Chapter 10 Sinusoidal Steady-State Analysiscontents.kocw.net/document/Chapter-10.pdf · 2013. 12. 27. · When the source is sinusoidal, we often ignore the transient/natural response

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Chapter 10Sinusoidal Steady-State

Analysis

10.1 Characteristics of Sinusoids

10.2 Forced Response to Sinusoidal Functions

10.3 The Complex Forcing Functions

10.4 The Phasor

10.5 Impedance and Admittance

10.6 Nodal and Mesh Analysis

10.7 Superposition, Source Transformations and Thevenin’s Theorem

10.8 Phasor Diagrams



Sinusoids

• the amplitude of the wave is Vm

• the argument is ωt• the radian or angular frequency is ω• note that sin() is periodic

sin

• the period of the wave is T• the frequency f is 1/T: units Hertz (Hz)

1,

22

• The new wave (in red) is said to lead the original (in green) by θ.

• The original sin(ωt) is said to lag the new wave by θ.

• θ can be in degrees or radians, but the argument of sin() is always radians.

sin

sin



• Converting Sines to Cosines

sin sin 180°

cos cos 180°

∓sin cos 90°

cos sin 90°

cos 5 10°

sin 5 10° 90°

sin 5 100°

sin 5 30° by130°

sin 5 260°

by230°



The Steady-State Response : the condition that is reached after the transient or natural response has died out

When the source is sinusoidal, we often ignore the transient/natural response and consider only the forced or “steady-state” response.The source is assumed to exist forever: −∞<t<∞

cos

Let cos sin

sin cos cos sin cos

⇒ sin cos 0

⇒ 0, 0 ⇒ ,

cos sin



• More Compact and User-Friendly Form

Let cos instead of cos sin

cos cos sin sin cos sin

cos , sin

cos sin

⇒ sincos tan ⇒ tan

cos cos tan by tan



Example 10.1 Find the current iL in the circuit.

100100 25 10 cos 10

8 cos 10

25 ∥ 100 20

cos tan

8

20 10 30 10cos 10 tan

3020

222 cos 10 56.3°


10.3 The Complex Forcing Function

With purely resistive circuit, it is no more difficult to analyze with sinusoidal sources thanwith dc sources.

It turns out that if the transient response is of no interest to us, there is an alternativeapproach for obtaining the sinusoidal steady-state response of any linear circuit.

cos sinEuler’s identity

cos sin ⇒ cos sin



• An Algebraic Alternative to Differential Equations

→

→

→

⇒

tan

cos tan

Let cos



Example 10.2 Find the voltage on the capacitor.

3 0

→ 3 2 0

let

3 2 5 0→ 1 10 3

⇒3

1 10

∠ tan V

31 10

31 10

cos 5 tan 10

29.85 cos 5 84.3


The term ejωt is common to all voltages and currents and can be ignored in all intermediate steps, leading to the phasor:

The phasor representation of a current (or voltage) is in the frequency domain

cos cos 0° → ∠0°

cos → ∠

complex value notation

,

∠

10.4 Phasor

Phasor

Example 10.3 Transform the time-domain voltage v(t)=100cos(400t-30) into the frequency domain.

100∠ 30°


10.4 Phasor

The Resistor

In the frequency domain, Ohm’s Law takes the same form:

8 cos 100 50° , 4Ω

2 cos 100 50°

8∠ 50°4 2∠ 50°

complex voltage and current

cos sin

cos sin

∠ ∠ in phase


10.4 Phasor

8∠ 50°, 100 / , 4

8∠ 50°100 4 0.02∠ 50°

1∠ 90° 0.02∠ 50°

⇒ 0.02∠ 140°

The Inductor

Differentiation in time becomes multiplication in phasor form: (calculus becomes algebra!)

∶

I lag V by 90

cos 2 sin 2- cos sin

Example 10.4


10.4 Phasor

Differentiation in time becomes multiplication in phasor form: (calculus becomes algebra!)

The Capacitor

:

I leads V by 90

Calculus (hard but real) Algebra (easy but complex)


10.4 Phasor

• Kirchhoff’s Laws Using Phasors

1 2 1 2

1 2 1 2

: ( ) ( ) ( ) 0 0: ( ) ( ) ( ) 0 0

N N

N N

KVL t t tKCL i t i t i t

V V VI I I

→

Let ∠0°∠0°

∠ tan

Example 10.5 Determine Is and is(t) if sources are operate at =2 rad/s and IC=228 A.

12 2∠28°

0.5∠ 90° 2∠28° 1∠ 62°12

12∠ 62°

12∠ 62° 2∠28°

32∠ 62°

⇒ 1.5 cos 2 62°



• Define impedance as Z=V/I, i.e. V=IZ

ZR=R ZL=jωL ZC=1/jωC

• Impedance is the equivalent of resistance in the frequency domain.

• Impedance is a complex number (unit ohm).• Impedances in series or parallel can be combined

using “resistor rules.”

• the admittance is Y=1/Z

YR=1/R YL=1/jωL YC=jωC

• if Z=R+jX; R is the resistance, X is the reactance (unit ohm Ω)

• if Y=G+jB; G is the conductance, B is the susceptance: (unit siemen S)

, | |∠

resistancereactance

1 1

conductance

susceptance



Example 10.6 Determine the equivalent impedance of the network working on =5 rad/s.

6 ∥ 0.46 0.46 0.4 0.02655 0.3982

10 0.02655 0.39820.02655 8.602

10 ∥ 0.02655 8.60210 0.02655 8.60210 0.02655 8.602 4.255 4.929 6.511∠49.20°

200 → 1 1

5 0.2 1 500 → 1 1

5 0.5 0.4

2 → 10



Example 10.7 Find the current i(t).

40∠ 90°

1.5 1000 ∥ 1000 2000

1.51 1 21 1 2 1.5

21

2 1.5 2.5∠36.87°kΩ

40∠ 90°2.5∠36.87° 0.016∠ 126.9°

16 cos 3000 126.9°



Example 10.8 Find the node voltages v1(t) and v2(t).

1∠0° 1 0 5 10 5 10

0.5∠ 90° 0.5 1 10 5 5 10

0.2 0.2 0.1 1,0.1 0.1 0.1 0.5

1 2 2.24∠ 63.4°

2 4 4.47∠116.6°

2.24 cos 63.4° ,4.47 cos 116.6°



Example 10.9 Obtain the expression for the time-domain currents i1 and i2.

5001 1

10 500 10 2

4 10 4 10 4

10 3 4 0,2 2 4 0

14 813 1.24∠29.7°,

20 3013 2.77∠56.3°

1.24 cos 10 29.7° ,2.77 cos 10 56.3°


10.7 Superposition, Source Transformations, and Thévenin’s Theorem

Example 10.10 Use superposition to find V1.

5 ∥ 105 105 10

4 2

5 ∥ 105 105 10

10

10 ∥ 510 510 5

2 4

4 2

4 210 2 4

4 2 10 2 4 1 0

4 286 8 2 2

1∠0° 1 0

0.5∠ 90° 0.5

4 2

4 22 4

2 4 10 4 2 0.5

6 86 8

1

2 2 1 1 2



Example 10.11 Determine the Thévenin equivalent seen by the –j10 impedance to find V1.

4 2 1 0 2 4 0.5 4 2 1 2 6 3

4 2 2 4 6 2

6 36 2 1060 30100

0.6 0.3

1 0 1 0.6 0.3 0.4 0.3

4 2 4 2 0.4 0.3 1 2



Example 10.12 Determine the power dissipated by the 10 resistor.

0.410 0.4 2 0

⇒ 79.23∠ 82.03°

79.23 cos 5 82.03°

1.6671.667 10 0.667 5 0

⇒ 811.7∠ 76.86°

811.7 cos 3 76.86°

10 10 79.23 cos 5 82.03° 811.7 cos 3 76.86°



The arrow for the phasor V on the phasor diagram is a photograph, taken at ωt = 0, of a rotating arrow whose projection on the real axis is the instantaneous voltage v(t).

6 8 10∠53.1°3 4 5∠ 53.1°

6 8 3 4 9 4

9.849∠23.96°

1 1

⇒ 1.414∠45° 10∠53.1° 14.14∠98.1°



50 10 50 10

50 5010 5 ,

10 10 10 ,

50 50 10 5

10∠ 90°

5 0.2∠0° 0.2 0,1∠0°

10∠ 90° 0.1∠90°

Let 1∠0° 1 0

Let ∠0°



Example 10.13 Construct a phasor diagram showing IR, IL, and IC.

Let 1∠0° 1 0

0.2 0.2∠0° 0.2

0.1 0.1∠ 90° 0.1

0.3 0.3∠90° 0.3

⇒

0.2∠0° 0.1∠ 90°

0.2 0.1 0.224∠ 26.6°

⇒ 0.3 0.2 0.1

0.2 0.2 0.283∠45°

Homework : 10장 Exercises 7의 배수 문제

회로이론-І 2013년2학기 이문석

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Chapter 10 Sinusoidal Steady-State Analysiscontents.kocw.net/document/Chapter-10.pdf · 2013. 12. 27. · When the source is sinusoidal, we often ignore the transient/natural response

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