Representation of Sinusoidal Signal a Phasor and

Solution of Current in R-L-C Series Circuits

 

In the last lesson, two points were described:

 

1. How a sinusoidal voltage waveform (ac) is generated?

 

2. How the average and rms values of the periodic voltage or current waveforms, are computed?

 

Some examples are also described there. In this lesson, the representation of sinusoidal (ac) voltage/current signals by a phasor is first explained. The polar/Cartesian (rectangular) form of phasor, as complex quantity, is described. Lastly, the algebra, involving the phasors (voltage/current), is presented. Different mathematical operations -addition/subtraction and multiplication/division, on two or more phasors, are discussed.

Keywords: Phasor, Sinusoidal signals, phasor algebra

After going through this lesson, the students will be able to answer the following questions;

1. What is meant by the term, 'phasor' in respect of a sinusoidal signal?

2. How to represent the sinusoidal voltage or current waveform by phasor?

3. How to write a phasor quantity (complex) in polar/Cartesian (rectangular) form?

4. How to perform the operations, like addition/subtraction and multiplication/division on two or more phasors, to obtain a phasor?

This lesson forms the background of the following lessons in the complete module of single ac circuits, starting with the next lesson on the solution of the current in the steady state, in R-L-C series circuits.

Symbols

i or i(t) Instantaneous value of the current (sinusoidal form)

I Current (rms value)

Maximum value of the current mI

−I Phasor representation of the current

φ Phase angle, say of the current phasor, with respect to the reference phasor

Same symbols are used for voltage or any other phasor.

Representation of Sinusoidal Signal by a Phasor

A sinusoidal quantity, i.e. current, tItimωsin)(=, is taken up as an example. In Fig. 13.1a, the length, OP, along the x-axis, represents the maximum value of the current, on a certain scale. It is being rotated in the anti-clockwise direction at an angular speed, mIω, and takes up a position, OA after a time t (or angle, tωθ=, with the x-axis). The vertical projection of OA is plotted in the right hand side of the above figure with respect to the angle θ. It will generate a sine wave (Fig. 13.1b), as OA is at an angle, θ with the x-axis, as stated earlier. The vertical projection of OA along y-axis is OC = AB =

Version 2 EE IIT, Kharagpur

θθsin)(mIi=, which is the instantaneous value of the current at any time t or angle θ. The angle θ is in rad., i.e. tωθ=. The angular speed, ω is in rad/s, i.e.fπω2=, where is the frequency in Hz or cycles/sec. Thus, f

ftItIIimmmπωθ2sinsinsin===

So, OP represents the phasor with respect to the above current, i.

The line, OP can be taken as the rms value, 2/mII=, instead of maximum value, I_m . Then the vertical projection of OA, in magnitude equal to OP, does not represent exactly the instantaneous value of I, but represents it with the scale factor of 707.02/1=. The reason for this choice of phasor as given above, will be given in another lesson later in this module.

Generalized case

The current can be of the form, )(sin)(αω−=tItim as shown in Fig. 13.1d. The phasor representation of this current is the line, OQ, at an angle,α (may be taken as negative), with the line, OP along x-axis (Fig. 13.1c). One has to move in clockwise direction to go to OQ from OP (reference line), though the phasor, OQ is assumed to move in anti-clockwise direction as given earlier. After a time t, OD will be at an angle θ with OQ, which is at an angle (αωαθ−=−t), with the line, OP along x-axis. The vertical projection of OD along y-axis gives the instantaneous value of the current,

)(sin)(sin2αωαω−=−=tItIim.

Phasor representation of Voltage and Current

The voltage and current waveforms are given as,

θsin2Vv=, and )(sin2φθ+=Ii

It can be seen from the waveforms (Fig. 13.2b) of the two sinusoidal quantities - voltage and current, that the voltage, V lags the current I, which means that the positive maximum value of the voltage is reached earlier by an angle, φ, as compared to the positive maximum value of the current. In phasor notation as described earlier, the voltage and current are represented by OP and OQ (Fig. 13.2a) respectively, the length of which are proportional to voltage, V and current, I in different scales as applicable to each one. The voltage phasor, OP (V) lags the current phasor, OQ (I) by the angleφ, as two phasors rotate in the anticlockwise direction as stated earlier, whereas the angleφ is also measured in the anticlockwise direction. In other words, the current phasor (I) leads the voltage phasor (V).

Mathematically, the two phasors can be represented in polar form, with the voltage phasor () taken as reference, such as , and . −V00∠=−VV

In Cartesian or rectangular form, these are,

000jVVV+=∠=−, and , φφφsincosIjIII+=∠=−

where, the symbol, jis given by 1−=j.

Of the two terms in each phasor, the first one is termed as real or its component in x-axis, while the second one is imaginary or its component in y-axis, as shown in Fig. 13.3a. The angle,φ is in degree or rad.

Phasor Algebra

Before discussing the mathematical operations, like addition/subtraction and multi-plication/division, involving phasors and also complex quantities, let us take a look at the two forms - polar and rectangular, by which a phasor or complex quantity is represented. It may be observed here that phasors are also taken as complex, as given above.

Representation of a phasor and Transformation

A phasor or a complex quantity in rectangular form (Fig. 13.3) is,

yxajaA+=−

where and are real and imaginary parts, of the phasor respectively. xaya

In polar form, it is expressed as

aaaAjAAAθθθsincos+=∠=−

where A and aθare magnitude and phase angle of the phasor.

From the two equations or expressions, the procedure or rule of transformation from polar to rectangular form is

axAaθcos= and ayAaθsin=

From the above, the rule for transformation from rectangular to polar form is

22yxaaA+=and ()xyaaa/tan1−=θ

The examples using numerical values are given at the end of this lesson.

Addition/Subtraction of Phasors

Before describing the rules of addition/subtraction of phasors or complex quantities, everyone should recall the rule of addition/subtraction of scalar quantities, which may be positive or signed (decimal/fraction or fraction with integer). It may be stated that, for the two operations, the quantities must be either phasors, or complex. The example of phasor is voltage/current, and that of complex quantity is impedance/admittance, which will be explained in the next lesson. But one phasor and another complex quantity should not be used for addition/subtraction operation.

For the operations, the two phasors or complex quantities must be expressed in rectangular form as

yxyxbjbBajaA+=+=−−;

If they are in polar form as

baBBAAθθ∠=∠=−−;

In this case, two phasors are to be transformed to rectangular form by the procedure or rule given earlier.

The rule of addition/subtraction operation is that both the real and imaginary parts have to be separately treated as

()()yxyyxxcjcbajbaBAC+=±+±=±=−−−

where ()()yyyxxxbacbac±=±=;

Say, for addition, real parts must be added, so also for imaginary parts. Same rule follows for subtraction. After the result is obtained in rectangular form, it can be transformed to polar one. It may be observed that the six values of , and - parts of the two phasors and the resultant one, are all signed scalar quantities, though in the example, and are taken as positive, resulting in positive values of . Also the phase angle sa'sb'sc'sa'sb'sc's'θ may lie in any of the four quadrants, though here the angles are in the first quadrant only.

This rule for addition can be extended to three or more quantities, as will be illustrated through example, which is given at the end of this lesson.

The addition/subtraction operations can also be performed using the quantities as phasors in polar form (Fig. 13.4). The two phasors are and . The find the sum , a line AC is drawn equal and parallel to OB. The line BC is equal and parallel to OA. Thus, . Also, )(OAA−)(OBB−)(OCC−−−−+=+=+==BAOBOAACOAOCC

OAOBBCOBOC+=+=

To obtain the difference , a line AD is drawn equal and parallel to OB, but in opposite direction to AC or OB. A line OE is also drawn equal to OB, but in opposite direction to OB. Both AD and OE represent the phasor ()(ODD−−−B). The line, ED is equal to OA. Thus, . Also −−−−=−=+==BAOBOAADOAODDOAOBEDOEOD+−=+=. The examples using numerical values are given at the end of this lesson.

Multiplication/Division of Phasors

Firstly, the procedure for multiplication is taken up. In this case no reference is being made to the rule involving scalar quantities, as everyone is familiar with them. Assuming that the two phasors are available in polar from as and . Otherwise, they are to be transformed from rectangular to polar form. This is also valid for the procedure of division. Please note that a phasor is to be multiplied by a complex quantity only, to obtain the resultant phasor. A phasor is not normally multiplied by another phasor, except in special case. Same is for division. A phasor is to be divided by a complex quantity only, to obtain the resultant phasor. A phasor is not normally divided by another phasor. aAAθ∠=−bBBθ∠=−

To find the magnitude of the product , the two magnitudes of the phasors are to be multiplied, whereas for phase angle, the phase angles are to added. Thus, −C

 ()baBAcBABABACCθθθθθ+∠⋅=∠⋅∠=⋅=∠=−−−)(

where and BAC⋅=bacθθθ+=

Please note that the same symbol, is used for the product in this case. −C

To divide .by −A−B to obtain the result ., the magnitude is obtained by division of the magnitudes, and the phase is difference of the two phase angles. Thus, −D

()babadBABABADDθθθθθ−∠⎟⎠⎞⎜⎝⎛=∠∠==∠=−−−

where and BAD/=badθθθ−=

If the phasors are expressed in rectangular form as

yxajaA+=− and yxbjbB+=−

where ()()xyayxaaaaA/tan;122−=+=θ

The values of −Bare not given as they can be obtained by substituting for . sb'sa'

To find the product,

()()()()xyyxyyxxyxyxcbabajbababjbajaBACC++−=+⋅+=⋅=∠=−−−θ

Please note that .The magnitude and phase angle of the result (phasor) are, 12−=j

()()[]()()BAbbaababababaCyxyxxyyxyyxx⋅=+⋅+=++−=22222122 , and

⎟⎟⎠⎞⎜⎜⎝⎛−+=−yyxxxyyxcbabababa1tanθ

The phase angle,

()()()(⎟⎟⎠⎞⎜⎜⎝⎛−+=⎥⎥⎦⎤⎢⎢⎣⎡⋅−+=⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛=+=−−−−yyxxxyyxxyxyxyxyxyxybacbababababbaabbaabbaa1111tan//1//tantantanθθθ

The above results are obtained by simplification.

To divide by −A−B to obtain as −D

yxyxyxbjbajaBAdjdD++==+=−−−

To simplify, i.e. to obtain real and imaginary parts, both numerator and denominator, are to be multiplied by the complex conjugate of −D−B, so as to convert the denominator into real value only. The complex conjugate of −B is

Version 2 EE IIT, Kharagpur

byxBbjbBθ−∠=+=*

In the complex conjugate, the sign of the imaginary part is negative, and also the phase angle is negative.

()()()()⎟⎟⎠⎞⎜⎜⎝⎛+−+⎟⎟⎠⎞⎜⎜⎝⎛++=−⋅+−⋅+=+=−2222yxyxxyyxyyxxyxyxyxyxyxbbbabajbbbababjbbjbbjbajadjdD

The magnitude and phase angle of the result (phasor) are,

()()[]()()()BAbbaabbbabababaDyxyxyxyxxyyyxx=++=+−++=2222222122 , and

⎟⎟⎠⎞⎜⎜⎝⎛+−=−yyxxyxxydbabababa1tanθ

The phase angle,

⎟⎟⎠⎞⎜⎜⎝⎛+−=⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛=−=−−−yyxxyxxyxyxybadbababababbaa111tantantanθθθ

The steps are shown here in brief, as detailed steps have been given earlier.

Example

The phasor, in the rectangular form (Fig. 13.5) is, −A

42sincosjajaAjAAAyxaaa+−=+=+=∠=−θθθ

where the real and imaginary parts are 4;2=−=yxaa

To transform the phasor, into the polar form, the magnitude and phase angle are −A

Version 2 EE IIT, Kharagpur

radaaaaAxyayx034.2565.11624tantan472.44)2(112222=°=⎟⎠⎞⎜⎝⎛−=⎟⎟⎠⎞⎜⎜⎝⎛==+−=+=−−θ

Please note that aθis in the second quadrant, as real part is negative and imaginary part is positive.

Transforming the phasor, into rectangular form, the real and imaginary parts are −A

0.4565.116sin472.4sin0.2565.116cos472.4cos=°⋅==−=°⋅==ayaxAaAaθθ

Phasor Algebra

Another phasor,−B in rectangular form is introduced in addition to the earlier one, −A

°∠=+=−45485.866jB

Firstly, let us take the addition and subtraction of the above two phasors. The sum and difference are given by the phasors, and respectively (Fig. 13.6). −C−D

°∠=+=+++−=+++−=+=−−−2.6877.10104)64()62()66()42(jjjjBAC

°−∠=−−=−+−−=+−+−=−=−−−0.166246.828)64()62()66()42(jjjjBAD

It may be noted that for the addition and subtraction operations involving phasors, they should be represented in rectangular form as given above. If any one of the phasors

is in polar form, it should be transformed into rectangular form, for calculating the results as shown.

If the two phasors are both in polar form, the phasor diagram (the diagram must be drawn to scale), or the geometrical method can be used as shown in Fig 13.6. The result obtained using the diagram, as shown are the same as obtained earlier.

[(OC) = 10.77, ; and ( OD) = 8.246, −C°=∠2.68COX−D°=∠0.166DOX]

Now, the multiplication and division operations are performed, using the above two phasors represented in polar form. If any one of the phasors is in rectangular form, it may be transformed into polar form. Also note that the same symbols for the phasors are used here, as was used earlier. Later, the method of both multiplication and division using rectangular form of the phasor representation will be explained.

The resultant phasor, i.e. the product of the two phasors is −C

()1236565.161945.3745565.116)485.8472.4(45485.8565.116472.4jBAC+−=°∠=°+°∠×=°∠×°∠=⋅=−−−

The product of the two phasors in rectangular form can be found as

1236)1224()2412()66()42(jjjjC+−=−+−−=+⋅+−=−

The result () obtained by the division of by −D−A−B is

()5.0167.0565.71527.045565.116485.8472.445485.8565.116472.4jBAD+=°∠=°−°∠⎟⎠⎞⎜⎝⎛=°∠°∠==−−−

The above result can be calculated by the procedure described earlier, using the rectangular form of the two phasors as

5.0167.072361266)1224()2412()66()66()66()42(664222jjjjjjjjjBAD+=+=++++−=−⋅+−⋅+−=++−==−−−

The procedure for the elementary operations using two phasors only, in both forms of representation is shown. It can be easily extended, for say, addition/multiplication, using three or more phasors. The simplification procedure with the scalar quantities, using the different elementary operations, which is well known, can be extended to the phasor quantities. This will be used in the study of ac circuits to be discussed in the following lessons.

The background required, i.e. phasor representation of sinusoidal quantities (voltage/current), and algebra - mathematical operations, such as addition/subtraction and multiplication/division of phasors or complex quantities, including transformation of phasor from rectangular to polar form, and vice versa, has been discussed here. The study of ac circuits, starting from series ones, will be described in the next few lessons.

Problems

13.1 Use plasor technique to evaluate the expression and then find the numerical value at t = 10 ms.

()()()()00dit = 150 cos100t-45+500 sin100t+cos100t-30dt⎡⎤⎣⎦

13.2 Find the result in both rectangular and polar forms, for the following, using complex quantities:

 

a) 5-j1215 53.1∠°

 

b) ()5-j12+15 -53.1∠°

 

c) 2 30- 4 2105 450∠°∠°∠°

 

d) 15 0+. 2 21032 -45⎛⎞∠°∠°⎜⎟∠°⎝⎠

 

List of Figures

Fig. 13.1 (a) Phasor representation of a sinusoidal voltage, and (b) Waveform

Fig. 13.2 (a) Phasor representation of voltage and current, and (b) Waveforms

Fig. 13.3 Representation of a phasor, both in rectangular and polar forms

Fig. 13.4 Addition and subtraction of two phasors, both represented in polar form

Fig. 13.5 Representation of phasor as an example, both in rectangular and polar forms

Fig. 13.6 Addition and subtraction of two phasors represented in polar form, as an example