Lecture 4: Resonance Prof. Niknejad › ~ee105 › fa03 › handouts › lectures › L… · Fall...
31
Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 4 Lecture 4: Resonance Prof. Niknejad
Transcript of Lecture 4: Resonance Prof. Niknejad › ~ee105 › fa03 › handouts › lectures › L… · Fall...
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4
Lect
ure
4: R
eson
ance
Prof
. Nik
neja
d
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Lect
ure
Out
line
Som
e co
mm
ents
/que
stio
ns a
bout
Bod
e pl
ots
Seco
nd o
rder
circ
uits
:–
Serie
s im
peda
nce
and
reso
nanc
e–
Vol
tage
tran
sfer
func
tion
(ban
dpas
sfilt
er)
–B
ode
plot
s for
seco
nd o
rder
circ
uits
Dep
artm
ent o
f EEC
SU
nive
rsity
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alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Goo
d Q
uest
ions
…
Why
doe
s the
Bod
e pl
ot o
f a si
mpl
e po
le o
r zer
o al
way
s hav
e a
slop
e of
20
dB/d
ecre
gard
less
of t
he
brea
k fr
eque
ncy?
You
’ve
been
slop
py w
ith si
gns,
wha
t’s th
e de
al?
Why
doe
s the
arc
tang
ent p
lot l
ook
so fu
nny?
Why
do
we
fact
or th
e tra
nsfe
r fun
ctio
n in
to te
rms
invo
lvin
g jω
?
Dep
artm
ent o
f EEC
SU
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alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Bod
e Pl
ot Q
uest
ion
#1
Not
e th
at th
e sl
ope
of a
pol
e or
zer
o is
inde
pend
ent
of th
e br
eak
poin
t:
On
a lo
g-lo
g sc
ale,
all
stra
ight
line
s hav
e th
e sa
me
slop
e …
the
slop
e ge
ts tr
ansl
ated
into
an
inte
rcep
t sh
ift!
ωτ
ωτ
ωτ
log
20lo
g20
log
201
log
20+
=≈
+j
On
a lo
g sc
ale,
this
term
has
a
fixed
slo
pe o
f 20
dB/d
ecad
eTh
is te
rm is
a c
onst
ant
Dep
artm
ent o
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SU
nive
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EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Bod
e Pl
ot Q
uest
ion
#2W
hy d
o yo
u so
met
imes
use
a p
ositi
ve si
gn a
nd
othe
r tim
es a
neg
ativ
e si
gn in
the
trans
fer f
unct
ion?
The
plus
sign
is ri
ght!
For
“pa
ssiv
e”ci
rcui
ts, t
he
pole
s are
all
in th
e LH
P (le
ft-ha
lf pl
ane)
. A
sim
ple
RC
circ
uit h
as:
Oth
erw
ise
the
circ
uit h
as a
neg
ativ
e re
sist
or!
We
can
synt
hesi
ze n
egat
ive
resi
stan
ce w
ith a
ctiv
e ci
rcui
ts…
)1(
)1)(
1()
1()
1)(1(
)(
)(
22
21
0pm
pp
znz
zK
jj
jj
jj
jG
Hωτ
ωτ
ωτ
ωτ
ωτ
ωτ
ωω
++
±+
++
=LL
Whi
ch o
ne is
righ
t?
0>
=RC
τ
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Bod
e Pl
ot Q
uest
ion
#3
I kno
w w
hat a
n ar
ctan
look
s lik
e an
d it
look
s no
thin
g lik
e w
hat y
ou sh
owed
us!
Line
ar S
cale
Log
Sca
le
Dep
artm
ent o
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alifo
rnia
, Ber
kele
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EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Log
Scal
es
Log
scal
es m
ove
forw
ard
non-
unifo
rmly
…
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
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EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Why
jω?
Whe
n w
e fa
ctor
our
tran
sfer
func
tions
, why
do
we
alw
ays l
ike
to p
ut th
ings
in te
rms o
f jω
, as o
ppos
ed
to sa
y ω
?R
ecal
l tha
t we
are
tryin
g to
find
the
resp
onse
of a
sy
stem
to a
n ex
pone
ntia
l with
imag
inar
y ar
gum
ent:
Rea
l sin
usoi
dal s
tead
y-st
ate
requ
ires t
he a
rgum
ent
to b
e im
agin
ary.
We
mus
t the
refo
re o
nly
cons
ider
th
e tra
nsfe
r fun
ctio
ns fo
r suc
h va
lues
…If
this
doe
sn’t
mak
e se
nse,
han
g in
ther
e!
LTI S
yste
mH
tj eω
)(
)(
φω
ω+t
j ej
H
Dep
artm
ent o
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SU
nive
rsity
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alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Seco
nd O
rder
Circ
uits
The
serie
s res
onan
t circ
uit i
s one
of t
he m
ost
impo
rtant
ele
men
tary
circ
uits
:
The
phys
ics d
escr
ibes
not
onl
y ph
ysic
al L
CR
ci
rcui
ts, b
ut a
lso
appr
oxim
ates
mec
hani
cal
reso
nanc
e (m
ass-
sprin
g, p
endu
lum
, mol
ecul
ar
reso
nanc
e, m
icro
wav
e ca
vitie
s, tra
nsm
issi
on li
nes,
build
ings
, brid
ges,
…)
Dep
artm
ent o
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SU
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rsity
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alifo
rnia
, Ber
kele
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EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Serie
s LC
R Im
peda
nce
With
pha
sora
naly
sis,
this
circ
uit i
s rea
dily
an
alyz
ed
Z
RC
jL
jZ
++
=ω
ω1
−
+=
++
=LC
Lj
RR
Cj
Lj
Z21
11
ωω
ωω
01
1]
Im[
2=
−
=LC
LZ
ωω
LC12=
ω
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Res
onan
ce
Res
onan
ce o
ccur
s whe
n th
e ci
rcui
t im
peda
nce
is
pure
ly re
al
Imag
inar
y co
mpo
nent
s of i
mpe
danc
e ca
ncel
out
For a
serie
s res
onan
t circ
uit,
the
curr
ent i
s m
axim
um a
t res
onan
ce
+ V R −
+ V
L –
+ V
C –
+ V s −V CV L
V RV s
0ω
ω>
V L
V C
V R V s 0ω
ω<
V L V C
V R V s
0ω
ω=
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Serie
s R
eson
ance
Vol
tage
Gai
n
Not
e th
at a
t res
onan
ce, t
he v
olta
ge a
cros
s the
in
duct
or a
nd c
apac
itor c
an b
e la
rger
than
the
volta
ge in
the
resi
stor
:
s
ss
L
VjQ
Lj
RVL
jZ
VL
jI
V
×=
==
=0
00
0)
(ω
ωω
ω
+ V R −
+ V
L –
+ V
C –
s
ss
C
VjQ
Lj
RVjL
ZV
Cj
IV
×−
=
−=
==
00
00
)(
1ω
ωω
ω
RZR
CLR
CLCR
CRL
Q0
0
01
11
1=
==
==
ωω
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Seco
nd O
rder
Tra
nsfe
r Fun
ctio
nSo
we
have
:
To fi
nd th
e po
les/
zero
s, le
t’s p
ut th
e H
in c
anon
ical
fo
rm:
One
zer
o at
DC
freq
uenc
y ca
n’t c
ondu
ct D
C d
ue
to c
apac
itor
RC
jL
j
RVV
jH
s+
+=
=
ωω
ω1
)(
0+ V o −
RCj
LCC
Rj
VVj
Hs
ωω
ωω
+−
==
20
1)
(
Dep
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ent o
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SU
nive
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rnia
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CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Pole
s of
2nd
Ord
er T
rans
fer F
unct
ion
Den
omin
ator
is a
qua
drat
ic p
olyn
omia
l:
LRj
jLC
LRj
RCj
LCC
Rj
VVj
Hs
ωωω
ωω
ωω
++
=+
−=
=2
20
)(
11
)(
LRj
jLR
jj
Hω
ωω
ωω
++
=2
2 0)
()
(LC1
2 0≡
ω
Qj
j
Qj
jH
02
2 0
0
)(
)(
ωω
ωω
ωω
ω+
+=
RLQ
0ω
≡
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Find
ing
the
pole
s…
Let’s
fact
or th
e de
nom
inat
or:
Pole
s are
com
plex
con
juga
te fr
eque
ncie
sTh
e Q
para
met
er is
cal
led
the
“qua
lity-
fact
or”
or Q
-fac
tor
This
is a
n im
porta
nt p
aram
eter
:R
e
Im
0)
(2 0
02
=+
+ω
ωω
ωQ
jj
22
−±
−=
−±
−=
Qj
Q41
12
42
00
2 0
2 00
ωω
ωω
ωω
∞→
→
0R
Q
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Res
onan
ce w
ithou
t Los
s
The
trans
fer f
unct
ion
can
be p
aram
eter
ized
in te
rms
of lo
ss.
Firs
t, ta
ke th
e lo
ssle
ss c
ase,
R=
0:
Whe
n th
e ci
rcui
t is l
ossl
ess,
the
pole
s are
at r
eal
freq
uenc
ies,
so th
e tra
nsfe
r fun
ctio
n bl
ows u
p!A
t thi
s res
onan
cefr
eque
ncy,
the
circ
uit h
as z
ero
imag
inar
y im
peda
nce
and
thus
zer
o to
tal i
mpe
danc
eEv
en if
we
set t
he so
urce
equ
al to
zer
o, th
e ci
rcui
t ca
n ha
ve a
stea
dy-s
tate
resp
onse
Re
Im
02 0
2 00
42
ωω
ωω
ωj
Q
±=
−±
−=
∞→
2
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Mag
nitu
de R
espo
nse
The
resp
onse
pea
kine
ssde
pend
s on
Q
QjQ
j
LRjLR
jj
H0
22 0
0
002
2 0
00
)(
ωω
ωω
ωω
ωωω
ωω
ωωω
ω+
−=
+−
=
1=
Q
10=
Q
100
=Q
0ω
1)
(0
02 0
2 0
2 0
0=
+−
=
Qj
Qj
jH
ωω
ωω
ω
ω
1)
(0=
ωjH
0)0(=
H
ω∆
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
How
Pea
ky is
it?
Let’s
find
the
poin
ts w
hen
the
trans
fer f
unct
ion
squa
red
has d
ropp
ed in
hal
f:
()
21)
(2
02
22 0
20
2=
+−
=
Q
Qj
Hω
ωω
ω
ωω
ω
21
1/1
)(
2
0
22 0
2=
+
−
=
Q
jH
ωω
ωω
ω
1/
2
0
22 0
=
−
Qω
ωω
ω
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Hal
f Pow
er F
requ
enci
es (B
andw
idth
)
We
have
the
follo
win
g:
1/
2
0
22 0
=
−
Qω
ωω
ω1
/0
22 0
±=
−Q
ωω
ωω
Four
sol
utio
ns!
02 0
02
=−ω
ωω
ωQ
m
ab
ba
>±
±=
+
±
±=
2 0
2
00
42
ωω
ωω
0000
<−
−>
+−
<−
+>
++
ba
ba
ba
ba
Take
pos
itive
freq
uenc
ies:
Q1
0
=∆ ωω
Q0ω
ωω
ω=
−=
∆−
+
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Mor
e “N
otat
ion”
Ofte
n a
seco
nd-o
rder
tran
sfer
func
tion
is
char
acte
rized
by
the
“dam
ping
”fa
ctor
as o
ppos
ed
to th
e “Q
ualit
y”fa
ctor
0)
(0
22 0
=+
+Q
jj
ωω
ωω
01 ωτ=
0)
(1
2=
++
Qj
jωτ
ωτ
02)
()
(1
2=
++
ζωτ
ωτ
jj
ζ21=
Q
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Seco
nd O
rder
Circ
uit B
ode
Plot
Qua
drat
ic p
oles
or z
eros
hav
e th
e fo
llow
ing
form
:
The
root
s can
be
para
met
eriz
ed in
term
s of t
he
dam
ping
ratio
:
01
2)(
)(
2=
++
ζωτ
ωτ
jj
dam
ping
ratio
22
)1(
12)
()
(1
ωτ
ωτ
ωτ
ζj
jj
+=
++
⇒=
Two
equa
l pol
es
1
)1)(
1(1
2)(
)(
12
21
2
−±
−=
++
=+
+⇒
>
ζζ
ωτ
ωτ
ωτ
ζωτ
ωτ
ζ j
jj
jj
Two
real
pol
es
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Bod
e Pl
ot: D
ampe
d C
ase
The
case
of ζ
>1 a
nd ζ
=1 is
a si
mpl
e ge
nera
lizat
ion
of si
mpl
e po
les (
zero
s).
In th
e ca
se th
at ζ
>1, t
he
pole
s (ze
ros)
are
at d
istin
ct fr
eque
ncie
s. Fo
r ζ=1
, th
e po
les a
re a
t the
sam
e re
al fr
eque
ncy:
22
)1(
12)
()
(1
ωτ
ωτ
ωτ
ζj
jj
+=
++
⇒=
22
1)
1(ωτ
ωτ
jj
+=
+
ωτ
ωτ
jj
+=
+1
log
401
log
202
Asy
mpt
otic
Slo
pe is
40
dB/d
ec
()
()
()
ωτ
ωτ
ωτ
ωτ
jj
jj
+∠
=+
∠+
+∠
=+
∠1
21
1)
1(2
Asy
mpt
otic
Pha
se S
hift
is 1
80°
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Und
erda
mpe
dC
ase
For ζ
<1, t
he p
oles
are
com
plex
con
juga
tes:
For ω
τ<<
1, th
is q
uadr
atic
is n
eglig
ible
(0dB
)Fo
r ωτ>
> 1,
we
can
sim
plify
:
In th
e tra
nsiti
on re
gion
ωτ~
1, t
hing
s are
tric
ky!
20
22
2
11
01
2)(
)(
ζζ
ζζ
ωτ
ζωτ
ωτ
−±
=−
±−
=
=+
+
jj
jj
ωτ
ωτ
ζωτ
ωτ
log
40)
(lo
g20
12)
()
(lo
g2
2=
≈+
+j
jj
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Und
erda
mpe
dM
agPl
ot
ζ=1
ζ=0.
01 ζ=0.
1ζ=
0.2
ζ=0.
4 ζ=0.
6ζ=
0.8
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Und
erda
mpe
dPh
ase
The
phas
e fo
r the
qua
drat
ic fa
ctor
is g
iven
by:
For ω
τ< 1
, the
pha
se sh
ift is
less
than
90°
For ω
τ= 1
, the
pha
se sh
ift is
exa
ctly
90°
For ω
τ> 1
, the
arg
umen
t is n
egat
ive
so th
e ph
ase
shift
is a
bove
90°
and
appr
oach
es 1
80°
Key
poi
nt: a
rgum
ent s
hifts
sign
aro
und
reso
nanc
e
()
−
=+
+∠
−2
12
)(
12
tan
12)
()
(ωτ
ωτζ
ζωτ
ωτ
jj
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Phas
e B
ode
Plot
ζ=0.
010.
10.
20.
40.
60.
8ζ=
1
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Bod
e Pl
ot G
uide
lines
In th
e tra
nsiti
on re
gion
, not
e th
at a
t the
bre
akpo
int:
From
this
you
can
est
imat
e th
e pe
akin
essi
n th
e m
agni
tude
re
spon
se
Exam
ple:
forζ=
0.1,
the
Bod
e m
agni
tude
plo
t pea
ks b
y 20
lo
g(5)
~14
dB
The
phas
e is
muc
h m
ore
diff
icul
t. N
ote
for ζ
=0, t
he p
hase
re
spon
se is
a st
ep fu
nctio
nFo
r ζ=1
, the
pha
se is
two
real
pol
es a
t a fi
xed
freq
uenc
yFo
r 0<ζ
<1, t
he p
lot s
houl
d go
som
ewhe
re in
bet
wee
n!Qj
jj
j1
21
2)(
)(
12)
()
(2
2=
=+
+=
++
ζζ
ζωτ
ωτ
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Ener
gy S
tora
ge in
“Ta
nk”
At r
eson
ance
, the
ene
rgy
stor
ed in
the
indu
ctor
and
ca
paci
tor a
ret
LIti
Lw
ML
02
22
cos
21))
((21
ω=
=
tC
It
CIC
di
CC
tv
Cw
MM
C
02
2 02
02
22 0
2
22
sin
21si
n21
)(
121
))(
(21
ωω
ωω
ττ
==
==
∫
LI
tC
tL
Iw
ww
MM
CL
s2
02
2 00
22
21)
sin
1co
s(
21=
+=
+=
ωω
ω
LI
WW
MS
L2
max
,21
==
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Ener
gy D
issi
patio
n in
Tan
k
Ener
gy d
issi
pate
d pe
r cyc
le:
The
ratio
of t
he e
nerg
y st
ored
to th
e en
ergy
di
ssip
ated
is th
us:
0
22
21ωπ⋅
=⋅
=R
IT
Pw
MD
ππ
ω
ωπ2
212
2121
0
0
2
2
QRL
RI
LI
ww
M
M
DS=
=⋅
=
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
Phys
ical
Inte
rpre
tatio
n of
Q-F
acto
rFo
r the
serie
s res
onan
t circ
uit w
e ha
ve re
late
d th
e Q
fact
or
to v
ery
fund
amen
tal p
rope
rties
of t
he ta
nk:
The
tank
qua
lity
fact
or re
late
s how
muc
h en
ergy
is st
ored
in
a ta
nk to
how
muc
h en
ergy
loss
is o
ccur
ring.
If Q
>>
1, th
en th
e ta
nk p
retty
muc
h ru
ns it
self
…ev
en if
yo
u tu
rn o
ff th
e so
urce
, the
tank
will
con
tinue
to o
scill
ate
for s
ever
al c
ycle
s (on
the
orde
r of Q
cyc
les)
Mec
hani
cal r
eson
ator
s can
be
fabr
icat
ed w
ith e
xtre
mel
y hi
gh Q
DS
wwQ
π2=
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
4P
rof.
A. N
ikne
jad
thin
-Film
Bul
k A
cous
tic R
eson
ator
(FB
AR
)R
F M
EMS
Agi
lent
Tech
nolo
gies
(IEE
EIS
SCC
200
1)Q
> 1
000
Res
onat
es a
t 1.9
GH
z
Can
use
it to
bui
ld lo
w p
ower
osc
illat
or
C0
Cx
Rx
L x
C1
C2
R0
Thin
Pie
zoel
ectri
c Fi
lm
Pad
