Download ECEN 301 Lecture #13

Imaginary
JS-H 1: 16
16 But, exerting all my powers to call upon God to deliver me
out of the power of this enemy which had seized upon me, and
at the very moment when I was ready to sink into despair and
abandon myself to destruction—not to an imaginary ruin, but
to the power of some actual being from the unseen world, who
had such marvelous power as I had never before felt in any
being—just at this moment of great alarm, I saw a pillar of light
exactly over my head, above the brightness of the sun, which
descended gradually until it fell upon me.
ECEN 301
Discussion #13 – Phasors
1
Lecture 13 – Network Analysis with
Capacitors and Inductors
Phasors
A “real” phasor is NOT the same thing used in Star Trek
ECEN 301
Discussion #13 – Phasors
2
Euler’s Identity
• Appendix A reviews complex numbers
Im
Complex exponential (ejθ) is a
point on the complex plane
j
ejθ
sinθ
-1
1
θ
cosθ
-j
1
Re
Euler’s equation
e j  cos   j sin 
e j  1  cos   j sin   cos 2   sin 2   1
Ae j  A cos   jA sin 
 A
ECEN 301
Discussion #13 – Phasors
3
Phasors
• Rewrite the expression for a general sinusoid signal:
A cos(t   )  Re{ Ae
magnitude
j (t  )
}
Angle (or argument)
Complex phasor notation for the simplification:
A cos(t   )  A  Ae
j
NB: The ejwt term is implicit (it is there but not written)
ECEN 301
Discussion #13 – Phasors
4
Frequency Domain
Graphing in the frequency domain: helpful in order
to understand Phasors
1.5
3.50
0.0
0.00
π
X(jw)
x(t)
π
2.00
4.00
0.00
-600.0
-1.5
-ω
0.0
Time domain
ECEN 301
600.0
w
time
cos(ω0t)
ω
π[δ(ω – ω0) + δ(ω – ω0)]
Frequency domain
Discussion #16 – Frequency Response
5
Gamma rays
Xray
0.3 mm
3 mm
30 mm
0.3 mm
3 mm
30 mm
0.3 m
3m
30 m
300 m
3 km
30 km
300 km
3000 km
1014 Hz
1013 Hz
1012 Hz
1011 Hz
1010 Hz
109 Hz
108 Hz
107 Hz
106 Hz
105 Hz
104 Hz
103 Hz
102 Hz
30Å
1017 Hz
1015 Hz
3Å
1018 Hz
300Å
0.3Å
1019 Hz
1016 Hz
0.03Å
…
1020 Hz
The Electromagnetic Spectrum
Ultraviolet
Infrared
Visible
light
Microwaves
…
Radio waves
wavelength
Phasors
1. Any sinusoidal signal can be represented by either:
•
•
Time-domain form: v(t) = Acos(ωt+θ)
Frequency-domain form: V(jω) = Aejθ = A θ
2. Phasor: a complex number expressed in polar form
consisting of:
•
•
Magnitude (A)
Phase angle (θ)
3. Phasors do not explicitly include the sinusoidal
frequency (ω) but this information is still important
ECEN 301
Discussion #13 – Phasors
7
Phasors
• Example 1: compute the phasor voltage for the equivalent voltage vs(t)
• v1(t) = 15cos(377t+π/4)
• v2(t) = 15cos(377t+π/12)
v1(t)
+
~
–
v2(t)
+
~
–
vs(t)
+
~
–
ECEN 301
Discussion #13 – Phasors
8
Phasors
• Example 1: compute the phasor voltage for the equivalent voltage vs(t)
• v1(t) = 15cos(377t+π/4)
• v2(t) = 15cos(377t+π/12)
v1(t)
v2(t)
+
~
–
+
~
–
1.
Write voltages in phasor notation
V1 ( j )  15e j / 4
 15

4
V
V2 ( j )  15e j /12
vs(t)
 15
+
~
–
ECEN 301
Discussion #13 – Phasors

12
V
9
Phasors
• Example 1: compute the phasor voltage for the equivalent voltage vs(t)
• v1(t) = 15cos(377t+π/4)
• v2(t) = 15cos(377t+π/12)
v1(t)
+
~
–
v2(t)
+
~
–
1.
2.
V1 ( j )  15

Write voltages in phasor notation
Convert phasor voltages from polar to
rectangular form (see Appendix A)
V
4
Convert to rectangula r :
vs(t)
+
~
–
ECEN 301
V2 ( j )  15

V
12
Convert to rectangula r :
 
 
V1 ( j )  15 cos   j15 sin  
4
4
 10.61  j10.61 V
Discussion #13 – Phasors
 
 
V2 ( j )  15 cos   j15 sin  
 12 
 12 
 14.49  j 3.88 V
10
Phasors
• Example 1: compute the phasor voltage for the equivalent voltage vs(t)
• v1(t) = 15cos(377t+π/4)
• v2(t) = 15cos(377t+π/12)
v1(t)
+
~
–
1.
2.
v2(t)
+
~
–
3.
Write voltages in phasor notation
Convert phasor voltages from polar to
rectangular form (see Appendix A)
Combine voltages
VS ( j )  V1 ( j )  V2 ( j )
 25.10  j14.49
vs(t)
+
~
–
ECEN 301
Discussion #13 – Phasors
11
Phasors
• Example 1: compute the phasor voltage for the equivalent voltage vs(t)
• v1(t) = 15cos(377t+π/4)
• v2(t) = 15cos(377t+π/12)
v1(t)
+
~
–
v2(t)
+
~
–
1.
2.
3.
4.
Write voltages in phasor notation
Convert phasor voltages from polar to
rectangular form (see Appendix A)
Combine voltages
Convert rectangular back to polar
VS ( j )  25.10  j14.49
Convert to polar :
r  (25.10) 2  (14.49) 2
 28.98
 14.49 

25
.
10


  tan 1 
vs(t)
+
~
–
ECEN 301


6
VS ( j )  28.98
Discussion #13 – Phasors

6
12
Phasors
• Example 1: compute the phasor voltage for the equivalent voltage vs(t)
• v1(t) = 15cos(377t+π/4)
• v2(t) = 15cos(377t+π/12)
v1(t)
+
~
–
v2(t)
+
~
–
1.
2.
3.
4.
5.
Write voltages in phasor notation
Convert phasor voltages from polar to
rectangular form (see Appendix A)
Combine voltages
Convert rectangular back to polar
Convert from phasor to time domain
VS ( j )  28.98

6


vS (t )  28.98 cos 377t  
6

vs(t)
+
~
–
ECEN 301
NB: the answer is NOT
simply the addition of
the amplitudes of v1(t)
and v2(t) (i.e. 15 + 15),
and the addition of their
phases (i.e. π/4 + π/12)
Bring ωt back
Discussion #13 – Phasors
13
Phasors
• Example 1: compute the phasor voltage for the equivalent voltage vs(t)
• v1(t) = 15cos(377t+π/4)
• v2(t) = 15cos(377t+π/12)
v1(t)
+
~
–
v2(t)
+
~
–
VS ( j )  28.98
vs(t)
+
~
–
ECEN 301

6


vS (t )  28.98 cos 377t  
6

Discussion #13 – Phasors
Im
14.49
π/6 Vs(jω)
Re
25.10
14
Phasors of Different Frequencies
Superposition of AC signals: when signals do not have
the same frequency (ω) the ejωt term in the phasors
can no longer be implicit
i (t )  i1 (t )  i2 (t )
I
I ( j )  I1 ( j1 )  I 2 ( j2 )
+
i1(t)
i2(t)
v
–
Load
 A1e j 0 e j1t  A2e j 0e j2t
 A1e j 0  A2 e j 0
NB: ejωt can no longer be implicit
ECEN 301
Discussion #13 – Phasors
15
Phasors of Different Frequencies
Superposition of AC signals: when signals do not have
the same frequency (ω) solve the circuit separately
for each different frequency (ω) – then add the
individual results
R2
i1(t)
ECEN 301
R1
+
–
vs(t)
Discussion #13 – Phasors
16
Phasors of Different Frequencies
•
Example 2: compute the resistor voltages
• is(t) = 0.5cos[2π(100t)] A
• vs(t) = 20cos[2π(1000t)] V
• R1 = 150Ω, R2 = 50 Ω
+ R2 –
i1(t)
ECEN 301
+
R1
–
+
–
vs(t)
Discussion #13 – Phasors
17
Phasors of Different Frequencies
• Example 2: compute the resistor voltages
• is(t) = 0.5cos[2π(100t)] A
• vs(t) = 20cos[2π(1000t)] V
• R1 = 150Ω, R2 = 50 Ω
+ R2 –
i1(t)
+
R1
–
1.
Since the sources have different frequencies
(ω1 = 2π*100) and (ω2 = 2π*1000) use
superposition
•
first consider the (ω1 = 2π*100) part of
the circuit
•
When vs(t) = 0 – short circuit
+ R2 –
i1(t)
ECEN 301
Discussion #13 – Phasors
+
R1
–
+ v (t)
– s
18
Phasors of Different Frequencies
• Example 2: compute the resistor voltages
• is(t) = 0.5cos[2π(100t)] A
• vs(t) = 20cos[2π(1000t)] V
• R1 = 150Ω, R2 = 50 Ω
i1(t)
+
R1|| R2
–
1.
Since the sources have different frequencies
(ω1 = 2π*100) and (ω2 = 2π*1000) use
superposition
•
first consider the (ω1 = 2π*100) part of
the circuit
I s ( j )  0.50
VI 1 ( j )  VI 2 ( j )  I s  R1 || R2
 Is 
R1 R2
R1  R2
(50)(150)
(50)  (150)
 18.750
 0.50 
ECEN 301
Discussion #13 – Phasors
19
Phasors of Different Frequencies
• Example 2: compute the resistor voltages
• is(t) = 0.5cos[2π(100t)] A
• vs(t) = 20cos[2π(1000t)] V
• R1 = 150Ω, R2 = 50 Ω
1.
+ R2 –
+
R1
–
+
–
Since the sources have different frequencies
(ω1 = 2π*100) and (ω2 = 2π*1000) use
superposition
•
first consider the (ω1 = 2π*100) part of
the circuit
•
Next consider the (ω2 = 2π*1000) part
of the circuit
R1
VV 1 ( j )  Vs 
R1  R2
 200 
i1(t)
ECEN 301
Vs ( j )  200
vs(t) V ( j )  200
s
+ R2 –
+
+ v (t)
s
R1
–
–
(150)
(50)  (150)
 150
Discussion #13 – Phasors
VV 2 ( j )  Vs 
R2
R1  R2
 200 
(50)
(50)  (150)
 50
 5
20
Phasors of Different Frequencies
• Example 2: compute the resistor voltages
• is(t) = 0.5cos[2π(100t)] A
• vs(t) = 20cos[2π(1000t)] V
• R1 = 150Ω, R2 = 50 Ω
+ R2 –
i1(t)
+
R1
–
+
–
1.
Since the sources have different frequencies
(ω1 = 2π*100) and (ω2 = 2π*1000) use
superposition
•
first consider the (ω1 = 2π*100) part of the
circuit
•
Next consider the (ω2 = 2π*1000) part of
the circuit
•
Add the two together
vs(t) V1 ( j )  VI 1 ( j )  VV 1 ( j )
 18.750  150
v1 (t )  18.75 cos[ 2 (100t )]  15 cos[ 2 (1000t )]
V2 ( j )  VI 2 ( j )  VV 2 ( j )
 18.750  50
v1 (t )  18.75 cos[ 2 (100t )]  5 cos[ 2 (1000t )]
ECEN 301
Discussion #13 – Phasors
21
Impedance
Impedance: complex resistance (has no physical significance)
• Allows us to use network analysis methods such as node voltage, mesh
current, etc.
• Capacitors and inductors act as frequency-dependent resistors
i(t)
vs(t)
+
~
–
i(t)
R
+
vR(t)
–
vs(t)
+
~
–
i(t)
C
+
vC(t)
–
vs(t)
+
~
–
L
+
vL(t)
–
I(jω)
Vs(jω) +~
–
ECEN 301
+
Z VZ(jω)
–
Discussion #13 – Phasors
22
Impedance – Resistors
Impedance of a Resistor:
• Consider Ohm’s Law in phasor form:
I(jω)
+
Z VZ(jω)
–
Vs(jω) +~
–
Im
Phasor
v (t )
i (t )  s
R
A
 cos(t )
R
VZ ( j )  A0
I Z ( j ) 
A
0
R
Phasor domain
I
V
Re
ECEN 301
vs (t )  A cos(t )
VZ ( j )
Z R ( j ) 
R
I Z ( j )
Discussion #13 – Phasors
NB: Ohm’s
Law works the
same in DC and
AC
23
Impedance – Inductors
Impedance of an Inductor:
• First consider voltage and current in the time-domain
i(t)
vs(t)
vL (t )  L
diL (t )
 vS (t )
dt
1
iL (t )   vL ( )d
L
1
  A cos( )d
L
A

sin( t )
L
ECEN 301
+
~
–
L
+
vL(t)
–
vS (t )  vL (t )  A cos(t )
NB: current is shifted
90° from voltage
Discussion #13 – Phasors
A
sin( t )
L
A



cos t  
L 
2
iL (t ) 
24
Impedance – Inductors
Impedance of an Inductor:
• Now consider voltage and current in the phasor-domain
i(t)
vs(t)
+
~
–
I(jω)
L
+
vL(t)
–
Phasor
Vs(jω) +~
–
+
Z VZ(jω)
–
vS (t )  vL (t )  A cos(t ) Phasor
VZ ( j )  A0
A

A
I
(
j

)



iL (t ) 
sin( t )
Z
L
2
L
A


V ( j )

cos t  
Z L ( j )  Z
 jL
L 
2
I Z ( j )
ECEN 301
Discussion #13 – Phasors
Phasor domain
Im
V
Re
-π/2
I
25
Impedance – Capacitors
Impedance of a capacitor:
• First consider voltage and current in the time-domain
i(t)
vs(t)
1
iC ( )d  vS (t )
C
dv (t )
iC (t )  C C
dt
d
 C [ A cos(t )]
dt
 C[ A sin( t )]
+
~
–
I(jω)
C
+
vC(t)
–
Phasor
Vs(jω) +~
–
vC (t ) 
Phasor
+
Z VZ(jω)
–
VZ ( j )  A0
I Z ( j )  CA

2


 CA cos t  
2

ECEN 301
Discussion #13 – Phasors
26
Impedance – Capacitors
Impedance of a capacitor:
• Next consider voltage and current in the phasor-domain
I(jω)
Vs(jω) +~
–
+
Z VZ(jω)
–
VZ ( j )  A0
I Z ( j )  CA
ECEN 301

2
VZ ( j )
Z L ( j ) 
I Z ( j )
1



C
2
j

C
1

j C
Discussion #13 – Phasors
Phasor domain
Im
I
V
π/2
 je
j
Re

2

1
j
27
Impedance
Impedance of resistors, inductors, and capacitors
I(jω)
Im
ωL
Phasor domain
Vs(jω) +~
–
ZL
π/2
-π/2
R
ZR
ZC
+
Z VZ(jω)
–
Re
+
R
–
+
L
–
+
C
–
-1/ωC
Z R ( j)  R
ECEN 301
Z L ( j)  jL
Discussion #13 – Phasors
Z C ( j ) 
1
jC
28
Impedance
Impedance of resistors, inductors, and capacitors
I(jω)
Im
ωL
Phasor domain
Vs(jω) +~
–
ZL
π/2
-π/2
R
ZR
ZC
Re
+
Z VZ(jω)
–
Impedence in general :
Z ( j )  R( j )  jX ( j )
-1/ωC
Not a phasor but a
complex number
ECEN 301
AC resistance
Discussion #13 – Phasors
reactance
29
Impedance
Practical capacitors: in practice capacitors contain a real component
(represented by a resistive impedance ZR)
• At high frequencies or high capacitances
• ideal capacitor acts like a short circuit
• At low frequencies or low capacitances
• ideal capacitor acts like an open circuit
Practical Capacitor
Ideal Capacitor
Im
Im
+
C
–
Re
+
R
–
+
C
–
R
-π/2
Re
ZR
ZC
-π/2
ZC
-1/ωC
-1/ωC
NB: the ratio of ZC to ZR is highly frequency dependent
ECEN 301
Discussion #13 – Phasors
30
Impedance
Practical inductors: in practice inductors contain a real component
(represented by a resistive impedance ZR)
• At low frequencies or low inductances ZR has a strong influence
• Ideal inductor acts like a short circuit
• At high frequencies or high inductances ZL dominates ZR
• Ideal inductor acts like an open circuit
• At high frequencies a capacitor is also needed to correctly model a practical inductor
Ideal Inductor
Practical Inductor
Im
ωL
+
L
–
ZL
π/2
Re
+
L
–
+
R
–
Im
ZL
ωL
π/2
R
Re
ZR
NB: the ratio of ZL to ZR is highly frequency dependent
ECEN 301
Discussion #13 – Phasors
31
Impedance
• Example 3: impedance of a practical capacitor
• Find the impedance
• ω = 377 rads/s, C = 1nF, R = 1MΩ
+
R
–
+
C
–
ECEN 301
+
Z
–
Discussion #13 – Phasors
32
Impedance
• Example 3: impedance of a practical capacitor
• Find the impedance
• ω = 377 rads/s, C = 1nF, R = 1MΩ
Z  Z R || Z C
R  (1 / jC )
R  (1 / jC )
R

1  jCR

+
R
–
+
C
–
+
Z
–
106

1  j (377)(10 9 )(106 )
106

1  j 0.377
 9.36 105 (0.36) 
ECEN 301
Discussion #13 – Phasors
33
Impedance
• Example 4: find the equivalent impedance (ZEQ)
• ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH
R1
L
R2
ECEN 301
ZEQ
C
Discussion #13 – Phasors
34
Impedance
• Example 4: find the equivalent impedance (ZEQ)
• ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH
Z EQ1  Z R 2 || Z C
R1

R2 (1 / jC )
R2  (1 / jC )
L

R2
1  jCR2
ZEQ
50
1  j (10 4 )(10  10 6 )(50)
50

1  j5
 1.92  j 9.62
 9.81(1.37) 

R2
ECEN 301
C
Discussion #13 – Phasors
35
Impedance
• Example 4: find the equivalent impedance (ZEQ)
• ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH
Z EQ  Z R1  Z L  Z EQ1
R1
 R1  jL  9.81(1.37)
L
ZEQ
ZEQ1
Z EQ1  9.81(1.37) 
ECEN 301
 100  j (10 4 )(10  2 )  1.92  j 9.62
 101.92  j 90.38
 136.20.723 
NB: at this frequency (ω) the circuit has an inductive
impedance (reactance or phase is positive)
Discussion #13 – Phasors
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