Download Chapter 5 Steady-State Sinusoidal Analysis

Electrical Engineering and Electronics II
Chapter 5
Steady-State Sinusoidal
Analysis
Scott
2008.10
•Main Contents
1. Identify the frequency, angular frequency, peak
value, rms value, and phase of a sinusoidal signal.
2. Solve steady-state ac circuits using phasors and
complex impedances.
3. Compute power for steady-state sinusoidal ac
circuits.
•Main Contents
4. Find Thévenin equivalent circuits for steady-state
ac circuits.
5. Determine load impedances for maximum power
transfer.
6. Solve balanced three-phase circuits.
•The importance of steady-state sinusoidal analysis

Electric power transmission and distribution by
sinusoidal currents and voltages

Sinusoidal signals in radio communication

All periodic signals are composed of sinusoidal
components according to Fourier analysis
5.1 Sinusoidal Currents and Voltages
•Parameters of Sinusoidal Currents and Voltages
•Vm is the peak value, unit is volt
•ω is the angular frequency, unit is radians
per second
•f is the frequency，unit is Hertz (Hz) or
inverse second.
•θ is the phase angle, unit is radian or degree.
•Parameters of Sinusoidal Currents and Voltages
1
Frequency
f 
T
Angular frequency   2
T
  2f
To uniformity, This textbook expresses sinusoidal
functions by using cosine function rather than the sine
function. 。
sin z   cosz  90


•Root-Mean-Square Values or Effective Values
Vrms
1

T
Pavg
T
 v t dt
2
T
I rms
0
2
rms
V

R
1
2



i
t
dt

T 0
Pavg  I
Average Power
2
rms
R
•RMS Value of a Sinusoid
Vrms
Vm

2
The rms value for a sinusoid is the peak value
divided by the square root of two.
However, this is not true for other periodic
waveforms such as square waves or triangular waves.
v(t )  100 cos(100t )
A voltage given by v(t )  100 cos(100t )is applied to a 50Ω resistance. Find
the rms value of the voltage and the average power delivered to the
resistance.
Vrms
Vm

 70.71V
2
V 2 rms 70.712
Pavg 

 100 W
R
50
v 2 (t )
p(t ) 
 200 cos 2 (100 t ) W
R
Complex Numbers
3
forms of complex numbers
 Arithmetic
operations of
complex numbers
•Rectangular form
A  a  jb


Imaginary unit
j
Real part
a  Re[A]
Im
A
b
|A|
1

0
Re
a
b  Im[ A]
2
2
2
2
2
 Magnitude
a b  A  A  a b
b
 Angle
  arctan
a
2
 Conjugate A  a  j b

Imaginary part
A  A AA  A
•Polar form
A | A | 
Im
A
b
|A|

0
Re
a
2
a | A | cos 
b
  arctan
a
b | A | sin 
A  a b
2

Conversion between Rectangular and polar
•Conversion
A | A | 
A  A cos   j A sin 
A  a  jb
 A  cos   jsin  
b
A  a  b  arctan
a
2
2
•Exponential form
 Euler’s
Identity
Im
A
b
|A|
j
e  cos   j sin 

0
A  A cos   j sin  
 Ae
j
Re
a
•Three forms
Rectangular form
A  a  jb
Polar form
A | A | 
Exponential form
A | A | e
j
•Arithmetic Operations
Given
 Identity
 Adding
A  a  jb
A B
B  c  jd
a  c, b  d
and subtracting
C  A  B  e1  je 2
e1  a  c
e2  b  d
•Arithmetic Operations
A  a  jb
Given
B  c  jd
 Product
j A
 Be
e
j( A  B )
C  A B  A e
 Ce
j C
e
C  AB
j B
j C
C   A   B
•Arithmetic Operations
B  c  jd
j A
A
e
 Dividing
D  A B 
j B
Be
A j( A  B )
j D
D  De 
e
B
A
D  AB 
B
j D
j( A  B )





D
A
B
e e
Given
A  a  jb
5.2 PHASORS
•Phasor Definition
Time function : v1 t   V1 cosωt  θ1 
Sinusoidal steady-state analysis is greatly facilitated if
currents and voltages are represented as vectors or
Phasors.
peak
Phasor: V1  V11
•rms Phasor or effective phasor.
•In this book, if phasors are not labeled as rms, then they are
peak phasors.
•Phasors as Rotating Vector
Angular velocity
正弦量可以表示为在复
平面complex plane上
按逆时针方向旋转的相
量的实部real part。
•Sinusoids can be visualized as the real-axis projection of
vectors rotating in the complex plane.
θ
•The phasor for a sinusoid is a ‘snapshot’ of the
corresponding rotating vector at t = 0.
•Phase Relationships
•To determine phase relationships from a
phasor diagram, consider the phasors to
rotate counterclockwise.
•Then when standing at a fixed point, if V1 arrives
first followed by V2 after a rotation of θ , we say that
V1 leads V2 by θ .
•Alternatively, we could say that V2 lags V1 by θ .
(Usually, we take θ≤180o as the smaller angle between
the two phasors.)
•Phase Relationships
•Phase Relationships
•Adding Sinusoids Using Phasors
Step 1: Determine the phasor for each term
Step 2: Add the phasors using complex
arithmetic.
Step 3: Convert the sum to polar form.
Step 4: Write the result as a time function.
•Adding Sinusoids Using Phasors
•Example:
v1  t   20 cos t  45  V
v2  t   10sin t  60  V
Find vs  v1  v2  ?
Solution:
V1  20  45 V
V2  10  30 V
•Adding Sinusoids Using Phasors
Vs  V1  V2
 20  45  10  30
 14.14  j14.14  8.660  j5
 23.06  j19.14
 29.97  39.7 V
 vs  t   29.97 cos t  39.7  V
•Adding Sinusoids Using Phasors
•Exercise
1.v1 (t )  10cos(t )  10sin(t )
2.i1 (t )  10cos(t  30 )  5sin(t  30 )
o
o
3.i2 (t )  20sin(t  90o )  15cos(t  60o )
•Answers
1.v1 (t )  14.14 cos(t  45o )
2.i1 (t )  11.18cos(t  3.44o )
3.i2 (t )  30.4 cos(t  25.3o )
5.3 COMPLEX IMPEDANCES
•By using phasors to represent sinusoidal
voltages and currents, we can solve sinusoidal
steady-state circuit problems with relative ease.
•Sinusoidal steady-state circuit analysis is
virtually the same as the analysis of resistive
circuits except for using complex arithmetic.
•Sinusoidal response of resistance
•
Relationship between voltage and current
i  I m cos(t  )
i
v  Ri  RI m cos(t  )
u
R
 Vm cos(t  )
then：
Vrms  RI rms
U
u
I
i
R
Vrms Vm

I rms I m
0

t
•Sinusoidal response of resistance
•
2.Phasor relation and diagram
Substituting for current and
voltage phasors
i  I m cos(t  )
I  I m 
Irms  I rms 
R
Vrms V

I rms I
I
i
v
R
R
V
Vrms  RI rms  RI rms 
V  RI  RI m 
Phasor diagram
I

V
0O
•Sinusoidal response of resistance

I  I m 0  I m Plot the phasor
diagram
0
V  RI m0  RI m
0
R
V

I
phasors expression in
Ohm’s Law

I
V
•Sinusoidal response of resistance
•
Power
Instantaneous power
p  vi
 2Vrms cos t 2I rms cos t
 2Vrms I rms cos 2 t  Vrms I rms  Vrms I rms cos 2t
No doubt, p≥0,Resistance always absorbs energy.
•Sinusoidal response of resistance
•
Power
p  2V rms I
2
cos
t  V rms I rms (1  cos 2t )
rms
2VrmsIrms
p
VrmsIrms
t
0
i
v
•Sinusoidal response of resistance
•
Power
Average Power
——Average value of instantaneous power in a period
1 T
1 T
P   pdt   Vrms I rms (1  cos 2t )dt
T 0
T 0
2
V
 Vrms I rms  I rms 2 R  rms
R
Average Power represents the consumed power, is also named as
Real power.
•Sinusoidal response of Inductance
•
Relationship between voltage and current
i  I m sin t
di
v  L   LI m cos t  Vm sin(t  900 )
dt
u、i have the same
frequency，u leads i
by 90o
i
u
0
π
2π
ωt
•Sinusoidal response of Inductance
•
Effective Value
i  I m sin t
Vm   LI m
v  Vm sin(t  900 )
Vrms   LI rms
Reactance电抗－感抗
Definition
U rms U m
XL 

  L  2 fL
I rms
Im
•Sinusoidal response of Inductance
•
Phasor relations
i
I  I0 A

I
v  Vm cos(t  900 )
0
I
V

V  Vm 90  jX L I m  jX L I m0  jX L I
0

V  jX L0I
i  I m cos t
L
v
Phasor diagram：


jωL
Z  j X L  j L 
V

I
Complex
impedance
•Sinusoidal response of Inductance
•
Power
v  Vm sin(t  900 )
i  I m sin t
p  pL  vi  Vm I m sin  t  sin( t  90)
 Vm I m sin  t  cos  t
 Vrms I rms sin 2 t
VrmsIrms
p
u
i
0
t
•Sinusoidal response of Inductance
Power
•
u
i
t
0
i
i
i
i
u
u
u
u
UI

0
吸
收
能
量
释
放
能
量

p

吸
收
能
量
释
放
能
量

di
i increases，  0，
dt
p(t )  0,WL increases
——magnitude of magnetic
field increases，inductance
absorbs energy
di
i decreases，  0，
dt
p(t )  0,WL decreases
t
——magnitude of magnetic
field decreases，inductance
supplys energy
•Sinusoidal response of Inductance
•
Power
Average Power
1 T
P   pdt
T 0
1 T
  Vrms I rms sin 2tdt  0
T 0
•Inductance is an energy-storage element rather
than an energy-consuming element
•Sinusoidal response of Inductance
Reactive Power
The power flows back and forth to inductances and
capacitances is called reactive power Q, it is the peak
instantaneous power。
2
VLrms
2
Q  VLrms I Lrms 
 X L I Lrms
XL
Unit: Var (乏)
Reactive power is important because it causes power
dissipation in the lines and transformers of a power
distribution system. Specific Charge is executed by
electric-power companies for reactive power.
•Sinusoidal response of capacitance
•
Relationship between voltage and current
i
v  Vm sin t
dv
i  C  CU m sin(t  900 )
dt
 I m sin(t  900 )
u
I m  CVm  2 fCVm
Current leads voltage
by 90°
C
v
i
0

2
•Sinusoidal response of capacitance
• Relationship
between voltage and current
I m  CVm  2 fCVm
i
 I rms  CVrms  2 fCVrms
v
Definition
Reactance
电抗－容抗
Vrms U m
1
1
XC 



I rms I m C 2 fC
C
•Sinusoidal
response
of
capacitance
i

I
v
• Phasor

C
U
-jωC
V  Vm 0 V
0
v  Vm sin t

i  I m sin(t  90 ) A I  I m 900 A
0

Vm 0
Vm
Vm




j


j
X
C

0
I

90
j
I
I
m
m
m
I
U
0
diagram


U
I j
XC

U
•Sinusoidal response of capacitance
•Power
•Instantaneous power
p  vi  Vm sin tI m sin(t  900 )
 Vrms I rms sin 2t
UrmsIrms
p
u
i
0
t
•Sinusoidal response of capacitance
•Power
u
i
t
0
i
i
i
i
u
u
u
u
UI

0
吸
收
能
量
释
放
能
量

p

吸
收
能
量
释
放
能
量

u increases ，Wc increases ，
du
 0，p(t )  0
dt
——capacitance charges，it
absorbs energy
u decreases ，Wc decreases ，
du
 0，p(t )  0
t dt
——capacitance discharges，
it supplys energy
•Sinusoidal response of capacitance
•Power
Average Power
1
P
T

T
0
1
pdt 
T

T
0
Vrms I rms sin 2tdt  0
•Capacitance is an energy-storage element rather
than an energy-consuming element.
•Sinusoidal response of capacitance
•Power
Reactive Power
• Assuming the same current acts on the inductance and
capacitance respectively, the initial phase is 0，then
i  I m sin t
vC  Vm sin(t  90)
p  vi  Vm sin tI m sin(t  900 )
 Vrms I rms sin 2t
QC  VCrms ICrms   I
2
Crms
XC
It is given that the frequency of sinusoidal source is 50Hz,
the rms or effective value is 10V, capacitor is 25μF, determine
the rms value of current. If the frequency is 5000Hz, then what
is the rms value of current now？

Solution
when f =50Hz
1
1
XC 

 127.4
6
2 fC 2  3.14  50  (25 10 )
I rms
U rms
10


 0.078 A  78mA
X C 127.4
When f =5000Hz
X C 
1
1

 1.274
6
2 fC 2  3.14  5000  (25 10 )
I rms
U rms
10


 7.8 A
XC
1.274
i
u
C
The higher frequency is under fixed rms value of voltage, the
bigger the rms value of current flowing through capacitor.
•Summary
ele
me
nt
circuit
Relationship
between u
and i
Ohm’s
Law
i
R
v
R
v  Ri U  RI
Complex
impedance
U
R 
I
Phasors
diagram


I U

i
L
v

di 
U
U  jX L I jX L 
L vL
dt
I
i
C
v

U
dv 
U   jX C I  jX C  
C iC
I
dt
U

I

I

U
•Inductance
•Capacitance
•Resistance
In phase
•True or False
1.
v  220 sin (ω t  45 )V
j45
45 
V  220 e V ？
3.
I RMS  4 e
Complex
j30 
A
 4 2 sin ( ω t  30 )A？
4.
I RMS  10 60 A
i  10 sin ( ω t  60  )A
？
2.
Peak Value
VRMS  100  15V
minus
VRMS
  100V ？
VRMS  100 e
j15
V？
•Complex Impedances
Twoterminal
circuit of
zero
sources
I
+
V
-
Complex Impedance
V V Ψ v V
Z 
 (Ψ v  Ψ i )  Z 
I I Ψ i I
Magnitude
V
Z 
I
Angle
  Ψ v Ψ i
•Complex Impedances of R L C
VR
R : ZR 
R
IR
| Z R | R
 0
VL
 j L  j X L | Z L |  L
L : ZL 
IL

1
| ZC |
C

 
2
C : Z  VC  1   j X
C
C
I C j C

2
•The Laws of Phasor Form
(KCL)：
i  0

I
 0
(KVL)：
v  0
V  0
v  iR
V  IZ
•Complex Impedances in Series
I
+
I
+
V1
V
–
+
V2
–
–
Z1
+
Z
V
–
Z2
Equivalent Impedance
Voltage-Divider
Zeq =Z1+Z2
Z1
V1 
V
Z1  Z 2
Z2
V2 
V
Z1  Z 2
•Complex Impedances in Parallel
I
+ I1
V
I
I2
+
Z1 Z2
Z
V
–
–
1
1 1
 
Equivalent Impedance Z
Z1 Z 2
eq
Current-Divider
I1 
Z2 
I
Z1  Z 2
I2 
Z1 
I
Z1  Z 2