Lecture 6: Resonance II - University of California,...

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Department of EECS University of California, Berkeley

EECS 105 Spring 2004, Lecture 6

Lecture 6: Resonance II

Prof. J. S. Smith


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.

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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



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

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Second Order Circuits

In the last lecture we introduced the resonant circuit:



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



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



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 ++=

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We then used the quadratic formula to find solutions for s, now in standard form:

→ss 0

20

2 20 ζωω ++=

( )1202,1 −±−= ζζωs



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

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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)(

−+

−

−+=



Solutions come in 3 kinds:

Depending on the damping factor, there are three types of solutions:

⎪⎩

⎪⎨

⎧

>

<=

111

ζUnderdamped

Critically Damped

Overdamped

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Overdamped Case

Time constants are real and negative

0)1(12

12 <⎩⎨⎧

=−±−=ss

s ζζτ

21

1

==

=

ζ

τ

ddV

1>ζ



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

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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



Underdamped (cont)

So we have:

( ) ( )( )jbtAjbtAeVtv atddC −++= expexp)( *

( )[ ]jbtAeVtv atddC expRe2)( +=

( )φω ++= tAeVtv atddC cos2)(

2

11ss

VA dd

+=

2

11ss

Vdd

+∠=φ

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Underdamped Peaking

For ζ < 1, the step response overshoots:

5.1

1

==

=

ζ

τ

ddV



Extremely Underdamped

01.1

1

==

=

ζ

τ

ddV

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Frequency domain analysis

We then looked at a Phasor analysis:

RCj

LjZ ++=ω

ω 1

Z



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 −+=

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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

−



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)(

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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



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∆

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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



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:

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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



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°

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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



Underdamped Mag Plot

ζ=1

ζ=0.01

ζ=0.1ζ=0.2ζ=0.4

ζ=0.6ζ=0.8

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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



Phase Bode Plot

ζ=0.010.10.20.40.60.8

ζ=1

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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



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

==

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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 ==⋅

=



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=

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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



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.

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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

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2)()(2)()()()(

pppp

zzzzq

jjjjjKjH

ωωζωωωωωζωωωωω

++++++

⋅=



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)

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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



We will use bilinear transfer functions for:

Small signal modeling of transistor circuits

Feedback and filtering in amplifiers

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