Physics 121 Fall 2002

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Transcript Physics 121 Fall 2002

Physics 121 - Electricity and Magnetism
Lecture 15E - AC Circuits & Resonance II
No Y&F reference for slides on Complex analysis
• Circuit Analysis using Complex Exponentials
–
–
–
–
Imaginary and Complex Numbers
Complex Exponential Phasors and Rotations
Phasors as Solutions of Steady State Oscillator Equations
Phasor representation applied to current versus voltage in R.
L. C.
– Series LCR Circuit using Complex Phasors
– Parallel LCR Circuit using Complex Phasors
– Transient Solution of Damped Oscillator using Complex
Phasors
Copyright R. Janow – Fall 2015
Summary: AC Series LCR Circuit
vR
E
R
L
vL
C
vC
Circuit
Element
Symbol
Resistance
or Reactance
Phase of
Current
Phase
Angle
Amplitude
Relation
Resistor
R
R
In phase
with VR
0º (0 rad)
VR = ImR
Capacitor
C
XC=1/wdC
Leads VC
by 90º
-90º (-p/2)
VC = ImXC
Inductor
L
XL=wdL
Lags VL
by 90º
+90º (p/2)
VL = ImXL
E(t)  Emax cos(wDt  F)
i(t)  Im cos(wDt)
Em
Im
F
Pav  Prms  ErmsI rms cos(F)
VL-VC
Z
VR
XL-XC
sketch shows
XL > XC
Em  Im | Z |
R
wDt
| Z |  [ R2  (XL  XC )2 ]1/2
tan(F ) 
VL  VC
X  XC
 L
VR
R
Copyright R. Janow – Fall 2015
Complex Exponentials in Circuit Analysis
z2  1 has roots   1
i 
1
z2   1 has roots   i
is “pure imaginary”
• Roots of a quadratic equation az2 + bz +c = 0
b
are complex if b2<4ac:
Re{z}   ,
2a
1
Im{z}  
4ac  b2
2a
b
1
z

b2  4ac
2a 2a
• Complex number in rectangular form:
z  Re(z)  i Im( z)
REVERSE SIGN OF
IMAGINARY PART
• Re{z} and Im{z} are both real numbers
• Complex conjugate: z*  Re(z)  i Im( z)
• Addition: z1  z2  Re(z1)  Re(z2 )  i { Im(z1)  Im(z1)}
z  z *  2 Re(z)
z  z *  2 Im( z)
• EE’s use j instead of i (i is for current)
Copyright R. Janow – Fall 2015
Representation using the Complex Plane
Imaginary axis (y)
|z|
q
x
“Argand diagram”
z  Re(z)  i Im( z)
x  Re(z)
y
y  Im{z}
 z  x  iy
• Polar form:
Real axis (x)
r |z|
x  r cos(q)
y  r sin(q)
 z  r cos(q)  i r sin(q)
• Magnitude2: r 2  zz *  (x  iy) (x  i y)  x2  y2
• Argument:
q  tan-1 ( y/x )
• Picture also displays complex-valued functions
Copyright R. Janow – Fall 2015
Complex Exponentials: Euler’s Formula
• Taylor’s Series definition of exponential function:
2
3
4
5
6
u u u
u u
eu  1  u 




 for all u (including complex)
2! 3! 4! 5! 6!
• Evaluate for u = iq:
i2q2 i3q3 i4 q4 i5q5 i6q6
i
q
e  1  iq 




 ... for all q
2!
3!
4!
5!
6!
q2 q4 q6
q3 q5
i
q
use i2n+1 = i(-1)2n
e  {1 


 ...}  i { q 

 ...}
2! 4! 6!
3! 5!
• Recognize series definitions of sine and cosine:
q2 q 4 q6
cos(q)  1 


 ...
2! 4! 6!
• Hence:
q3 q5
sin(q)  q 

 ... (radians)
3! 5!
eiq  cos(q)  i sin(q)
Above is a complex number of magnitude 1, argument q:
• Special
Properties:
1/i   i
e0  1  ei2p  ei2np n  1,2,...
e ip   1  (e  ip )*
eip   1
eip / 2  i
e ip / 2   i  (e ip / 2 )*
Copyright R. Janow – Fall 2015
Rules for Complex Quantities
Euler’s Formula for magnitude r:
z  r eiq  r cos(q)  i r sin(q)
Polar form of arbitrary complex z
z *  r e  iq  r cos(q)  i r sin(q) Complex conjugate of z
z1  z2  (x1  x2 )  i(y1  y2 )
Sum:
| z1  z2 |  { (x 1  x2 )2  (y1  y2 )2 }1/ 2
 iq1
 iq2
i (q  q2 )
Product:
z1 z2  r1 e
r2 e
 r1 r2 e 1
z z *  | z |2
| 1/ z |  1/ | z |
•Magnitude of a product = product of individual magnitudes
•Argument of a product = sum of individual arguments
•Two complex entities are equal if and only if their amplitudes
(magnitudes) are equal and their arguments (phases) are equal
Periodicity 2p:
ei (  2np)  ei [ e 2pi ]n  ei [1n ]  ei 
DeMoivre’s Theorem:
zn  { r ei q }n  rnei nq  rn {cos(nq)  isin(nq)}
i (q  2pk) / n
z1/n  r1/ne
for integer n & k = 0,1….n-1
Copyright R. Janow – Fall 2015
Phasors, Rotation Operators, Evolution Operators:
e+i
rotates complex number z
by angle + in complex plane
ze
 i
Im
z

z'  z ei  r ei qei   r ei (q  )
Time Evolution: Let  = w(t-t0)
iw(t f  t0 ) evolves z from t0 to tf for tf > t0
e
q

ze
Re
 i
Represent sinusoidally varying real quantity a(t) as a vector in the complex
plane, rotating (counterclockwise) at angular frequency w.
Define A(t)
•
•
•
•
 Amaxei (wt  ) then Re {A(t)}  Amaxcos(wt  )  a(t)
A(t) is called a “time domain phasor”
Re {A} = is the measurable instantaneous value of A(t)
Time dependent exponentials like eiwt are sometimes called evolution operators
Cancelling common factors of eiwt leaves a “frequency domain phasor” A (often
simply called a “phasor”):
A  A max ei 
Copyright R. Janow – Fall 2015
Phasor Definitions, continued
A phasor represents a sinusoidal, steady state signal whose amplitude Amax,
phase , and frequency w are time invariant.
peak value of a(t), real
time domain phasor, complex
a(t)  Re { (Amax ei ) eiwt }
instantaneous value,
time domain, real
frequency domain phasor A, complex,
rotated by  from real axis
Phasor Transform P:
forward - time to frequency domain
inverse - frequency to time domain
e.g. : A  Amax ei 
a(t)  P -1 { A } e.g. : a(t)  Re{Aeiwt }
A  P { a(t) }
Advantage of phasor transform: For sinusoidal signals, differential equations
(time domain) become algebraic equations (frequency domain) as common
factors of eiwt cancel).
Process: Replace real variables in time dependent analysis problem with
variables written using complex exponentials. Cancel common factors of eiwt
and solve remaining algebraic problem. Then return to real solution in time
Copyright R. Janow – Fall 2015
domain.
Time domain phasors are alternative solutions to sines and
cosines in differential equations representing oscillations.
Recall:
d(e f (a) )
df(a)
f
(
a
)
 e
da
da
chain rule:
d(eia )
 i eia
da
a can be complex
First and Second Derivatives: set a = wt+, amplitude Amax
dA
d
d

[ Amaxei (wt  ) ]  i A (wt  )  iwA
dt
dt
dt
2
d A d
dA
2

[
i
w
A
]

i
w


w
A
2
dt
dt
dt
Example: Simple
Harmonic Oscillator
d2x
2


w
x w0  k/m
0
2
dt
x2 (t)  x0sin(wt  ' )
Trig Solutions: x1(t)  x0cos(wt  )
Complex Solutions:
x 3 (t)  x 0ei(wt )
x 4 (t)  x * (t)  x 0ei(wt )
note : cos(x)  (x n  xn *)/2 also
a solution
Copyright R. Janow – Fall 2015
Complex Exponential Representation applied to Passive
Circuit Elements: Revisit AC Voltage vs. Current in L, C, and R
Apply complex voltage & current
(time domain phasors):
Resistor:
ε(t) - vR(t )  0
vR (t)  i(t) R 
magnitudes:
exponents:
Note: now using j = sqrt(-1)
i(t)
jω t
j(w t  F)
D
I Re
equate to E e D
m
m
ε
Em  Im R
jwDt  jwDt  jF  F  0
jw t )
j(w t  F)
di
d jw t)
 L Im [e D ]  ImLjwDe D equate to Eme D
dt
dt
magnitudes:
ImwDL  Em
L  wDL
Em  ImL
jw t
exponents: jwD t  jF
e
 j e D e jF  j  F  p / 2
Capacitor:
vC(t ) 
vR( t )
ε(t) - vL(t )  0
Inductor:
vL(t )  L
i(t)  Ime jwDt
E (t)  Eme j( wDt F)
i(t)
ε
L
vL(t)
ε(t) - vC(t )  0
I
I
jw t
jw t
j(w t F)
q(t) 1
  i(t)dt  m  e D dt  m e D equate to Eme D
C
C
C
jwDC
Im / wDC  Em C  1/ wDC Em  ImC
exponents: e jwD t  jF   j e jwD t e jF   j  F  p / 2
magnitudes:
i(t)
ε
C
vC ( t )
Copyright R. Janow – Fall 2015
Summary: Complex Exponential Representation of AC
Voltage vs. Current in L, C, and R
Applied voltage: E (t)  Eme
j( wD t F )
& current: i(t)  Ime
jwD t
time domain
Phase angle factor ejF rotates E CCW from current phasor
Passive circuit elements:
Resistor:
Inductor:
Capacitor:
VR / IRm  R
VL / ILm  L  wDL
VC / ICm   C  1
wDC
VR& IRm in phase
VL leads ILm by p/2
VC lags ICm by p/2
Phasor rotations of voltage drops relative to currents:
e0 =1
e+jp/2 = +j
e-jp/2 = -j
Voltage drop phasors relative to currents: time domain
vR (t)  IRm R e
jωDt
vL(t )  ILmLe
jwDt
e jp / 2
vC(t )  ICmCe
jwDt
e jp / 2
Frequency domain (factor out time dependence)
vR (t)  IRm R
vL(t )  ILmLe jp / 2
vC(t )  ICm Ce jp / 2
Define: Complex impedance z = E(t)/i(t) and |z| = Vmax/Imax
Impedances of simple circuit elements (complex):
zR  R
zL   jL
zC   jC
Magnitudes of Impedances |z| = [zz*]1/2 (real)
| zR | R
| zL |  L
Copyright
| z | R.Janow
 – Fall 2015
C
C
Complex Impedance z
Applied voltage: E (t)  Eme
j( wD t F )
& current: i(t)  Ime
jwD t
time domain
Definition: complex impedance z (or simply impedance) is the ratio of the
(complex) voltage phasor to the (complex) current phasor (in time or frequency
domain).
jF
E (t)  i(t) z
| z |  Em / Im
 z  | z |e
Definition: Phase angle F measures rotation of the applied voltage referenced
to the current in the branch. It is also the angle between z and the real axis.
The rotation operator is ejF.
Sketch shows positive Im{z}.
jF
e  cos(F)  j sin(F) z  Re{z}  j Im{z}
F is positive, implying that
Em leads Im.
Re{z}  | z | cos(F)
Im{z}  | z | sin(F)
tan(F) 
Im{z} sin(F)

Re{z} cos(F)
1/z occurs when analyzing parallel branches:
1
z*
e jF
cos(F)
sin(F)
2
zz* | z | 




j
z
|z|
|z|
|z|
| z |2
cos(F) Re{z}
Re{1 / z} 

|z|
| z |2
sin(F)
Im{z}
Im{1/z}  

|z|
| z |2
Im
Z
Im(z)
Em
F
Re(z)
tan(F)  
Im
Re
Im{1/z}
Re{1/z}
Copyright R. Janow – Fall 2015
Impedances of Series or Parallel Collections of Circuit Elements
Impedances of individual passive circuit elements:
zR  R
zL   jL
zC   jC
The kth sub-circuit (arbitrary complexity)
consists of R, L, and/or C basic elements:
zk
Assertion: Follow series or parallel resistor addition rules to
compute equivalent impedances (complex)
Series branch formula:
z ser 
zk
z2
z1
n
 zk
complex
k 1
Parallel sub-circuit formula:
1
n
1
 
complex
zpara k 1 zk
z1
z2
zk
Copyright R. Janow – Fall 2015
Revisit series LCR circuit using complex phasors
AC voltage: E (t)  Eme j( wDt F)
time domain phasors
Currents are the same everywhere in an essential branch,
same phase, same magnitude. Im  IRm  ILm  ICm
vR
E
jw t
Current: i(t)  Ime D
R
L
C
vL
Kirchhoff Loop rule for series LRC circuit:
E(t)  vR (t)  vL (t)  vC (t)
Substitute the time domain voltage phasors for vR, vL, vC.
vC
Eme j(wDtF)  Im R e
jωDt
 ImLe jwDte jp / 2  ImCe jwDte jp / 2 (1)
Divide equation (1) by i(t) ( = Imejwt )
E(t)
 z  | z | e jF  R  Le  jp / 2   Ce jp / 2
i(t)
replace : e jp / 2  j
e jp / 2   j
z  R  j(L  C ) (1)
Magnitude of Z:
| z |
z z *  [ R 2  (L  C )2 ]1/2
VL
Em
Im
F
wDt
VR
VC
Sketch shows F positive for
VL>VC  XL>XC
A t = 0 sketch would show
phasors
in frequency domain
Copyright R. Janow – Fall 2015
Revisit the series LCR circuit, Continuation
Impedance z (Equation 1) can also be found by summing the impedances of
the 3 basic circuit elements in series LCR circuit (invoke Series Formula*)
z  R  j( L  C )  | z | e jF (1)
Re{z}  R
Im
Z
Im{z}  XL  XC
z is not a phasor, as it is time independent.
Previous “phasor diagrams” showed z rotating
with Em.
Phase angle F for the circuit:
tan(F) 
Power
factor:
V  VC
Im{z} L  C

 L
Re{z}
R
VR
cos(F) 
VR
R

Em | z |
XL-XC
F
R
Re
Sketch shows F >0 for XL > XC
 VL > VC. F is positive when Im{z}
is positive
* The equivalent (complex) impedance
for circuit elements in series
(arbitrarily many) is the sum of the
individual (complex) impedances.
Resonance: As before, z is real for XL= XC (w2 = 1/LC). |z| is minimized.
Current amplitude Im is maximized at resonance
Phase angle in terms of admittance 1/z:
1 z*
2
2


Re{z}
/
|
z
|

j
Im{z}
/
|
z
|
z | z |2
tan(F) 
Im{ z}
Im{1/z}
 
Re{z}
Re{1/z}
Copyright R. Janow – Fall 2015
Parallel LCR circuit using complex phasors
All steady state voltages and currents
oscillate at driving frequency wD
AC voltage:
E (t)  Eme j( wDt F)
Instantaneous Current: i(t)  Ime jwDt
i
a
iL
iR
E
vR
iC
vL
R
L
vC
C
b
Kirchhoff Loop Rule (time domain):
E(t)  vR (t)  0
E(t)  vC (t)  0
E(t)  vL (t)  0
• 2 essential nodes “a” & “b”
• 4 essential branches,
• Not all independent
Instantaneous voltages across parallel branches have the same magnitude and
Use voltage as reference
phase:
 E(t)  vR (t)  vL (t)  vC (t)
instead of current
Common voltage phase, all branches:
e
Common voltage peak, all branches:
Em  VRm  VLm  VCm
j(wD t  F)
Kichhoff Current Rule (node a or b):
i(t)  iR (t)  iL (t)  iC (t) (1)
Current Amplitudes in each parallel branch (reference now to voltage drop):
Em
IRm 
R
ILm 
L  wDL
Em
L
C  1 wD C
ICm 
Em
C
Copyright R. Janow – Fall 2015
Parallel LCR circuit, Continuation #1
Instantaneous currents in each branch lead, lag,
or are in phase with the (reference) voltages:
iR (t)  IRme
iL (t)  ILme
j(wD t F)
j(wD t F)  jp / 2
iC (t)  ICme
e
j(wD t F)  jp / 2
e
Im
Em, VR, VL, VC
• in phase with E(t)
F
• lags E(t) by p/2
IRm
wDt
Im
• leads E(t) by p/2
ICm
ICm- ILm
Current amplitude addition rule is Pythagorean
Im  [ I2Rm  ( ICm  ILm )2 ]1/2
ILm
Re
Sketch shows F < 0 (applied
voltage lags current) for
ICm > ILm  XL>XC
Substitute currents into junction rule equation (1):
jw t
j(w t F)
j(w t F)  jp / 2
j(w t F)  jp / 2
i(t)  Ime D  IRme D
 ILme D
e
 ICme D
e
(2)
Cancel ejwt factor and multiply by e-jF
(current phasor – frequency domain)
Ime jF  IRm  ILme jp / 2  ICme jp / 2 
Em
R

Em  jp / 2
E
e
 m e jp / 2 (2.1)
L
C
Copyright R. Janow – Fall 2015
Parallel LCR circuit, continuation #2
Recall, impedance
“admittance”
Note that:
voltage phasor
z 
 z e jF 
current phasor
1 current phasor
1  jF


e
z voltage phasor
z
e jp / 2   j
e jp / 2   j
Substitute, then divide (Eq. 2.1) above by Em
Em jF
e
Im
Im  jF

e
Em
|z| 
Em
Im
1
z*

z | z |2
1
1
1
1

 j[

] (3)
z
R
C
L
Admittance 1/z in Equation 3 is also the sum of the reciprocal impedances of
the 3 basic circuit elements in parallel LCR circuit (invoke Parallel Formula)
Find |1/z|: multiply 1/z by complex conjugate (1/z)* and take square root:
1
1 1
1
1 

   j[

]
C
L 
| z |2 z z *  R
1
1
1
1 

]
  j[
C
L 
 R
1/2
2

2 
1
1
1  
1
    

 R 

 
|z|
 C L  


(4)
Represent z in terms of 1/z:
1
z * Re{z}
Im{ z}
1
1



j

Re{
}

j
Im{
}
2
z zz * | z |2
z
z
|z|
1
1
Re{z}  | z |2 Re{ }  | z |2
R
z
1
1
Re{ } 
R
z
1
1
1
Im{ }  [

]


z
C
L
1
1
1
Im{z}   | z |2 Im{ }   | z |2 [

]


z
C
L
Copyright R. Janow – Fall 2015
Parallel LCR circuit, Continuation #3
Im
Phase angle F:
1/z
1/XC-1/XL
Im{z}
Im{1 / z}
tan(F) 
 
Re{z}
Re{1 / z}
1
1
tan(F)  R (

) (5)
XL XC
tan(F) 
ILm  ICm
IR

F
E
1/R
Sketch shows F negative for 1/XC
> 1/XL  IC > IL  XL > XC
1/XL  1 / XC
1/R
For XL > XC: 1/XL – 1/XC < 0
-Tan (F) and F are positive in Series LCR circuit (see above),
Voltage Em leads current Im
- Tan(F) and F are negative in Parallel LCR circuit
Current Im leads applied voltage Em
Converse for XC > XL
F = 0 at resonance (XL = XC) in both Series and Parallel circuits
Copyright R. Janow – Fall 2015
Parallel LCR circuit, Continuation #4
Resonance: Minimizes
E
Im  m
|z|
as function of frequency
1/2
2

2 
1
1
1  
 1

    

 R 
|z|


L  
 C


(4)
Minimum of 1/|z| when XL = XC,
i.e. when w2 = 1/LC
Same resonant frequency as series LCR, but current is minimized
instead of maximized at resonance
Lim
1
1

|z|
R
Lim
1
1

also as w  wres,  z  R
z
R
as w  wres  | z |  R
At resonance, the current amplitudes ILm and ICm in the L & C
branches are equal, but are 180o apart in phase. These
cancel at all times at nodes a and b of the circuit.
Copyright R. Janow – Fall 2015
Using complex exponentials to solve a differential equation
Revisit Damped Oscillator: After “step response” to
switch at ‘a’ saturates, turn switch to ‘b’. Decaying
natural oscillations start when damping is weak.
dUtot
 i2 (t)R
dt
• R dissipates potential energy
• not a steady state system
• solutions not simple sinusoids
d2Q(t) R dQ

 w20Q(t)  0
2
dt
L dt
Trial solution… Q(t)  Q 0e j( wt ) (2)
 Assume complex frequency
a
+
E
w0  1/LC (1)
R
b
L
C
second order equation
for Q(t) – see Lect. 13
… but if w is real, solution oscillates forever
w  wx  jwy
wx and wy assumed real
dQ(t)
d2Q(t)
j( wt   )
2 j( wt  )
2
 Q 0 jwe
 jwQ(t)

Q
(
j
w
)
e


w
Q(t)
0
2
dt
dt
R
Substitute into Equation (1):  w2 Q  j w Q  w20Q(t)  0
L
“characteristic
R
Cancel common factors of Q:
 w2  j w  w20  0 (3)
equation”
L
Derivatives:
Phasor-like trial solution (2) turned differential equation into polynomial equation
Copyright R. Janow – Fall 2015
Using complex exponentials to solve a differential equation, #2
2
2
2
Expand: w  wx  2 jwx wy  wy
Equation (3) becomes 2 separate equations for real and imaginary terms
R
R
 2 wx wy  wx  0 (4.1)
Im{Eq 3} 
wy 
L
2L
R
2
2
2
Re{Eq 3}   wx  wy  wy  w0  0 (4.2)
L
2
R 
R 2

2
2
wx  w20  2L
Substitute wy:  wx  
  w0  0
 2L 
( 
Shifted natural frequency wx is
For real wx (underdamping):
damping
 Q(t)  Q0 e
w20 
real for
Q( t )  Q 0e
j(wx  jwy ) t
oscillation

R
t
2L
e
j(ωx t  )
For critical damping, wx = 0:
j
Q(t)  Q0 e e

R
t
2L
w20
imaginary for
(6)
w20 
( 
< ( 
underdamping
R 2
2L
overdamping
2
j j wy t jω x t
R 2
2L
e j  Q0e
Re Q(t)  Q0 e
( 
R 2
2L
(5)
e

e
R
t
2L
cos(wx t  ) (7)
No oscillations, decay only:
R
t
2L
Copyright R. Janow – Fall 2015
Re Q(t)  Q0 cos() e

(8)
Using complex exponentials to solve a differential equation, #3
( 
wx becomes imaginary. Equations 4.1 & 4.2 invalid
For overdamping: w0 < 2L
Return to Eq. 3.0. Assume frequency is
2
2
w

j
w
'
w
'
real
w


(
w
'
)
y
y
y
pure imaginary ( no oscillation)
2
Eq. 3.0 becomes:
Solution:
R
 (w y ' ) 2 
2
R
wy '  w20  0
L
R
w  jwy ' / -  j
j
2L
quadratic, real coefficients
(  w
R 2
2L
2
0
(9) two roots
both pure imaginary
Both roots lead to decay w/o oscillation
+ root: w+ implies damping faster than e-Rt/2L
- root: w- implies damping slower than e-Rt/2L but not growth
Most general solution: linear combination of Q+ and Q-, each of form of Eq. 2.0
Q( t ) 
1 
2 





j
Q (t )  Q (t )  Q 0e e
Recall definition, hyperbolic cosine:
ReQ(t) 
Rt / 2L 1
2

 
e


1 x
e  ex
2
Q 0 cos() e-Rt/2L cosh (
 R 


 2L 
 R 2 2

  w0 t
 2L 

 e
 R 2 2
 
  w0 t
 2L 




 cosh(x)
2 w2 t ) (10)
0
Correctly reduces to critically damped Eq. 8.0
Copyright R. Janow – Fall 2015
Copyright R. Janow – Fall 2015