Upload
others
View
12
Download
0
Embed Size (px)
Citation preview
1
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6
Lecture 6: Resonance II
Prof. J. S. Smith
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Announcements
The labs start this weekYou must show up for labs to stay enrolled in the course.The first lab is available on the web site, it is an introduction to pspicePspice is now available from the course web site.
2
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Context
In the last lecture, we discussed – second order transfer functions, circuits which have
resonances– Power into a load, and reactive power– The concept of the Q of a resonator
In this lecture, we will review and complete the discussion of circuits with resonance, in standard form, and the Bode plot for the magnitude and phase of a circuit with an underdamped resonance
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Next Lecture
In the next lecture, we will start discussing the physical building blocks of integrated circuits– Metals– Insulators– Semiconductors
Reading assignment: Chapter 1 in the text
3
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Second Order Circuits
In the last lecture we introduced the resonant circuit:
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
And we derived the time domain DE for this circuit:
This is the equation for a damped harmonic oscillator which is being driven. This we can write in a standard form:
Where→
)()()()(2
2
tvdt
tdvRCtvtvdtdLC s
CCC =++
LCtv
dttdv
Qtvtv
dtd
LCtv
dttdv
LCRCtv
LCtv
dtd
sCCC
sCCC
)()()()(
)()()(1)(
0202
2
2
2
=++
=++
ωω
RLQ 0ω=
LC12
0 =ω
4
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
The behavior can also be characterized by a damping factor instead of Q
In which case the equation can be written:
Q21
=ζ
LCtv
dttdvtvtv
dtd sC
CC)()(2)()( 0
202
2
=++ ζωω
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Transient solution
To find the transient solution, for t>0, we tried a solution of the form:
– Where s is a complex number
Giving us a second order equation for s:
ddst
C VAetv +=)(
ss 020
2 20 ζωω ++=
5
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
We then used the quadratic formula to find solutions for s, now in standard form:
→ss 0
20
2 20 ζωω ++=
( )1202,1 −±−= ζζωs
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
General Case
Solutions are real or complex conj depending on if ζ > 1 or ζ < 1
⎩⎨⎧
=−±−=2
12 )1(1ss
s ζζτ
( ) ( )tsBtsAVtv ddC 21 expexp)( ++=
0)0( =++= BAVv ddC
( ) ( ) 0expexp0)()0(02211
0
=+⇒===
=t
t
C tsBstsAsdt
tdvCi
ddVBABsAs−=+=+ 021
6
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Solving for the coefficients:
Solve for A and B:
2
1
2
1
2
1
2
1
2
1 11
)1(
1ss
Vss
ss
VssV
AVB
ss
VAdddddd
dddd
−=
−
+−−=−−=
−
−=
ddVAssA −=−+
2
1
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−−= tsts
ddC esse
ssVtv 21
2
1
2
11
11)(
( ) ( )ts
ss
Vss
ts
ss
VVtvdd
ddddC 2
2
1
2
1
1
2
1exp
1exp
1)(
−+
−
−+=
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Solutions come in 3 kinds:
Depending on the damping factor, there are three types of solutions:
⎪⎩
⎪⎨
⎧
>
<=
111
ζUnderdamped
Critically Damped
Overdamped
7
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Overdamped Case
Time constants are real and negative
0)1(12
12 <⎩⎨⎧
=−±−=ss
s ζζτ
21
1
==
=
ζ
τ
ddV
1>ζ
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Critically Damped
Time constants are real and equal
τζζ
τ1)1(1 2 −=−±−=s
( )ττ
ζ
//
11)(lim tt
ddC teeVtv −−
→−−=
11
1
==
=
ζ
τ
ddV
1=ζ
8
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Underdamped
Now the s values are complex conjugate
jbasjbas
−=+=
2
1
( ) ( )tjbaBtjbaAVtv ddC )(exp)(exp)( −+++=
( ) ( )( )jbtBjbtAeVtv atddC −++= expexp)(
B
ss
Vss
ssV
ss
VAdd
dddd =−
=−
=−
−=
2
1
2
1
1
2*2
*1
*
111
( ) ( )( )jbtAjbtAeVtv atddC −++= expexp)( *
1<ζ
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Underdamped (cont)
So we have:
( ) ( )( )jbtAjbtAeVtv atddC −++= expexp)( *
( )[ ]jbtAeVtv atddC expRe2)( +=
( )φω ++= tAeVtv atddC cos2)(
2
11ss
VA dd
+=
2
11ss
Vdd
+∠=φ
9
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Underdamped Peaking
For ζ < 1, the step response overshoots:
5.1
1
==
=
ζ
τ
ddV
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Extremely Underdamped
01.1
1
==
=
ζ
τ
ddV
10
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Frequency domain analysis
We then looked at a Phasor analysis:
RCj
LjZ ++=ω
ω 1
Z
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Power into an Impedance Z
Letting the phase angle of the current be zero we found that the power into a circuit was:
The first term is the average power into the circuit, the second is the unavoidable ripple in the power, since this is AC, but the imaginary part of Z yields power which goes in at one part of the cycle, and then is returned at a later point. This is called reactive power.
( ))2sin()2cos(21 2 tZtZZIP irr ωω −+=
11
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
The Phasor results
At resonance the circuit impedance is purely real The reactive power for the Cap is supplied from the inductor, and visa versa.Imaginary components of impedance cancel outFor a series resonant circuit, the current is maximum at resonance
+VR
−
+ VL – + VC –
+Vs
−
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Second Order Transfer Function
So we have:
To find the poles/zeros, we put the H in canonical form:
RCj
Lj
RVVjH
s ++==
ωω
ω 1)( 0+Vo
−
RCjLCCRj
VVjH
s ωωωω+−
== 20
1)(
12
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. SmithFor a complex conjugate pair of poles in the denominator:
Putting the denominator in standard form, and factoring it:
If Q>1/2 Poles are complex conjugate frequencies
Re
Im
0)( 20
02 =++ ωωωωQ
jj
22 −±−=−±−=Q
jQQQ 4
11242 0
020
200 ωωωωωω
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Magnitude Response
How peaked the response is depends on Q
So at the peak (Z):
Qj
Qj
LRj
LRj
jH022
0
0
0
0220
0
0
)( ωωωω
ωω
ωωωωω
ωωω
ω+−
=+−
=
1=Q
10=Q
100=Q
0ω
1)(0
020
20
20
0 =+−
=
Qj
Qj
jH ωωωω
ω
ω
1)( 0 =ωjH
0)0( =H
ω∆
13
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
How Peaky is it?
Let’s find the points when the magnitude of the transfer function squared has dropped in half:
( ) 21)( 2
02220
20
2 =
⎟⎟⎠
⎞⎜⎜⎝
⎛+−
⎟⎟⎠
⎞⎜⎜⎝
⎛
=
Q
QjH
ωωωω
ωωω
21
1/
1)( 2
0
220
2 =
+⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
Q
jH
ωωωω
ω
1/
2
0
220 =⎟⎟
⎠
⎞⎜⎜⎝
⎛ −Qωωωω
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Half Power Frequencies (Bandwidth)
We have the following:
1/0
220 ±=−
Qωωωω
020
02 =−ωωωωQ
m
abbaQQ
>±±=+⎟⎟⎠
⎞⎜⎜⎝
⎛±±= 2
0
200
42ωωωω
Q0ωωωω =−=∆ −+
1/
2
0
220 =⎟⎟
⎠
⎞⎜⎜⎝
⎛ −Qωωωω
Q1
0
=∆ωω
Four solutions!
0000
<−−>+−<−+>++
babababa
Take positive frequencies:
14
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Second Order Circuit Bode Plot
Quadratic poles or zeros have the following form:
The roots can be parameterized in terms of the damping ratio:
02)()( 002 =++ ωζωωω jj
20
200
2 )(2)()(1 ωωωωωωζ jjj +=++⇒=
damping ratio
Two equal poles
1
)1)(1(2)()(1
2
0
21200
2
−±−=
++=++⇒>
ζζωω
ωτωτωζωωωζ
j
jjjj
Two real poles
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Bode Plot: Damped Case
The case of ζ >1 and ζ =1 is a simple generalization of simple poles (zeros). In the case that ζ >1, the poles (zeros) are at distinct frequencies. For ζ =1, the poles are at the same real frequency:
22 )1(12)()(1 ωτωτωτζ jjj +=++⇒=22 1)1( ωτωτ jj +=+
ωτωτ jj +=+ 1log401log20 2
( ) ( ) ( )ωτωτωτωτ jjjj +∠=+∠++∠=+∠ 1211)1( 2
AsymptoticSlope is 40 dB/dec
Asymptotic Phase Shift is 180°
15
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Underdamped Case
For ζ <1, the poles are complex conjugates:
For ω/ω0 << 1, this quadratic is negligible (0dB)For ω/ω0 >> 1, we can simplify:
In the transition region ω/ω0 ~ 1, things are tricky!
22
0
200
2
11
02)()(
ζζζζωω
ωζωωω
−±=−±−=
=++
jj
jj
0
2
00
2
0
log40)(log2012)()(log20ωω
ωωζ
ωω
ωω
=≈++ jjj
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Underdamped Mag Plot
ζ=1
ζ=0.01
ζ=0.1ζ=0.2ζ=0.4
ζ=0.6ζ=0.8
16
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Underdamped Phase
The phase for the quadratic factor is given by:
For ω/ω0 < 1, the phase shift is less than 90°For ω/ω0 = 1, the phase shift is exactly 90°For ω/ω0 > 1, the argument is negative so the phase shift is above 90° and approaches 180°Key point: argument shifts sign around resonance
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−=⎟⎟
⎠
⎞⎜⎜⎝
⎛++∠ −
2
0
01
0
2
0 )(1
2tan12)()(
ωω
ζωω
ζωω
ωω jj
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Phase Bode Plot
ζ=0.010.10.20.40.60.8
ζ=1
17
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Bode Plot GuidelinesIn the transition region, note that at the breakpoint:
From this you can estimate the peakiness in the magnitude response Example: for ζ=0.1, the Bode magnitude plot peaks by 20 log(5) ~14 dBThe phase is much more difficult. Note for ζ=0, the phase response is a step functionFor ζ=1, the phase is two real poles at a fixed frequencyFor 0<ζ<1, the plot should go somewhere in between!
Qjjjj 1212)()(12)()( 2
0
2
0
==++=++ ζζζωω
ωω
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Energy Storage in “Tank”
At resonance, the energy stored in the inductor and capacitor are
tLItiLw ML 0222 cos
21))((
21 ω==
tC
ItC
IC
diC
CtvCw
MM
C
02
20
2
02
220
2
22
sin21sin
21
)(121))((
21
ωω
ωω
ττ
==
⎟⎠⎞
⎜⎝⎛== ∫
LItC
tLIwww MMCLs2
02
20
022
21)sin1cos(
21
=+=+= ωω
ω
LIWW MSL2
max, 21
==
18
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Energy Dissipation in Tank
Energy dissipated per cycle:
The ratio of the energy stored to the energy dissipated per cycle is thus:
0
2 221
ωπ
⋅=⋅= RITPw MD
ππω
ωπ 22
12
21
21
0
0
2
2Q
RL
RI
LI
ww
M
M
D
S ==⋅
=
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Physical Interpretation of Q-FactorFor the series resonant circuit we have related the Q factor to very fundamental properties of the tank:
The tank quality factor relates how much energy is stored in a tank to how much energy loss is occurring.If Q >> 1, then the tank pretty much runs itself … even if you turn off the source, the tank will continue to oscillate for several cycles (on the order of Q cycles)Mechanical resonators can be fabricated with extremely high Q
D
S
wwQ π2=
19
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
thin-Film Bulk Acoustic Resonator (FBAR)RF MEMS
Agilent Technologies(IEEE ISSCC 2001)Q > 1000Resonates at 1.9 GHz
Can use it to build low power oscillator
C0
Cx Rx Lx
C1 C2R0
Pad
Thin Piezoelectric Film
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Crystals
To establish accurate clocks, we need oscillators with narrow resonances, and therefore high QSince mechanical resonators can have very high Q’s, because they can be built to loose very little energy on each cycle, most clocks are built around a mechanical resonance.
20
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Features of bilinear transfer functions
For bilinear transfer functions, we can put them into the form:
Where the roots in the numerator are called zeros, and the roots of the denominator are called poles.
Since the roots come from a real polynomial (in jω),they are either real, or come in complex conjugate
pairs. We can write those that come in pairs:
L
L
))()(())()(()()(
321
321
ppp
zzzq
jjjjjjjKjH
ωωωωωωωωωωωωωω
++++++
⋅=
( )( )LL
LL2
0000002
1
2000000
21
2)()(2)()()()(
pppp
zzzzq
jjjjjKjH
ωωζωωωωωζωωωωω
++++++
⋅=
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Feature by Feature:
Overall factor KDC poles or ZerosOverall decade for decade shift with frequencyEach pole contributes:– ≈1 far below its corner frequency– At the corner frequency, the response is down by 3db– 1/jω far above its corner frequency (-90° phase shift)
Each zero contributes:– ≈1 below its corner frequency– At the corner frequency, the response is up 3 db– jω above its corner frequency(90° phase shift)
21
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
Feature by Feature:Each complex conjugate pair of poles contribute:– ≈1 far below their corner frequency– 1/(jω)2 far above their corner frequency (-180° phase
shift)– At the resonance, the response peaks up ≈ by a factor Q– Log
Each complex conjugate pair of zeros contribute:– ≈1 below their corner frequency– (jω)2 far above their corner frequency (180° phase shift)– At the resonance, the response dips down ≈ by a factor Q– Log
( ) ( )2r 10H j 20 log 2 1ω = − ⋅ ζ − ζ
( ) ( )2r 10H j 20 log 2 1ω = + ⋅ ζ − ζ
Q21
=ζ
Department of EECS University of California, Berkeley
EECS 105 Spring 2004, Lecture 6 Prof. J. S. Smith
We will use bilinear transfer functions for:
Small signal modeling of transistor circuits
Feedback and filtering in amplifiers