lecture 5: resonance

1

Department of EECS University of California, Berkeley

EECS 105 Spring 2004, Lecture 5

Lecture 5: Resonance

Prof. J. S. Smith


EECS 105 Spring 2004, Lecture 5 Prof. J. S. Smith

Context

In the last lecture, we discussed simple zeros and poles of a transfer function, and creating approximate Bode plots for hand plotting of phase and magnitude characteristics.Is this lecture, we will discuss second order transfer functions, circuits which have resonances.

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Second Order Circuits

The series resonant circuit is one of the most important elementary circuits:

This model is not only useful for physical LCR circuits, but also approximates mechanical resonances, molecular resonance, microwave cavities, transmission lines, buildings, bridges, …



Time Domain analysis

The differential equations for this circuit are:

dttdiLtv L

L)()( =

rtitv rr ⋅= )()(

dttdvCti C

c)()( =

dttdvRCtv

dtdt

tdvCdLv

tvtvtvtv

CC

C

s

RCLs

)()(

)()()()()(

++⎟⎠⎞

⎜⎝⎛

=

++=

3



So the differential equation for the circuit is:

Let’s see how this circuit responds to a step input, zero before time t=0, and Vdd for t>0

First of all, note that the steady state solution is

dttdvRCtvtv

dtdLCtv C

CCs)()()()( 2

2

++=

dds Vtv =∞→ )(



Transient solution

To find the transient solution, for t>0, let’s try a solution of the form:

– Where s is a complex number

Now we substitute this into our D. E.:

Giving us a second order equation for s:

ddst

C VAetv +=)(

ststdd

stdd

CCCs

RCsAeAeVAeLCsVdt

tdvRCtvtvdtdLCtv

+++=

++=

2

2

2 )()()()(

RCsLCs ++= 10 2

4



We can use the quadratic formula to find solutions for s: →

If s is real, then the circuit is overdamped, and the voltage will change exponentially to its steady state value.If s is complex, the circuit is underdamped, and the solution will oscillate around the steady state value before settling down to it.

RCsLCs ++= 10 2

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛±=

LCLR

LRs 1

22

2

2,1



Underdamped caseFor the underdamped case:

We will need to take sums of the complex exponentials to get real solutions, solutions are of the form:

Where: and

and A and are determined by the boundary conditions

2

2,1 21

2⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛±=

LR

LCj

LRs

)sin()( 1 φωα ++= − tAeVtv tddC

2

21

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

LR

LCω

LR

2=α

φ

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EECS 105 Spring 2004, Lecture 5 Prof. J. S. SmithResponse of underdamped circuit to step

If the circuit is moderately damped:

0.5 1 1.5 2 2.5 3

0.25

0.5

0.75

1

1.25

1.5

1.75

2



Underdamped response to step

And if very underdamped:

0.5 1 1.5 2 2.5 3

0.25

0.5

0.75

1

1.25

1.5

1.75

2

6



Underdamped Oscillations:

In the very underdamped case (R small), the ringing dies exponentially in a time:

And each oscillation takes a time:

So the number of oscillations of ringing that will occur is approximately:

RL21 =−α

πππω

2212)2(

12

1 LCL

RLC

≈⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

−

−

CL

RLC

RLN

ππ1

22 =≈



Frequency domain analysisWith phasor analysis, this circuit is readily analyzed, for example, the input impedance:

You can also write the expression for the voltage across any component

RCj

LjZ ++=ω

ω 1

⎟⎠⎞

⎜⎝⎛ −+=++=

LCLjRR

CjLjZ 2

111ω

ωω

ω

Z

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Series LCR Impedance

For example, the voltage across the capacitor:

SCCc VRCj

LjCj

ZIV1

11−

⎟⎟⎠

⎞⎜⎜⎝

⎛++==

ωω

ω

Z



Low frequency behavior

At low frequencies, the characteristic of this circuit is dominated by the capacitor

The inductor looks like a shortat low frequencies

The ω in the denominator of theterm for the capacitor makes it the major contribution

RCj

LjZ ++=ω

ω 1

8



High frequency behavior

At high frequencies, the characteristics of this circuit are dominated by the inductor

The capacitor looks like a short at low frequencies

The ω proportionality of theterm for the inductor makes it the major contribution

RCj

LjZ ++=ω

ω 1



Near resonanceNear resonance, energy will oscillate between the capacitor and the inductor

Notice that the terms for thecapacitor and the inductor haveopposite sign, so they can add upto zero impedance at one frequency At that frequency (ω0=[LC] -1/2) , energy is perfectlyoscillating between the inductor and the capacitor,→The only impedance left at that frequency is the resistor.

RCj

LjZ ++=ω

ω 1

9



At resonance

LC12 =ω

01]Im[ =⎟⎟⎠

⎞⎜⎜⎝

⎛+=

CjLjZ

ωω



Power sidebar

The power into a circuit is just the voltage times the current, but remember this is not linear, so we’ll go back to complete notation:

ZIV

eIeIti

eVeVtv

tjtj

tjtj

=

+=

+=

−

−

ˆˆ

)*ˆˆ(21)(

)*ˆˆ(21)(

ωω

ωω

10



Convert from phasors, then multiply

Power:

( )( ) ( ))*ˆ*ˆ()ˆˆ(

41)ˆ*ˆ(*)ˆˆ(

41

)*ˆ*ˆ()ˆ*ˆ(*)ˆˆ()ˆˆ(41

)*ˆ*ˆ()ˆ*ˆ(

)*ˆˆ()ˆˆ(41

)*ˆˆ)(**ˆˆ(41

)*ˆˆ)(*ˆˆ(41)()(

22

22

tjtj

tjtj

tjtjtjtj

tjtjtjtj

tjtjtjtj

tjtjtjtj

eIVeIVIVIVP

eIVIVIVeIVP

eIeVeIeV

eIeVeIeVP

eIeVeZIeVP

eIeIeVeVtitvP

ωω

ωω

ωωωω

ωωωω

ωωωω

ωωωω

−

−

−−−

−

−−

−−

+++=

+++=

⎟⎟⎠

⎞⎜⎜⎝

⎛

++

+=

++=

++=⋅=



Expression for Power from ZPower:

Notice that the second and fourth terms are just the complex conjugate of the first and third, so we can write:

The second term averages to zero, so the first term gives us the average power

( ) ( )( ) ( ))*ˆ**ˆ()ˆˆ(

41)ˆ**ˆ(*)ˆˆ(

41

)*ˆ*ˆ()ˆˆ(41)ˆ*ˆ(*)ˆˆ(

41

22

22

tjtj

tjtj

eIZIeIZIIZIIZIP

eIVeIVIVIVP

ωω

ωω

−

−

+++=

+++=

{ } { }tjeIZIREIZIREP ω2ˆˆ21*ˆˆ

21

+=

11



Reactive PowerLet’s take the phase angle of the current to be zero, to make it simpler:

Now look at the second term:

The first term is unavoidable ripple in the power, since this is AC, but the imaginary part of Z is power which goes in at one part of the cycle, and then is returned at a later point. This is called reactive power.

( ){ }

( ))2sin()2(cos(21

)2sin()2cos(21

22

22

tZtZIP

tjtZREIP

ir ωω

ωω

−=

+=

{ }⎟⎠⎞

⎜⎝⎛ += tj

r ZeREZIP ω22

21

21



Resonance

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

−

VL

VC

VR

Vs

0ωω <

VL

VC

VRVs

0ωω =

VC

VL

VR

Vs

0ωω >

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Series Resonance Voltage Gain

At resonance, the voltage across the inductor, and across the capacitor, can be larger than the voltage of the voltage source:

Remember:

+VR

−

+ VL – + VC –

s

ssL

VjQ

LjRVLj

ZVLjIV

⋅=

=== 000

0 )(ωω

ωω

s

ssC

VjQ

LjRV

jL

ZV

CjIV

⋅−=

−=== 00

00 )(1 ωω

ωω

RZ

RCL

RCLC

RCRLQ 0

0

0 1111=====

ωω



Second Order Transfer FunctionSo we have:

To find the poles/zeros, let’s put the H in canonical form:

One zero at DC frequency can’t conduct DC due to capacitor

RCj

Lj

RVVjH

s ++==

ωω

ω 1)( 0+Vo

−

RCjLCCRj

VVjH

s ωωωω+−

== 20

1)(

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Poles of 2nd Order Transfer Function

Denominator is a quadratic polynomial:

LRjj

LC

LRj

RCjLCCRj

VVjH

s ωω

ω

ωωωω

++=

+−==

22

0

)(11)(

LRjj

LRj

jHωωω

ωω

++=

220 )(

)(LC12

0 ≡ω

Qjj

Qj

jH022

0

0

)()( ωωωω

ωω

ω++

=R

LQ 0ω≡



Finding the poles…

Let’s factor the denominator:

Poles are complex conjugate frequencies

Re

Im

0)( 20

02 =++ ωωωωQ

jj

22 −±−=−±−=Q

jQQQ 4

11242 0

020

200 ωωωωωω

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Resonance without Loss

The transfer function can be parameterized in terms of loss. First, take the lossless case, R=0:

When the circuit is lossless, the poles are at realfrequencies, so the transfer function blows up!At this resonance frequency, the circuit has zero imaginary impedance and thus zero total impedanceEven if we set the source equal to zero, the circuit can have a steady-state response

Re

Im

020

200

42ωωωωω j

QQQ

±=⎟⎟⎠

⎞⎜⎜⎝

⎛−±−=

∞→

2



Magnitude Response

How peaked the response is depends on Q

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 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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More “Notation”

Often a second-order transfer function is characterized by the “damping” factor as opposed to the “Quality” factor

0)( 0220 =++

Qjj ωωωω

0)(1 2 =++Q

jj ωτωτ 0

1ω

τ =

02)()(1 2 =++ ζωτωτ jj

ζ21

=Q



Second Order Circuit Bode Plot

Quadratic poles or zeros have the following form:

The roots can be parameterized in terms of the damping ratio:

012)()( 2 =++ ζωτωτ jj

22 )1(12)()(1 ωτωτωτζ jjj +=++⇒=

damping ratio

Two equal poles

1

)1)(1(12)()(12

212

−±−=

++=++⇒>

ζζωτ

ωτωτζωτωτζ

j

jjjj

Two real poles

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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°



Underdamped Case

For ζ <1, the poles are complex conjugates:

For ωτ << 1, this quadratic is negligible (0dB)For ωτ >> 1, we can simplify:

In the transition region ωτ ~ 1, things are tricky!

22

2

11

012)()(

ζζζζωτ

ζωτωτ

−±=−±−=

=++

jj

jj

ωτωτζωτωτ log40)(log2012)()(log20 22 =≈++ jjj

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Underdamped Mag Plot

ζ=1

ζ=0.01

ζ=0.1ζ=0.2ζ=0.4

ζ=0.6ζ=0.8



Underdamped Phase

The phase for the quadratic factor is given by:

For ωτ < 1, the phase shift is less than 90°For ωτ = 1, the phase shift is exactly 90°For ωτ > 1, the argument is negative so the phase shift is above 90° and approaches 180°Key point: argument shifts sign around resonance

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

=++∠ −2

12

)(12tan12)()(ωτωτζζωτωτ jj

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Phase Bode Plot

ζ=0.010.10.20.40.60.8

ζ=1



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)()( 22 ==++=++ ζζζωτωτ

20



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

==



Energy Dissipation in Tank

Energy dissipated per cycle:

The ratio of the energy stored to the energy dissipated is thus:

0

2 221

ωπ

⋅=⋅= RITPw MD

ππω

ωπ 22

12

21

21

0

0

2

2

QR

L

RI

LI

ww

M

M

D

S ==⋅

=

21



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=



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

lecture 5: resonance

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