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Electric Circuit Theory

Nam Ki Min

010-9419-2320 [email protected]

Sinusoidal Steady-State Analysis

Chapter 9

Nam Ki Min

010-9419-2320 [email protected]

CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 [email protected]

Chapter 9 Sinusoidal Steady-State Analysis 3 9.1 The Sinusoidal Source

Sinusoidal Source

Sinusoidal voltage source

• Produces a voltage that varies sinusoidally with time.

Sinusoidal current source

Sinusoidal source or sinusoidally time-varying excitation or sinusoid. • Produce a signal that has the form of the sine or cosine function.

• Produces a current that varies sinusoidally with time.

𝑣

𝑣

𝑒

𝑒

𝑣𝑠

𝑖 𝑖𝑠



Alternating Circuit

A sinusoidal current is usually referred to as alternating current (ac).

Circuits driven by sinusoidal current or voltage sources are called ac circuits.

𝜋 2

−𝜋 2

𝜋 0

𝜔 𝑣(𝑡)

AC Waveforms

Sinusoidal wave



Why Sinusoids?

Nature itself is characteristically sinusoidal. We experience sinusoidal variation in the motion of a pendulum, the vibration of a string, the economic fluctuations of the stock market, and the natural response of underdamped second-order systems’

A sinusoidal signal is easy to generate and transmit.

It is the form of voltage generated throughout the world and supplied to homes, factories, laboratories, and so on.

It is the dominant form of signal in the communications and electric power industries.

Through Fourier analysis, any practical periodic signal can be represented by a sum of sinusoids. Sinusoids, therefore, play an important role in the analysis of periodic signals.

A sinusoid is easy to handle mathematically.

𝜋 2

−𝜋 2

𝜋 0

𝜔 𝑣(𝑡)



Sinusoidal Voltage

𝑣 𝑡 = 𝑉𝑚 sin𝜔𝑡

𝑣 𝑡 = 𝑉𝑚 cos𝜔𝑡

𝑉𝑚 cos𝜔𝑡 𝑉𝑚 sin𝜔𝑡

𝑣 𝑡

𝜔𝑡 𝜙 = 90°

𝑣 𝑡 = 𝑉𝑚 sin(𝜔𝑡 ± 90°) = ±𝑉𝑚 cos𝜔𝑡

𝑣 𝑡 = 𝑉𝑚 cos(𝜔𝑡 ± 90°) = ∓𝑉𝑚 sin𝜔𝑡

𝑉𝑚

A sinusoid can be expressed in either sine or cosine form.

When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes.

This is achieved by using the following trigonometric identities:

sin(𝐴 ± 𝐵) = sin 𝐴 cos𝐵 ± cos𝐴 sin 𝐵

cos(𝐴 ± 𝐵) = cos𝐴 cos𝐵 ∓ sin𝐴 sin𝐵



Sinusoidal Voltage

𝑣 𝑡 = 𝑉𝑚 cos(𝜔𝑡 + 𝜙)

• 𝑉𝑚: amplitude

• 𝜔 ∶ angular frequency in radians/s

• 𝜔𝑡: argument

• 𝑇 ∶ period in second

𝑇 =2𝜋

𝜔

𝑣 𝑡 + 𝑇 = 𝑉𝑚 cos𝜔 𝑡 + 𝑇 = 𝑉𝑚cos 𝜔 𝑡 +2𝜋

𝜔

= 𝑉𝑚 cos𝜔𝑡

= 𝑉𝑚 cos(𝜔𝑡 + 2𝜋)

= 𝑣(𝑡)

• 𝑓 ∶ frequency in Hz

𝑓 =1

𝑇 𝜔 = 2𝜋𝑓

A more general expression:

• 𝜙: phase angle

(9.1)



Converting Sine to Cosine

𝑉𝑚 cos𝜔𝑡

𝑉𝑚 sin𝜔𝑡

𝑣 𝑡

𝜔𝑡 𝜙 = 90°

𝑉𝑚

𝑣 𝑡 = 𝑉𝑚 sin(𝜔𝑡 + 90°) = 𝑉𝑚 cos𝜔𝑡

𝑣 𝑡 = 𝑉𝑚 cos(𝜔𝑡 − 90°) = 𝑉𝑚 sin𝜔𝑡

𝑣 𝑡 = 𝑉𝑚 cos(𝜔𝑡 ± 180°) = −𝑉𝑚 cos𝜔𝑡

𝑣 𝑡 = 𝑉𝑚 sin(𝜔𝑡 ± 180°) = −𝑉𝑚 sin𝜔𝑡

𝑣 𝑡 = 𝑉𝑚 cos(5𝑡 + 10°) = 𝑉𝑚 sin(5t + 90° + 10°)

= 𝑉𝑚 sin(5t + 100°)

𝑣(𝑡) = 𝑉𝑚 sin(5t − 260°)

= 𝑉𝑚 sin(5t − 180° − 80°)

= −𝑉𝑚 sin(5t + 90 − 170°)

= −𝑉𝑚 cos(5t − 180 + 10°)

= 𝑉𝑚 cos(5t + 10°)

= −𝑉𝑚 sin(5t − 80°)

= −𝑉𝑚 cos(5t − 170°)



Phase Relation of a Sinusoidal Wave



Phase Shift (difference)

The two waves (A versus B) are of the same amplitude and frequency, but they are out of step with each other.

In technical terms, this is called a phase shift (difference).

𝜙𝐴

𝜙𝐵 𝜙 = 0

𝑣 𝑡 = 𝑉𝑚 sin(𝜔𝑡 + 𝜙𝐴)

𝑣 𝑡 = 𝑉𝑚 sin(𝜔𝑡 + 𝜙𝐵)

𝑣 𝑡 = 𝑉𝑚 cos(𝜔𝑡 + 𝜙𝐴)

𝑣 𝑡 = 𝑉𝑚 cos(𝜔𝑡 + 𝜙𝐵)

𝜔

𝜔 𝜙𝐴 𝜙𝐵

𝑣 𝑡 𝑣 𝑡



Phase Shift (difference)

Examples of phase shifts

The sinusoids are said to be in phase.

The sinusoids are said to be out of phase.

B “lags” A

A “lags” B

Leading

Lagging


Chapter 9 Sinusoidal Steady-State Analysis 12 9.2 The Sinusoidal Response

Sinusoidal Response

A sinusoidal forcing function produces both a natural (or transient) response and a forced (or steady-state) response, much like the step function, which we studied in Chapters 7 and 8.

The natural response of a circuit is dictated by the nature of the circuit, while the steady-state response always has a form similar to the forcing function.

However, the natural response dies out with time so that only the steady-state response remains after a long time.

When the natural response has become negligibly small compared with the steady-state response, we say that the circuit is operating at sinusoidal steady state.

It is this sinusoidal steady-state response that is of main interest to us in this chapter.


Chapter 9 Sinusoidal Steady-State Analysis 13

𝑑𝑖

𝑑𝑡+𝑅

𝐿𝑖 =

𝑉𝑚𝐿cos(𝜔𝑡 + 𝜙)

9.2 The Sinusoidal Response

► Example ◄

For the circuit in Fi.9.5, find current

𝑣𝑠(𝑡) = 𝑉𝑚 cos(𝜔𝑡 + 𝜙)

𝐿𝑑𝑖

𝑑𝑡+ 𝑅𝑖 = 𝑉𝑚 cos(𝜔𝑡 + 𝜙)

𝑝(𝑡) =𝑅

𝐿

𝑔(𝑡) =𝑉𝑚𝐿cos(𝜔𝑡 + 𝜙)

𝜇 𝑡 = exp 𝑅

𝐿𝑑𝑡 = 𝑒

𝑅𝐿𝑡

(9.7)

(9.8)

(3)

(1)

(2)

(4)

𝜇 𝑥 = exp 𝑝(𝑥)𝑑𝑥



𝑖 𝑡 =1

𝑒𝑅𝐿𝑡 𝑒

𝑅𝐿𝑡 𝑉𝑚𝐿cos(𝜔𝑡 + 𝜙)𝑑𝑡 +

𝐾

𝑒𝑅𝐿𝑡

9.2 The Sinusoidal Response

► Example ◄

𝜇 𝑡 = 𝑒𝑅𝐿𝑡

𝑔(𝑡) =𝑉𝑚𝐿cos(𝜔𝑡 + 𝜙) =

1

𝑒𝑅𝐿𝑡

𝑉𝑚𝐿 𝑒

𝑅𝐿𝑡 cos𝜔𝑡 cos𝜙 − sin𝜔𝑡 sin𝜙 𝑑𝑡 +

𝐾

𝑒𝑅𝐿𝑡

=1

𝑒𝑅𝐿𝑡

𝑉𝑚𝐿

cos𝜙 𝑒𝑅𝐿𝑡 cos𝜔𝑡 𝑑𝑡 − sin𝜙 𝑒

𝑅𝐿𝑡 sin𝜔𝑡 𝑑𝑡 +

𝐾

𝑒𝑅𝐿𝑡

cos(𝜔𝑡 + 𝜙) = cos𝜔𝑡 cos𝜙 − sin𝜔𝑡 sin𝜙

exp 𝑎𝑥 cos 𝑏𝑥 𝑑𝑥 =𝑒𝑎𝑥

𝑎2 + 𝑏2(𝑎 cos 𝑏𝑥 + 𝑏 sin 𝑏𝑥)

exp 𝑎𝑥 sin 𝑏𝑥 𝑑𝑥 =𝑒𝑎𝑥

𝑎2 + 𝑏2(𝑎 sin 𝑏𝑥 − 𝑏 cos 𝑏𝑥)

(5)

(6)


Chapter 9 Sinusoidal Steady-State Analysis 15 9.4 The Passive Circuit Elements in the Frequency Domain

=1

𝑒𝑅𝐿𝑡

𝑉𝑚𝐿

cos𝜙 𝑒𝑅𝐿𝑡 cos𝜔𝑡 𝑑𝑡 − sin𝜙 𝑒

𝑅𝐿𝑡 sin𝜔𝑡 𝑑𝑡 +

𝐾

𝑒𝑅𝐿𝑡

exp 𝑎𝑥 cos 𝑏𝑥 𝑑𝑥 =𝑒𝑎𝑥

𝑎2 + 𝑏2(𝑎 cos 𝑏𝑥 + 𝑏 sin 𝑏𝑥)

exp 𝑎𝑥 sin 𝑏𝑥 𝑑𝑥 =𝑒𝑎𝑥

𝑎2 + 𝑏2(𝑎 sin 𝑏𝑥 − 𝑏 cos 𝑏𝑥)

► Example ◄

=1

𝑒𝑅𝐿𝑡

𝑉𝑚𝐿

cos𝜙𝑒𝑎𝑥

𝑎2 + 𝑏2 (𝑎 cos 𝑏𝑡 + 𝑏 sin 𝑏𝑡) − sin𝜙

𝑒𝑎𝑥

𝑎2 + 𝑏2 (𝑎 sin 𝑏𝑡 − 𝑏 cos 𝑏𝑡) +

𝐾

𝑒𝑅𝐿𝑡

𝑎 cos 𝑏𝑥 + 𝑏 sin 𝑏𝑥 = 𝑎2 + 𝑏2𝑎

𝑎2 + 𝑏2cos 𝑏𝑥 +

𝑏

𝑎2 + 𝑏2sin 𝑏𝑥 = 𝑎2 + 𝑏2 cos𝜃 cos 𝑏𝑥 + sin 𝜃 sin 𝑏𝑥

𝑎 =𝑅

𝐿

𝑏 = 𝜔 𝑒𝑎𝑥 = 𝑒

𝑅𝐿𝑡

=𝑉𝑚𝐿

cos𝜙1


1


𝐾

𝑒𝑅𝐿𝑡

= 𝑎2 + 𝑏2 cos(𝜃 − 𝑏𝑥)

𝑎 sin 𝑏𝑥 − 𝑏 cos 𝑏𝑥 = = 𝑎2 + 𝑏2 sin(𝑏𝑥 − 𝜃) 𝑎2 + 𝑏2 cos 𝜃 sin 𝑏𝑥 − sin 𝜃 cos 𝑏𝑥

(7)

(8)



► Example ◄

𝑎 cos 𝑏𝑥 + 𝑏 sin 𝑏𝑥 = 𝑎2 + 𝑏2𝑎

𝑎2 + 𝑏2cos 𝑏𝑥 +

𝑏

𝑎2 + 𝑏2sin 𝑏𝑥 = 𝑎2 + 𝑏2 cos𝜃 cos 𝑏𝑥 + sin 𝜃 sin 𝑏𝑥

= 𝑎2 + 𝑏2 cos(𝜃 − 𝑏𝑥)

𝑎 sin 𝑏𝑥 − 𝑏 cos 𝑏𝑥 = = 𝑎2 + 𝑏2 sin(𝑏𝑥 − 𝜃) 𝑎2 + 𝑏2 cos 𝜃 sin 𝑏𝑥 − sin 𝜃 cos 𝑏𝑥

=𝑉𝑚𝐿

cos𝜙1

𝑎2 + 𝑏2𝑎2 + 𝑏2 cos(𝜃 − 𝑏𝑡 − sin𝜙

1

𝑎2 + 𝑏2𝑎2 + 𝑏2 sin(𝑏𝑡 − 𝜃) +

𝐾

𝑒𝑅𝐿𝑡

=𝑉𝑚𝐿

1

𝑎2 + 𝑏2cos𝜙 cos(𝜃 − 𝜔𝑡) − sin𝜙 sin(𝜔𝑡 − 𝜃) +

𝐾

𝑒𝑅𝐿𝑡

=𝑉𝑚𝐿

1

𝑅𝐿

2

+ 𝜔2

cos𝜙 cos(𝜔𝑡 − 𝜃) − sin𝜙 sin(𝜔𝑡 − 𝜃) +𝐾

𝑒𝑅𝐿𝑡

𝑎 =𝑅

𝐿

𝑏 = 𝜔

=𝑉𝑚𝐿

cos𝜙1


1


𝐾

𝑒𝑅𝐿𝑡

(9)

(10)



► Example ◄

=𝑉𝑚𝐿

1

𝑅𝐿

2

+ 𝜔2

cos𝜙 cos(𝜔𝑡 − 𝜃) − sin𝜙 sin(𝜔𝑡 − 𝜃) +𝐾

𝑒𝑅𝐿𝑡

𝑖(𝑡) = 𝑉𝑚1

𝑅2 + (𝜔𝐿)2cos(𝜔𝑡 + 𝜙 − 𝜃) +

𝐾

𝑒𝑅𝐿𝑡

𝑎 =𝑅

𝐿

𝑏 = 𝜔

0 = 𝑉𝑚1

𝑅2 + (𝜔𝐿)2cos(𝜙 − 𝜃) + 𝐾 𝐾 = −𝑉𝑚

1

𝑅2 + (𝜔𝐿)2cos(𝜙 − 𝜃)

𝑖(𝑡) =𝑉𝑚

𝑅2 + 𝜔𝐿 2cos(𝜔𝑡 + 𝜙 − 𝜃) −

𝑉𝑚

𝑅2 + 𝜔𝐿 2cos(𝜙 − 𝜃) 𝑒−

𝑅𝐿𝑡

(10)

(11)

(12)

(13) (9.9)



𝑖 𝑡 = −𝑉𝑚

𝑅2 + 𝜔𝐿 2cos(𝜙 − 𝜃) 𝑒−

𝑅𝐿𝑡 +

𝑉𝑚

𝑅2 + 𝜔𝐿 2cos(𝜔𝑡 + 𝜙 − 𝜃) (9.9)

Sinusoidal Response

Transient(Natural) Component

• The natural response of a circuit is

dictated by the nature of the circuit.

• The natural response dies out with time.

Steady-state Component

(Forced Response)

• The steady-state solution is a sinusoidal function.

• The frequency of the response signal is identical to

the frequency of the source signal.

• The maximum amplitude of the steady-state

response, in general, differs from the maximum

amplitude of the source.

• The phase angle of the response signal, in general,

differs from the phase angle of the source.

When the natural response has become

negligibly small compared with the steady-

state response, we say that the circuit is

operating at sinusoidal steady state.

It is this sinusoidal steady-state response

that is of main interest to us in this chapter.


Chapter 9 Sinusoidal Steady-State Analysis 19 9.3 The Phasor

Definition

A phasor is a complex number that represents the amplitude and phase angle of a sinusoid.

𝑣 𝑡 = 𝑉𝑚 cos(𝜔𝑡 + 𝜙) → 𝐕 = 𝑉𝑚𝑒𝑗𝜙 = 𝑉𝑚∠𝜙

phasor representation

When a phasor is used to describe an AC quantity, the length of a phasor represents the amplitude of the wave while the angle of a phasor represents the phase angle of the wave relative to some other (reference) waveform.



Complex Number

A complex number z can be written in rectangular form as

𝑧 = 𝑥 + 𝑗𝑦

𝑗 = −1

Representation of a complex number z in the complex plane

𝑥: the real part of z

the imaginary part of z 𝑦:

The complex number z can also be written in polar or exponential form as

𝑧 = 𝑟𝑒𝑗𝜙

𝑟 = 𝑥2 + 𝑦2

𝜙 = 𝑡𝑎𝑛−1𝑦

𝑥

𝑧 = 𝑟∠𝜙

The relationship between the rectangular form and the polar form is

𝑧 = 𝑥 + 𝑗𝑦 = 𝑟(cos𝜙 + 𝑗 sin𝜙) = 𝑟∠𝜙

9.3 The Phasor



Complex Number

Addition and subtraction of complex numbers

𝑧1 = 𝑥1 + 𝑗𝑦1 = 𝑟1∠𝜙1

Multiplication and division

𝑧1𝑧2 = 𝑟1𝑟2∠𝜙1 + 𝜙2

Reciprocal

Square Root

𝑧2 = 𝑥2 + 𝑗𝑦2 = 𝑟2∠𝜙2

𝑧1 + 𝑧2 = (𝑥1 + 𝑥2) + 𝑗(𝑦1 + 𝑦2)

𝑧1 − 𝑧2 = (𝑥1 − 𝑥2) + 𝑗(𝑦1 − 𝑦2)

𝑧1𝑧2=𝑟1𝑟2∠𝜙1 − 𝜙2

1

𝑧=

1

𝑥 + 𝑗𝑦 =

1

𝑥2 + 𝑦2∠ − 𝜙

1

𝑧=

1

𝑟∠𝜙 =1

𝑟 ∠ − 𝜙

𝑧 = 𝑟∠𝜙/2

9.3 The Phasor




Complex Conjugate

Complex Conjugate

𝑧 = 𝑥 + 𝑗𝑦

𝑧∗ = 𝑥 − 𝑗𝑦 = 𝑟∠ − 𝜙 = 𝑟𝑒−𝜙

−𝑗 =1

𝑗 ←

1

𝑗=

1

1∠90°=1

1∠ − 90° = −𝑗

9.3 The Phasor




The Phasor Representation of the Sinusoid v(t)

The idea of phasor representation is based on Euler’s identity. In general,

𝑒±𝑗𝜃 = cos 𝜃 + 𝑗 sin 𝜃 (9.10)

We can regard cos 𝜃 and sin 𝜃 as the real and imaginary parts of 𝑒𝑗𝜃; we may write

cos 𝜃 = Re{𝑒𝑗𝜃}

sin 𝜃 = Im{𝑒𝑗𝜃}

(9.11)

(9.12)

We write the sinusoidal voltage function given by Eq.(9.1) in the form suggested by Eq.(9.11)

(9.1) 𝑣 𝑡 = 𝑉𝑚 cos(𝜔𝑡 + 𝜙)

= Re{𝑉𝑚𝑒𝑗 𝜔𝑡+𝜙 }

= Re{𝑉𝑚𝑒𝑗𝜙𝑒𝑗𝜔𝑡}

𝑉𝑚𝑒𝑗𝜙 : a complex number that carries the amplitude and phase angle

of the given sinusoidal function.

We define the phasor representation or phasor transform of the given sinusoidal function as

𝑣 𝑡 = Re{𝐕𝑒𝑗𝜔𝑡} 𝐕 = 𝑉𝑚𝑒𝑗𝜙 (9.15)

(9.14)

9.3 The Phasor



The Phasor Representation of the Sinusoid v(t)

Phasor representation

𝐕 = 𝑉𝑚𝑒𝑗𝜙 (9.15)

𝐕 = 𝑉𝑚∠𝜙

𝐕 = 𝑉𝑚(cos𝜙 + 𝑗 sin𝜙) (9.16)

One way of looking at Eqs. (9.15) and (9.16) is to consider the plot of the 𝐕𝑒𝑗𝜔𝑡 on the complex plane.

As time increases, the 𝐕𝑒𝑗𝜔𝑡 rotates on a circle of radius 𝑉𝑚 at an angular velocity ω in the counterclockwise direction, as shown in Fig. (a). In other words, the entire complex plane is rotating at an angular velocity of ω.

We may regard 𝑣(𝑡) as the projection of the 𝐕𝑒𝑗𝜔𝑡 on the real axis, as shown in Fig.(b).

The value of the 𝐕𝑒𝑗𝜔𝑡 at time t = 0 is the phasor 𝐕 of the sinusoid 𝑣(𝑡). The 𝐕𝑒𝑗𝜔𝑡 may be regarded as a rotating phasor.

Thus, whenever a sinusoid is expressed as a phasor, the term 𝒆𝒋𝝎𝒕 is implicitly present. It is therefore important, when dealing with phasors, to keep in mind the frequency ω of the phasor; otherwise we can make serious mistakes.

𝐕

𝐕𝑒𝑗𝜔𝑡

9.3 The Phasor



Phasor Transform : Summary

Eqs.(9.14) through (9.16) reveal that to get the phasor corresponding to a sinusoid, we first express the sinusoid in the cosine form so that the sinusoid can be written as the real part of a complex number. Then we take out the time factor 𝑒𝑗𝜔𝑡, and whatever is left is the phasor corresponding to the sinusoid.

By suppressing the time factor, we transform the sinusoid from the time domain to the phasor domain. This transformation is summarized as follows:

= Re{𝑉𝑚𝑒𝑗𝜙𝑒𝑗𝜔𝑡} 𝑣 𝑡 = 𝑉𝑚 cos(𝜔𝑡 + 𝜙)

= Re{𝐕𝑒𝑗𝜔𝑡}

𝐕 = 𝑉𝑚𝑒𝑗𝜙

= 𝑉𝑚∠𝜙 (1)

Time-domain representation Phasor or frequency-domain representation

𝑉𝑚 sin(𝜔𝑡 + 𝜙) = 𝑉𝑚 cos(𝜔𝑡 + 𝜙 − 90°)

9.3 The Phasor



Inverse Phasor Transform

Equation (1) states that to obtain the sinusoid corresponding to a given phasor V, multiply the phasor by the time factor 𝑒𝑗𝜔𝑡 and take the real part.

𝐕 = 115∠ − 45°


𝜔 = 500 rad/s

= 𝑉𝑚 cos(𝜔𝑡 + 𝜙)

𝑣 𝑡 = 115 cos(500𝑡 − 45°)

𝑣(𝑡) = Rm{𝑉𝑚𝑒𝑗𝜙𝑒𝑗𝜔𝑡} = 𝑉𝑚𝑒

𝑗𝜙

𝐕 = 𝑗8𝑒−𝑗20°

= (1∠90°)(8∠ − 20°)

= 8∠90° − 20°

= 8∠70°° 𝑣 𝑡 = 8 cos(𝜔𝑡 + 70°)

𝐕 = 𝑗 5 − 𝑗12 = 12 + 𝑗5

= 122 + 52∠𝜙 𝜙 = tan−15

12= 22.62°

𝑣 𝑡 = 13 cos(𝜔𝑡 + 22.62°) = 13∠22.62°

9.3 The Phasor



Phasor Diagram

Since a phasor has magnitude and phase (“direction”), it behaves as a vector and is printed in boldface.

For example, phasors 𝐕 = 𝑉𝑚∠𝜙 and 𝐈 = 𝐼𝑚∠ − 𝜃 are graphically represented in Figure.

Such a graphical representation of phasors is known as a phasor diagram.

9.3 The Phasor



Phasor Diagram

The relationship between two phasors at the same frequency remains constant as they rotate; hence the phase angle is constant.

Consequently, we can usually drop the reference to rotation in the phasor diagrams and study the relationship between phasors simply by plotting them as vectors having a common origin and separated by the appropriate angles.

Finally, we should bear in mind that phasor analysis applies only when frequency is constant; it applies in manipulating two or more sinusoidal signals only if they are of the same frequency.

Phasor Diagram

𝝓

𝜙

𝐈

𝐕

𝐕

𝐈 𝜙

𝑣(𝑡) = 𝑉𝑚 cos(𝜔𝑡 + 𝜙2)

𝑖(𝑡) = 𝐼𝑚 cos(𝜔𝑡 + 𝜙1)

𝐕 = 𝑽𝒎 ∠𝜙2

𝐈 = 𝑰𝒎 ∠𝜙1 𝝓 = 𝝓𝟐 −𝝓𝟏

9.3 The Phasor



< The V- I Relationship for a Resistor>

Time Domain

• Current

𝑖(𝑡) = 𝐼𝑚 cos(𝜔𝑡 + 𝜙)

• Voltage

𝑣 𝑡 = 𝑖𝑅 = 𝑅𝐼𝑚 cos(𝜔𝑡 + 𝜙)

Phasor form

𝐈 = 𝐼𝑚∠𝜙

𝐕 = 𝑅𝐈 = 𝑅𝐼𝑚∠𝜙

Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain.

This shows that the voltage-current relation for the resistor in the phasor domain continues to be Ohm’s law, as in the time domain.



< The V- I Relationship for a Resistor>

Phasor Diagram


𝐕 = 𝑅𝐈 = 𝑅𝐼𝑚∠𝜙

𝐈 = 𝐼𝑚∠0°

𝐕 = 𝑅𝐈 = 𝑅𝐼𝑚∠0°



< The V- I Relationship for an Inductor>

Time Domain

𝑖(𝑡) = 𝐼𝑚 cos(𝜔𝑡 + 𝜙)

𝑣 𝑡 = 𝐿𝑑𝑖

𝑑𝑡= −𝜔𝐿𝐼𝑚 sin(𝜔𝑡 + 𝜙) 𝑣

𝑖

= −𝜔𝐿𝐼𝑚 cos(𝜔𝑡 + 𝜙 − 90°)

= 𝜔𝐿𝐼𝑚 cos(𝜔𝑡 + 𝜙 + 90°)




Phasor form


𝐕 = 𝜔𝐿𝐼𝑚∠𝜙 + 90°

𝑣(𝑡) = 𝜔𝐿𝐼𝑚 cos(𝜔𝑡 + 𝜙 + 90°)

= (1∠90°)(𝜔𝐿𝐼𝑚∠𝜙)

𝑗 = 1∠90° = 𝑗𝜔𝐿𝐼𝑚∠𝜙

= 𝑗𝜔𝐿𝐈

Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain.




𝑑𝑖

𝑑𝑡 = 𝜔𝐼𝑚 cos(𝜔𝑡 + 𝜙 + 90°)

= Re{𝜔𝐼𝑚 𝑒𝑗𝜔𝑡𝑒𝑗𝜙𝑒𝑗90° }

= Re{𝑗𝜔𝐼𝑚 𝑒𝑗𝜔𝑡𝑒𝑗𝜙 }

= Re{𝑗𝜔𝐈 𝑒𝑗𝜔𝑡 }

𝑑𝑖

𝑑𝑡

𝐈 = 𝐼𝑚∠𝜙 𝑖(𝑡) = 𝐼𝑚 cos(𝜔𝑡 + 𝜙)

𝑗𝜔𝐈

Differentiating a sinusoid is equivalent to

multiplying its corresponding phasor by 𝑗𝜔.

𝑣 𝑡 = 𝐿𝑑𝑖

𝑑𝑡= 𝐿𝑗𝜔𝐈 = 𝑗𝜔𝐿𝐈



Phasor Diagram

9.4 The Passive Circuit Elements in the Frequency Domain


𝐈 = 𝐼𝑚∠0°

= 𝑗𝜔𝐿𝐈

Voltage leads current by 90°in an inductor

Current lags voltage by 90° in an inductor

𝐕 = 𝜔𝐿𝐼𝑚∠90°


𝐕 = 𝜔𝐿𝐼𝑚∠𝜙 + 90°



< The V- I Relationship for a Capacitor>

Time Domain

𝑣(𝑡) = 𝑉𝑚 cos(𝜔𝑡 + 𝜙)

𝑖 𝑡 = 𝐶𝑑𝑣

𝑑𝑡= −𝜔𝐶𝑉𝑚 sin(𝜔𝑡 + 𝜙)

= −𝜔𝐶𝑉𝑚 cos(𝜔𝑡 + 𝜙 − 90°)

= 𝜔𝐶𝑉𝑚 cos(𝜔𝑡 + 𝜙 + 90°)

𝑣

𝑖




Phasor form


𝐈 = 𝜔𝐶𝑉𝑚∠𝜙 + 90°

𝑖(𝑡) = 𝜔𝐶𝑉𝑚 cos(𝜔𝑡 + 𝜙 + 90°)

= (1∠90°)(𝜔𝐶𝑉𝑚∠𝜙)

𝑗 = 1∠90° = 𝑗𝜔𝐶𝑉𝑚∠𝜙

= 𝑗𝜔𝐶𝐕

𝐕 =1

𝑗𝜔𝐶𝐈

Voltage-current relations for a capacitor in the: (a) time domain, (b) frequency domain.




Phasor Diagram


𝐈 = 𝜔𝐶𝑉𝑚∠𝜙 + 90°

𝐕 = 𝑉𝑚∠0°

𝐈 = 𝜔𝐶𝑉𝑚∠90°

= 𝑗𝜔𝐶𝐕



Summary

Summary of voltage-current relationships



< Impedance, Reactance, and Admittance>

Impedance

We obtained the voltage-current relations for the three passive elements as

𝐕 = 𝑅𝐈

𝐕 =1

𝑗𝜔𝐶𝐈

𝐕 = 𝑗𝜔𝐿𝐈

𝐕

𝐈= 𝑅

𝐕

𝐈= 𝑗𝜔𝐿

𝐕

𝐈 =

1

𝑗𝜔𝐶

From these three expressions, we obtain Ohm’s law in phasor form for any type of element as

𝐙 =𝐕

𝐈 𝐕 = 𝐙𝐈 or

Z is a frequency-dependent quantity known as impedance, measured in ohms.




Impedance

The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms.

The impedance represents the opposition which the circuit exhibits to the flow of sinusoidal current. Although the impedance is the ratio of two phasors, it is not a phasor, because it does not correspond to a sinusoidally varying quantity.

The impedances of resistors, inductors, and capacitors can be readily obtained from Eq. (9.39). Table 9.1 summarizes their impedances.




Resistance and Reactance

As a complex quantity, the impedance may be expressed in rectangular form as

𝐙 =𝐕

𝐈= 𝑅 + 𝑗𝑋

𝑅 : Resistance

𝑋 : Reactance

Inductive and capacitive reactance

𝐙 = 𝑅 + 𝑗𝑋

𝑋<0 :

𝑋>0 :

𝐙 = 𝑅 − 𝑗𝑋

Inductive reactance

Capacitive reactance

𝐙 = 𝑅

𝐙 = 𝑗𝜔𝐿

𝐙 =1

𝑗𝜔𝐶= −𝑗

1

𝜔𝐶

Resistor :

Inductor :

Capacitor:




Impedance in Polar Form

𝐙 = 𝑅 + 𝑗𝑋

𝐙 = 𝑍∠𝜃 𝑍 = 𝑅2 + 𝑋2 𝜃 = tan−1

𝑋

𝑅

𝐙

𝑅

𝑗𝑋

𝜃

𝑅 = 𝑧 cos 𝜃

𝑋 = 𝑍 sin 𝜃

Admittance

It is sometimes convenient to work with the reciprocal of impedance, known as admittance.

The admittance Y is the reciprocal of impedance, measured in siemens (S).

𝐘 =1

𝐙=𝐈

𝐕

As a complex quantity, we may write Y as

𝐘 = 𝐺 + 𝑗𝐵

𝐺 : Conductance

𝐵 : Susceptance

𝐘 =1

𝑅 + 𝑗𝑋 =𝑅 − 𝑗𝑋

𝑅2 + 𝑋2

𝐺 =𝑅

𝑅2 + 𝑋2 𝐵 = −

𝑋

𝑅2 + 𝑋2




𝐘 = 𝐺 + 𝑗𝐵

𝐺 : Conductance

𝐵 : Susceptance

𝐘 =1

𝑅 + 𝑗𝑋 =𝑅 − 𝑗𝑋

𝑅2 + 𝑋2

𝐺 =𝑅

𝑅2 + 𝑋2 𝐵 = −

𝑋

𝑅2 + 𝑋2



► Example ◄

(a) Convert the circuit to the frequency domain.

(b) Find an admittance for resistor and capacitor

𝐕s = 10 ∠0° V

𝐙R = 5 Ω

𝐙c = −𝑗1

𝜔𝐶= −𝑗

1

4 × 0.1= −𝑗2.5 Ω

𝐘R =1

5= 0.2 S

𝐘c = 𝑗𝜔𝐶 = 𝑗4 × 0.1 = 𝑗0.4 S

Voltage phasor

Impedance

Admittance


Chapter 9 Sinusoidal Steady-State Analysis 45 9.5 Kirchhoff’s Laws in the Frequency Domain

KVL in the frequency Domain

For KVL, let 𝑣1, 𝑣2, ⋯ , 𝑣𝑛 be the voltages around a closed loop. Then

𝑣1 + 𝑣2 +⋯+ 𝑣𝑛 = 0

In the sinusoidal steady state, each voltage may be written in cosine form, so that Eq. (9.36) becomes

(9.36)

𝑉𝑚1 cos(𝜔𝑡 + 𝜃1) + 𝑉𝑚2 cos(𝜔𝑡 + 𝜃2) + ⋯+ 𝑉𝑚𝑛 cos(𝜔𝑡 + 𝜃𝑛) = 0 (9.37)

We now Euler’s identity to write Eq.(9.37) as

Re 𝑉𝑚1𝑒𝑗𝜃1𝑒𝑗𝜔𝑡} + Re{𝑉𝑚2𝑒

𝑗𝜃2𝑒𝑗𝜔𝑡} + ⋯+ Re{𝑉𝑚𝑛𝑒𝑗𝜃𝑛𝑒𝑗𝜔𝑡} = 0 (9.38)

or

Re (𝑉𝑚1𝑒𝑗𝜃1 + 𝑉𝑚2𝑒

𝑗𝜃2 +⋯+ 𝑉𝑚𝑛𝑒𝑗𝜃𝑛)𝑒𝑗𝜔𝑡} = 0

If we let 𝐕𝑘 = 𝑉𝑚𝑘𝑒𝑗𝜃𝑘, then

Re (𝐕1 + 𝐕𝟐 +⋯+ 𝐕𝒏)𝑒𝑗𝜔𝑡} = 0

Since 𝑒𝑗𝜔𝑡 ≠ 0,

𝐕1 + 𝐕𝟐 +⋯+ 𝐕𝒏 = 0

(9.39)

(9.40)

(9.41)

indicating that Kirchhoff’s voltage law holds for phasors.

+ 𝑣1 − + 𝑣2 −

+ 𝑣𝑛 −


Chapter 9 Sinusoidal Steady-State Analysis 46 9.5 Kirchhoff’s Laws in the Frequency Domain

KCL in the frequency Domain

By following a similar procedure, we can show that Kirchhoff’s current law holds for phasors. If we let 𝑖1, 𝑖2, ⋯ , 𝑖𝑛 , in be the current leaving or entering a closed surface in a network at time t, then

𝑖1 + 𝑖2 +⋯+ 𝑖𝑛 = 0

If 𝐈1, 𝐈𝟐, ⋯ , 𝐈𝑛, are the phasor forms of the sinusoids 𝑖1, 𝑖2, ⋯ , 𝑖𝑛 , then

(9.42)

(9.43) 𝐈1 + 𝐈𝟐 +⋯+ 𝐈𝒏 = 0

which is Kirchhoff’s current law in the frequency domain.

► Example ◄

Find 𝑖(𝑡) in the circuit

𝐕s = 10 ∠0° V

KVL :

𝐙R = 5 Ω 𝐙c = −𝑗2.5 Ω

𝐕s = 𝐕R + 𝐕C = 𝐈𝐙R + 𝐈𝐙C = 5𝐈 + −j2.5 𝐈

𝐈 =10∠0°

5 − 𝑗2.5=10(5 + 𝑗2.5)

52 + 2.52= 1.6 + 𝑗0.8 = 1.789∠26.57° A

𝑖1

𝑖2

𝑖𝑛


Chapter 9 Sinusoidal Steady-State Analysis 47 9.6 Series, Parallel, and Delta-to-Wye Simplifications

Series Connection

showing that the total or equivalent impedance of series-connected impedances is the sum of the individual impedances.

This is similar to the series connection of resistances.

𝐕ab = 𝐕𝟏 + 𝐕𝟐 +⋯+ 𝐕𝒏

= 𝐈𝐙1 + 𝐈𝐙𝟐 +⋯+ 𝐈𝐙𝐧

= 𝐈(𝐙1 + 𝐙𝟐 +⋯+ 𝐙𝐧)

𝐙ab =𝐕ab𝐈= 𝐙1 + 𝐙𝟐 +⋯+ 𝐙𝐧



► Example ◄

For the circuit in Fi.9.5, find (a) current, (b) impedance, and (c) reactance.

𝑣𝑠(𝑡) = 𝑉𝑚 cos(𝜔𝑡 + 𝜙)

• Phasor form

𝐙 = 𝑅 + 𝑗𝜔𝐿 = 𝑍∠𝜃

𝐕𝑠 = 𝑉𝑚∠𝜙

𝐈 =𝐕𝑠𝐙=

𝑉𝑚∠𝜙

𝑅2 + (𝜔𝐿)2∠𝜃

𝑍 = 𝑅2 + (𝜔𝐿)2

𝜃 = tan−1𝜔𝐿

𝑅

Voltage :

Impedance :

𝐙

𝑅

𝑗𝜔𝐿

𝜃

𝑅 = 𝑧 cos 𝜃

𝑋 = 𝑍 sin 𝜃 Current : =

𝑉𝑚∠𝜙 − 𝜃

𝑅2 + (𝜔𝐿)2

𝑖 𝑡 =𝑉𝑚

𝑅2 + 𝜔𝐿 2cos(𝜔𝑡 + 𝜙 − 𝜃)



Parallel Connection

This indicates that the equivalent admittance of a parallel connection of admittances is the sum of the individual admittances.

𝐈 = 𝐈𝟏 + 𝐈𝟐 +⋯+ 𝐈𝒏

= 𝐕1

𝐙𝟏+1

𝐙2+⋯+

1

𝐙n =

𝐕

𝒁𝑎𝑏

1

𝒁𝑎𝑏=1

𝐙𝟏+1

𝐙2+⋯+

1

𝐙n

𝒀𝑎𝑏 = 𝐘1 + 𝐘2 +⋯+ 𝐘𝑛



Delta-to-Wye Transformation

Z1 =𝑍𝑏𝑍𝑐

𝑍𝑎 + 𝑍𝑏 + 𝑍𝑐 , (9.51)

Z2 =𝑍𝑐𝑍𝑎

𝑍𝑎 + 𝑍𝑏 + 𝑍𝑐 , (9.52)

Z3 =𝑍𝑎𝑍𝑏

𝑍𝑎 + 𝑍𝑏 + 𝑍𝑐 , (9.53)

Za =𝑍1𝑍2+ 𝑍2𝑍3+ 𝑍3𝑍1

𝑍1 , (9.54)

Zb =𝑍1𝑍2+ 𝑍2𝑍3+ 𝑍3𝑍1

𝑍2 , 9.55

Zc =𝑍1𝑍2+ 𝑍2𝑍3+ 𝑍3𝑍1

𝑍3 , (9.56)

electric circuit theoryelearning.kocw.net/kocw/document/2015/korea_sejong/... · 2016-09-09 ·...

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