Transcript Week 4-AC - khurramkamal

• Sinusoidal Waveform
• Phasor Method
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Single-Phase System
 Alternating signal is a signal that varies with respect to time
 Alternating signal can be categories into ac voltage and ac
current
 This voltage and current have positive and negative value
There are many waveform in alternating circuit such as
(i)
Sinus waveform
(ii)
cosines waveform
(iii)
square waveform
(iv)
triangular waveform
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■ Sinus and cosines waveform is an important because
it is a basic waveform for electric supply in
transmission system.
Example of waveform is shown in figure below:
■
Vm/Im
Vm
Time(t)
- Vm
V/I
V/I
t
t
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■ There are four type
of alternating
waveform:
– Symmetry
– Non Symmetry
– Periodic
– Non Periodic
Example of signal is shown below:
V/I
t
Symmetry and periodic
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Sinus and cosines
waveform
■ Voltage and current value is represent
■
■
■
by vertical axis and time represent by
horizontal axis.
In the first half, current or voltage will
increase into maximum positive value
and come back to zero.
Then in second half, current or voltage
will increase into negative maximum
voltage and come back to zero.
One complete waveform is called one
cycle.
Vm/Im
Vm
Tim
- Vm
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There are several specification in sinusoidal waveform:
1. Period
2. Frequency
3. instantaneous value
4. peak value
5. peak to peak value
6. average value
7. effective value
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Period
■ Period is defines as the amount of time is take to go
through one cycle
■ It symbol as T for time
■ Period for sinusoidal waveform is equal for each
cycle.
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Frequency
■ Frequency is defines as
number of cycles in one
seconds.
■ It symbol as f and the units
is Hertz ( Hz)
■ It can derives as
1
f 
T
Signal with lower frequency
Vm/Im
Vm
1.0
Time(t)
- Vm
Signal with higher frequency
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V/I
Instantaneous value
10V
■ Instantaneous value is
■
■
magnitude value of
waveform at one specific
time.
Symbol for Instantaneous
value of voltage is v(t) and
current is i(t).
Example of Instantaneous
value for voltage is shown:
0.25
0.5
0.75
1.0
1.5
Time(s)
-10V
v (0.25)  10V
v (0.5)  0V
v (0.75)  10V
v (1.0)  0V
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■
■
■
Peak Value
Peak value is a maximum
value from reference axis into
maximum value of waveform.
For one complete cycle, there
are two peak value that is
positive peak value and
negative peak value.
It symbol is
V/I
10V
Peak positive
value
0.25
-10V
0.5
0.75
1.0
1.5
Time(s)
Peak negative
value
Vp
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■
■
Peak to peak value
Peak to peak value is a
maximum amplitude
value of waveform that
been calculate from
negative peak value into
positive peak value
It symbol is
V/I
10V
Peak to peak
value
0.25
0.5
0.75
1.0
1.5
-10V
Vp p
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Average value
■ Average value is average value for all instantaneous
■
value in half or one complete waveform cycle
It can be calculate in two ways:
1. Calculate the area under the graph:
Average value = area under the function in a period
period
2. Use integral method
T
1
average _ value   v(t )dt
T0
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Effective value
■ The most common method of specifying the
■
■
amount of sine wave of voltage or current by
relating it into dc voltage and current that will
produce the same heat effect.
It is called root means square value, rms
The formula of effective value for sine wave
waveform is
vm
Vrms 
 0.7071vm
2
im
I rms 
 0.7071im
where Im & Vm are peak values
2
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A more general expression for the sinusoid (as shown in
the figure):
v(t) = Vm sin (wt + q)
where q is the phase angle
v(t)
Vm
q
V1 = Vm sin wt




wt
-Vm
V2 = Vm sin wt + q)
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Phase angle
■ Phase angle is a shifted angle waveform from the
■
■
■
■
reference origin
Phase angle is been represent by symbol θ or Φ
Units is degree ° or radian
Two waveform is called in phase if its have a same
phase degree or different phase is zero
Two waveform is called out of phase if its have a
different phase or different phase is not zero
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A sinusoid can be expressed in either sine or cosine form.
When comparing two sinusoids, it is expedient to express both
as either sine or cosine with positive amplitudes.
We can transform a sinusoid from sine to cosine form or vice
versa using this relationship:
- sin ωt = sin (ωt ± 180o)
- cos ωt = cos (ωt ± 180o)
cos ωt = sin (ωt + 90o)
sin ωt = cos (ωt - 90o)
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The sine and cosine may be useful in manipulation of
sinusoidal functions:
sin(    )  sin  cos   cos  sin 
cos(   )  cos  cos   sin  sin 
sin(    )  sin  cos   cos  sin 
cos(   )  cos  cos   sin  sin 
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Phasor Diagram
■ Sinusoids are easily expressed in terms of phasors
■ Phasor is a complex number that represent
■
■
magnitude and angle for a sine wave
Phasor diagram is a vector line that represent
magnitude and phase angle of a sine wave
The magnitude of the phasor is equal to rms value
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Phasor Diagram
■ For example, if given a sine wave waveform
v(t )  Vm sin( wt   0 )V
■ It can be represent by a phasor diagram
 Vm 
V 
  0
 2
V
V rms

o
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Phasor Diagram
I
q
o
V

o
V1
From the phase diagram above, it can be conclude that:
i) I leading V for θ° degree or V lagging I for θ° degree
ii) V leading V1 for Φ° degree or V1 lagging V for Φ° degree
iii) I leading V1 for (Φ° + θ° ) degree or V1 lagging I for (Φ° + θ° ) degree
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Phasor Diagram
– Adding phasors is equivalent to adding the
corresponding time function for each phasor
– One way is to dissolve the phasor to complex
numbers and then adding then up according to the
real & imaginary values
 Each phasor can be represented by a complex
number. Break each phasor into real and imaginary
parts.
o V1 = V1cos(1) + jV1sin(1)
o V2 = V2cos(2) + jV2sin(2)
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Phasor Diagram
 So, the sum of the two phasors can be computed by
adding the real and the imaginary parts separately,
giving:
o = V1cos(1) + V2 cos(2) + j[V1sin(1)+ V2 sin(2)]
 Then, we can note the real part and the imaginary part
are the real and imaginary parts of the sum.
o = Vsum,real + jVsum,imaginary
 And, the phasor for the sum voltage can also be
represented with a magnitude-angle representation.
o = Vsumcos(sum) + jVsumsin(sum)
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Phasor Diagram
■ whenever you deal with complex numbers or
variables. If we have two phasors that we are adding,
we visualize the situation as shown below.
■Vsum = V1 + V2
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Phasor Diagram
■ We can add the two phasors any way possible. That
includes doing it graphically by hand, breaking the
phasors into components and summing the real and
imaginary components - as we did above - or any other
way you can imagine to sum two vector-like quantities.
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Phasor Diagram
■ It is now clear that all the laws & rules applicable to DC
■
can be generalised and used in AC circuits analysis.
However, due to the alternating property of current &
voltage, analysing it requires phasor method. In other
words, apart of amplitude, we do need to consider the
frequency of voltage or current.
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Phasor Diagram
■ To achieve this, using phasor and adopting complex
■
number in calculation we can analyse the steady state
AC circuits.
Thus with these methods of analysis, the power supplied
to the load in the circuit can be obtained.
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