(α−120 o )=3sin ( (α+90 o )−cos (α+30 o )=−coswork6000/A00 elec eng fundamentals...
Transcript of (α−120 o )=3sin ( (α+90 o )−cos (α+30 o )=−coswork6000/A00 elec eng fundamentals...
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“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
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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 +=+∫
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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−
=
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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.
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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.
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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)
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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]
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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
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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=
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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)
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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)
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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
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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)
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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.
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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
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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
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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
++
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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
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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
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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
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= ( ) ( ) ( )
−+ ∫∫
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
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= ( ) ( ) ( ) ( ) ( )
−+ ∫∫
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
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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.
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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
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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
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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