Showing posts with label Basics. Show all posts
Showing posts with label Basics. Show all posts



Previously we already discussed basics of single phase transformer and three phase transformer. We also discussed about three phase transformer connections or Vector Groups. Now, here we will discuss about auto-transformers. Auto transformers have several applications. But first we will develop some basic concepts of autotransformer. Before proceeding further one should have some basic knowledge of transformer.

Auto transformers can be made in two ways. In one way it can be realized by additively connecting the primary and secondary windings of the two winding transformer. In other way the autotransformers can be thought of built as a single unit with one continuous winding. We will start from first approach and move to second.

Single Phase Autotransformer

Let us first consider a normal single phase two winding transformer. The schematic diagram of a  transformer is shown in Fig-A(i). We will obtain an autotransformer from this usual transformer.

Let the primary side and secondary side voltage ratings are respectively V1, I and V2, I respectively. Number of turns in primary and secondary side are N1 and Nrespectively. Now let us connect the transformers as shown in Fig-A(ii) with additive polarity. For the analysis purpose for better visibility we rearrange the windings in Fig-A(ii) as shown in Fig-B. Of course here we have shown a load connected across secondary

For the purpose of simplifying the analysis the transformer is considered as ideal. (In case of power transformer the ideal transformer analysis gives quite accurate result). The autotransformer analysis can be very simple if you recall two important concepts of a transformer.

  • The voltage developed in the windings are dependent on the flux linkages. The windings are wound on the same magnetic core so they link the same flux. Hence
                                                            V/ N1= V/ N2

             So whenever voltage V1 exist across primary winding, then voltage V2 will be induced across the
          secondary winding irrespective of changes in connections.
  •  Similarly the magnetic circuit demands that mmf should be balanced. It implies the primary side ampere turn should equal the secondary side ampere turn. Hence
                                                            I. N1= I. N2

It means current  I2 that flows  in secondary winding is associated with current  Iin primary winding according to above mmf balance formula.

In Fig-B the primary  of autotransformer is taken across both the windings where as secondary is across N2 winding.
The autotransformer is so loaded that the secondary current is I1+I2 . It makes the current flowing in the windings as I1 and I2   which are the rated values.
using KCL and KVL if the primary side voltage and current of the auto transformer is Vp and Ip and secondary side voltage and current of the auto transformer is Vs and Is then,

                                                            V= V1+V
                                                             Ip = I
                                                            Vs = V
                                                             Is = I1+I2.

Now the capacity of the autotransformer is (V1+V2).I1 or (I1+I2).V2

Using the voltage ratio and mmf balance formula it is quite easy to show that the autotransformer capacity formula can be simplified as

                  The capacity of autotransformer = Sa = (V+ V2).I= (I+ I2).V2 = V1 I1 (1 + N/ N1)
                                                                                          = V2 I2 (1 + N/ N1)   

                       Where as the capacity of our original transformer = S =V1 I1 = V2 I2

                                                                          So,   Sa = S(1 + N/ N1)

But (1 + N2 /N1) is always greater than 1.
Hence by forming an autotransformer the capacity of the resulting autotransformer is always more than the original isolated winding type transformer. In the so formed autotransformer it should be remembered that the voltage and current through the windings remains as before i.e the rated values. From the above formula it is clear that larger value of N2/N1  gives larger capacity of the autotransformer. 

In this case the voltage ratio of autotransformer =  (V1+V2)/ V2= V1 / V +1 = (N1 / N) + 1

                                                                Hence   V/ Vs =  (N1 / N) +1 =   (N1 / N) +1

From this formula it is clear that if  N2 /N1 is made large to increase the capacity then the voltage ratio between primary and secondary of autotransformer approaches 1. For this reason auto transformers are advantageous for use in power network when the voltage ratio between both sides is near unity. It is used in grid substations as interconnecting transformers (ICT). The autotransformers are used to interconnect two  different voltage levels. For example interconnection of 400kV and 220kV, 735kV and 345kV and 765kV and 400kV etc.. The voltage ratio should be less than 3:1 for more advantageous use.

Again look at the formula,
                                                     I. N1= I. N2

As we said  N2 /N1 should be large for adavantageous use.

                                      To achieve it, N2  should be much greater than N1

                                      It implies that  I2   will be much less than I1

The current I2  flows in the common winding of the autotransformer. Hence the auto transformer can be designed with N2 number of turns made of conductor of smaller cross section area, so resulting in a big saving.The autotransformers are cheaper and lighter in comparison to two winding transformers.

Just looking at the sketch you may think that instead of connecting two windings in additive ways. Why should not it be made of single winding and one terminal brought out from the middle as per requirement. Yes this is true and the autotransformer can be thought of made of a single winding having a part of winding common to primary and secondary .

Sometimes this method is used to obtain a variable secondary voltage. This case it is so designed that the middle contact can smoothly slides over the coil. It is commonly used in the academic electrical laboratories. This is usually called as Variac(Fig-C) or Dimmerstat. There are some other terminologies adopted by different manufacturers.  In this design it is not possible to adopt conductors of two different cross sectional area as in case of ICT where turns ratio is fixed(due to fixed voltage ratio) between primary and secondary. 

Autotransformers are also used for voltage regulation in distribution networks, for starting of induction motors and as lighting dimmers. Autotransformers are also used in electric traction.

One main disadvantage about autotransformer is that the primary and secondary are electrically connected.  So the electrical disturbance i.e high voltage transients from one side can be easily transmitted to the other side.
The other disadvantage is that the impedance of the autotransformer is considerably low, so the short circuit current will be more. More over an open circuit in common winding results in full primary side voltage across the load which is harmful.

But in several cases the advantages outweigh the disadvantages.

Three Phase Autotransformer

First thing is that the theory of single phase autotransformer is the basis of three phase autotransformer. Three single phase autotransfor bank can be used for forming a three phase transformer or a single unit  three phase autotransformer can be built. The three phase autotransformers (see Fig-D) are connected in star-star(Wye-Wye). If the autotransformers are connected as Delta-Delta, then phase difference between primary and secondary exist which is not desired (See Vector Groups).  

In three phase Y-Y connected power autotransformers an additional delta connected winding is used to take care of zero sequence currents (for unbalanced systems), and third harmonic currents.

Although we discussed here is for one particular case still we revealed the general approach. If you wish to connect load across the other winding, then you can proceed in a similar way.

Moreover the above analysis is for step down case. You can easily analyze for step up case by interchanging the position of source and load. In this case the direction of I1+I2 is reversed so also the directions of Iand I2. It should be recalled again that the change of direction of current in series winding is associated with change of direction of current in common winding to satisfy  mmf balance.

Complex Power and Power Triangle


In the last article we discussed the basics of Electrical Power and related terminologies. Here we will develop the concept of  Complex Power and Power Triangle. These two are very important concepts used frequently by power engineers. As this article requires the knowledge of previous article so it is advised to at least have a look  at the last article.

From the previous article the following points are clear.
  • The instantaneous power p is composed of real power and reactive power.
  • The time average value of instantaneous power is the real power (true power) is |V | | I | cos φ
  • The instantaneous reactive power oscillates about the horizontal axis, so its average value is zero
  • The maximum value of the reactive power is |V | | I | sin φ

It should be remembered that real power is the average value and the reactive power is maximum value.

Complex Power

In power system analysis the concept of Complex Power is frequently used to calculate the real and reactive power.

This is a very simple and important representation of real and reactive power when voltage and current phasors are known. Complex Power is defined as the product of Voltage phasor and conjugate of current phasor. See Fig-A

Let voltage across a load is represented by phasor V  and  current through the load is I.

If S is the complex power then,

                 S = V . I*

V is the phasor representation of voltage and I* is the conjugate of current phasor. 

So if  V is the reference phasor then V can be written as |V| ∠0.
(Usually one phasor is taken reference which makes zero degrees with real axis. It eliminates the necessity of introducing a non zero phase angle for voltage)

Let current lags voltage by an angle φ, so  I = | I | ∠-φ
(current phasor makes -φ degrees with real axis)

                            I*=  | I | ∠φ
                           S = |V|  | I | ∠(0+φ) =  |V|  | I | ∠φ

(For multiplication of phasors we have considered polar form to facilitate calculation)

Writting the above formula for S in rectangular form we get

                           S =  |V|  | I | cos φ  +  j  |V|  | I | sin φ 

             The real part of complex power S is |V| | I | cos φ which is the real power or average power and the imaginary part  |V| | I | sin φ is the reactive power. 

             So,              S = P + j Q

             Where        P = |V| | I | cos φ    and    Q = |V| | I | sin φ

It should be noted that S is considered here as a complex number. The real part P is average power which is the average value, where as imaginary part is reactive power which is a maximum value. So I do not want to discuss further and call S as phasor. If you like more trouble I also advise you to read my article about phasor  or some other articles on phasor and complex numbers.

Returning to the main point, from the above formula it is sure that P is always more than zero. Q is positive when φ is positive or current lags voltage by φ degrees. This is the case of inductive load. We previously said that inductance and capacitance do not consume power. The power system engineers often say about reactive power consumption and generation. It is said that inductive loads consume reactive power and capacitors produce reactive power. This incorrect terminology creates confusion.

The fact is that most of the loads are inductive and they unnecessarily draw more current from source. Although in each cycle both inductance and capacitance draw power from the source and return same amount of power to the source but the behavior of inductance and capacitance are opposing to each other. When capacitors are connected in parallel to inductive load the power requirement of inductive load is supplied by capacitor in half cycle and in next half cycle the reverse happens. Depending upon the values of capacitor this power requirement of inductance in the load may be fully or partially satisfied. If partially satisfied the rest will be drawn from the distant source. By properly selecting the capacitance the maximum value of reactive power (Q) drawn from the distant source (or returned to the distant source) is reduced. This reduction in reactive power results in reduction of line current so the reduction of losses in transmission line and improvement in voltage at load end.

Power Triangle

Returning to the complex power formula, P, Q and S are represented in a power triangle as shown in figure below.

S is the hypotenuse of the triangle, known as Apparent Power. The value of apparent power is |V|| I |

or    |S| = |V|| I |

 It is measured in VoltAmp or VA.

P is measured in watt and Q is measured in VoltAmp-Reactive or VAR. In power systems instead of these smaller units larger units like Megawatt, MVAR and MVA is used.

The ratio of real power and apparent power is the power factor of the load.

power factor = Cos φ = |P| / |S|

                      = |P| / √(P 2+Q 2)

The reactive power Q and apparent power S are also important in power system analysis. As just shown above the control of reactive power is important to maintain the voltage within the allowed limits. Apparent power is important for rating the electrical equipment or machines.

Total Power of Parallel Circuits

In real world the loads are usually connected in parallel. Here we will show the total power consumed by parallel branches. See Figure-C. It has two branches.

First we have to draw the individual power triangles for each branch. Next the power triangles are arranged back to back keeping real power in positive x direction as shown. The total power consumed is obtained by connecting starting point O to the tip of last triangle. This is actually the result of addition of complex numbers.

If   S= P+j Q1
     S= P+j Q2
     Then,  S = S1+  S 

   or    S = (P1+  P ) + j  (Q1+  Q)

         P = P1+  P
        Q = Q1+  Q

In the above diagram S1  P1 ,   Q1 and φ 1 correspond to branch1 and S P2 ,   Q2 and φ  correspond to branch2.
S, P, Q and φ correspond to total power consumption as seen by the generator

Instantaneous, Average, Real and Reactive Power


We discussed and developed some important concepts of transmission lines in last few articles. Last time we discussed about long transmission lines. Here we discuss a simple but important basic concept Electric Power. This will refresh our knowledge before we move further. 

Electric Power has same meaning as mechanical power but here the power or energy that we are concerned is in Electrical form. We often encounter terms like instantaneous, average, total, real, reactive, apparent and complex power or simply power. What they mean? how are they related ? That we will discuss here and in next article.   

DC Circuit

As long as our analysis is restricted to Direct Current(DC) circuit the power consumed by the resistance load is the product of voltage across the resistance and current flowing through the resistance. It is really simple.

P = V . I

The power consumed by the load is the product of voltage across the load and current drawn by the load (Fig-A). Or the Power supplied by the DC source (battery/cell) is the product of voltage across the cell and current supplied by the cell. Both are equal in our example figure(considering ideal battery of zero internal resistance). The law of energy conservation implies power supplied by the source must be same as power consumed by the circuit. In DC circuit case instantaneous power is same as average power.
AC Circuit

In AC circuit analysis, what is this power that we talk about. The main problem is that the AC voltage and current varies sinusoidally with time. Moreover the presence of circuit reactive elements like Inductor and capacitor shift the current wave with respect to voltage wave (angle of phase difference). 

Power is rate at which energy is consumed by load or produced by generator. Whether it is DC circuit or AC circuit, the value of instantaneous power is obtained by multiplying instantaneous voltage with instantaneous current. If at any instant of time t the voltage and current values are represented by sine functions as

         v = Vm  sin ωt  

          i = Im  sin (ωt-φ)

Vm and Im  are the maximum values of the sinusoidal voltage and current. Here ω=2 π f
f is the frequency and ω is the angular frequency of rotating voltage or current phasors. It should be clear that for a power system f is usually 50 or 60 Hz
φ is the phase difference between the voltage and current.

As we said the instantaneous power is the product of instantaneous voltage and current, if we name instantaneous power as p then

p = v.i =  Vm  sin ωt  .  Im  sin (ωt-φ)
         or  p = Vm Im  sin ωt  sin (ωt-φ)

Applying trigonometric formula 2.sin A.sin B = cos(A-B) - cos (A+B)  we get

It can be written as

This is the equation of instantaneous power

In the Fig-C is drawn all the three waves corresponding to v, i and p. Graphically also we can get the value of instantaneous power (p) at any instant of time t by simply multiplying the value of current i and voltage v at that particular instant t. (You can verify that in the diagram p is negative when either v or i is negative otherwise p is positive. See the points where p is zero). In the graph we have shown horizontal axis as angle φ instead of time t for easy visualization. It should be clear that both way it is correct.

Clearly the instantaneous power p is composed of two terms. The first term is constant because for a given load the phase angle φ is fixed. It does not change unless the load is changed.  The second term  is varying with time sinusoidally due to the presence of the term cos (2ωt-φ). Look that the instantaneous power frequency is twice the frequency of voltage or current.

So the instantaneous power in a single phase circuit varies sinusoidally.

The instantaneous power,  p = constant term + sinusoidal oscillating term.

In one complete period the average of oscillating term is zero.

Then what is the average power within a given time, say one Time Period of the wave?

It is the constant term.

Here is another way to think about the average power.
Just observe that the instantaneous power is negative for a small time. For any time interval you just find the total +ve area A+ (above horizontal-axis (blue line) and below p curve) and total -ve area A- (below horizontal axis and  above p curve). The net area is obtained by subtracting A- from A+. By dividing this net area ( by the time interval T we get the average power(P). You can do this using calculus. What you will ultimately get is only the first term in the above formula for instantaneous power p.

In still another way it is easier to realize that the formula for instantaneous power p has a constant term  (Vm.Im / 2) cos φ and the other sinusoidal term (Vm.Im / 2) cos (2 wt - φ). Actually p is the oscillating power which oscillates about the average constant term  (Vm.Im / 2) cos φ .

So the average power is

The above formula can be written as




V and I are the phasor representation of RMS values* of voltage and current sinusoids. The symbols |V| and |I| are  the magnitudes of phasors V and I. (See at the buttom for definition of RMS value).

This above formula is your favorite formula for useful power that we are most concerned about. This average power formula is used to find the power consumed by the load. The monthly electric energy bill at home is based on this power. The engineers and technicians in power or electrical industry simply use the term power instead of average power. So whenever we simply call power it means average power.

Of course the instantaneous power is oscillating in nature. As we already said it does not oscillates about the horizontal-axis rather about the average power P (cyan color horizontal line). 

P will be zero when cos φ =0 or  φ  = 90 degree, that is when the phase angle between voltage and current waves is 90 degrees. It is only when the load is pure inductive or capacitive. In this case the second term only remains in the instantaneous power formula.

From the above figure for some time the power becomes negative that means the load supply energy to source for this period. This is due to the presence of reactive element in load.

The above formula for instantaneous power can be written in another form. This form actually is an attempt to distinguish the oscillating reactive power from the instantaneous power formula.   Rearranging the terms in equation for instantaneous power above we get

                      p = |V| | I | cos φ (1-cow2ωt) - |V| | I | sin φ sin2ωt

In this equation the first term |V| | I | cos φ (1-cow2ωt) is oscillatory whose average value is |V| | I | cos φ. We already talked about this average power.

The second term |V| | I | sin φ sin2ωt which is also oscillatory but with zero average value. The maximum value of this term is |V| | I | sin φ. This is the so called Reactive power. So Reactive power is the maximum value of a oscillatory power that is repeatedly drawn from the source and again returned to the source within each cycle. So the average of this reactive power is zero.

The average power P is called as Real Power. It is also sometimes called active power.

              Real power = P = |V| | I | cos φ

It is usually written as P = VI cos φ. But it should be remembered that V and I are the rms values of voltage and current. For example when we say single phase 220 volt AC it means the rms value of voltage is 220 volts ( it is not maximum value of voltage sinusoid)

              Reactive power = Q = |V| | I | sin φ

Real power is measured in Watt and the reactive power is measured in VAR (VoltAmpereReactive). In power sector these units are too small so real power is measured in Megawatt (MW) and reactive power in Megavar (MVAR). The letter R at the end denotes reactive power.

Many times students and practicing engineers are confused about the average power (often simply called power). They think that what they get by multiplying RMS voltage and RMS current is RMS power. No that is wrong. There is no RMS power. RMS power has no meaning or not defined. (Also see definition of RMS value, below at the end). It is average power or real power or true power.

Power In Three phase Balanced System

Let us consider a three phase balanced system. A three phase balanced system is analysed considering only one phase and neutral return. This is called per phase analysis. So the above analysis for single phase is true for balanced three phase case. Let the total power here is Pt. Then we get total three phase power as thrice of single phase case.

P= 3 |V| | I | cos φ

It should be remembered that |V| and | I |  are the per phase values. and φ is the phase angle of load in per phase analysis.

The above formula for balanced three phase system can be written as

P= √3 |Vl| | Il | cos φ

In the above formula Vl and  Il are line voltage and current (Fig-D). This equation is independent of type of three phase load connection i.e delta or star connected load. You have to know the line voltage, line current and phase angle φ as above. This form is very convenient and used often in power calculation.

There is one main difference between the single phase and total three phase power. The instantaneous single phase power is pulsating. In the balanced three phase case, each phase instantaneous power is pulsating but the three pulsating power waves are 120 degrees displaced from each other. At any instant of time the total of these three instantaneous power waves is a constant which is 3 |V| | I | cos φ. So the total power consumed in three phase balanced system is not pulsating. Non-pulsating power also imply the desired non-pulsating torque in case of three phase rotating machines. In large 3-phase motors this is really desired.

*RMS value of AC Sinusoids

The value of AC voltage or current that produces the same heating (or same energy) that is produced if DC voltage or current numerically equal to RMS value of AC is applied instead of AC. This concept helps make the formula for power similar for both DC and AC circuits.

You should read the next article about Power Triangle and Complex Power

Operators j and a in Electrical Engineering

We have already discussed about phasors and its simple properties. Perhaps now it is the time that we want to explore a little more. Every effort is made to keep it as simple as our previous article. 

Before proceeding further I want to clarify that here we are mainly concerned about phasor multiplication and 'j' and 'a' operators.This article will also help us better appreciate the use of symmetrical components ( for analysis of unbalanced 3-phase systems)  and subsequently other phenomena  in transformer and AC circuits.

We know that phasor in the form x+j y is drawn as an arrow from origin to (x, y) point.

Till now I represented the phasor in x+j y form also called rectangular form. A phasor can also be represented in polar form. In the polar form we also need two parameters, these are length of phasor (r) and angle(phi) it makes with the +ve horizontal axis . See the Figure-A.

Phasor Multiplication

I have already discussed the use of j in phasor representation.
We know that j is equal to square root of -1.

or    j = sqrt(-1) so j.j = -1

Now consider two phasors A = 2 + j 3 and B = -1 + j 2

Let us multiply A and B

 A.B = (2+j 3) . (-1 + j2) = -2 + j 4 -j 3 + j.j (3.2) = -2 + j 1 - 6 = -8 + j 1 

Directly multiply each of real and imaginary parts from A with each of B. It is simple!

It is even easier to multiply in polar form. See the example below. As illustrated in figure-A,  we represent below the phasors A and B in the polar form. For phasor A, 4 is its length and it makes 20 degrees with x-axis. Similarly B is of length 3 units and it makes 40 degrees with +ve horizontal axis

( 20, 40 and 60 are angles in degree)

Representing in polar form, the multiplication has become extremely easy.
Just multiply the lengths and add the angles to get the new phasor. 
You can convert it back to the rectangular form.

                                      A.B = 12(cos 60 + sin 60)   
j and a Operators

What we will get, if a phasor is multiplied with j?

for example
      if    A = 3 + j 4
Then    j A =j(3 + j 4) = j 3 + j.j 4 = -4 + j 3   ( As j.j =-1)

Now draw the phasor -4 + j 3. It will be observed that the angle between 3 + j 4 and -4 + j 3 is 90 degrees.

Any phasor when multiplied by j  will rotate the original phasor by 90 degrees in anticlockwise direction. Now if the resultant phasor is again multiplied by j then the phasor is again rotated by 90 degrees in anticlockwise direction, so on.
In our example j(-4 + j 3) = -j4 -3 = -(3 + j 4), which is in opposite direction to (p + j q). So clearly the phasor has again undergone 90 degrees anticlockwise rotation. See the figure. Every time we apply j, we rotate the phasor counterclockwise by 90 degrees.

Now let us consider about another operator ' a ' (standard symbol). It has the capacity to rotate a phasor counterclockwise by 120 degrees. applying ' a ' twice the phasor is rotated by 240 degrees, by applying thrice the original phasor is rotated 360 degrees or one complete rotation, so the original phasor.

It is clear that as the phasor is rotated 120 degrees (magnitude remains the same) then in polar form

                              a = 1/120deg  

in rectangular form   a = 1.cos 120 + j 1.sin 120
                          or  a = -0.5 + j 0.866

see the Fig-C how a phasor A is rotated by 120 degrees when applied with operator a.

I colored them red green and blue to recall our balanced three phase system.

Clearly we are able to get the phasors B and C  by applying the operator a repeatedly on phasor A.
Otherwise we can say that, the balanced system of A-B-C sequence can be equally represented in terms of 'a' and A only. The operator a will be used more in our article symmetrical components.

Fourier Series in Electrical Engineering

The Fourier Series deals with periodic waves and named after J. Fourier who discovered it.

The knowledge of Fourier Series is essential to understand some very useful concepts in Electrical Engineering.Fourier Series is very useful for circuit analysis, electronics, signal processing etc. . The study of Fourier Series is the backbone of Harmonic analysis. We know that harmonic analysis is used for filter design, noise and signal analysis. etc.. Harmonic analysis is also very important in power system studies. In power network, harmonics are mainly generated by non-linear elements and switching equipment. Although it is a applied mathematics topic but like our previous article here also we will try to minimise the maths and depict in a simpler way.

The Fourier Series deals with periodic waves and named after J. Fourier who discovered it.

Fourier Series 

The Fourier series is concerned with periodic waves. The periodic wave may be rectangular, triangular, saw tooth or any other periodic form(single valued). Here I will also call this periodic wave as signal wave.Any periodic signal wave can be represented as a sum of a series of sinusoidal waves of different  frequencies and amplitudes.

Otherwise we can also say that a series of sinusoidal waves of different frequencies and amplitudes add up to give a periodic wave of non-sinusoidal form.

Let us consider a periodic signal wave v . According to the definition of Fourier series, this periodic signal wave can be written as sum of sinusoidal waves as below.

Now the question is how can we find the coefficients

I am not going to show you the details of how I obtained the formulas, but you can remember the three general  formulas as shown below for obtaining the coefficients. Many of you may not require to find the harmonics by using the formulas below, but those who are interested can use these formulas. Of course, students are required to remember these formulas.

                                                         n = 1, 2, 3, 4 .....

Putting the values of n we obtain different coefficients. For example if n=1 we get a1 and b1.

Some knowledge about the properties of the Fourier series will immensely help you. In most cases signal waves maintain symmetry. Depending on the symmetry of the wave we may not be always required to find all the sine and cosine terms coefficients. So now I will guide you through some important properties that you should remember so that just looking at the signal wave you can immediately say which coefficients should be present in the series.

Let us first of all talk a little about harmonics.

What is harmonic ?

Every periodic wave has a Time period(T) which is one complete cycle. The whole signal is the repetition of this period. Frequency (N) of the wave is the number of complete cycles in one second.

clearly  N = 1 / T

Observe the general Fourier series, it has a component (a1 sin wt), this sinusoid has the same frequency as the actual signal wave and it is called the fundamental component. The next component is (a2 sin 2wt), its frequency is twice that of fundamental (or original signal wave), this sinusoid is called second harmonic. So also the third, fourth etc.

As an example if the fundamental wave has a frequency of 60 Hz, then the frequency of second harmonic is 120 (2* 60) Hz, third harmonic 180 Hz, 9th harmonic will be 540 Hz etc. The frequency of nth harmonic is (n.60).

Properties of Fourier Series

We will now consider some important and very useful properties of Fourier series
  • In the fourier Series the constant term a0 will not appear if the signal wave average value in one period is zero. (one period is T which is equal to 2PI)

               Looking at the figure it is clear that area bounded by the Square wave above and below t-axis are
               A1 and A2 respectively. Here A1=A2, so the average is zero. The wave has a0 equal to zero. You
               can also confirm by using the formula above. But shifting the same wave in vertical direction as
               redrawn in Fig-B  will automatically introduce a0 in Fourier series.

              Yes by integrating using the formula above you will find a0. By shifting I have actually disturbed  the
              symmetry around horizontal axis and introduced the average value a0.  Here in this figure a0 is exactly how
              much I shifted the  whole wave in vertical direction. This ais also called  DC component of the signal
    • The Fourier series having sine and cosine terms can be combined. 
             We know that if,    p =  a1 sin x +b1 cos x        then it can be written in the form
                                          p = r sin (x + phi)               phi is the angle displacement from sin x.
             Using school trigonometry r & phi are found from a1 and b1
             (Remember that same harmonics are combined in this form. sin x with cos 2x or cos x with sin 2x etc.
             are not combined in above form).

    • Considering the square wave in the above Fig-A we will get only terms with sine function. There will be  no cosine terms because the coefficients a1, a2 etc will be zero. This you can verify from above formula also. The Fourier series is:

             As the above square wave maintains symmetry about origin so it will be composed of sine waves only.
            (sine waves are symmetric about origin). In Fig-C, I have reproduced the square wave. Here I have  
            calculated only three terms of the Fourier series ' v ', the fundamental, third harmonic and fifth harmonic. 
            The blue curve is the sum of these three terms of the Fourier series of the square wave shown. You can 
            see how just by considering three terms of the Fourier series, the blue curve approximate the square 
            wave. In Fig-D the sum of Fourier series(in blue) is drawn by calculating more coefficients( 9 ). Observe
            how the blue curve approximates the suare wave so that I had to remove the square wave from the figure for clarity. Taking more terms the curve will be even smoother.

    • Again consider the square wave. I have now chosen the vertical axis by shifting pi/2 to the right  and redrawn in Fig-E. Now applying the above equations for getting the coefficients, we will find that the a0 and the sine term coefficients b1, b2...etc has vanished and only cosine terms are there in the series. 
              The Fourier series is:

             It is very important to compare Figure-C and Figure-E. In both cases the square waves are identical.
             Only the vertical axes are chosen differently in each cases. Comparing the individual harmonics and their
             sum (blue)  in both the figures it is clear that actually choosing the position of vertical axis does not
             change the component harmonics but what previously you were getting as sine waves have now
             become cosine waves automatically. This is only due to the choice of vertical axis. 
             Both are sinusoidal waves and both are correct. If some intermediate position of axis is chosen, we will
             get both sine and cosine terms. The sine and cosine terms can be combined as above to get only either
             phase shifted sine(or cosine) terms for each harmonic.

             So the choice of axis does not change the shape or number of harmonics, only mathematically the series
             is different.

             So it is important to realize that for a particular shape of signal wave the harmonics are fixed, whether
             you want to show them as series of sine terms or cosine terms or both, it does not matter. Of course in
            exam you may not be allowed to choose axis. 
    • If the signal wave has quarter wave symmetry, then the Fourier expansion will contain only odd harmonics like 3rd, 5th etc.. To test quarter wave symmetry look at the signal wave at Fig-A. Draw an imaginary vertical line passing through T/4. For the +ve half wave if one side of this vertical line is mirror image of other side then the wave shows quarter wave symmetry.
             Note: A sine wave is symmetric with respect to origin (odd function) and cosine is symmetric with
                       respect to y-axis (even function). Combination of only sine waves will be symmetric with respect
                      to origin (odd function) and combination of cosine waves will remain symmetric with respect to
                      y-axis (even function)

    Linear and Nonlinear Systems in Electrical Engineering

     Like any physical system, the Electrical Systems also work based on some well defined principles. For the study of the characteristics of any system, device or element, the block diagram notation (as shown below ) is used universally.

    In the figure the element or system is represented by a rectangular block. Naturally the device will receive some input and give the output. The arrow is shown to represent the direction of input and output signal flow.

    Basically any system can be  reduced to this simple form for studying its behavior.  When the block diagram represents a single element or simple device then instead of 'input' and 'output' we better use the terms 'cause' and 'effect'  respectively (even the terms 'excitation' and 'response' are equally used).  For example the cause may be voltage applied across the element and effect is the current that flows in the element. Here we will mainly use the terms 'input', 'output' and 'cause', 'effect'. I feel that the terms excitation and response are more specialized. Here we also use different terms like element, device and system interchangeably to satisfy people of different mindset. It should be remembered that these terms are not exactly of the same meaning. For example different elements or devices may be connected in a particular way to form a system etc.

    The simplest way of studying the behaviour of a element or system is by applying a series of inputs and observing the respective outputs. That is what is our approach here.

    Linear System

    When the input(cause) and corresponding output(effect) are tied by the rules of linearity, then the device or system behavior is said to be linear, otherwise nonlinear.

    Then what is exactly this linearity ?

    Suppose we apply a series of inputs say x1, x2, x3 and x4. When input is x1 the output obtained is y1. Similarly when inputs are x2, x3 and x4 the respective outputs are y2, y3 and y4.

    Now a graph is drawn taking points P1(x1, y1), P2(x2, y2), P3(x3, y3), P4(x4, y4). See the points marked with small circles in X-Y plane. In the X axis we take input values and in Y- axis output values.

    Now join the points by smooth free hand curve, the result is figure-B(ii). The more points we take, the more accurate will be the curve.
    Depending on the device behavior the curve of input versus output can be of any shape. some more cases are shown in fig-C(i) to (iv).

    Each of the curves may fit to some mathematical linear or nonlinear equation. For many devices the  relationship between input and output can be found directly analytically.

    Remember that the graph (or curve) is drawn with input( X axis) versus output (Y axis) ( not input vs. time or output vs. time).

    If a system or device or element satisfy the following two properties then it is called linear.

    • Superposition
              Suppose when  x1 input is applied the  y1 output is obtained,

                      and when x2 input is applied the y2 output is obtained.

              Now, if input applied is (x1+x2),  then the output obtained will be y1+y2 .
                    (equivalently we say that if x1 and x2 are applied simultaneously then out put will be the sum of the
                               outputs obtained individually)
    • Homogeneity
              when  x1 is the input  applied the  y1 output is obtained,

              Then if  (kx1) input is applied, then output obtained will be ky1. Here k is any real number.
    By examining all the input(X) vs output(Y) curves one can be sure that above two properties will be satisfied by only curve c(ii).
    The equation of this curve is in the form y = mx. Here the constant 'm' may be any real number.

    It is easy to remember that the input vs. output characteristic curve of a linear device should be straight line passing through the origin (center). Only in this case the above two laws of linearity are satisfied.

    All the other curves in fig-C are non-linear and they do not satisfy the above two properties of linearity (check it). It should be realized that even the straight line in fig-C(i) does not represent a linear system. Students are often confused and consider any straight line relationship between the input and output as linear. But it is not. One can also verify that the above two properties are not satisfied by every straight line relationship.

    In maths the linear equation is represented by the form y = mx + c, which is a straight line whose orientation depends on the values of m and c. The constants 'm' and 'c' can be any real numbers. In case of physical systems, all represented by these linear equations are not strictly linear systems. For the system to be linear, c must be zero.

    Linear and Nonlinear elements/systems in Electrical engineering

    Now is the time to consider few examples. The best example of a linear element is an ordinary resistance. If the voltage applied across the resistance is 'cause' and current flowing through the resistance as 'effect'. Then the graph of voltage versus current is a straight line (Ohms Law) passing through the origin. see fig-D. You can verify that it satisfies the linearity laws.

    The next example is the simple Diode circuit (Fig-E). Look at the voltage versus current curve here. Although it passes through the origin it is not a straight line, hence one can immediately say that it's a nonlinear element.

    Now we will consider the example of a magnetic circuit.

    Consider the magnetic core in a transformer. Due to the current 'i' applied in the coil surrounding the core, flux     'phi' is established in the magnetic core. A curve is drawn between field intensity and flux density. The arrow direction in the curve indicates that two different values of flux density obtained for same value of field intensity, one is for when the excitation current is increasing and other when the excitation current is decreasing. The o-a-b path is only at  beginning of magnetisation. This type of behavior in magnetic circuit of transformer is called hysteresis. The closed loop formed by the curve is called hysteresis loop. Now just looking at the diagram one can be sure that the field intensity vs. flux density curve is non-linear.

    This non-linearity in transformer characteristic gives rise to harmonics, which requires some techniques to handle.

    Nonlinearities are often encountered in electrical devices and systems. Most of the elements show non-linearities to some degree beyond certain input range. There are some methods to deal with the nonlinearities. We can also say that every element or system behaves like a linear element or system within a small range of input variation. For small variations of input around the operating point, the curve can be approximated by a small straight line. The middle of this small straight line is to be treated as origin. See Fig-G. In the figure the portion a-b can be treated as linear and Q as operating point.

    The truth is that in real world sometimes even the systems with much pronounced nonlinear characteristics is analysed using linear techniques. this is because the variation of system input and output is small enough around the operating point.

     'x' and 'y' are small changes around the operating point that is our new origin 'Q'. Again remember that the curve with respect to 'x' & 'y' coordinates is linear (within a-b range) while with respect to 'X' and 'Y' coordinates it is still nonlinear.

    (While solving problems considering small changes around the new origin 'Q', we only concern about 'x', 'y' and 'Q', forgetting original 'X', 'Y' and 'O')

    Let us consider one more example.
    (This portion requires little more knowledge in curve tracing and electronics)

    Consider the characteristics of a transistor (see Fig-H). The output 'Ic' versus 'Vce' (collector to emitter voltage)   is drawn for different values of base current 'Ib'.

       Ic = collector current
    Vce = collector to emitter voltage
       Ib = base current

    Examining the transfer curve (between Ib(cause) and Ic(effect)) it is clear that the curve of Ib versus Ic is non-linear. Close examination reveals that the curve seems to behave linearly in part of the curve (point a to point b). Then our aim is to fix 'Q' in such a position so that the variations ib and ic around Q lies within this linear region.

    ( Ib and Ic are total values but ib and ic are variations around Q)

    Just look at the point Q that we fixed in the curve. If the circuit is allowed to operate around this point Q, so that its input value say Ib varies in small values, then this variation with respect to this operating point Q is linear  variation.
    It should be noted that the overall variation with respect to origin 'O' is not linear. The linearity is with respect to point Q.

    Then why do we need this linear variation? Yes, in the linear portion the output wave shape is not distorted. That means the shape of the wave is preserved. Operating in somewhat linear range gives birth to very little noise.

    This is one important reason why we need the biasing in the transistor and set the quiescent point Q.

    In the linear range the superposition law works.
    If sinusoidal input signal is applied to the transistor, then actually the transistor operates at base value plus sinusoidal input.

    So actually whether you apply input (ib) or not, IB,IC and VCE are always present as long as Vcc battery is connected as shown. These are the quiescent point values due to biasing, which helps the transistor operate in linear range. 
    From the total collector current Ic,  the output ic is filtered out using a capacitor or signal transformer.

    So, we applied ib and got ic. However ic is several times ib and so we got ic as amplified version of ib.

    The linearisation method is used in many situations. This technique is also helpful in power system small signal stability studies and other small signal oscillation studies.

    Phase Sequence in Electrical Systems

    In Electrical Systems, sometimes without identification of phase sequence it is impossible to proceed further. Both students and practicing engineers often find it confusing.  There are some important situations where identification of phase sequence is a must. These are when:

    • One synchronous generator is to be synchronized to the grid.
    • Two systems are to operate in parallel.
    • Two transformers are to operate in parallel.
    • Connecting two different lines originating from the same source.
     In a three phase system the voltage or current sinusoid attain peak values periodically one after another. The sinusoid are displaced 120 degrees from each other. So also phasors representing the three sinusoids for voltage or current waves of three lines are phase displaced by 120 degrees.

    Now the question arises what is the sequence ? In which order the voltage or current waves attain the peak values cyclically. In the diagram just look at the ABC Anti Clock Wise phase sequence. Here the phasors are rotating in anti clock wise direction. An imaginary viewer(see figure-A, left ) will encounter first phase A, then B, then C again A, then wise.The sequence is ABCABCABC.......  or ABC Anti Clock Wise sequence .

    You might imagine about the possibility of phase sequence ACB Anti Clock Wise. Yes it can be! In this case the phasors rotating the same anti clock wise direction the imaginary viewer(see figure-A, right )  will encounter first phase A, then C, then B. So here the sequence continues like ACBACBACB......  .

    You might think that in anti clock wise rotation, do the other sequences possible?

    you may think why not BCA or BAC or CBA or ......?

    From the above ABC sequence if you start from B then you can see that BCA is nothing but the same ABC sequence. Similarly BAC and CBA sequence are the same as ACB, only we started from other phase.

    Note:  If you studied permutation and combination maths then it is easier to appreciate the case.

    Hence it is clear that for anti clock wise rotation there are two possible phase sequences ABC or ACB.

    Why anti clock wise ? yes it is the convention mostly used. Just recall the school maths when you always measured the trigonometric angle starting from positive x-axis in anticlockwise direction and called it positive angle and in clock wise direction the angle is negative. Accordingly the sine wave is drawn. This is the reason why anti clock wise rotation is so prevalent.

    However the Clock Wise ABC and ACB phase sequence can also be used. The phase sequence identification is purely a convention. It helps in identifying the sequence in which three phase voltage or current attain  the peak values.

    The Anti Clock Wise ABC is equivalent to Clock Wise ACB. Just analyze by placing the imaginary viewer and rotating the phases in respective directions.

    When any two of the three phase conductors connecting to the three phase induction motor is interchanged the phase sequence of the supply to motor is changed. This results in the rotation of motor in opposite direction. Actually this is the principle used in mechanical phase sequence detectors. The same direction of rotation means same phase sequence. These days solid state sequence detectors are increasingly used.

    In some regions of the world other letters may be used for phase sequence, like L1L2L3 or RYB. 
    It is really confusing when synchronization of two different systems are considered.

    In real world, before synchronization or paralleling, the two sides phase sequence is identified by using the same sequence detector. If the same direction of rotation is observed for both sides by the detector, then they are marked accordingly for same sequence. Of course only same sequence is not sufficient. It is also ascertained that terminals of same phase are connected together.