Sinusoidal Steady-State Analysis Chapter 10

Transcription

Chapter 10
Sinusoidal 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
회로이론-І 2013년2학기 이문석
1
Sinusoids
sin
sin
•
•
•
•
•
•
the amplitude of the wave is Vm
the argument is ωt
the radian or angular frequency is ω
note that sin() is periodic
the period of the wave is T
the frequency f is 1/T: units Hertz (Hz)
1
sin
•
•
•
,
2
2
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.
2
•
Converting Sines to Cosines
sin
sin
180°
cos
cos
180°
∓sin
cos
cos
90°
sin
90°
cos 5
sin 5
sin 5
sin 5
10°
10° 90°
100°
30°
sin 5
by130°
260°
by230°
3
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
sin
⇒
⇒ cos
cos
sin
cos
sin
cos
cos
0,
0
cos
Let
sin
0
,
⇒ sin
4
•
More Compact and User-Friendly Form
Let
instead of
cos
cos
cos
cos
sin
,
cos
sin
cos
sin
cos
sin
sin
sin
sin
cos
⇒
cos
tan
cos
⇒
tan
tan
by tan
5
Example 10.1 Find the current iL in the circuit.
25 ∥ 100
20
100
10 cos 10
100 25
8 cos 10
cos
tan
8
20
10
222 cos 10
30 10
56.3° 회로이론-І 2013년2학기 이문석
cos 10
tan
30
20
6
10.3 The Complex Forcing Function
With purely resistive circuit, it is no more difficult to analyze with sinusoidal sources than
with dc sources.
It turns out that if the transient response is of no interest to us, there is an alternative
approach for obtaining the sinusoidal steady-state response of any linear circuit.
Euler’s identity
cos
sin
⇒
cos
cos
sin
sin
7
•
An Algebraic Alternative to Differential Equations
Let
cos
→
→
→
⇒
tan
cos
tan
8
Example 10.2 Find the voltage on the capacitor.
let
3
→
3
→
⇒
0
3
2
0
2 5
1
10
3
3
1
10
∠
0
tan
V
3
1
10
3
cos 5
1
10
29.85 cos 5
tan
10
84.3
9
10.4 Phasor
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
cos
→
0° →
∠0°
∠
complex value notation
,
∠
Phasor
Example 10.3 Transform the time-domain voltage v(t)=100cos(400t-30) into the frequency
domain.
100∠
30°
10
10.4 Phasor
The Resistor
In the frequency domain, Ohm’s Law takes the same form:
8 cos 100
50° ,
2 cos 100
8∠
50°
4
2∠
4Ω
50°
50°
complex voltage and current
cos
cos
∠
∠
sin
sin
in phase
11
10.4 Phasor
The Inductor
Differentiation in time becomes multiplication in phasor form: (calculus becomes algebra!)
cos
∶ sin
2
cos
-
2
sin
I lag V by 90
Example 10.4
8∠
50°,
8∠ 50°
100 4
⇒
1∠
90° 0.02∠
0.02∠
100
/ ,
0.02∠
4
50°
50°
140°
12
10.4 Phasor
The Capacitor
Differentiation in time becomes multiplication in phasor form: (calculus becomes algebra!)
:
I leads V by 90
Calculus (hard but real)
Algebra (easy but complex)
13
10.4 Phasor
•
Kirchhoff’s Laws Using Phasors
KVL : 1 (t )  2 (t )     N (t )  0  V1  V2    VN  0
KCL : i1 (t )  i2 (t )    iN (t )  0  I1  I 2    I N  0
Let
→ ∠0°
∠0°
∠
tan
Example 10.5 Determine Is and is(t) if sources are operate at =2 rad/s and IC=228 A.
1
2
⇒ 회로이론-І 2013년2학기 이문석
2∠28°
0.5∠
90° 2∠28°
1
2
1
∠ 62°
2
1
∠ 62°
2
1.5 cos 2
1∠
62°
2∠28°
3
∠
2
62°
62°
14
•
Define impedance as Z=V/I, i.e. V=IZ
ZR=R
•
•
•
•
•
,
ZC=1/jωC
| |∠
reactance
resistance
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
•
ZL=jωL
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)
1
1
susceptance
conductance
15
Example 10.6 Determine the equivalent impedance of the network working on =5 rad/s.
200
→
2 → 6 ∥ 0.4
1
5 0.2
1
1
500
→
1
1
5 0.5
0.4
10
6
6
10 ∥ 0.02655
0.4
0.4
0.02655
8.602
0.3982
10 0.02655
8.602
10 0.02655
8.602
10 0.02655
0.02655
8.602
4.255
4.929
0.3982
6.511∠49.20°
16
Example 10.7 Find the current i(t).
40∠
90°
1.5
1000 ∥ 1000
2000
1 1
2
2
1.5
1.5
1
1
2
1
2
1.5 2.5∠36.87°kΩ
40∠ 90°
2.5∠36.87°
16 cos 3000
0.016∠
126.9°
126.9° 17
Example 10.8 Find the node voltages v1(t) and v2(t).
1∠0°
0.5∠
1
90°
0
5
0.5
0.2
0.2
0.1
0.1
10
1
0.1
0.1
5
10
10
5
5
1
1,
0.5
2.24 cos
4.47 cos
10
2
2
2.24∠
4
63.4°
4.47∠116.6°
63.4° ,
116.6°
18
Example 10.9 Obtain the expression for the time-domain currents i1 and i2.
500
4
1
10
10
4
1
500
10
2
10
4
10
2
3
4
2
4
14
8
1.24∠29.7°,
13
20
30
2.77∠56.3°
13
1.24 cos 10
2.77 cos 10
0,
0
29.7° ,
56.3°
19
10.7 Superposition, Source Transformations, and Thévenin’s Theorem
Example 10.10 Use superposition to find V1.
5∥
10
5
5
4
10
10
2
10 5
10
5
2
4
10 ∥
5
5∥
5 10
5
10
10
10
1∠0°
1
0.5∠
90°
4
2
4
2
4
6
28
8
4
2
4
2
0
0.5
6
6
8
8
10
4
2
2
4
4
10
2
1
0
4
0.5
1
2
4
2
2
2
2
4
10 2
2
1
1
2
20
Example 10.11 Determine the Thévenin equivalent seen by the –j10 impedance to find V1.
4
4
6
2 1
2
1
3
4
6
2
2
6
1
0
4
2
0
2
2
2
4
0.5
4
3
6
2
10
60
30
100
0.6
0.3
1
4
0.6
2 0.4
0.3
0.3
0.4
1
0.3
2
21
Example 10.12 Determine the power dissipated by the 10  resistor.
0.4
⇒
10
79.23∠
0.4
82.03°
79.23 cos 5
⇒
0
82.03° 1.667
5
1.667
10
0.667
811.7∠ 76.86°
811.7 cos 3
2
10
82.03°
10 79.23 cos 5
0
76.86° 811.7 cos 3
76.86°
22
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).
3
6
8
⇒
5∠
53.1°
10∠53.1°
1
4
6
8 3
4
9
4
9.849∠23.96°
1
1.414∠45° 10∠53.1°
14.14∠98.1°
23
50
50
Let
∠0°
10
10
50
10
50
5 ,
10
,
10
50
10
50
5
10
10∠
Let
1∠0°
1
90°
0
5
0.2∠0° 0.2
1∠0°
10∠ 90°
0,
0.1∠90°
24
Example 10.13 Construct a phasor diagram showing IR, IL, and IC.
Let
1∠0°
1
0
⇒
0.2∠0°
0.2
0.1∠
0.1
0.2
0.1
0.3
0.2∠0°
0.2
0.1∠
90°
0.3∠90°
0.224∠
0.3
⇒
0.2
0.2
90°
0.2
26.6°
0.1
0.283∠45°
0.1
0.3
Homework : 10장 Exercises 7의 배수 문제
25
이 자료에 나오는 모든 그림은 McGraw·hill 출판사로부터 제공받은 그림을 사
용하였음.
All figures at this slide file are provided from The McGraw-hill company.

Sinusoidal Steady-State Analysis Chapter 10

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