Transcript Document

Chapter 5
Steady-State Sinusoidal Analysis
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 ac
4. Find Thévenin and Norton equivalent circuits.
5. Determine load impedances for maximum power
transfer.
5. Steady-State Sinusoidal Analysis
* In most circuits, the transient response (i.e., the
complimentary solution) decays rapidly to zero, the steadystate response (i.e., the forced response or the particular
solution) persists.
* In this chapter, we learn efficient methods for finding the
steady-state responses for sinusoidal sources.
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5. Steady-State Sinusoidal Analysis
5.1 Sinusoidal Currents and Voltages
5.1.1 Phase and Phase Angle
* Consider the sinusoidal voltage as shown,
v ( t )  Vm cosω t  θ  where
V m is the Peak value of the voltage
ω is the angular frequency in rad/s
θ is the phase angle (usually in degree)
Since the angle increases by 2π per cycle
we have : ω T  2 π , T is the period in s,
the frequency in Hz (or s - 1 ) : f 
ω
1
T
2π
 2π f
T
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5. Steady-State Sinusoidal Analysis – 5.1 Sinusoidal Current and Voltage
5.1.1 Phase and Phase Angle
* We usually use cosine function to model a sinusoidal signal.
In case there is a sine function, we can use the following
conversion:
sin(z)  cos(z - 90)
For example:
v x t   10 sin200t  30 
v x t   10 cos200t  30  90 
 10 cos200t  60 
and thus we can say that the phase angle of v x (t) is - 60 
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5. Steady-State Sinusoidal Analysis – 5.1 Sinusoidal Current and Voltage
5.1.2 Root-Mean-Square (RMS) Values (or Effective Values)
Consider applyinga periodicvoltage v(t) with period T to a resistance R.
v 2 t 
The power delivered to the resistance is given by pt  
,
R
the energy delivered per period is : E T 
 pt  dt
T
0
The average powe per period is :
 1 T 2

ET
1 T
1 T v t 
Pav 
  pt  dt  
dt  
v ( t )dt  / R

0
0
0
T
T
T
R
 T

By defining the root - mean - squre (rms) value of the periodicvoltage v(t) as :
2
2
Vrms 
1
T

T
0
v t  dt , we have Pavg
2
Vrms

R
2
Similarly,if we define the rms value of the periodiccurrent as :
I rms 
1
T

T
0
i 2 t  dt , we have Pavg  I rms R
2
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5. Steady-State Sinusoidal Analysis – 5.1 Sinusoidal Current and Voltage
5.1.3 RMS Value of a Sinusoid
Consider a sinusoidal voltage given by :
v t   Vm cosω t  θ 
Vrms
1

T

0
2
Vm

2T
2
Vm

2T
Vrms 
Vm
2
T
1
v (t)dt 
T
2

T
0
Vm cos2 ω t  θ  dt
2
 1  cos2ω t  2θ dt
T
0
1
1






T

sin
2
ω
T

2
θ

sin
2
θ


2ω
2ω


 0.7071Vm Note : residential power : 110Vrms (Vm  155.5V),60 Hz
Similarly, I rms 
Im
2
 0.7071I m
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5. Steady-State Sinusoidal Analysis – 5.1 Sinusoidal Current and Voltage
Example 5.1 – Power delivered to a resistance by a sinusoidal source
Given : a v(t)  100cos(100πt) V
is applied to a 50 Ω resistance
Find : Vrms , the average power Pavg
and pt 
f  ω/2 π  50 H z , T  1/f  20 ms
Vrms  Vm / 2  70.71V
2
Vrms
( 70.71 ) 2
Pavg 

 100W
R
50
v 2 t  1002 cos2 100π t 
P t  

R
50
 200cos2 100π t  W
 100  100 cos200πt  W
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5. Steady-State Sinusoidal Analysis
5.2 Phasors
5.2.1 Definition
* A phasor is a vector in complex number plane that represents
the magnitude and phase of a sinusoid.
* In ac circuit analysis, voltages and currents are usually
represented as phasors.
For a sinusoidal voltageof the form v1 t   V1 cosω t  θ1 
we definethe phasoras : V1  V1 θ1
If the sinusoid is of the form v 2 t   V2 sinω t  θ 2 
we can convertit to v 2 t   V2 cosω t  θ 2  90, therefore
V2  V2 ( θ 2  90 )
Sim ilarly, for i1 t   I 1 cosω t  θ1  and i2 t   I 2 sinω t  θ 2 
we can write I 1  I 1 θ1 and I 2  I 2 ( θ 2  90 )
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5. Steady-State Sinusoidal Analysis – 5.2 Phasors
5.2.1 Definition
* Euler’s Identity:
complexexponential e jθ  cosθ  j sinθ
Ae jθ  A cosθ  jA sinθ
Ae j( ωt  θ )  Acos( ωt  θ )  jAsin( ωt  θ )
Acos( ωt  θ )  Re (Ae j( ωt  θ ) )
* In phasor application:
In circuit application, Ae j( ωt  θ ) is simplified to the form Ae jθ
by deleting the term Ae jwt .
As a result,
A cos( ωt  θ ) is presented as Re(Ae jθ ) or simply its phasor form
A  Aθ  A cos( ωt  θ )
For example, the sinusoidal voltage
v 1 (t)  V1 cos( ωt  θ 1 ) is presented as V1  V1 θ 1
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5. Steady-State Sinusoidal Analysis – 5.2 Phasors
5.2.4 Phasor vs. Sinusoids
* The phasor is simply a “snapshot” of a
rotating vector at t=0.
com plexexponential e jθ  cosθ  j sinθ
Ae jθ  A cosθ  jA sinθ
Ae j( ωt θ )  Acos(ωt  θ )  jAsin( ωt  θ )
Acos(ωt  θ )  Re (Ae j( ωt θ ) )
Consider the sinusoidal voltage as shown :
vt   Vm cos t   
It' s just the real part of a rotating vector :

vt   Re Vm e j  t  

Vm e j  t    Vm (  t   ) and
at t  0, v(0)  Vm cos  Vm 
The phasor for v(t) is defined as :
V  Vm 
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5. Steady-State Sinusoidal Analysis – 5.2 Phasors
5.2.2 Phasor Summation
Consider the summation of three voltages
v(t)  10 cos( ωt)  5 sin( ωt  60 )  5 cos( ωt  90 )
We first rewrite all the functions to cosine form
v t   10 cosω t   5 cosω t  30   5 cosω t  90 

 
 

v t   Re 10e
 5e
 5e
 Re 10  5 e
 5 e e 
 Re 10e j ω t  Re 5 e j ( ω t  30 )  Re 5 e j ( ω t  90 )
j ( ω t  30 )
jωt
 j 30

j ( ω t  90 )
j 90
jωt
In Euler' s Identity (or in polar form) :

v t   Re 100  5  30  590  e j ω t

Consider the complex number :
100  5  30  590  10  4.33  j 2.50  j 5
 14.33  j 2.5  14.549.90  14.54 e j 9.90

v t   Re 14.54 e

 v t   Re 14.54 e j 9.90 e j ω t
j ( ω t  9.90 )

 14.54 cosω t  9.90 

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5. Steady-State Sinusoidal Analysis – 5.2 Phasors
5.2.2 Phasor Summation
Now we use Phasor notation to simplify our calculation:
vt   10 cos t   5 sin t  60  5 cos t  90
vt   10 cos t   5 cos t  30  5 cos t  90
V  V1  V2  V3
 100  5  30  590
 10  4.33  j 2.50  j 5
 14.33  j 2.5
 14.549.90
 V  14.549.90
 vt   14.54 cos t  9.90
Note: In using phasors to add sinusoids, all of the terms must
have the same frequency.
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5. Steady-State Sinusoidal Analysis – 5.2 Phasors
Example 5.2 – Using phasors to Add Sinusoids
Suppose that v1 t   20 cos t  45
v 2 t   10 sin t  60
Find v s (t)  v1 (t)  v 2 (t)
The phasors are : V1  20  45 and V2  10  30
V s  V1  V2
 20  45  10  30
 14.14  j14.14  8.660  j 5
 23.06  j19.14
 29.97  39.7
v s t   29.97 cos t  39.7
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5. Steady-State Sinusoidal Analysis – 5.2 Phasors
5.2.3 Fundamental Phasor Operations
If V1  V1  1 , V2  V2  2 , then
Multiplication : V1  V2  V1V2  1   2
Division :
V1 V1

 1 -  2
V 2 V2
Reciprocal:
1
1
1

  -1
V1 V1  1 V1
SquareRoot :
V1  V1  1 /2
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5. Steady-State Sinusoidal Analysis – 5.2 Phasors
5.2.5 Phase Relationships
Considerthe voltages:
v1 t   3 cos t  40 and v 2 t   4 cos t  20
They can be represented as : V1  340 and V2  4   20
and we m ay say that :
V1 leads V2 by 60 or V2 lags V1 by 60 .
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5. Steady-State Sinusoidal Analysis – 5.3 Complex
Impedances
5.3.1 Impedance
* Impedance means “complex resistance”.
* The impedance concept is equivalent to stating that capacitors and
inductors act as “frequency-dependent resistors.”
* By using impedances, we can solve sinusoidal steady-state circuit
with relatively ease compared to the methods of Chapter 4.
* Except for the fact that we use complex arithmetic, sinusoidal steadystate analysis is the same as the analysis of resistive circuits.
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5. Steady-State Sinusoidal Analysis – 5.3 Complex Impedances
5.3.2 Inductance
Consider a current i L t   I m sin t    flow through an inductor:
diL t 
  L I m cos t   
dt
In their phasor form s : I L  I m (   90 ) and V L   L I m   Vm 
v L t   L
Note : the current lags the voltage by 90
Rewrite voltage as : V L   L90  I m (   90 )  j  L  I L
Define the im pedanceof the inductance as : Z L  j  L   L90
We have : V L  Z L I L
This is Ohm 's law in phasor form .
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5. Steady-State Sinusoidal Analysis – 5.3 Complex Impedances
5.3.3 Capacitance
Consider a voltage v C (t)  Vm cos( ωt  θ ) across a capacitor
dvC
i C (t)  C
 -ωCV m sin( ωt  θ )  ωCV m cos( ωt  θ  90 )
dt
In their phasor forms : VC  Vm θ , I C  ωCV m θ  90   I m θ  90 
Note : the current leads voltage by 90 
1
1
Rewrite voltage as VC  Vm θ 
 - 90  I m θ  90 
 - 90  I C
ωC
ωC
1
1
Define the impedanceof capacitance as : Z C 

 - 90 
jωC ωC
we have : VC  Z C I C
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5. Steady-State Sinusoidal Analysis – 5.3 Complex Impedances
* ELI the ICE man.
* Impedance that are pure imaginary are called reactance.
5.3.4 Resistance
Definethe im pedanceof resistanceas Z R  R
we have : VR  R I R
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5. Steady-State Sinusoidal Analysis – 5.3 Complex Impedances
20
5. Steady-State Sinusoidal Analysis – 5.3 Complex Impedances
Quiz - Exercises 5.6, 5.7, 5.8
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5. Steady-State Sinusoidal Analysis – 5.3 Complex Impedances
Quiz - Exercises 5.6, 5.7, 5.8
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5. Steady-State Sinusoidal Analysis – 5.3 Complex Impedances
Additional Example: represent the circuit shown in the
frequency domain using impedances and phasors.
23
5. Steady-State Sinusoidal Analysis – 5.3 Complex Impedances
Additional Example: represent the circuit shown in the
frequency domain using impedances and phasors.
24
5. Steady-State Sinusoidal Analysis – 5.4 Circuit Analysis
5.4 Circuit Analysis
* Impedance circuit analysis is the same as resistive circuit
analysis, we can directly apply KCL, KVL, nodal analysis,
mesh analysis,…
KVL equation:
v1 t   v2 t   v3 t   0  V1  V2  V3  0
KCL equations:
i1 (t)  i2 (t) - i3 (t)  0  I 1  I 2 - I 3  0
* The above phasor approach can only apply for steady state
with sinusoids of the same frequency.
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5. Steady-State Sinusoidal Analysis – 5.4 Circuit Analysis
Example 5.3 – Steady-State AC Analysis of a Series Circuit
* Find the steady-state current, the phasor voltage across each
element, and construct a phasor diagram.
Vs  10030, Z L  j  L  j150 , Z C   j
1
  j 50
C
Z eq  R  Z L  Z C  100  j150  j 50  100  j100  141.445
Vs
10030

 0.707  15
Z 141.445
i t   0.707 cos500t  15
I
VR  R  I  100 0.707  15  70.7   15
VL  j  L  I  15090  0.707  15  106.175
VC   j
1
 I  50  90  0.707  15  35.4  105
C
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Example 5.4 – Series/Parallel Combination of Complex Impedances
* Find the voltage across the capacitor, the phasor current
through each element, and construct a phasor diagram
V s  10 - 90 , Z L  j  L  j100 , Z C   j
Z RC 
1
  j100
C
1
1
10


 70.71  45
1 R  1 Z C 1 100  1 (  j100 ) 0.0141445
Z RC  50  j 50
VC  V s
Z RC
70.71  45
 10  90
 10  180
Z L  Z RC
j100  50  j 50
vC t   10 cos1000t  180  10 cos1000t 
I
Vs
10  90
10  90


 0.1414  135
Z L  Z RC
j100  50  j 50 50  j 50
VC 10  180

 0.1  180
R
100
V
10  180 10  180
IC  C 

 0.1  90
ZC
 j100
100  90
IR 
27
5. Steady-State Sinusoidal Analysis – 5.4 Circuit Analysis
Example 5.5 – Steady-State AC Node-Voltage Analysis
* Find the voltage at node 1 using nodal analysis
Write KCL at node1 and node2
V1 V1 - V2

 2 - 90
10
- j5
V2 V2  V1

 1.50 
j10
 j5
0.1  j0.2 V1  j0.2V2   j 2
 j0.2V1  j0.1V2  1.5
Solve for V1 :
V1  16.129.7 or v1 t   16.1 cos100t  29.7
28
5. Steady-State Sinusoidal Analysis – 5.4 Circuit Analysis
Exercise 5.11 – Steady-State AC Mesh-Current Analysis
* Solve for the mesh currents
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5. Steady-State Sinusoidal Analysis – 5.4 Circuit Analysis
Additional Example :
Deter min e B and L
when i(t)  Bcos (3t - 51.87 )A
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5. Steady-State Sinusoidal Analysis – 5.4 Circuit Analysis
Additional Example :
31
5. Steady-State Sinusoidal Analysis – 5.4 Circuit Analysis
Additional Example :
32
5. Steady-State Sinusoidal Analysis – 5.4 Circuit Analysis
Additional Example : Commercial Airliner Door Bridge Circuit
33
5. Steady-State Sinusoidal Analysis – 5.4 Circuit Analysis
Additional Example :
34
5. Steady-State Sinusoidal Analysis – 5.4 Circuit Analysis
Additional Example :
35
5. Steady-State Sinusoidal Analysis – 5.4 Circuit Analysis
Additional Example :
36
5. Steady-State Sinusoidal Analysis – 5.4 Circuit Analysis
Quiz – Exercise 5.10
* Find the phasor voltage and phasor current at each element
37
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
5.5 Power in AC Circuit
5.5.1 Voltage, Current and Impedance
In the network, v(t)  Vm cos(t) or V  Vm 0 
the im pedance Z  Z   R  jX
where Z  R 2  X 2 ,   tan-1 (X/R)
The phasorcurrent is I 
V Vm 0

 I m 
Z
Z 
Vm
where I m 
Z
(Note : If V  Vm  v , then
I  I m  v -   I m  i , where  v -  i   )
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5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
5.5.2 Voltage, Current and Power for a Resistive Load
If the load is pure resistive,
Z  R0, θ  0
v t   Vm cosω t 
i t  
Vm
cos( ω t)  I m cosω t 
R
pt   v t  i t   Vm I m cos2 ω t 
1 1
 cos2ω t)
2 2
V I
The average power P  m m  Vrms I rms
2
 Vm I m (
(1) Current is in phase with voltage.
(2) Energy flows continuously from
source to load.
39
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
5.5.3 Voltage, Current and Power for an Inductive Load
If the load is pure inductance,
Z  ωL90  , θ  90 
v t   Vm cosω t 
i t   I m cosω t  90 
 I m sinω t 
pt   v t  i t 
 Vm I m cosω t sinω t 
Vm I m

sin2ω t   Vrms I rms sin2ωt
2
The average power P  0
(1) Current lags the voltage by 90 degree (ELI)
(2) Half of the power is positive, energy is delivered to the inductance
and stored in the magnetic field; the other half of the power is
negative, the inductance returns energy to the source.
(3) The average power is zero, and we say reactive power flows back40
and forth in-between the source and the load.
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
5.5.4 Voltage, Current and Power for a Capacitive Load
If the load is pure capacitance,
1
Z
 - 90  , θ  -90
ωC
v t   V m cosω t 
i t   I m cosω t  90 
  I m sinω t 
pt   v t  i t 
 V m I m cosω t  sinω t 
Vm I m
sin2 ω t   V rms I rms sin 2 ωt
2
The average power P  0

(1) Current leads the voltage by 90 degree (ICE)
(2) The average power is zero: reactive power flows back and forth inbetween the source and the load.
(3) Reactive power is negative (positive) for a capacitance
(inductance).
41
(4) Reactive power in inductance and in capacitance cancel each other.
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
5.5.5 Power Calculation for a General (RLC) Load
For generalRLC load,Z  R  jX  Z  , - 90    90
vt   Vm cos t 
it   I m cos t   
pt   Vm I m cos t  cos t   
 Vm I m cos  cos2  t   Vm I m sin  cos t  sin t 
Vm I m
Vm I m

cos 1  cos2 t  
sin  sin2 t 
2
2
  0 : pure resistive, only the 1st term rem ains;
  90 : pure reactive, only the second term rem ains.
Active (Real) load due to R
Reactive load due to L, C
42
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
5.5.5 Power Calculation for a General (RLC) Load
vt   Vm cos t 
vt   Vm cosω t  θ v 
i t   I m cos t   
p t   Vm I m cos t  cos t   
p(t)  Vm I m cos  cos2  t 
 Vm I m sin   cos t  sin  t 
Vm I m
p t  
cos 1  cos2 t 
2
Vm I m

sin   sin 2 t 
2
i t   I m cosω t  θ i , θ  θ v - θ i
 I m cos( ω t  θ v - θ )
p t   Vm I m cosω t  θ v  cosω t  θ v  θ 
p t   Vm I m cosθ  cos2 ω t  θ v 
 Vm I m sinθ  cosω t  θ v  sinω t  θ v 
Vm I m
p t  
cosθ 1  cos2( ω t  θ v )
2
V I
 m m sinθ  sin2( ω t  θ v )
2
Since the term s involving cos(2(t   v )) and sin(2(t   v ))
have average values of zero, the average(real) power P is :
Vm I m
P
cos , using Vrms  Vm / 2 and I rms  I m / 2 ,
2
we have P  Vrms I rms cos 
43
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
5.5.5 Power Calculation for a General (RLC) Load
Vm I m
V I
cosθ 1  cos2ω t   m m sinθ sin2ω t 
2
2
V I
Average Power: P  m m cosθ   Vrms I rms cosθ  W
2
pt  
Power Factor: PF  cosθ , θ  θv  θi is called the power angle
Power factor is often stated as percentage, e.g.,
90% lagging (i.e., current lags voltage, inductive load)
60% leading (i.e., current leads voltage, capacitive load)
Vm I m
Q
sinθ   V
I
sinθ  VAR (Volt Am peresReactive)
rm s rm s
Reactive Power:
2
The last term in power formula is the power flowing back and forth
between the source and the energy-storage elements. Reactive
power is its peak power.
Apparent Power: S  Vrms I rms VA (Volt - Ampere)
P 2  Q 2  Vrms I rms  cos2 θ   Vrms I rms  sin 2 θ   Vrms I rms 
Note: 5kW load is different from 5kVA load.
44
2
2
2
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
5.5.5 Power Calculation for a General (RLC) Load
45
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
5.5.6 Impedance triangle and Power Triangle
* The impedance triangle:
* The Power triangle:
Apparent power, average (real) power, and reactive power form a
triangle.
S  Vrms I rms
VA (Volt - Am pere)
P 2  Q 2  Vrms I rms  cos2    Vrms I rms  sin 2    Vrms I rms 
2
2
2
46
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
5.5.6 Impedance triangle and Power Triangle
pt  
Z  Z θ  R  j X , cosθ  
The average power P 
Vm I m
V I
cosθ 1  cos2ω t   m m sinθ sin2ω t 
2
2
R
X
, sinθ  
Z
Z
Vm I m
V I
R
cosθ  m m 
2
2
Z
2
I
2
(Note : Vm  I m Z )
 m R  I rms R
2
V I
2
Similarly, the reactive power Q  m m sinθ  I rms X
2
(X  0, inductive load; X  0, capacitiveload)
2
V
Also : P  Rrms where V Rrms is the voltage acrtoss the resistance
R
2
V Xrms
Q
where V Xrms is the voltage across the reactance
X
2
2
2
Vm I m R Vm R
Vm R 2 1 VRm 1 VRrms
P




2
2
2 Z
2 Z
2 Z R
2 R
R
2
47
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
Example 5.6 – AC Power Calculation
(1) Find the power and reactive power taken from the source
θ  θ v  θ i  90  ( 135 )  45
Vs ,rms 
Vs
I rms 
I
2


10
2
 7.071V
0.1414
 0.1 A
2
2
The power : P  Vs ,rms I rms cosθ 
 7.071 0.1 cos45  0.5W
The reactive power Q  Vs ,rms I rms sinθ 
 7.071 0.1 sin45  0.5VAR
48
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
Example 5.6 – AC Power Calculation
(2)The reactive power delivered to the inductor:
Q L  I rms X L  ( 0.1 ) 2 ( 100 )  1.0VAR
2
(3)The reactive power delivered to the capacitor:
2
 0 .1 
2
QC  I C ,rms X C  
 ( 100 )  0.5VAR
 2
(4)The (real) power delivered to the resistance :
2
( Note : Q  QL  QC )
 IR 
 0.1 
2


PR  I R ,rms R  
R
 100  0.5W

 2
 2
Of course,the power absorbedby the capacitance and inductanceis 0
PL  0 , PC  0
2
All of the power delivedrd by the sourceis absorbedby the resistance.
49
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
Additional Example:
50
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
Additional Example:
51
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
Additional Example:
52
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
Additional Example:
53
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
Additional Example:
54
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
Additional Example:
55
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
Example 5.7 – Using Power Triangles
Find power, reactive power, and power factor for the source
and the phasor currents as shown.
We first find the power and reactive power for each load, then
sum over to obtain the power and reactive power for the
source.
56
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
Example 5.7 – Using Power Triangles
For load A, we have : cos A   0.5
PA  Vrms I Arms cos A   104 ( 0.5 )  5 kW
Q A   Vrms I Arms   PA
2
   5000
  104
2
2
2
 8.660kVAR
For load B,  B  arccos(0.7 )  45.57
QB  PB tan B   5000tan45.57
QB  5.101kVAR
The power and reactive power delivered by the source :
P  PA  PB  5  5  10kW
Q  Q A  QB  8.660  5.101  3.559kVAR
57
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
Example 5.7 – Using Power Triangles
The power and reactive power delivered by source:
P  PA  PB  10kW , Q  Q A  QB  3.559kVAR
i
The power angle: θ  tan1 Q/P  19.59
v
The power factor : cosθ   0.9421 or 94.21%leading.
The apparentpower : S  Vrms I rms  P 2  Q 2  10.61kVA
The effective sourcevoltage: Vrms  V / 2  1kV
The effective current : I rms  Vrms I rms /Vrms  10.61A
I m  I  2 I rms  15 A
θ i  θ v  θ  30  ( 19.59 )  49.59, I  I θ i  1549.59
58
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
5.5.7 Power-Factor Correction
59
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
Example 5.8 – Power-Factor Correction
A 50kW load operates from a 60-Hz 10kV-rms line with a power
factor of 60% lagging. Compute the capacitance that must
be placed in parallel with the load to achieve a 90% lagging
power factor.
θ L  cos1 ( 0.6 )  53.13
QL  P tanθ   66.67kVAR
θ new  cos1 ( 0.9 )  25.84
50kW
Qnew  P tanθ new   24.22kVAR
PF  0.6
QC  Qnew  QL  42.45kVAR
2
 
2
Vrms
104
XC 

 2356Ω
QC
 42450
ω  2π 60  377.0
1
1
C

 1.126μF
ω X C 377  2356
QL  66.67kVA
Qnew  22.42kVA
P  50kW
60
5. Steady-State Sinusoidal Analysis – 5.5 Power in AC Circuit
Quiz - Exercise 5.12 – Power in AC Circuits
.
61
5. Steady-State Sinusoidal Analysis – 5.6 Thevenin and Norton
5.6 Thevenin and Norton Equivalent Circuits
* A two terminal circuit composed of sinusoidal sources (of the
same frequency), resistances, capacitances, and
inductances can be simplified to Thevenin or Norton
equivalent circuit.
5.6.1 Thevenin Equivalent Circuits
Vt  Voc
Voc Vt
Zt 

I sc
I sc
* The Thevenin impedance can also be
obtained by zeroing sources.
5.6.2 Norton Equivalent Circuits
Zt 
Voc Vt

I sc I sc
I n  I sc
62
5. Steady-State Sinusoidal Analysis – 5.6 Thevenin and Norton
Example 5.9 – Thevenin and Norton Equivalents
We find Z t by zeroing sources
1
1 100  1 (  j100 )
1

0.0141445
 70.71  45
 50  j 50
Apply short circuit at term inals
V
1000
IR  s 
 10 A
100
100
I sc  I R  I s
Zt 
 1  190
 1.414  45 A
Vt  I sc Z t
 1.414  45  70.71  45
 100  90 V
63
5. Steady-State Sinusoidal Analysis – 5.6 Thevenin and Norton
Additional Example:
64
5. Steady-State Sinusoidal Analysis – 5.6 Thevenin and Norton
65
5. Steady-State Sinusoidal Analysis – 5.6 Thevenin and Norton
5.6.1 – Maximum Power Transfer
Assum eZ t  Rt  jX t , Z L  R L  jX L
I  Vt /(Z t  Z L )  Vt /[(R t  RL )  j(X t  X L )]
I m  Vtm / Z  Vtm /[(R t  RL ) 2  (X t  X L ) 2 ] 1/2
Avg.output power P  I
P  Pmax when
2
rms
Vtm2 RL
1
RL 
2 [(R t  RL ) 2  (X t  X L ) 2 ]
P
P
 0 and
0
R L
X L
 RL  Rt and X L  -X t  Z L  Rt - X t
i.e., when Zload  Z t
*
we have m ax.power output
Z total  Z t  Z load  Rt  j X t  Rt  j X t  2Rt
If Z load can only be resistive, then when
Z load  Rload  Z t we havem ax.output power.
66
5. Steady-State Sinusoidal Analysis – 5.6 Thevenin and Norton
Example 5.10 – Maximum Power Transfer
Since Z t  50  j 50
(a) we will havem ax.output power when
Z load  Z t  50  j 50
*
Vt
100  90
Ia 

 1  90
Z t  Z load 50  j 50  50  j 50
2
 1 
2
P  I arms Rload  
 ( 50 )  25W
 2
(b) If the load has to be resistive,
Rload  Z t  50  j 50  70.71
Ib 
Vt
100  90

Z t  Z load 50  j 50  70.71
100  90

 0.765  67.50
130.66  22.50
2
P  I brms Rload
2
 0.7653

 70.71  20.71W
2 

67
5. Steady-State Sinusoidal Analysis – 5.6 Thevenin and Norton
Quiz – Exercise 5.14 and Exercise 5.15
68
5. Steady-State Sinusoidal Analysis – 5.6 Thevenin and Norton
Quiz – Exercise 5.14
69
5. Steady-State Sinusoidal Analysis – 5.6 Thevenin and Norton
Quiz – Exercise 5.15
70
5. Steady-State Sinusoidal Analysis – 5.6 Thevenin and Norton
Quiz – Exercise 5.15
71
5. Steady-State Sinusoidal Analysis – SUMMARY
72
5. Steady-State Sinusoidal Analysis – SUMMARY
73
5. Steady-State Sinusoidal Analysis – SUMMARY
74
Chapter 6
Frequency Response, Bode Plots,
and Resonance
1. State the fundamental concepts of Fourier analysis.
2. Determine the output of a filter for a given input
consisting of sinusoidal components using the
filter’s transfer function.
3. Use circuit analysis to determine the transfer
functions of simple circuits.
4. Draw first-order lowpass or highpass filter circuits
and sketch their transfer functions.
5. Understand decibels, logarithmic frequency scales,
and Bode plots.
6. Calculate parameters for series and parallel
resonant circuits.
6. Frequency Response – 6.1 Fourier Analysis, Filters, Transfer Functions
6.1 Fourier Analysis, Filters, and Transfer Functions
6.1.1 Fourier Analysis
* Most real-world information-bearing electrical signals are not
sinusoidal.
* Fourier theorem tells that a non-sinusoidal signal can be
expressed by the summation of sinusoidal functions in the
form:.
a0  a1cos(0 t)  b1 sin(0 t )  a2 cos(20 t)  b2 sin( 2  t )  ...
 an cos(n 0 t)  bn sin(n0 t)  ...
76
6. Frequency Response – 6.1 Fourier Analysis, Filters, Transfer Functions
6.1.1 Fourier Analysis
* All real-world signals are sums of sinusoidal components
having various frequencies, amplitudes, and phases.
* The square wave is a special example:
v sq (t) 
4A
4A
4A
sin(ω0 t) 
sin(3ω0 t) 
sin(5ω0 t)  ...
π
3π
5π
where ω0  2π T
* Most of the real-world signals are
confined to finite range of frequency.
* It is important to learn how the circuits
respond to components having
different frequencies.
77
6. Frequency Response – 6.1 Fourier Analysis, Filters, Transfer Functions
6.1.2 Filters
* Filters process the sinusoidal components of an input signal
differently depending of the frequency of each component.
Often, the goal of the filter is to retain the components in
certain frequency ranges and reject components in other
frequency ranges.
78
6. Frequency Response – 6.1 Fourier Analysis, Filters, Transfer Functions
6.1.3 Filters and Transfer Functions
* Since the impedances of inductances and capacitances
change with frequency, RLC circuits provide one way to
realize electrical filters.
1
1
Z L  L90  2fL90, ZC 
 - 90 
 - 90
C
2f
* The transfer function of a two-port filter is defined as:
H (f )  H(f)) H(f) 
Vout
Vin
where H(f) is the magnitude
 H(f)  Vout - Vin is the phase
79
6. Frequency Response – 6.1 Fourier Analysis, Filters, Transfer Functions
Example 6.1 – Using Transfer Function to Find Output
For the transfer functions shown, find the output signal,
given the input: vin ( t )  2 cos( 2000t  40 )
The frequencyof the input signalis f  1000 Hz
Vout
 H ( 1000)  330 
Vin
Vin  240  Vout  H ( 1000)* Vin  330  240  6 70
vout (t )  6 cos(2000t  70)
80
6. Frequency Response – 6.1 Fourier Analysis, Filters, Transfer Functions
Example 6.2 – Multi-input components, Superposition Principle
The input involves two components:
vin ( t )  2cos( 2000t )  cos( 4000t  70 )
Vin 1
Vin 2
We use superposition principle:
Vout 1  H ( 1000)  Vin 1  330  20  6 30
Vout 2  H ( 2000) Vin 2  260  1  70  2  10
vout ( t )  vout 1 ( t )  vout 2 ( t )
vout ( t )  6 cos( 2000t  30 )  2 cos( 4000t  10 )
81
6. Frequency Response – 6.1 Fourier Analysis, Filters, Transfer Functions
Example 6.2 – Multi-input components, Superposition Principle
vin ( t )  2cos( 2000t )  cos( 4000t  70 )
vout ( t )  6 cos( 2000t  30 )  2 cos( 4000t  10 )
82
6. Frequency Response – 6.2 First-Order Low-Pass Filters
＊ Ideal Filters
83
6. Frequency Response – 6.2 First-Order Low-Pass Filters
6.2 First Order Low-Pass Filters
A low-pass filter is designed to pass low-frequency
components and reject high-frequency components. In
other words, for low frequencies, the output magnitude is
nearly the same as the input; while for high frequencies, the
output magnitude is much less than the input.
6.2.1 Transfer Function
Considerthe first - order low - pass filter
as shown,the input signalis a sinusoidal
having a phasorVin , we have
Vin
I
R  1 j2πfC
Vout 
1
Vin
Vin
1
I 


j2πfC
j2πfC R  1 j2πfC 1  j2πfRC
Vout
1
H(f) 

Vin 1  j 2πfRC
1
We define f B 
the" break" frequency,
2πRC
the" half - power" frequency
H(f) 
1
1  j( f f B )
84
6. Frequency Response – 6.2 First-Order Low-Pass Filters
6.2.2 Magnitude and Phase Plots of the Transfer Function
f
1
1
H ( f )   arctan( )
H( f ) 
H( f ) 
2
fB
1  j( f f B )
1  ( f fB )
As f  0, H ( f )  1 low - frequencycom ponentspassed,
also H(f)  0
As f  f B , H(f)  0 high - frequencycom ponentsrejected,
also H(f)  90
As f  f B , H(f) 
1
2
, Vout rms 
1
2
2
Vin rms , since P  Vrms
 Half Power
85
6. Frequency Response – 6.2 First-Order Low-Pass Filters
Example 6.3 – Calculation of RC Low-pass Output
vin ( t )  5 cos( 20t )  5 cos( 200t )  5 cos( 2000t )
Vin 1  50, f 1  10, Vin 2  50, f 2  100,
Vin 3  50, f 3  1000
1
1
fB 

 100Hz
6
2RC 2 * (1000 2 ) *10 *10
1
H( f ) 
 H ( 10 )  0.9950  5.71
1  j( f f B )
H ( 100)  0.7071  45
H ( 1000)  0.0995  84.29
Vout 1  H ( 10 ) Vin1  4.975  5.71
vout 1 ( t )  4.975cos( 20πt  5.71 )
Vout 2  H ( 100) Vin2  3.53545
vout 2 ( t )  3.535cos( 200πt  45 )
Vout 3  H ( 1000) Vin3  0.4975  84.29
vout 3 ( t )  0.4975cos( 2000πt  84.29 )
86
6. Frequency Response – 6.2 First-Order Low-Pass Filters
Example 6.3 – Calculation of RC Low-pass Output
vin ( t )  5 cos( 20πt )  5 cos( 200πt )  5 cos( 2000πt )
Vin 1  50, f 1  10, Vin 2  50, f 2  100,
Vin 3  50, f 3  1000
vout ( t )  4.975cos( 20πt  5.71 )
vout 1 (t)
 3.535cos( 200πt  45 )
vout 2 (t)
 0.4975cos( 2000πt  84.29 )
vout 3 (t)
87
6. Frequency Response – 6.2 First-Order Low-Pass Filters
Quiz – Exercise 6.4: Another First-Order Low-Pass Filter
Showthat the transfer functionis
V
1
H(f)  out 
Vin
1  j(f/f B )
where f B  R/2L
This is also a low-pass filter
88
6. Frequency Response – 6.3 Decibels and the Cascade Connection
6.2 Decibels and the Cascade Connections
6.3.1 Decibels
* We usually express the ratio of voltage (or power) amplitude
in decibels.
H( f )  H ( f ) dB  20 log H ( f )
for voltage
H( f )  H( f ) db  10log H( f ) for power
89
6. Frequency Response – 6.3 Decibels and the Cascade Connection
6.3.2 Cascade two-Port Networks
Vout Vout 2 Vout 1 Vout 2 Vout 1 Vout 2
H( f ) 





Vin
Vin 1
Vin 1 Vout 1
Vin1 Vin2
H ( f )  H1 ( f )  H 2 ( f )
H( f ) dB  H1 ( f ) dB  H 2 ( f ) dB
90
6. Frequency Response – 6.4 Bode Plots
6.4 Bode Plots
H( f ) 
1
1  ( f f B )2
91
6. Frequency Response – 6.4 Bode Plots
6.4 Bode Plots
break frequency f B
H( f ) 
 f 
H ( f )   arctan 
 fB 
1
1  ( f f B )2
H ( f ) dB  20 log
1
1 ( f fB )
2
For f  f B H(f)  0 dB
For f  f B /10,  H(f)  0
For f  10 f B ,  H(f)  90
 f 

For f  f B H ( f ) dB  20 log
 fB 
92
6. Frequency Response – 6.5 First-Order High-Pass Filters
6.5 First-Order High-Pass Filters
6.5.1 Transfer Function
V
j( f f B )
H ( f )  out 
Vin 1  j ( f f B )
1
fB 
2RC
H( f ) 
f fB
1  f fB 
2
 f 
H ( f )  90   arctan  
 fB 
93
6. Frequency Response – 6.5 First-Order High-Pass Filters
6.5.2 Bode Plots
H( f ) 
f fB
1  f fB 
2
H ( f ) dB
  f 2 
 f 
  10 log1  
 
 20 log
  f B  
 fB 
 f 
For f  f B , H ( f ) dB  20 log 
 fB 
For f  f B , H  f  dB  0
For f  f B /10, H  f   90
For f  10 f B , H  f   0
94
6. Frequency Response – 6.5 First-Order High-Pass Filters
Exercise 6.13– Another First-Order High-Pass Filter
Showthat the transfer function
of the circuit is :
H( f ) 
Vout
j( f f B )

Vin 1  j( f f B )
where f B  R/2L
95
6. Frequency Response – 6.5 First-Order Filters
First-Order Low-Pass Filters
First-Order High-Pass Filters
96
6. Frequency Response – 6.6 Series Resonances
6.6.1 Resonant Circuits (Second Order)
* The resonance circuits forms the basis of second-order filters
that have better performance than the first-order filters.
* When a sinusoidal source of the proper frequency is applied
to a resonant circuit, voltage much larger then the source
can appear.
97
6. Frequency Response – 6.6 Series Resonances
6.6.1 Resonant Circuits (Second Order)
The im pedanceseen by the sourceis
1
Z s ( f )  j 2 fL  R  j
2fC
When Z s (f) is purely resistive,the frequency
f 0 is called the" resonant frequency"
2f 0 L 
1
2f 0 C
 f 
0
1
2 LC
The quality factor is defined as the ratio of the reactanceof the inductance
at the resonant frequencyto the resistance: Qs  2f 0 L/R (  1/2f 0 CR )

 f
f 0 

Now the im pedancecan be rewritten as : Z s ( f )  R 1  jQs 

f 
 f0

98
6. Frequency Response – 6.6 Series Resonances
6.6.1 Resonant Circuits (Second Order)
With resonant frequency f 0 
1
2 LC
and quality factor Qs  2f 0 L/R (  1/2f 0 CR )
The im pedanceseen by the sourceis

 f
f 0 
1

 R 1  jQs   
Z s ( f )  j 2 fL  R  j
f 
2fC
 f0

99
6. Frequency Response – 6.6 Series Resonances
6.6.2 Series Resonant Circuits as Band-Pass Filter

 f
f 0 
Z s ( f )  R 1  jQs   
f 
 f0


VR
Vs
Vs R
I

Z s  f  1  jQs ( f f 0  f 0 f )
VR  RI 
-
Vs
VR
1


1  jQs ( f f 0  f 0 f )
Vs 1  jQs ( f f 0  f 0 f )

f f0 2 
2
H f  
 1  Qs ( - ) 
Vs
f0 f 

 Band - Pass Filter
VR
-1/2
freq.naer f 0 pass, others rejected
100
6. Frequency Response – 6.6 Series Resonances
6.6.2 Series Resonant Circuits as Band-Pass Filter

f f0 2 
2
H f  
 1  Qs ( ) 
Vs
f0 f 

VR
-1/2
The half - power frequency:
occurs at H  f   1/ 2
denotedas : f H and f L
Band - width is defined as :
B  f H  f L  f 0 /Qs
f H  f0 
B
B
, f L  f0 2
2
101
6. Frequency Response – 6.6 Series Resonances
Example 6.5 – Series Resonant Circuit
Resonant frequency: f 0 
Quality Factor : Qs 
1
2 LC
 1000Hz
2f 0 L
 10
R
f 0 1000

 100Hz
Qs
10
B
B
 f 0   1050 Hz , f L  f 0 -  950Hz
2
2
Band - width : B 
fH
102
6. Frequency Response – 6.6 Series Resonances
Example 6.5 – Series Resonant Circuit
At resonant frequency f 0  1000Hz
Z L  j 2f 0 L  j1000
Z c  -J/2f 0 C  - j1000
Z s  R  Z L  Z c  100
I
Vs 10

 0.010
Zs
100
VR  RI  100 0.010  10
VL  Z L I  j1000 0.010  1090
VC  Z C I   j1000 0.010  10  90
103
6. Frequency Response – 6.7 Parallel Resonances
6.7 Parallel Resonance
With resonant frequency f 0 
1
2π LC
and quality factor Qp  R/(2πf 0 L)  (2πf 0 CR)
104
6. Frequency Response – 6.8 Ideal and Second-Order Filters
6.8 Ideal and Second-Order Filters
6.8.1 Ideal Filters
105
6. Frequency Response – 6.8 Ideal and Second-Order Filters
6.8.1 Ideal Filters
106
6. Frequency Response – 6.8 Ideal and Second-Order Filters
6.8.2 Second-Order Low-Pass Filter
Vout
 jQs ( f 0 f )
H( f ) 

Vin 1  jQs ( f f 0  f 0 f )
f0 
1
2π LC
2πf 0 L
Qs 
R

1
2πf 0 CR
In design a filter, we want
the gain to be approxim ately
constantin the passband
 chooseQs  1
107
6. Frequency Response – 6.8 Ideal and Second-Order Filters
6.8.2 Second-Order High-Pass Filter
108
6. Frequency Response – 6.8 Ideal and Second-Order Filters
6.8.2 Second-Order Band-Pass Filter
109
6. Frequency Response – 6.8 Ideal and Second-Order Filters
6.8.2 Second-Order Band-Reject (Notch) Filter
110
6. Frequency Response – 6.8 Ideal and Second-Order Filters
Example 6.7 – Filter Design
Design a second-order filter with L=50mH that passes
components higher in frequency than 1kHz, rejects
components lower than 1kHz.
We need a high-pass filter.
To obtain a approximately
constant transfer function
In the pass-band, we choose
Qs  1 and select f 0  1kHz
since f 0 
1
2 LC
1
we have C 
 0.507F
2
2
( 2 ) f 0 L
2f 0 L
and R 
 314.1
Qs
111
6. Frequency Response – 6.8 Ideal and Second-Order Filters
* The Popular Sallen-Key Filters
112
6. Frequency Response – 6.8 Ideal and Second-Order Filters
* Higher-order Filters using Cascade of 2nd-order Filters
113
6. Frequency Response – 6.8 Ideal and Second-Order Filters
* Higher-order Filters using Cascade of 2nd-order Filters
114