(v

dx

dvu

dx

duv

v

u

dx

d−

=




ELECTRICAL ENGINEERING FUNDAMENTAL REVISITED

1) CAPCITANCE

dt

dvCi =

where i is current, C is capacitance, v is voltage and t is time. Unit of

capacitance is Farad, (coulombs per volt) but this is a very large unit, and

usually micro-farad (1Fµ = 10-6F) or pico-farad (1pF = 10-12F) are used.

Capacitances in the femto-farad (1fF = 10-15F) range are responsible for

protection of computer chips.

It can be shown that for 2 capacitances in parallel, the equivalent

capacitance is given by:

21 CCCeq +=

Likewise, it can also be shown that for 2 capacitances in series, the

equivalent capacitance is given by:

21 11

1

CCCeq +

=

In an ac circuit, capacitive reactance decreases with frequency. To calculate

capacitive reactance in Ω

fCXc π2

1−=

The minus sign meaning current through a capacitive reactance leads the

voltage by 90o.




2) INDUCTANCE

dt

diLv =

where v is voltage, i is current, t is time and L is inductance, henries (H)

which is equivalent to volt seconds per ampere. Typically, we deal with

inductances ranging from a fraction of a microhenry ( Hµ ) to several tens

of henries.

It can be shown that for 2 inductance in parallel, the equivalent inductances

is given by:

21 11

1

LLLeq +

=

Similarly, it can also be shown that for 2 inductances in series, the

equivalent inductance is given by:

21 LLLeq +=

In an ac circuit, inductive reactance increases with frequency. To calculate

inductive reactance in terms of Ω

fLXL π2=

Opposite to capacitive reactance, current through an inductive reactance

lags the voltage by 90o.




Example

A 20V(rms), 1020Hz power source is connected in series with a 400mH inductor and

a 0.5 Fµ capacitor, calculate the current.

Answer:

The impedance is

Z = ( )( ) ( ) 63

105.010202

11040010202 −

−

××−×

ππ = (2563-312) = 2251Ω

Current is 225120

=8.88mA(rms)

Note that the current lags the voltage by 90o because the inductance is larger than the

capacitance, and there is no resistance in this circuit.

Example

A 70.7V(rms), 79.6Hz power supply is connected in series with a 100Ω resistance, a

0.3H inductor and a 40Fµ capacitor. Calculate the circuit current.

Answer:

The reactance is

( )( ) ( ) 610406.792

13.06.792 −××

−π

π = 150-50 = 100Ω

The impedance is Z = 22 100100 + =141.4Ω

The current is =4.1417.70

0.5Amp(rms)

Note that the current lags the voltage because inductance is larger than capacitance

(lags by

−−

100

50150tan 1 =45o in this case)




Example

A 20V(rms), 1020Hz power supply is connected in series with a 100Ω resistance, a

400mH inductor and a 2.16Fµ capacitor. Calculate the circuit current.

Answer:

The reactance is

( )( ) ( ) 63

1016.210202

11040010202 −

−

××−×

ππ = 2563.5-72.2 = 2491.3Ω

The impedance is Z = ( )22 3.2491100 + =2493.3Ω

The current is =3.2493

208mA

Note that the current lags the voltage because inductance is larger than capacitance

(lags by

−−

1002.725.2563

tan 1 =87.7o in this case)

[Examples adopted from Freeman, Roger L. Fundamentals of Telecommunications]




3) Root mean square

Root mean square value of a periodic voltage v(t) is defined as

( )∫=T

rms dttvT

V0

21

where T is the period of the voltage.

Similarly, root mean square value of a periodic current is defined as

( )∫=T

rms dttiT

I0

21




4) Consider applying a periodic voltage v(t) with a period T to a resistance R.

The power delivered to the resistance is given by

( ) ( )R

tvtp

2

=

Furthermore, the energy delivered in one period is given by

( )∫=T

T dttpE0

The average power Pavg delivered to the resistance is the energy delivered in

one cycle divided by the period. Thus

( )∫==T

Tavg dttp

TT

EP

0

1

Thus ( )

∫=T

avg dtR

tv

TP

0

21

This can be rearranged as

( )R

dttvTP

T

avg

∫= 0

21

( )

R

dttvT

P

T

avg

2

0

21

=∫

Therefore

R

VP rms

avg

2

=

Similarly

RIP rmsavg2=




Consider a sinusoidal voltage given by

( ) ( )θω += tVtv m cos

Root mean square value of this sinusoidal voltage can be calculated by

( )∫ +=T

mrms dttVT

V0

22 cos1 θω

( )[ ]∫ ++=T

mrms dtt

T

VV

0

2

22cos12

θω

( )T

mrms tt

T

VV

0

2

22sin2

1

2

++= θωω

( ) ( )

−++= θω

θωω

2sin2

122sin

2

1

2

2

TTT

VV m

rms

Because the period is T, naturally πω 2=T ,

( ) ( )θω

θωω

2sin2

122sin

2

1 −+T

( ) ( )θω

θπω

2sin2

124sin

2

1 −+=

( ) ( )θω

θω

2sin2

12sin

2

1 −=

=0

Therefore

2m

rms

VV =

where Vm is the peak value of the sinusoidal voltage.

(For a mathematical treatment of sinusoidal analysis and calculation of phasor angles,

please read any textbook in electrical engineering such as HAMBLEY, A.R. Electrical

Engineering Principles and applications)




LAPLACE TRANSFORM RE-VISITED

The Laplace transform technique is widely used in engineering calculations because it

is a very convenient way of manipulating and solving differential equations. Laplace

transformation is a process of changing the independent variable from time t to a

complex variable s ( ωσ js += referred to as the Laplace operator) Following a

commonly used convention, a function of time such as f(t) or x(t) will be written in

lower case, and its Laplace transform F(s) or X(s) will be written in capital letters.

The symbol LLLL will be used to denote ‘the transform of’ a function in the time

domain into a function in the s-domain so that

( )[ ] ( )∫∞

−

+

==0

)( dttfetfsF stL

with initial condition of f(t) assumed to be zero at t < 0.

( ) ( )[ ] ( )∫∞+

∞−

− ==j

j

stdsesFj

sFtfσ

σπ2

11L

Laplace transform allows linear differential equations in the time domain to be

converted to algebraic equations in the s-domain which are in general much easier to

handle.

Important properties of the Laplace Transform and its inverse

1) Laplace transform is a linear transformation

LLLL [A f(t)] = A F(s)

and

LLLL [f1(t) ± f2(t)] = F1(s) ± F2(s)

so

LLLL [a1f1(t) ± a2f2(t)] = a1F1(s) ± a2F2(s)




2) Derivative

)0()( +−=

fssF

dt

dfL

where f(0+) is the initial value of f(t) evaluated as the one-sided limit of f(t) as t

approaches zero from positive values.

Likewise,

[ ] )0(')0()(

)0('

2

2

++

+

−−=

−

=

ffssFs

fs

dt

fd

dt

dfL

L

Therefore )0(')0()(22

2++ −−=

fsfsFs

dt

fdL

3) Integral

s

sFtf

t )()(

0

=

∫L

If there is an impulse function at t = 0,

( )s

dttf

s

sFtf t

t+=∫

∫ +=

0

0

)()(L

4) Shifting Theorem – real translation in the time domain

The Laplace transform of the function f(t-T) (i.e. delayed by a dead time T) is

[ ] )()( sFeTtf sT−=−L

for T > 0

and f(t-T) = 0 wherever t≤ T




5) Initial value theorem

If f(t) contains neither impulses nor higher-order singularities at t = 0,

)(lim)(lim)0(0

ssFtffst ∞→=→

+ ==

for t > 0

6) Final value theorem

If )(lim tft ∞→

exists, then

)(lim)(lim)(0

ssFtffst →∞→

==∞

provided that f(t) contains neither impulses nor higher-order singularities at t = 0.

The final value theorem enables the final value of f(t), i.e. when t tends to infinity, to

be evaluated directly from a knowledge of F(s). It is useful both directly and as a

means of checking that a time solution f(t) is obviously in error or not.

7) Complex translation in the s-domain

LLLL [e-atf(t)] = F(s+a)




EXAMPLE

Laplace transform of the differential equation

026532

2

3

3

4

4

=++−+ ydt

dy

dt

yd

dt

yd

dt

yd

is

s4Y + 3s3Y – 5s2Y + 6sY + 2Y = 0

i.e.

(s4 + 3s3 – 5s2 + 6s + 2)Y = 0

provided that all initial conditions are zero.




Table of Laplace transform

Time domain s-domain Remarks

f(t) LLLL [f(t)] = F(s)

1(t) s

1 Unit step

a(t) s

a Integrator – step function of magnitude a

f(t-kT) e-skTF(s)

Function delayed by a constant dead time kT

For LLLL [f(t)] = F(s)

r1(t-kT) kTse

s

r − Step function of magnitude r, delayed

till time t = kT




Time domain s-domain Remarks

t 2

1

s Unit ramp

b(t-kT) 2s

be kTs−

Ramp of slope b delayed by time kT

τ

τte−1

1

1

+sτ

First order lag where τ is the time

constant




Time domain s-domain

( )[ ]1

1sin1

2

2

<

−−

−

ζ

ζωζ

ω ζω

where

te ntn n

22

2

2 nn

n

ss ωζωω

++

( )[ ] 1

cos1sin1

11 12

2

<

+−−

− −−

ζ

ζζωζ

ζω

where

te ntn

( )22

2

2 nn

n

sss ωζωω

++




Fourier analysis re-visited

Although sinusoidal mathematical functions are used to represent electrical signals,

most real world information-bearing signals are not sinusoidal. However, we can

construct any waveform by adding sinusoids that have the proper amplitudes,

frequencies, and phases. For example, in the following diagram, the waveform shown

on the left hand side is the sum of the sinusoids shown on the right hand side.

[Adopted from HAMBLEY, A.R. Electrical Engineering Principles and applications]

The waveform on the left hand side looks relatively simple because it is composed of

only three components.

Fourier analysis is a mathematical technique for finding the amplitudes, frequencies,

and phases of the components of a given waveform. All real world signals are sums of

sinusoidal components. A periodic function ( )te ω , of any shape having a radian

frequency of ω , and repetitive every π2 radians, can be expressed as a summation

of harmonic terms:

( )teω = ( ) ( )( )∑∞

=++

1

0 sincos2 n

nn tnbtnaa ωω

= ( )( )∑∞

=++

1

0 sin2 n

nn tnca ψω

It follows that

22nnn bac += = peak value of the nth harmonic

= −

n

nn b

a1tanψ = phase displacement of the nth harmonic




The coefficients can be found by the following expressions:

( )∫=π

ωωπ

2

0

0

21

2tdte

a= time value average of the periodic function = dc term value

( ) ( )∫=π

ωωωπ

2

0

cos1

tdtntean

( ) ( )∫=π

ωωωπ

2

0

sin1

tdtntebn




Fourier series of a square wave

[Adopted from HAMBLEY, A.R. Electrical Engineering Principles and applications]

The above diagram shows a square wave with peak to peak value of 2A. It can be

represented by its harmonic terms as follows:

( )tvsq ω = ( )( )∑∞

=++

1

0 sin2 n

nn tnca ψω

Now, 20a

is obviously 0 because time value average of the square wave is 0.

( ) ( )∫=π

ωωωπ

2

0

1 cos1

tdttva sq




= ( ) ( ) ( )

−+ ∫∫

π

π

π

ωωωωπ

2

0

cos1cos tdttdtA

= ( ) ( )[ ]ππ

π ωωπ

2

0sinsin tt

A − =0

Therefore 01 =a

( ) ( )∫=π

ωωωπ

2

0

1 sin1

tdttvb sq

= ( ) ( ) ( )

−+ ∫∫

π

π

π

ωωωωπ

2

0

sin1sin tdttdtA

= ( ) ( )[ ]ππ

π ωωπ

2

0coscos tt

A +− = πA4

Therefore πA

b4

1 =

So πA

bac42

12

11 =+= , 0tan1

111 =

= −

b

aψ

Also,

( ) ( )∫=π

ωωωπ

2

0

2 2cos1

tdttva sq = ( ) ( ) ( )∫π

ωωωπ

2

0

22cos21

tdttvsq = 0

( ) ( )∫=π

ωωωπ

2

0

2 2sin1

tdttvb sq = ( ) ( ) ( )∫π

ωωωπ

2

0

22sin21

tdttvsq

= ( ) ( ) ( ) ( ) ( )

−+ ∫∫

π

π

π

ωωωωπ

2

0

22sin122sin2

tdttdtA

= ( ) ( )[ ]ππ

π ωωπ

2

02cos2cos

2tt

A +− = 0

So 02 =c , 02 =ψ

Furthermore,

( ) ( )∫=π

ωωωπ

2

0

3 3cos1

tdttva sq = ( ) ( ) ( )∫π

ωωωπ

2

0

33cos31

tdttvsq




= ( ) ( ) ( ) ( ) ( )

−+ ∫∫

π

π

π

ωωωωπ

2

0

33cos133cos3

tdttdtA

= ( ) ( )[ ]ππ

π ωωπ

2

03sin3sin

3tt

A − =0

( ) ( )∫=π

ωωωπ

2

0

3 3sin1

tdttvb sq = ( ) ( ) ( )∫π

ωωωπ

2

0

33sin31

tdttvsq

= ( ) ( ) ( ) ( ) ( )

−+ ∫∫

π

π

π

ωωωωπ

2

0

33sin133sin3

tdttdtA

= ( ) ( )[ ]ππ

π ωωπ

2

03cos3cos

3tt

A +− = π3

4A

So 3

43

Ac = , 03 =ψ

It can be similarly shown that all nψ are 0 for this square wave, and all even nc are

0 (that to say, there is no even harmonics for this square wave), and 5

45

Ac = ,

74

7

Ac = etc.

Summarizing, Fourier analysis shows that the above square wave can be written as an

infinite series of sinusoidal components:

( )

++++= LttttA

tvsq ωωωωπ

7sin7

15sin

5

13sin

3

1sin

4

in which T

πω 2= is called the fundamental angular frequency of the square wave.

The lower figure above shows several of the terms in this series and the result of

summing the first five terms. Clearly, even the sum of the first five terms only is

already a fairly good approximation to the square wave, and the approximation

becomes better as more components are added. Thus, the square wave is composed of

an infinite number of sinusoidal components. The frequencies of the components are




odd integer multiples of the fundamental frequency, the amplitudes decline with

increasing frequency, and the phases of all components are -90o.

The fact that all signals are composed of sinusoidal components is a fundamental idea

in electrical engineering. The frequencies of the components, as well as their

amplitudes and phases, for a given signal can be determined by theoretical analysis or

by instrument called a spectrum analyzer.




Bessel functions Bessel functions of the first kind of order n and argument β :

Generating function and definition:

( ) ( )∑∞

−∞=

=n

tjnn

tj mm eJe ωωβ βsin

( ) ( )∫

−

−=π

π

λλβ λπ

β deJ njn

sin

2

1

( ) ( ) ( )( )∑

∞

=

+

+−=

0

2

!!

21

k

nkk

n nkkJ

ββ

Properties of ( )βnJ :

1) ( ) ( ) ( )ββ nn

n JJ 1−=−

2) ( ) ( ) ( )ββ

ββ nnn Jn

JJ2

11 == +−

3) ( ) 12 =∑∞

−∞=nnJ β




Selected values of ( )βnJ

n\ β 0.1 0.2 0.5 1 2 5 8 10

0 0.997 0.990 0.938 0.765 0.224 -0.178 0.172 -0.246

1 0.050 0.100 0.242 0.440 0.557 -0.328 0.235 0.043

2 0.001 0.005 0.031 0.115 0.353 0.047 -0.113 0.255

3 0.003 0.020 0.129 0.365 -0.291 0.058

4 0.002 0.034 0.391 -0.105 -0.220

5 0.007 0.261 0.286 -0.234

6 0.001 0.131 0.338 -0.014

7 0.053 0.321 0.217

8 0.018 0.224 0.318

9 0.006 0.126 0.292

10 0.001 0.061 0.208

11 0.026 0.123

12 0.010 0.063

13 0.003 0.029

14 0.001 0.012

15 0.005

16 0.002




Fourier analysis of a sinusoidally pulse width modulated waveform is very complex,

however, the phase to neutral voltage can be expressed by means of Bessel functions

of the first kind of order n:

( )tVMv m

dc ωcos2

=

( )∑∞

=

+1

0 cos2

sin2

2

mcar

dc tmm

mMJV ωπππ

( ) ( )∑ ∑∞

=

∞±

±=

+

+

+1 1

cos2

sin22

m nmcar

ndc tntm

nm

m

mMJV ωωπ

π

π

where carω is the radian frequency of the carrier signal, and

mω is the radian frequency of the modulated signal,

M is the amplitude modulation ratio, i.e. car

m

V

V

dcV is the voltage of the DC link.

The first term in the above equation gives the fundamental frequency output

component, which is proportional to the modulation index.

The rms value of the fundamental component of the phase to phase voltage is

dcMV22

3

Part of the text and figures adopted from SHEPHER, W., ZHANG, L., Power

converter circuits and HAMBLEY, A.R. Electrical Engineering Principles and

applications

(α−120 o )=3sin ( (α+90 o )−cos (α+30 o )=−coswork6000/A00 elec eng fundamentals revisited... · MEBS 6000 2010 Utilities services M.Sc.(Eng) in Building Services Engineering

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