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PART TWO: AC Circuits

Chapter 5Sinusoids and phasors

THIS COURSE CAN BE FOUND AT:

http://users.utcluj.ro/~denisad

… PART II

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PHASORSREPRESENTATION OF SINUSOIDAL TIME FUNCTIONS BY VECTORS

AND COMPLEX NUMBERS

• A phasor is a complex number that represents the amplitude and a

phase of a sinusoid.

- Sinusoids are easily expressed in terms of phasors, which are more convenient to

work with that sine and cosine functions.

- The origin of the term phasor rightfully suggests that a (diagrammatic) calculus somewhat

similar to that possible for vectors is possible for phasors as well.

- An important additional feature of the phasor transform is that differentiation and integration

of sinusoidal signals (having constant amplitude, period and phase) corresponds to simple

algebraic operations on the phasors; the phasor transform thus allows the analysis

(calculation) of the AC steady state of RLC circuits by solving simple algebraic equations

(with complex coefficients) in the phasor domain instead of solving differential equations

(with real coefficients) in the time domain.

- The originator of the phasor transform was Charles Proteus Steinmetz working at General

Electric in the late 19th century.

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5.5. PHASORS

(Eq. 1)

(Eq. 2a)

(Eq. 2b)

(Eq. 3)

(Eq. 2a)

5.5. PHASORS

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5.5. PHASORS

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A) Geometrical (phasorial) representation.

2IOA =

+=

tAOx0

Vector:

- its projection onto the vertical

axis Oy represents the

instantaneous value of i(t) to the

scale chosen.

- Ox axis is called reference axis.

)sin(2 += tIi

KEEP IN MIND:

The phasor is not an electric current (voltage), it is only a symbol for it.

5.5. PHASORS

Mathematical operations in phasorial representation.

a) The multiplication of a sinusoid by a scalar „a”:

The resulting phasor has its peak value „a”

times bigger and the same phase angle.

b) The addition : 22112121 /2/2 +++=++→ tItIOAOAii

5.5. PHASORS

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c) The derivation :

)cos(2 += tIdt

di

++=

2sin2

tI

dt

di

dt

di

22

++→

tI

The derivation of a sinusoid corresponds to the multiplication of the

peak value by and counter-clockwise rotation of the phasor with π/2

5.5. PHASORS

d) The integration:

The integration of a sinusoid corresponds to the division of the peak

value by and clockwise rotation of the phasor with π/2

Important remark: in practical work, the Argand diagram is

simplified by omitting the axes.

−+=

2sin

2

t

Iidt

2/2

−+

→ t

I

5.5. PHASORS

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REVIEW: COMPLEX NUMBERS

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REVIEW: COMPLEX NUMBERS

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REVIEW: COMPLEX NUMBERS

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REVIEW: COMPLEX NUMBERS

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B) Analytical (or complex) representation.

)sin(2 += tIi

- The complex time function

(complex instantaneous value):

)(2 += tjeIi

)(2)sin(2 +→ =+= tjeIitIi

),sin(2)cos(2 +++= tIjtIi

]2[)( )( +== tjmm eIii

5.5. PHASORS

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Mathematical operations in complex representation.

a) The multiplication by a scalar „a”:

↔

b) The addition :

c) The derivation:

demonstration:

↔

↔

where:

)sin(2 += taIia iaeaI tj =+ )(2

2121 iiii ++→

dt

diij→

=+= )cos(2 tIdt

di

2)()

2(

22

j

tjtj

eeIeI = +++

,2 )( ijeIj tj == +

jjej

=+=2

sin2

cos2

5.5. PHASORS

d) The integration :

↔

where:

The simplified complex representation:

It is called complex effective value.

→ i

jidt

1

)2

sin(2

)cos(2

−+=+−= tI

tI

idt =−+ )

2(2

tj

eI

ij

eIj

eeI tj

jtj === +

−+

12

12 )(2)(

jjje

j 1

2sin

2cos2 =−=

−+

−=

−

→ =+= jIeItIi )sin(2

5.5. PHASORS

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5.6. THE CHARACTERISATION OF THE LINEAR CIRCUITS IN

COMPLEX PLANE

5.6.1. The complex impedance

is the resistive or in-phase component,

X = Z sinφ is the reactive or quadrature component.

2 sin( )u U t = + jU U e =

2 sin( )i I t = + jI I e =

1( , , , ,...)U

Z f R L CI

= =

( ) cos( ) sin( )jU U U UZ e j

I I I I

−= = = − + −

jXRjZZZeZ j +=+== sincos

cosZR =

Where: Z is the impedance

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22 XRZ +=

R

Xarctan=

−+=+=

CLjRjXRZ

1

For the RLC series circuit:

( )j

j

j

U Ue UI e

Z ZZe

−= = =

Im[ 2 ] 2 sin( )j t Ui I e t

Z

= = + −

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5.6.2. The complex admittance.

is the conductance (the real part of Y),

is the susceptance (the imaginary part of Y).

11

( ; , , ,...)I

Y g R L CU Z

= = =

( ) cos( ) sin( )j

j

j

I e I I IY e j

U U UU e

− −

= = = − − −

jBGjYYYeY j −=−== − sincos

cosYG =

sinYB =

22 BGY +=

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jBGXR

Xj

XR

R

XR

jXR

jXRZY −=

+−

+=

+

−=

+==

222222

11

( )j jI U Y UYe UYe − −= = =

22 BGY += the admittance triangle

- the current:

( ) Im[ 2 ] 2 sin( )j ti t Ie UY t = = + −

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- it should be noted that and because: 2Z

RG =

2Z

XB =

5.6.3. The complex power

Let and

- the complex power: S U I

=

( ) (cos sin )j j j j jS Ue Ie UIe UIe Se UI j P jQ − −= = = = = + = +

* *cos sinjS U I UIe UI jUI P jQ −= = = − = −

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References

[1] Charlews K. Alexander, Matthew N.O.Sadiku, Fundamentals of Electric

Circuits (Fifth Edition), published by McGraw-Hill, 2013

[2] Radu V. Ciupa, Vasile Topa, The Theory of Electric Circuits, published

by Casa Cartii de Stiinta, 1998

[3] Dan. D Micu, Laura Darabant, Denisa Stet et al., Teoria circuitelor

electrice. Probleme, published by UTPress, 2016

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