lecture 1: course overview and introduction to phasorsee105/fa03/handouts/lectures/lecture1.pdf ·...
TRANSCRIPT
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1
Lect
ure
1: C
ours
e O
verv
iew
and
In
trod
uctio
n to
Pha
sors
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
1P
rof.
A. N
ikne
jad
EEC
S 10
5: C
ours
e O
verv
iew
Phas
orsa
nd F
requ
ency
Dom
ain
(2 w
eeks
)In
tegr
ated
Pas
sive
s (R
, C, L
) (2
wee
ks)
MO
SFET
Phy
sics
/Mod
el (1
wee
k)PN
Junc
tion
/ BJT
Phy
sics
/Mod
el (1
.5 w
eeks
)Si
ngle
Sta
ge A
mpl
ifier
s (2
wee
ks)
Feed
back
and
Diff
Am
ps (1
wee
k)Fr
eq R
esp
of S
ingl
e St
age
Am
ps (1
wee
k)M
ultis
tage
Am
ps (2
.5 w
eeks
)Fr
eq R
esp
of M
ultis
tage
Am
ps (1
wee
k)
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
EEC
S 10
5 in
the
Gra
nd S
chem
e
Exam
ple:
Cel
l Pho
ne
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Tran
sist
ors
are
Bric
ks
Tran
sist
ors a
re th
e bu
ildin
g bl
ocks
(bric
ks) o
f the
m
oder
n el
ectro
nic
wor
ld:
Focu
s of c
ours
e:–
Und
erst
and
devi
ce p
hysi
cs–
Bui
ld a
nalo
g ci
rcui
ts–
Lear
n el
ectro
nic
prot
otyp
ing
and
mea
sure
men
t–
Lear
n si
mul
atio
ns to
ols s
uch
as S
PIC
E
Ana
log
“Am
p”
Dig
ital
Gat
e
MO
S C
ap
PN
Jun
ctio
nVar
iabl
eC
apac
itor
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
SPIC
E
* Ex
ampl
e ne
tlis
tQ1
1 2
0 n
pnmo
dR1
1 3
1k
Vdd
3 0
3v.t
ran
1u 1
00u
SPICE
SPICE
stim
ulus
resp
onse
netli
st
SPIC
E =
Sim
ulat
ion
Prog
ram
with
ICEm
phas
isIn
vent
ed a
t Ber
kele
y (r
elea
sed
in 1
972)
.DC
: Fi
nd th
e D
C o
pera
ting
poin
t of a
circ
uit
.TR
AN
: So
lve
the
trans
ient
resp
onse
of a
circ
uit (
solv
e a
syst
em o
f ge
nera
lly n
on-li
near
ord
inar
y di
ffer
entia
l equ
atio
ns v
ia a
dapt
ive
time-
step
solv
er)
.AC
: Fi
nd st
eady
-sta
te re
spon
se o
f circ
uit t
o a
sinu
soid
al e
xcita
tion
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
BSI
MTr
ansi
stor
s are
com
plic
ated
. A
ccur
ate
sim
requ
ires 2
D o
r 3D
num
eric
al si
m(T
CA
D) t
o so
lve
coup
led
PDEs
(qua
ntum
ef
fect
s, el
ectro
mag
netic
s, et
c)Th
is is
slow
…a
circ
uit w
ith o
ne tr
ansi
stor
will
take
hou
rs
to si
mul
atio
nH
ow d
o yo
u si
mul
ate
larg
e ci
rcui
ts (1
00s-
1000
s of
trans
isto
rs)?
Use
com
pact
mod
els.
In E
ECS
105
we
will
der
ive
the
so
calle
d “l
evel
1”
mod
el fo
r a M
OSF
ET.
The
BSI
M fa
mily
of m
odel
s are
the
indu
stry
stan
dard
m
odel
s for
circ
uit s
imul
atio
n of
adv
ance
d pr
oces
s tra
nsis
tors
.B
SIM
= B
erke
ley
Shor
t Cha
nnel
IGFE
T M
odel
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Ber
kele
y…
A g
reat
pla
ce to
stud
y ci
rcui
ts, d
evic
es, a
nd C
AD
!
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Rev
iew
of L
TI S
yste
ms
Sinc
e m
ost p
erio
dic
(non
-per
iodi
c) si
gnal
s can
be
deco
mpo
sed
into
a su
mm
atio
n (in
tegr
atio
n) o
f si
nuso
ids v
ia F
ourie
r Ser
ies (
Tran
sfor
m),
the
resp
onse
of a
LTI
syst
em to
virt
ually
any
inpu
t is
char
acte
rized
by
the
freq
uenc
y re
spon
se o
f the
sy
stem
:
Any
line
ar c
ircui
t W
ith L
,C,R
,Man
d de
p. s
ourc
es
Am
pS
cale
Pha
se S
hift
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Exam
ple:
Low
Pas
s Fi
lter (
LPF)
Inpu
t sig
nal:
W
e kn
ow th
at:
)co
s()
(t
Vt
vs
sω
=)
cos(
)(
0
φω
+⋅
=t
VK
tv
V
so
321
Am
p sh
ift
dtdvt
vt
v
dtdvRC
tv
tv
dtdvC
ti
Rti
tv
tv s
ss
00
00
0
0
)(
)(
)(
)()
(
)(
)(
)(
τ+
=
−==
−=
Pha
se s
hift
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
LPF
the
“har
d w
ay”
(con
t.)Pl
ug th
e kn
own
form
of t
he o
utpu
t int
o th
e eq
uatio
n an
d se
e if
it ca
n sa
tisfy
KV
L an
d K
CL
Sinc
e si
ne a
nd c
osin
e ar
e lin
early
inde
pend
ent f
unct
ions
:
)co
s(s
insi
n)
sin
(cos
cos
cos
sin
cos
cos
sin
)si
n(si
nsi
nco
sco
s)
cos(
)si
n()
cos(
cos
00
00
φω
τφ
ωφ
ωτ
φω
ω
φω
ωτ
φω
ω
+−
−=
+=
+−
=+
+−
+=
tV
tV
tV
yx
yx
yx
yx
yx
yx
tV
tV
tV ss
0co
ssi
n2
1=
+t
at
aω
ω
02
1≡
≡a
aIF
F
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
LPF:
Sol
ving
for r
espo
nse…
App
lyin
g lin
ear i
ndep
ende
nce
()
20
2/12
0
2000
1
00
00
)(
11)
(1
))
(1(
cos
)ta
n1(
cos
)si
n(c
ostan
tan
0si
nco
s0
cos
sin
ωτ
ωτ
ωτ
φ
φω
τφ
φω
τφ
ωτ
φ
ωτ
φφ
ωτ
φφ
ωτ
φ +=
=+
=+
=−
=−
−=
−=
=−
−=
−−
−
s
s
s
s
ss
VVV
V
VV
VV
VV
VV
VV
V
Phas
e R
espo
nse:
Am
plitu
de R
espo
nse:
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
LPF
Mag
nitu
de R
espo
nse
24
68
10
0.2
0.4
0.6
0.81
τω
1=
τω
/10
=
1.0≈
1≈
707
.0≈
20
)(
11ω
τ+
=sVV
Pass
band
of fi
lter
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
LPF
Phas
e R
espo
nse
24
68
10
-80
-60
-40
-200
o45
−
ωτ
10
tan−
−=
sVVp
o90−
≈ τω
/10
=
τω
1=
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
dB:
Hon
or th
e in
vent
or o
f the
pho
ne…
The
LPF
resp
onse
qui
ckly
dec
ays t
o ze
roW
e ca
n ex
pand
rang
e by
taki
ng th
e lo
g of
the
mag
nitu
de re
spon
sedB
= d
eciB
el(d
eci=
10)
0.1
110
100
-40
-30
-20
-100
=
sdB
sVV
VV0
0lo
g20
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Why
20?
Pow
er!
Why
mul
tiply
log
by “
20”
rath
er th
an “
10”?
Pow
er is
pro
porti
onal
to v
olta
ge sq
uare
d:
At b
reak
poin
t:
Obs
erve
: sl
ope
of si
gnal
atte
nuat
ion
is 2
0 dB
/dec
ade
in fr
eque
ncy
=
=
ss
VVVV
dB0
2
0lo
g20
log
10
dB3/1
dB
0−
=
→
=sVV
τω
dB60
/10
00
dB40
/10
0
dB
0
dB
0
−=
→=
−=
→=
s
s VVVV
τω
τω
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Why
intr
oduc
e co
mpl
ex n
umbe
rs?
They
act
ually
mak
e th
ings
eas
ier
One
insi
ghtfu
l der
ivat
ion
of
Con
side
r a se
cond
ord
er h
omog
eneo
us D
E
Sinc
e si
ne a
nd c
osin
e ar
e lin
early
inde
pend
ent,
any
solu
tion
is a
line
ar c
ombi
natio
n of
the
“fun
dam
enta
l”so
lutio
ns
ix e
=
=+
xxy
yy
cos
sin0
''
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Insi
ght i
nto
Com
plex
Exp
onen
tial
But
not
e th
at
is
als
o a
solu
tion!
That
mea
ns:
To fi
nd th
e co
nsta
nts o
f pro
p, ta
ke d
eriv
ativ
e of
this
eq
uatio
n:
Now
solv
e fo
r the
con
stan
ts u
sing
bot
h eq
uatio
ns:
ix ex
ax
aeix
cos
sin
21
+=
xa
xa
eiix
cos
sin
12
+−
=
01
det
sin
cos
cos
sin
21
21
≠−
==
=
−
Ab
aaA
eieaa
xx
xx
ixix
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
The
Rot
atin
g C
ompl
ex E
xpon
entia
l
So th
e co
mpl
ex e
xpon
entia
l is n
othi
ng b
ut a
poi
nt
traci
ng o
ut a
uni
t circ
le o
n th
e co
mpl
ex p
lane
:
xi
xeix
sin
cos
+=
2
ti
ti
ee
ωω
−+
ti eω
ti
eω
−
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Mag
ic:
Turn
Diff
Eq
into
Alg
ebra
ic E
q
Inte
grat
ion
and
diff
eren
tiatio
n ar
e tri
vial
with
co
mpl
ex n
umbe
rs:
Any
OD
E is
now
triv
ial a
lgeb
raic
man
ipul
atio
ns …
in fa
ct, w
e’ll
show
that
you
don
’t ev
en n
eed
to
dire
ctly
der
ive
the
OD
E by
usi
ng p
haso
rsTh
e ke
y is
to o
bser
ve th
at th
e cu
rren
t/vol
tage
re
latio
n fo
r any
ele
men
t can
be
deriv
ed fo
r com
plex
ex
pone
ntia
l exc
itatio
n
ti
ti
ei
edtd
ωω
ω=
ti
ie
id
eω
ωτ
ωτ
1=
∫
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Com
plex
Exp
onen
tial i
s Po
wer
ful
To fi
nd st
eady
stat
e re
spon
se w
e ca
n ex
cite
the
syst
em w
ith
a co
mpl
ex e
xpon
entia
l
At a
ny fr
eque
ncy,
the
syst
em re
spon
se is
cha
ract
eriz
ed b
y a
sing
le c
ompl
ex n
umbe
r H:
This
is n
ot su
rpris
ing
sinc
e a
sinu
soid
is a
sum
of c
ompl
ex
expo
nent
ials
(and
bec
ause
of l
inea
rity!
)
From
this
per
spec
tive,
the
com
plex
exp
onen
tial i
s eve
n m
ore
fund
amen
talLT
I Sys
tem
Ht
i eω)
()
(φ
ωω
+ti e
H
iee
tt
it
i
2si
nω
ω
ω−
−=
2co
st
it
ie
et
ωω
ω−
+=
)(ω
H)
(ωφ
Hp=
Mag
Res
pons
e
Pha
se R
espo
nse
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
LPF
Exam
ple:
The
“so
ft w
ay”
Let’s
exc
ite th
e sy
stem
with
a c
ompl
ex e
xp:
tj
tj
o
tj
sss
eV
eV
tv
eV
tv
dtdvt
vt
v
ωφ
ω
ω
τ
0)
(0
00
)()(
)(
)(
===
+=
+
use
j to
avoi
d co
nfus
ion
com
plex
real
tj
tj
tj
se
Vj
eV
eV
ωω
ωω
τ0
0⋅
⋅+
=
() τ
ω⋅
+=
jV
V s1
0 () τ
ω⋅
+=
jVV s
11
0Ea
sy!!!
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Mag
nitu
de a
nd P
hase
Res
pons
e
The
syst
em is
cha
ract
eriz
ed b
y th
e co
mpl
ex
func
tion
The
mag
nitu
de a
nd p
hase
resp
onse
mat
ch o
ur
prev
ious
cal
cula
tion:
() τ
ωω
⋅+
==
jVV
Hs
11
)(
0
ωτ
ω
ωτ
ω
1
20
tan
)(
)(
11
)(
−−
=
+=
=
H
VVH
s
p
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Why
did
it w
ork?
The
syst
em is
line
ar:
If w
e ex
cite
syst
em w
ith a
sinu
soid
:
If w
e pu
sh th
e co
mpl
ex e
xp th
roug
h th
e sy
stem
firs
t an
d ta
ke th
e re
al p
art o
f the
out
put,
then
that
’s th
e “r
eal”
sinu
soid
al re
spon
se
)](
Re[
])(R
e[]
Re[
xx
yL
L=
=
]R
e[co
s)
(t
js
ss
eV
tV
tv
ωω
==
]R
e[)
cos(
)(
)(
φω
φω
+=
+=
tj
oo
oe
Vt
Vt
v
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
And
yet
ano
ther
per
spec
tive…
Aga
in, t
he sy
stem
is li
near
:
To fi
nd th
e re
spon
se to
a si
nuso
id, w
e ca
n fin
d th
e re
spon
se to
and
an
dsu
m th
e re
sults
:
)(
)(
)(
21
21
xx
xx
yL
LL
+=
+=
ti
eω
−t
i eω
LTI S
yste
mH
ti eω
)(
1)
(φ
ωω
+ti e
H
LTI S
yste
mH
ti
eω
−)
(2
)(
φω
ω+
−−
ti e
H
LTI S
yste
mH
2
ti
ti
ee
ωω
−+
2)
()
(t
it
ie
He
Hω
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
1P
rof.
A. N
ikne
jad
Ano
ther
per
sepc
tive
(con
t.)
Sinc
e th
e in
put i
s rea
l, th
e ou
tput
has
to b
e re
al:
That
mea
ns th
e se
cond
term
is th
e co
njug
ate
of th
e fir
st:
Ther
efor
e th
e ou
tput
is:
2)
()
()
(t
it
ie
He
Ht
yω
ωω
ω−
−+
=
func
tion)
odd
()
()
(fu
nctio
n)ev
en(
)(
)(
φω
ωω
ω−
=−
=−
=−
HH
HH
pp
()
)co
s()
(2
)(
)(
)(
)(
φω
ω
ωφ
ωφ
ω
+=
+=
+−
+
tH
ee
Ht
yt
it
i
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
“Pro
of”
for L
inea
r Sys
tem
sFo
r an
arbi
trary
line
ar c
ircui
t (L,
C,R
,M, a
nd
depe
nden
t sou
rces
), de
com
pose
it in
to li
near
sub-
oper
ator
s, lik
e m
ultip
licat
ion
by c
onst
ants
, tim
e de
rivat
ives
, or i
nteg
rals
:
For a
com
plex
exp
onen
tial i
nput
x th
is si
mpl
ifies
to
:
∫∫∫∫∫
∫+
++
++
++
==
LL
xx
xx
dtdb
xdtd
bax
xy
22
21
)(L
LL
++
++
++
==
∫∫∫
tj
tj
tj
tj
tj
tj
ec
ec
edtd
be
dtdb
aee
ωω
ωω
ωω
21
22
21
)(L
LL
++
++
++
=2
21
22
1)
()
(ω
ωω
ωω
ωω
ωω
jec
jec
ej
be
jb
aey
tj
tj
tj
tj
tj
++
++
++
==
LL
22
12
21
)(
)(
ωω
ωω
ω
jcjc
jb
jb
ae
Hx
yt
j
y
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
“Pro
of”
(con
t.)
Not
ice
that
the
outp
ut is
als
o a
com
plex
exp
tim
es a
co
mpl
ex n
umbe
r:
The
ampl
itude
of t
he o
utpu
t is t
he m
agni
tude
of t
he
com
plex
num
ber a
nd th
e ph
ase
of th
e ou
tput
is th
e ph
ase
of th
e co
mpl
ex n
umbe
r
++
++
++
==
LL
22
12
21
)(
)(
ωω
ωω
ω
jcjc
jb
jb
ae
Hx
yt
j
))(
cos(
)(
]R
e[
)(
)(
)(
)(
22
12
21
ωω
ω
ω
ωω
ωω
ωω
ω
Ht
Hy
eH
ey
jcjc
jb
jb
ae
Hx
y
Hj
tj
tj
p
LL
p
+=
=
++
++
++
==
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Phas
ors
With
our
new
con
fiden
ce in
com
plex
num
bers
, we
go fu
ll st
eam
ahe
ad a
nd w
ork
dire
ctly
with
them
…w
e ca
n ev
en d
rop
the
time
fact
or
si
nce
it w
ill
canc
el o
ut o
f the
equ
atio
ns.
Exci
te sy
stem
with
a p
haso
r:R
espo
nse
will
als
o be
pha
sor:
For t
hose
with
a L
inea
r Sys
tem
bac
kgro
und,
we’
re
goin
g to
wor
k in
the
freq
uenc
y do
mai
n–
This
is th
e La
plac
edo
mai
n w
ith
ti eω
11
1~φj e
VV
=2
22~
φj eV
V= ωj
s=
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Cap
acito
r I-V
Pha
sorR
elat
ion
Find
the
Phas
orre
latio
n fo
r cur
rent
and
vol
tage
in a
ca
p:
dtt
dvC
ti
Cc
)(
)(
=)
(ti c)
(tv C
+ _
cc
tj
ct
jc
tj
ct
jc
tj
ct
jc
VC
jI
eC
Vj
eI
eC
Vj
edtd
CV
eV
dtdC
eI
ωω
ω
ωω
ωω
ωω
==
=
=
]R
e[]
Re[
]R
e[]
Re[
]R
e[)
(
]R
e[)
(t
jc
c
tj
cc
eV
tv
eI
ti
ωω
==
Dep
artm
ent o
f EEC
SU
nive
rsity
of C
alifo
rnia
, Ber
kele
y
EE
CS
105
Fall
2003
, Lec
ture
1P
rof.
A. N
ikne
jad
Indu
ctor
I-V
Phas
orR
elat
ion
Find
the
Phas
orre
latio
n fo
r cur
rent
and
vol
tage
in
an in
duct
or:
)(ti
)(tv
+ _
]R
e[)
(]
Re[
)(
tjt
j
Vet
vIe
tiωω
==
dttdi
Lt
v)
()
(=
IL
jV
LIe
jVe
LIe
je
dtdLI
Iedtd
LVe
tj
tj
tj
tj
tj
tj ωω
ω
ωω
ωω
ωω
==
=
=
]R
e[]
Re[
]R
e[]
Re[