Phasors Demystified

Phasors Demystified

Phasors Demystified
Swapnil Sunil Jain
Aug 7, 2006

Phasors Demystified

Suppose the following integro-differential equation is given in the time-domain

1

1

This is not the most general integro-differential equation but it has all the basic elements required for this discussion and hence the reader can easily extend this discussion for the more generalized case.

:

C

1

⁢

d

d

⁢

t

⁢

y

⁢

(

t

)

+

C

2

⁢

∫

-

∞

t

y

⁢

(

t

)

⁢

𝑑

t

+

C

3

⁢

y

⁢

(

t

)

=

x

⁢

(

t

)

(1)

where

y

⁢

(

t

)

and

x

⁢

(

t

)

are sinusoidal waveforms of the same frequency. Now, since

y

⁢

(

t

)

is a sinusoidal function it can be represented as

A

y

⁢

cos

⁡

(

ω

⁢

t

+

ϕ

y

)

and, similarly, x(t) can be represented as

A

x

⁢

cos

⁡

(

ω

⁢

t

+

ϕ

x

)

. Furthermore, using the properties of complex numbers we can write

y

⁢

(

t

)

=

A

y

⁢

cos

⁡

(

ω

⁢

t

+

ϕ

y

)

=

ℜ

⁡

(

A

y

⁢

e

j

⁢

ϕ

y

⁢

e

j

⁢

ω

⁢

t

)

x

⁢

(

t

)

=

A

x

⁢

cos

⁡

(

ω

⁢

t

+

ϕ

x

)

=

ℜ

⁡

(

A

x

⁢

e

j

⁢

ϕ

x

⁢

e

j

⁢

ω

⁢

t

)

Now if we define the quantities

Y

~

as

A

y

⁢

e

j

⁢

ϕ

y

and

X

~

as

A

x

⁢

e

j

⁢

ϕ

x

(where

Y

~

and

X

~

are called

phasors

), then we can write the above expression in a more

compact

form as

y

⁢

(

t

)

=

ℜ

⁡

(

Y

~

⁢

e

j

⁢

ω

⁢

t

)

x

⁢

(

t

)

=

ℜ

⁡

(

X

~

⁢

e

j

⁢

ω

⁢

t

)

Now, using the above expression for

y

⁢

(

t

)

and

x

⁢

(

t

)

we can rewrite our original integro-differential equation as

C

1

⁢

d

d

⁢

t

⁢

ℜ

⁡

[

Y

~

⁢

e

j

⁢

ω

⁢

t

]

+

C

2

⁢

∫

-

∞

t

ℜ

⁡

[

Y

~

⁢

e

j

⁢

ω

⁢

t

]

⁢

𝑑

t

+

C

3

⁢

ℜ

⁡

[

Y

~

⁢

e

j

⁢

ω

⁢

t

]

=

ℜ

⁡

[

X

~

⁢

e

j

⁢

ω

⁢

t

]

Moving the

derivative

and the

integral

inside the

ℜ

operator we get

C

1

⁢

ℜ

⁡

[

d

d

⁢

t

⁢

Y

~

⁢

e

j

⁢

ω

⁢

t

]

+

C

2

⁢

ℜ

⁡

[

∫

-

∞

t

Y

~

⁢

e

j

⁢

ω

⁢

t

⁢

𝑑

t

]

+

C

3

⁢

ℜ

⁡

[

Y

~

⁢

e

j

⁢

ω

⁢

t

]

=

ℜ

⁡

[

X

~

⁢

e

j

⁢

ω

⁢

t

]

⇒

C

1

⁢

ℜ

⁡

[

Y

~

⁢

j

⁢

ω

⁢

e

j

⁢

ω

⁢

t

]

+

C

2

⁢

ℜ

⁡

[

Y

~

⁢

e

j

⁢

ω

⁢

t

j

⁢

ω

]

+

C

3

⁢

ℜ

⁡

[

Y

~

⁢

e

j

⁢

ω

⁢

t

]

=

ℜ

⁡

[

X

~

⁢

e

j

⁢

ω

⁢

t

]

⇒

ℜ

⁡

[

Y

~

⁢

j

⁢

ω

⁢

C

1

⁢

e

j

⁢

ω

⁢

t

]

+

ℜ

⁡

[

Y

~

⁢

C

2

j

⁢

ω

⁢

e

j

⁢

ω

⁢

t

]

+

ℜ

⁡

[

Y

~

⁢

C

3

⁢

e

j

⁢

ω

⁢

t

]

-

ℜ

⁡

[

X

~

⁢

e

j

⁢

ω

⁢

t

]

=

0

⇒

ℜ

⁡

[

Y

~

⁢

j

⁢

ω

⁢

C

1

⁢

e

j

⁢

ω

⁢

t

+

Y

~

⁢

C

2

j

⁢

ω

⁢

e

j

⁢

ω

⁢

t

+

Y

~

⁢

C

3

⁢

e

j

⁢

ω

⁢

t

-

X

~

⁢

e

j

⁢

ω

⁢

t

]

=

0

⇒

ℜ

⁡

[

e

j

⁢

ω

⁢

t

⁢

(

Y

~

⁢

j

⁢

ω

⁢

C

1

+

Y

~

⁢

C

2

j

⁢

ω

+

Y

~

⁢

C

3

-

X

~

)

]

=

ℜ

⁡

[

0

]

+

j

⁢

ℑ

⁡

[

0

]

Equating the

real parts

above we get,

⇒

e

j

⁢

ω

⁢

t

⁢

(

Y

~

⁢

j

⁢

ω

⁢

C

1

+

Y

~

⁢

C

2

j

⁢

ω

+

Y

~

⁢

C

3

-

X

~

)

=

0

⇒

Y

~

⁢

j

⁢

ω

⁢

C

1

+

Y

~

⁢

C

2

j

⁢

ω

+

Y

~

⁢

C

3

-

X

~

=

0

(for

t

≠

-

∞

)

(2)

Hence, we have now arrive at the phasor domain expression for (1). You can see from the

analysis

above that we aren’t adding or losing any information when we transform equation (1) into the

”phasor domain”

and arrive at equation (2). One can easily get to (2) by using simple algebraic properties of real and

complex numbers

. Furthermore, since (2) can be derived readily from (1), in practice we don’t even bother to do all the intermediate steps and just skip straight to (2) calling this ”skipping of steps” as ”transforming the equation into the phasor domain.”

We can now continue the analysis even further and solve for

y

⁢

(

t

)

which is the whole motivation behind the use of phasors. Solving for

Y

~

in (2) we get

⇒

Y

~

=

X

~

(

j

⁢

ω

⁢

C

1

+

C

2

j

⁢

ω

+

C

3

)

Now, since

y

⁢

(

t

)

=

ℜ

⁡

[

Y

~

⁢

e

j

⁢

ω

⁢

t

]

we have

y

⁢

(

t

)

=

ℜ

⁡

[

X

~

⁢

e

j

⁢

ω

⁢

t

(

j

⁢

ω

⁢

C

1

+

C

2

j

⁢

ω

+

C

3

)

]

⇒

y

⁢

(

t

)

=

ℜ

⁡

[

A

x

⁢

e

j

⁢

ϕ

x

⁢

e

j

⁢

ω

⁢

t

(

j

⁢

ω

⁢

C

1

+

C

2

j

⁢

ω

+

C

3

)

]

The above equation makes sense because you can see that the output y(t) is given completely in terms of the variables

A

x

and

ϕ

x

(which depend only on the input sinusoid

x

⁢

(

t

)

) and the constants

C

1

,

C

2

and

C

3

—as we expected! So by converting the integro-differential equation (1) into the phasor domain (2), all the complicated integration and

differentiation

operations become simple manipulation of complex variables—which is why phasors are so useful!

Title

Phasors Demystified

Canonical name

PhasorsDemystified1

Date of creation

2013-03-11 19:26:47

Last modified on

2013-03-11 19:26:47

Owner

swapnizzle (13346)

Last modified by

(0)

Numerical id

1

Author

swapnizzle (0)

Entry type

Definition

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