ECEN 301 Discussion #13 – Phasors 1 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.
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
ECEN 301 Discussion #13 – Phasors 1
ImaginaryJS-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 2
Lecture 13 – Network Analysis with Capacitors and Inductors
PhasorsA “real” phasor is NOT the same thing used in Star Trek
ECEN 301 Discussion #13 – Phasors 3
Euler’s Identity
• Appendix A reviews complex numbers
Complex exponential (ejθ) is a point on the complex plane
A
jAAAe
je
je
j
j
j
sincos
1sincossincos1
sincos
22
1
1-1
j
-j
Re
Im
sinθ
cosθ
θ
ejθ
Euler’s equation
ECEN 301 Discussion #13 – Phasors 4
Phasors
• Rewrite the expression for a general sinusoid signal:
}Re{)cos( )( tjeAtA
Angle (or argument)magnitude
Complex phasor notation for the simplification:
jAeAtA )cos(
NB: The ejwt term is implicit (it is there but not written)
ECEN 301 Discussion #16 – Frequency Response 5
Frequency Domain
Graphing in the frequency domain: helpful in order to understand Phasors
-1.5
0.0
1.5
0.00 2.00 4.00
time
x(t)
0.00
3.50
-600.0 0.0 600.0
w
X(jw)
-ω ω
π π
cos(ω0t)π[δ(ω – ω0) + δ(ω – ω0)]
Time domain Frequency domain
The Electromagnetic Spectrum
Visiblelight
UltravioletXray
Gamma rays InfraredMicrowaves
Radio waveswavelength
0.03
Å
0.3Å
3Å 30Å
300Å
1020
Hz
0.3 m
m
3 m
m
30 m
m
0.3
mm
3 m
m
30 m
m
0.3
m
3 m
30 m
300
m
3 km
30 k
m
……
1019
Hz
1018
Hz
1017
Hz
1016
Hz
1015
Hz
1014
Hz
1013
Hz
1012
Hz
1011
Hz
1010
Hz
109 H
z
108 H
z
107 H
z
106 H
z
105 H
z
104 H
z
300
km10
3 Hz
3000
km
102 H
z
ECEN 301 Discussion #13 – Phasors 7
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 8
Phasors• Example 1: compute the phasor voltage for the equivalent voltage vs(t)
Superposition of AC signals: when signals do not have the same frequency (ω) the ejωt term in the phasors can no longer be implicit
i1(t) Load
+
v
–
I
i2(t)0
20
1
02
01
2211
21
21
)()()(
)()()(
jj
tjjtjj
eAeA
eeAeeA
jIjIjI
tititi
NB: ejωt can no longer be implicit
ECEN 301 Discussion #13 – Phasors 16
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
R1
R2
i1(t) vs(t)+–
ECEN 301 Discussion #13 – Phasors 17
Phasors of Different Frequencies
+R1
–
+ R2 –
i1(t) vs(t)+–
• Example 2: compute the resistor voltages
• is(t) = 0.5cos[2π(100t)] A
• vs(t) = 20cos[2π(1000t)] V
• R1 = 150Ω, R2 = 50 Ω
ECEN 301 Discussion #13 – Phasors 18
Phasors of Different Frequencies
+R1
–
+ R2 –
i1(t)
1. Since the sources have different frequencies (ω1 = 2π*100) and (ω2 = 2π*1000) use superposition • first consider the (ω1 = 2π*100) part of
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
)]1000(2cos[15)]100(2cos[75.18)(
015075.18
)()()(
1
111
tttv
jVjVjV VI
)]1000(2cos[5)]100(2cos[75.18)(
05075.18
)()()(
1
222
tttv
jVjVjV VI
+R1
–
+ R2 –
i1(t) vs(t)+–
ECEN 301 Discussion #13 – Phasors 22
ImpedanceImpedance: 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
vs(t)+–~ R
+vR(t)
–
i(t)
vs(t)+–~ C
+vC(t)
–
i(t)
vs(t)+–~ L
+vL(t)
–
i(t)
Vs(jω) +–~
+VZ(jω)
–
I(jω)
Z
ECEN 301 Discussion #13 – Phasors 23
Impedance – Resistors
Impedance of a Resistor:• Consider Ohm’s Law in phasor form:
0)(
0)(
R
AjI
AjV
Z
Z
Vs(jω) +
–~
+VZ(jω)
–
I(jω)
Z
)cos(
)()(
)cos()(
tR
AR
tvti
tAtv
s
s
Re
Im
I V
Phasor
Phasor domain
RjI
jVjZ
Z
ZR
)(
)()(
NB: Ohm’s Law works the same in DC and AC
ECEN 301 Discussion #13 – Phasors 24
Impedance – Inductors
Impedance of an Inductor:• First consider voltage and current in the time-domain
)sin(
)cos(1
)(1
)(
)()(
)(
tL
A
dAL
dvL
ti
tvdt
tdiLtv
LL
SL
L
vs(t)+–~ L
+vL(t)
–
i(t)
NB: current is shifted 90° from voltage
2cos
)sin()(
)cos()()(
tL
A
tL
Ati
tAtvtv
L
LS
ECEN 301 Discussion #13 – Phasors 25
Impedance – Inductors
Impedance of an Inductor:• Now consider voltage and current in the phasor-domain
2)(
0)(
L
AjI
AjV
Z
Z
vs(t)+–~ L
+vL(t)
–
i(t)
2cos
)sin()(
)cos()()(
tL
A
tL
Ati
tAtvtv
L
LS
Vs(jω) +–~
+VZ(jω)
–
I(jω)
Z
Phasor
Phasor
LjjI
jVjZ
Z
ZL
)(
)()(
Phasor domain
Re
Im
I
V
-π/2
ECEN 301 Discussion #13 – Phasors 26
Impedance – Capacitors
Impedance of a capacitor:• First consider voltage and current in the time-domain
2cos
)]sin([
)]cos([
)()(
)()(1
)(
tCA
tAC
tAdt
dC
dt
tdvCti
tvdiC
tv
CC
SCC
vs(t)+–~ C
+vC(t)
–
i(t)
Vs(jω) +–~
+VZ(jω)
–
I(jω)
Z
Phasor
Phasor
2)(
0)(
CAjI
AjV
Z
Z
ECEN 301 Discussion #13 – Phasors 27
Impedance – Capacitors
Impedance of a capacitor:• Next consider voltage and current in the phasor-domain
2)(
0)(
CAjI
AjV
Z
Z
Vs(jω) +–~
+VZ(jω)
–
I(jω)
Z
Cj
C
jC
jI
jVjZ
Z
ZL
1
2
1
)(
)()(
Phasor domain
Re
Im
I Vπ/2
jej
j 12
ECEN 301 Discussion #13 – Phasors 28
Impedance
LjjZL )(RjZR )(
+L–
+C–
Re
Im
-π/2
π/2R
-1/ωC
ωL
ZR
ZC
ZL
Phasor domain
+R–
CjjZC 1
)(
Vs(jω) +–~
+VZ(jω)
–
I(jω)
Z
Impedance of resistors, inductors, and capacitors
ECEN 301 Discussion #13 – Phasors 29
Impedance
)()()(
:generalin Impedence
jjXjRjZ
Re
Im
-π/2
π/2R
-1/ωC
ωL
ZR
ZC
ZL
Phasor domainVs(jω) +
–~
+VZ(jω)
–
I(jω)
Z
Impedance of resistors, inductors, and capacitors
AC resistance reactanceNot a phasor but a complex number
ECEN 301 Discussion #13 – Phasors 30
ImpedancePractical 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
+C–
+R–
+C–Re
Im
-π/2
-1/ωC
ZC
Re
Im
-π/2
R
-1/ωC
ZR
ZC
Ideal Capacitor Practical Capacitor
NB: the ratio of ZC to ZR is highly frequency dependent
ECEN 301 Discussion #13 – Phasors 31
ImpedancePractical 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
Re
Im
π/2
ωLZL
Re
Im
π/2 R
ωL
ZR
ZL
Ideal Inductor Practical Inductor
NB: the ratio of ZL to ZR is highly frequency dependent
+L–
+R–
+L–
ECEN 301 Discussion #13 – Phasors 32
Impedance• Example 3: impedance of a practical capacitor
• Find the impedance• ω = 377 rads/s, C = 1nF, R = 1MΩ
+R–
+C–
+Z–
ECEN 301 Discussion #13 – Phasors 33
Impedance• Example 3: impedance of a practical capacitor
• Find the impedance• ω = 377 rads/s, C = 1nF, R = 1MΩ
+R–
+C–
+Z–
)36.0(1036.9
377.01
10
)10)(10)(377(1
10
1
)/1(
)/1(
||
5
6
69
6
j
j
CRj
R
CjR
CjR
ZZZ CR
ECEN 301 Discussion #13 – Phasors 34
Impedance
• Example 4: find the equivalent impedance (ZEQ)
• ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH
ZEQ
R2 C
R1
L
ECEN 301 Discussion #13 – Phasors 35
Impedance
• Example 4: find the equivalent impedance (ZEQ)
• ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH
ZEQ
R2 C
R1
L
)37.1(81.9
62.992.1
51
50
)50)(1010)(10(1
50
1
)/1(
)/1(
||
64
2
2
2
2
21
j
j
j
CRj
R
CjR
CjR
ZZZ CREQ
ECEN 301 Discussion #13 – Phasors 36
Impedance
• Example 4: find the equivalent impedance (ZEQ)
• ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH
723.02.136
38.9092.101
62.992.1)10)(10(100
)37.1(81.924
1
11
j
jj
LjR
ZZZZ EQLREQ
ZEQ
R1
L
ZEQ1
)37.1(81.91EQZNB: at this frequency (ω) the circuit has an inductive impedance (reactance or phase is positive)