Chapter 8. Sinusoidal Sources and Phasors

Chapter 8. Sinusoidal Sources and Phasors

Chapter Contents

• 8.1 Properties of Sinosoids• 8.2 RLC Circuit Example• 8.3 Complex Sources• 8.4 Phasors• 8.5 I-V Laws for Phasors• 8.6 Impedance and Admittance• 8.7 Kirchhoff's Laws and Impedance

Equivalents• 8.8 Phasor Circuits

8.1 PROPERTIES OF SINUSOIDS

• The sine wave :

8.1 PROPERTIES OF SINUSOIDS

• The sinusoid is a periodic function. There is a smallest number s.t.

• Period and frequency :

- Period :

- Frequency : German physicist

Heinrich R. Hertz (1857-1894)

8.1 PROPERTIES OF SINUSOIDS

• Relation between frequency and radian frequency :

• A more general sinusoidal expression :

Ф : phase angle, or phase

8.1 PROPERTIES OF SINUSOIDS

- Phase angle Ф should be expressed in radians, but degrees are a very familiar measure for angle. Therefore, we may write

8.1 PROPERTIES OF SINUSOIDS

- Phase leading and lagging :

Ex.)

8.1 PROPERTIES OF SINUSOIDS

• Sine and cosine functions :

• Half-period shifts and full-period shifts :

8.1 PROPERTIES OF SINUSOIDS

Ex. 8.1)

8.1 PROPERTIES OF SINUSOIDS

• Linear combination of a sine wave and a cosine wave :

8.1 PROPERTIES OF SINUSOIDS

• Quadrature representation of the sinusoid

8.1 PROPERTIES OF SINUSOIDS

• Ex.) Let's find the forced mesh current in the series cct.

* By KVL,

8.2 RLC CIRCUIT EXAMPLE

+-

2H

15 cos 2 t [V] i

F101

• Trial forced solution : → eq.(1)

8.2 RLC CIRCUIT EXAMPLE

• Combining the two quadrature terms,

8.2 RLC CIRCUIT EXAMPLE

• An alternative method for treating ccts with sinusoidal sources =

replacing the given sources by complex sources.

• A complex number is a point in the complex plane.

(1) Rectancluar form :

8.3 COMPLEX SOURCES

(2) Polar form :

(3) Trigonometric form :

8.3 COMPLEX SOURCES

(4) Exponential polar form :

- Euler's identity :

- Therefore, the exponential form is

- Polar form of ejθ

8.3 COMPLEX SOURCES

Ex. 8.3) A= 4+j3

<Solution>• Polar form :- Since and ,

• Exponential polar form :

8.3 COMPLEX SOURCES

• Complex exponential function : ejωt (rotating phasor)

- cosωt : projection of this point onto the horizontal axis real part- sinωt : projection of this point onto the vertical axis imaginary

part

8.3 COMPLEX SOURCES

The complex exponential traces out unit circles in the

complex plane, beginning on the positive real axis at time and

moving counterclockwise, completing one full circle (or

period) every seconds.

• General scaled and phase-shifted complex exponential :

Vmej(ωt+ф)

- Euler's identity form :

8.3 COMPLEX SOURCES

• Application of complex numbers to electric ccts : superposition

3Hv g=f 1 ( t)+ j f 2 ( t) i=i 1 ( t)+ j i2 ( t)LinearCircuit

C

+-

3Hv g=f 2 ( t) i=i 2 ( t)LinearCircuit

C

+-

+-

3Hv g=f 1 ( t) i=i 1 ( t)LinearCircuit

C

8.3 COMPLEX SOURCES

* The forced response to will

be a complex exponential of the same

frequency , .

Ex. 8.4)

<Solution>Replacing the sinusoidal source by the complex source , by KVL,

• Substituting the trial solution , .

8.3 COMPLEX SOURCES

+-

2H

15 cos 2 t [V] i

F101

Ex. 8.5)

<Solution>• First we replace the real excitation by the complex excitation.

• Complex response i 1 satisfies

• Substituting the trial forced solution :

8.3 COMPLEX SOURCES

• The preceding results may be put in much more compact form by the use of quantities called phasors.

– The phasor method is credited to Charles Proteus Steinmetz (1865-1923), a famous electrical engineer with the General Electric Company.

• Consider the forced response of a cct to sinusoidal excitation at frequency ω. Each sinusoidal source may be expressed as a cosine.

• We replace by a complex exponential source.

8.4 PHASORS

• Phasor representation of

- In general, each current and each voltage will be of the form

We define and as phasors, i.e., the complex numbers that multiply in the expression for currents and voltages.

8.4 PHASORS

• Relation between the sinusoidal source and the phasor :

• Phasor transformation :

• Since , by using the polar exponential form,

The amplitude of the sinusoid is the magnitude of its phasor, and the phase angle of the sinusoid is the angle of its phasor.

8.4 PHASORS

Ex. 8.6)

<Solution>• Since , we use as the complex

exponential source.

• Substituting the trial forced solution into eq.(1) : eq.(1)

• Therefore, the forced solution :

8.4 PHASORS

+-

1H

v g i

• If the sinusoidal source is given in the sine form , we will first convert to the cosine form.

– Complex sources and source phasors for cosine and sine sources

8.4 PHASORS

Ex. 8.7)

<Solution>• Replacing the sources by exponential sources,

8.4 PHASORS

+-

3H

i

36 cos (2 t+ 30 ) Vo 2 sin (2 t - 15 ) Ao

+- 3H

i

2 e j(2 t - 105 ) Ao

36 e j(2 t+ 30 ) Vo

i1i2

• KVL equation on mesh-1,

where

8.4 PHASORS

+- 3H

i

2 e j(2 t - 105 ) Ao

36 e j(2 t+ 30 ) Vo

i1i2

• Substituting the trial forced solution :

8.4 PHASORS

+- 3H

i

2 e j(2 t - 105 ) Ao

36 e j(2 t+ 30 ) Vo

i1i2

• Relationships between phasor voltage and phasor current for

resistors, inductors, and capacitors are very similar to Ohm's

law for resistors.

• Consider a cct in which all currents and voltages are of the form .

We are interested in the forced response only.

(1) I-V relationship for resistor cct :

8.5 I-V LAWS FOR PHASORS

• Canceling the eωt factors,

- Since and ,

8.5 I-V LAWS FOR PHASORS

Ex. 8.8)

<Solution>• The phasor voltage and current :

• Current in time-domain :

8.5 I-V LAWS FOR PHASORS

v=Ri R =5

i

+

-

Time-domain

V =R I R =5

I

+

-

Frequency-domain

v ( t) v

i

t

30

(2) I-V relationship for inductor cct :

• Substituting the complex current and voltage into the time-domain relation,

• Therefore, we obtain the phasor relation :

8.5 I-V LAWS FOR PHASORS

• If the current in the inductor is given by , then the phasor current is

• therefore, the phasor voltage :

8.5 I-V LAWS FOR PHASORS

v=L di/dt L

i

+

-

Time-domain

V =j L I jL

I

+

-

Frequency-domain

• Therefore, the voltage in the time domain :

: The current lags the voltage by 90°.

8.5 I-V LAWS FOR PHASORS

(3) I-V relationship for capacitor cct :

• Substituting the complex current and voltage into the time-domain relation,

• Therefore, we obtain the phasor relation : Since

8.5 I-V LAWS FOR PHASORS

• If the capacitor voltage is given by , then the phasor voltage is

• therefore, the phase current :

8.5 I-V LAWS FOR PHASORS

V

I= jC V

+

-

jC1

v C

i=C dv/dt

+

-

• Therefore, the current in the time domain :

: The current leads the voltage by 90°.

8.5 I-V LAWS FOR PHASORS

v , i

v

i

t

V90

Ex. 8.9) Determine the current i(t) through a 1 [uF] capacitor when

is applied.

<Solution>

8.5 I-V LAWS FOR PHASORS

• In general cct with 2 accessible terminals, if the time-domain

voltage and current are

• then

8.6 IMPEDANCE AND ADMITTANCE

V

I

+

-

Phasorcircuit

• Impedance :

- Therefore, the I-V relation : ( Ohm's law )

- Magnitude and phase of impedance :

• Impedance is a complex #, being the ratio of two complex #'s, but it is not a phasor because of no corresponding sinusoidal time-domain function.

8.6 IMPEDANCE AND ADMITTANCE

• Impedance in rectangular form :

8.6 IMPEDANCE AND ADMITTANCE

Ex. 8.10) Determine the impedance in the polar and the rectangular forms.

<Solution>

8.6 IMPEDANCE AND ADMITTANCE

V

I

+

-

Phasorcircuit

• The impedances of resistors, inductors, and capacitors :

– Inductive and capacitive reactances : XL, XC

8.6 IMPEDANCE AND ADMITTANCE

• Admittance :

– Relation between components of Y and Z :

8.6 IMPEDANCE AND ADMITTANCE

Ex. 8.11)

<Solution>

• Admittances of a resistor, an inductor, and a capacitor :

8.6 IMPEDANCE AND ADMITTANCE

(1)Kirchhoff's Voltage Law• If a complex excitation is applied to a cct, complex voltages

appear across the elements in the cct.

• KVL around a typical loop : KVL holds for phasors

where

8.7 KIRCHHOFF'S LAWS AND IMPEDANCE EQUIVALENTS

+

-

V 1e j(t+)

V m e j(t+ )

V 2e j(t+) V N e j(t+)

(2) Kirchhoff's Current Law

• At any node with N connected branches : KCL hods for phasors

Where

8.7 KIRCHHOFF'S LAWS AND IMPEDANCE EQUIVALENTS

Ex. 8.12)

<Solution>

• Since a single phasor current I flows in each element,

- By KVL,

- Therefore, the equivalent impedance :

8.7 KIRCHHOFF'S LAWS AND IMPEDANCE EQUIVALENTS

V

+

-

IZ 1

+ -V 1

Z N

+ -V N

Z 2

+ -V 2

Z eq

• Equivalent admittance of parallel admittances :

– In the case of two parallel elements ( N=2 ),

8.7 KIRCHHOFF'S LAWS AND IMPEDANCE EQUIVALENTS

Ex. 8.13)

<Solution>• By KVL in the phasor cct,

8.7 KIRCHHOFF'S LAWS AND IMPEDANCE EQUIVALENTS

Ex. 8.14)

<Solution>• Using series-parallel impedance equivalents,

8.7 KIRCHHOFF'S LAWS AND IMPEDANCE EQUIVALENTS

• Phasor cct : Time-domain cct with the voltages and currents

replaced by their phasors and the elements identified by their

impedances.

The describing equation obtained from this cct is then the phasor

equation.

8.8 PHASOR CIRCUITS

Ex. 8.15)

<Solution>• The phasor cct is obtained by replacing the voltage source and the

currents by their phasors and the elements by their impedances.

• Impedance from the source terminals :

• Current through the source terminal :

• By current division,

8.8 PHASOR CIRCUITS

+-

II1

5 0 o V

j- j

Ex. 8.16)

<Solution>• By KCL,

8.8 PHASOR CIRCUITS

3 cos 4 t A

i

+-

+

-

F81-

v 1v 12

1-

aI

+-

+

-

V 1 21-V 1

- j

a

3 0 o VA

• Op-amp cct :

• Phasor analysis may be used to find the steady-state

response only with stable ccts.

8.8 PHASOR CIRCUITS

Ex. 8.17)

<Solution>(1) Natural response :• By KVL around the left loop,

• Substituting and differentiating,

• Characteristic equation and the natural response :

8.8 PHASOR CIRCUITS

4 cos 3 t A

1F

1H

v -

i

1

+-

+

v23 v

23V

(2) Forced response :• The corresponding phasor equivalent cct :

• By KVL,

• Substituting ,

8.8 PHASOR CIRCUITS

+j 3

V -

I

1

+-

+

V234 0 o A V

23

- j 13

V

• Report<Problems> 8.7, 8.9, 8.13(a)(b)(d), 8.14(a)(c),

8.20, 8.31, 8.34, 8.36, 8.41, 8.47 (10 problems)