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Mathematical representation of phasor in Complex form

by | Last updated Jul 5, 2021 | AC Circuits

The complex form of representing a phasor is used to solve complex sinusoidal equations. DC analysis requires simple mathematics to solve its equations. Since an AC waveform has magnitude and direction, it needs a complex number to solve AC sinusoidal equations.

A phasor, which is a vector quantity can be represented mathematically in four ways: Rectangular form, trigonometric form, exponential form and Polar form.

Rectangular form

The phasor quantities are algebraically expressed in terms of rectangular components. It is also called a Cartesian form of representation. In this form, a phasor can be divided into two components, namely a horizontal component and a vertical component.

For example, let us consider the phasor E, as shown in the below diagram. From the endpoint of the phasor, draw a horizontal line and vertical line towards the X-axis and Y-axis respectively. Now, the phasor E is said to have horizontal component ‘a’ and vertical component ‘b’.

Thus in rectangular form, the phasor can be mathematically expressed as

    \[ \boxed{ E = a + jb}\]

Here ‘j’ is a complex operator, which indicates that the component ‘b’ is perpendicular to component ‘a’. In this way of writing an expression is said to be a complex form of the phasor.

Complex Rectangular form

Mathematically, the component ‘a’ is called a real number and component ‘b’ is called an imaginary number. In electrical engineering, it is called active and reactive components respectively.

Both the real and imaginary parts of a complex number can be either positive or negative. Hence both the real and imaginary axis extended in respective positive and negative directions. It results in a complex plane with four quadrants called an Argand Diagram, as shown below.

Argand diagram

In the Argand diagram, there are four phasors each in four quadrants. In rectangular form, the four phasors are represented as,

    \[OE_1 = a_1 + jb_1, OE_2 = -a_2 + jb_2, OE_3 = -a_3 - jb_3, OE_4 = a_4 - jb_4 \]

Significance of j operator

In general, an operator such as +, -, *, / performs some operations on the numbers. Similarly, j is a complex operator, which indicates the location of the phasor that is rotating in anticlockwise direction. The value for ‘j‘ is given by j = \sqrt{-1} or j^2 = -1.

Let E be the phasor in horizontal X-axis, which rotates in the anticlockwise direction. When the j operator is joined with the phasor E, it becomes jE and is displaced through 900. When the j operator is again multiplied with jE, the phasor becomes j2E = -E. Now the phasor -E is rotated through an angle 1800.

Significance of j operator

Further application of j operator on -E, it becomes –jE and this phasor is displaced by 2700. It is directly opposite to the phasor jE. When the j operator is again applied to the phasor –jE, it becomes –j2E = E. Now the phasor completes 3600 rotation, returning back to its original position.

The significance of j operator can be summarized as below,

900 rotation; j = \sqrt{-1} = +j

1800 rotation; j^2 = {( \sqrt{-1} )}^2 = -1

2700 rotation; j^3 = {( \sqrt{-1} )}^3 = -j

3600 rotation; j^4 = {( \sqrt{-1} )}^4 = +1

Trigonometric form

Let us consider a phasor OB having the magnitude E and it is displaced by an angle \phi with respect to X-axis. Draw a vertical line from B towards the horizontal axis. Now OAB is a right-angled triangle.

By applying trigonometry to the triangle, we obtain,

    \[cos \phi = \frac{OA}{OB} = \frac{OA}{E} => OA = E cos\phi \]

    \[sin\phi = \frac{AB}{OB} = \frac{AB}{E} => AB = E sin\phi \]

Trigonometric form

It is observed that, the horizontal component of E is E cos\phi and its vertical component is E sin \phi. Thus in trigonometric form, we can represent the phasor as,

    \[ \boxed{ E = E cos \phi + j E sin \phi }\]

This is equivalent to the rectangular form E = a + jb, where a = E cos\phi and b = E sin \phi.

Exponential form

Euler’s formula gives the relationship between the exponential and trigonometry functions. It is given by,

    \[ e^{j\phi} = cos \phi + j sin \phi \]

Also, in Maclaurin series, the functions sin \phi , cos \phi and e^{j \phi} are expanded as,

    \[cos \phi = 1 - \frac{1}{2!} \phi^2 + \frac{1}{4!} \phi^4 - \frac{1}{6!} \phi^6 + ... --->(1)\]

    \[sin \phi= \phi - \frac{1}{3!} \phi^3 +\frac{1}{5!} \phi^5 - \frac{1}{7!} \phi^7 + ... --->(2)\]

    \[ e^{j\phi} = 1 + j \phi + \frac{1}{2!} {j \phi}^2 + \frac{1}{3!} {j \phi}^3 + \frac{1}{4!} {j \phi}^4 + \frac{1}{5!} {j \phi}^5 + \frac{1}{6!} {j \phi}^6 + \frac{1}{7!} {j \phi}^7 + ... \]

As we know, j^2 = -1, j^3 = -j, j^4 = +1, j^5 = +j, j^6 = -1, j^7 = -j..., the above expression for e^{j \phi} is re-written as,

    \[ e^{j\phi} = 1 + j \phi - \frac{1}{2!} \phi^2 - \frac{j}{3!}  \phi^3 + \frac{1}{4!} \phi^4 + \frac{j}{5!}  \phi^5 - \frac{1}{6!} \phi^6 - \frac{j}{7!} \phi^7 ... \]

    \[ e^{j\phi} = \left(1 - \frac{1}{2!} \phi^2 + \frac{1}{4!} \phi^4 - \frac{1}{6!} \phi^6 + ... \right) + j \left( \phi - \frac{1}{3!} \phi^3 +\frac{1}{5!} \phi^5 - \frac{1}{7!} \phi^7 + ... \right ) \]

Using equations (1) and (2), the above expression becomes,

    \[ e^{j\phi} = cos \phi + j sin \phi \]

Thus the trigonometric form of a phasor E = E(cos \phi + j sin \phi ) can be written in exponential form as,

    \[ \boxed{ E = E e^{j\phi} }\]

In the exponential form of a phasor, E represents the magnitude and \phi represents the phase angle with respect to the reference axis.

Polar form

In complex polar form, the phasor is represented with its magnitude and phase angle as,

    \[\boxed{ E = E \angle \pm \phi }\]

Here E is the magnitude of the phasor, \phi is the angle of the phasor with respect to X-axis.

Let us draw this phasor having the magnitude E, leading by angle \phi with respect to the horizontal axis. If the arrowhead of E is joined with a vertical line towards the horizontal axis, we get a right-angled triangle.

Polar form

By using the Pythagoras theorem, it is possible to determine the magnitude and angle of the complex number.

The Pythagoras theorem relates the sides of a right angled triangle in a simple way

    \[{E}^2 = x^2 + y^2 \]

    \[ \boxed{ E = \sqrt{x^2 + y^2} }\]

By applying trigonometry, we can also write,

    \[cos \phi = \frac{x}{E} => x = E cos\phi    .........(1)\]

    \[sin\phi = \frac{y}{E} => y = E sin\phi   ..........(2)\]

Dividing the equation(2) by eqution(1), we get,

    \[\frac{sin \phi}{cos \phi} = \frac{y}{x} => tan \phi = \frac{y}{x} \]

    \[\boxed{\phi = tan^{-1} \frac{y}{x}} \]

Summarizing the different forms to represent the phasor quantities,

Rectangular form: E = a \pm jb

Trigonometric form : E = E ( cos \phi \pm j sin \phi )

Exponential form: E = E e^{\pm j \phi}

Polar form: E = E \angle \pm \phi

Complex Arithmetic operation of phasor

Two or more phasor quantities can be added, subtracted, multiplied, divided using the rectangular, trigonometric, exponential and polar forms. Let us learn, how each operation are performed.

Phasor Addition

The rectangular form is sufficient to perform the addition and subtraction operation of phasor quantities. Consider the two phasors, A = a_1 + j b_1 and A = a_2 + j b_2 for which the operation is to be performed.

For addition, E = A + B = (a_1 + j b_1 ) + (a_2 + j b_2) = (a_1 + a_2) +j(b_1 + b_2)

The magnitude of resultant phasor is determined as,

    \[E = \sqrt{(a_1+a_2)^2 + (b_1 +b_2)^2}\]

Phase angle with respect to X-axis is,

    \[ \phi = tan^{-1} \left(\frac{b_1 +b_2}{a_1 + a_2} \right) \]

Phasor Subtraction

For subtraction, E = A - B = (a_1 + j b_1 ) - (a_2 + j b_2) = (a_1 - a_2) +j(b_1 - b_2)

Its magnitude is obtained as,

    \[E = \sqrt{(a_1-a_2)^2 + (b_1 -b_2)^2}\]

Phase angle with respect to X-axis is,

    \[ \phi = tan^{-1} \left(\frac{b_1 -b_2}{a_1 - a_2}\right) \]

Phasor Multiplication

Performing the phasor multiplication and phasor division by using the rectangular form is complex. But by using polar form or exponential form, the operation will be very simple.

Let us consider the two phasors, A = a_1 + j b_1 and B = a_2 + j b_2.

In polar form of representation,

    \[ A = a_1 + j b_1 = A \angle \phi_1\]

    \[B = a_2 + j b_2 = B \angle \phi_2 \]

The resultant phasor is given by,

    \[ A*B = A \angle \phi_1 + B \angle \phi_2 = AB \angle{(\phi_1 +\phi_2)}\]

Hence, it is observed that the product of any two phasors is given by another resultant phasor, whose magnitude is given by A*B and the phase angle is equal to the sum of angles of A and B.

Phasor Division

Similarly for division,

    \[ \frac{A}{B} = \frac{A \angle \phi_1}{B \angle \phi_2} = \frac{A}{B} \angle{(\phi_1 -\phi_2)}\]

Here, it is observed that the quotient of any two phasors results in another phasor, whose magnitude is given by (A/B) and the phase angle is equal to the angle of A minus the angle of B.



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