elen665 passiverlc i lect3
TRANSCRIPT
1
Passive RLC Networks*
References: - CMOS RF IC Design, Thomas Lee- Solid State Radio Engineering, Krau et al Wiley, 1980- Microwave Transistor Amplifier, Guillermo Gonzalez, Prentice Hall, 1997
RF circuits have a relatively large ratio of passive to active components.
Passive circuits are important for matching (which is important for efficient
power transfer). They can be used for boosting amplifier gain at high
frequencies. Furthermore, they are also used to filter unwanted signals.
* A significant part of these notes were provided by Dr. Sherif H. K. Embabi
2
1. Series RLC Network
( ) jXRZjLjRjZc
o +=θ∠=ω
−ω+=ω1
( ) RjZLC
o
o
=ω
ω==ω frequency) (resonance rad/sec 1
at
Qf
B
CRRL
Q
Q
o
o
o
*
=
ω=
ω=
π=
by given is B width,bandpower -half The
1cycleper dissipatedenergy
circuit in the storeengergy ousinstantanemax 2
51591
911520
==
µ=Ω=
QpFC
H.LR
3
2. Parallel RLC Tank
( )
At
frequency) (resonance 1
here w
1
1
GY
sec/radLC
LR
CRG
CQ
jQG
LjCjGjY
o
o
oo
o
o
o
=
ω=ω
=ω
ω=ω=
ω=
ω
ω−
ωω
+=
ω+ω+=ω
We also know that the half power bandwidth, B, is given by:
HzRCQ
fB o
π==
21
4
Note that at ωο
CL
LZZ oLc =ω==
This is know as the characteristicImpedance of the network.
Note that at ωο
QI
CRIZV
II
in
oinc
Lc
=
ω===
Which could be very high!
QRL
RQCKR
o
o
ω=
ω=Ω=1
Z
5
3. Practical Parallel LC Network
Parallel RLC rarely exist in real life.A more realistic network is the LC tank.
Real inductors are lossy (even more that capacitors). That is why a series resistor is added to the inductor.
( )( ) ( )
( ) ( )
( )1 can write we
define weIf
or
rad/sec 1
2
2
2
2222
+==
ω=
==ω==ω
−=ω
ω+ω
−ω+ω+
=ω
tss
st
s
sot
s
sto
s
sto
s
s
so
ss
s
ss
QRCR
LR
RL
Q
CRL
RjZL
CRGjY
LR
CL
LRL
CLR
RjY
This relation implies that a small resistance (Rs) can be transformed toa larger value (Rt).
6
The practical parallel LC circuit can be converted to a parallel RLC network.
( )
po
pp
tsp
s
sot
t
tsp
L
RCC
QRR
RL
QLL
ω==
+=
ω=
+=
toequal also is and 1
1
2
2
2
( )
+=
+=ω
=
1
1 1
2
2
2
QQCC
QRRRC
Q
sp
spO
Note: These conversion expressions are only valid close to ωo
Similarly:
7
Xs Rs
or
Xs Rs
Rs = Component Losses
Rp
Xp
or
Rp
Xp
A series-to-parallel transformation
Q = Qs which equals Qp which equals the Q of the component.
and, p
pp Q
RX =
If the Q of the component is greater than 10, then,and,
sp
sp
XX
RQR
≈
≈ 2
ppp
p
s
ss
s
ss
p
pp
QQR
X
RX
Q
RX
QX
RQ
===
===
2
8
4. The Maximum Power Transfer Theorem
For a given fixed source impedance Zs, what Load impedance ZL maximizes the power delivered to the load?
~VS
sss XjRZ += LLL XjRZ +=
The power delivered to Load impedance is only due to RL (reactive elements do not dissipate power). Hence,
( ) ( )22
22
deliveredsLsL
sL
L
R
XXRR
VRR
VP
+++==
VS and VL are r.m.s. voltage across RS and RL , respectively.Max power can be achieved if XL = -XS and RL = RS which implies that ZLand ZS are complex conjugates.
9
5. Impedance TransformationImpedance transformers are critical for RF circuit design to achieve max. power transfer.
(a) L-MatchingRS
LS
CRp upwardtransformer
RSC Lp
is equivalentto
( )
CL
ZCL
RRQ
CL
RQRR
QCL
LQ
QLL
RL
Q
QRQRRLR
CL
so
os
sp
s
ssp
o
so
ssps
so
ssps
s
so
1
1
1
1
1
1 1
2
22
22
2
2
222
=ω
=≅>>
=≅ω
−=ω∴
≅
+=
ω=
≅+=−=ω
ed. transformupwardbeen has that implies which , and 2sspps RRQR,RR =→
(characteristicimpedance Zo)
10
Example:Find a matching network that match a 50Ω load to a 1000Ω source resistance. (fo=1.5MHz)
Ω==
=×=
µ=⇒
×π=ω
ω=Ω=⇒Ω=⇒
=−=⇒
+=
==
229X 462.6pFC
1050
123
10512
21850
358941
1
20 ratioer transform
c
3
6
2
CL
RRrecall
H.L
.*If
LXXQ
.R
RQ
Q
R
R
ssp
s
o
sos
s
s
p
s
p
Ω=
+=++−
=−
−=
50
100010001000
10001000
Z
or
22
2
LEFT
LEFT
LEFTLEFTc
cc
c
c
R
jXRX
XjX*jXjX*
So by making RLEFT=50Ω and solving for Xc we can find C.
11
Another alternative matching circuit:
Notice that matching is only valid around ωο
The figure shows the fractional power reaching the Load vs. frequency
x−ω
2
12
Downward Transformer
Rp
L
CpRsRS
Cs
L
is equivalent to
( )p
ps
pps
CQ
QCC
Q
R
Q
RR
≅+
=
≅+
=
2
2
22
1
1
L-match can only specify two of the following:
( )
=−
−
ω−
22 1
or ratiotion transformaQ
Q
Qo
So if Q is selected the ratio is determined.
Since the transformation ratio is usually not vary large, Q must be relatively small.
13
(b) π-Matching
First the resistor Rp is downward transformed to RI , which is then upward transformed to Rin by the second L-Match
This way Q is decoupled from the transformation ratio.
It can be shown that
( )11
111
21 −+−=+ω
=R
R
RR
RLL
Q pino
For a given Q, Rin and Rp, find RI.Then get
po
ino
o`
RQ
C
RQ
C
QRLL
ω=
ω=
ω=+
12
11
121
14
How to Construct π-Matching Network
Based on the required (QTOT = Q1+Q2) find RI. From the Appendix
+
−
+−+
+
+=
2
224
1114 L
s
Ls
TOTTOT
LsI
RR
RR
RRR
RI RI
Rs
Vs
1pX
1sX 2sX
2pX RL
RI RI
RI
Vs
1sX 2sX
RI
'sX1
'sX2
11
1
22
11
−==
−==
I
L
LLp
I
s
sp
RRR
QR
X
RRQ
RX
IsIs RQX,RQX 2211 ==
21 ssTOT,s XXX +=
1
1
1
2
22
2
−=
+=
+=
I
L
LI
p
psc
RR
Q
QR
R
Q
RR
15
Appendix11 −+−=
I
L
I
sTOT R
RRR
Q
Find the expression for RI as a function of Q, Rs, and RL
122 22 +−+−
+=
I
Ls
I
Ls
I
Ls
RRR
RRR
RRR
Q
( )( ) ( )( )( ) ( ) ( )( ) ( )
( ) ( )( ) ( )( ) ( )( ) ( ) ( ) 024
044224
444222
42
22222
22224
222222
222
=−++−+
=−+++−++−+
++−=++−+++
++−=+−+
LsILsI
LsLsILsLsI
IILsLsLsILsI
IILsLsLsI
RRRRRQRQQ
RRRRRRRRRQRQQ
RRRRRRRRQRRRQR
RRRRRRRRQR
( ) ( ) ( )( )
( ) ( )
+−
+−+
++
=
−
+−+±+
+=
−+−++
±+
+=
2
22
22
22
2224222
4111
4
41
41
44442
14
Ls
LsLs
LsLsLs
LsLsLs
I
RRRR
QQRR
RRQ
RRRRQ
RRQQRRQQQQ
RRR
16
Example: Design π-matching network to match a 50Ω Load to a 1000Ω Rs,With a QTOT = 12.
X=99.5 X=20
~
Z=10+j0 Z=10+j0
1000 X=100.5 X=-25 5050Ω
~
X=119.5
1000Ω
Ω=
+−
+−+
++
= 104
1114
212
22Ls
Ls
TOTTOT
LsI RR
RRQQ
RRR
For Load Side
Ω==
Ω=⋅==−=
=
25
20215
5
22
22
2
QR
X
RQXQ
RR
Lp
Is
I
L
Xs2
RI=10Ω Xp2 50=RL
For the source side
Ω==
Ω=⋅=
=
=
5100
599
959
100
11
11
1
.QRX
.RQX
.QRR
sp
Is
I
s
17
The total Q for the π-matched is 11.95 (?)
Note the improvement in the selectivity.
18
(c) T-Matching
( )1
11
1211
−+−=+= L
ino R
R
RR
CCRQDetermine R1Then calculate
o
L
o
in
o
R`QL
R`QL
RQ
CC
ω=
ω=
ω=+
2
1
121 T-Matching is suitable when the source and
termination parasitics are inductive, since they can be absorbed by network.
Similarly for the π-Matching it is more suitable for capacitive parasitics.
19
TABLE 1 Parallel-Series Conversion Formulas for RC Networks
Cp Rp
Rs
Cs
Define: Define:
p
pp
pp
X
RQ
CX
=
ω=
1
s
ss
ss
RX
Q
CX
=
ω=
1
Parallel Equivalent of The Series Network
Series Equivalent ofThe Parallel Network
20
EXACT FORMULAS
APPROXIMATE FORMULAS
( )
+=
+=
+=
+=
+=+=
2
2
2
2
2
2
2
2
22
1
1
1
1
1 1
p
ppse
s
sspe
p
ppse
s
sspe
p
psesspe
Q
QCC
CC
Q
QXX
XX
Q
RRQRR
pspe
psespe
p
psesspe
ps
CCC
XXXX
Q
RRQRR
≈≈
≈≈
≈≈
≥≥
se
22
C
10 If 10 If
21
Cp Rp
Rs
Cs
TABLE 2 Parallel-Series Conversion Formulas for RL Networks
Define: Define:
p
pp
pp
XR
Q
LX
=
ω=
s
ss
ss
RX
Q
LX
=
ω=
Parallel Equivalent of The Series Network
Series Equivalent ofThe Parallel Network
22
EXACT FORMULAS
APPROXIMATE FORMULAS
( )
+=
+=
+=
+=
+=+=
1
1
1
1
1 1
2
2
2
2
2
2
2
2
22
p
ppse
s
sspe
p
ppse
s
sspe
p
psesspe
Q
QLL
LL
Q
QXX
XX
Q
RRQRR
pspe
psespe
p
psesspe
ps
LLL
XXXX
Q
RRQRR
≈≈
≈≈
≈≈
≥≥
se
22
L
10 If 10 If
23
TABLE 3 Design Formulas for the Resonant RL||C Circuit
Rt C
L
R
Quantity Exact Expression Units Approximate Expression101 ≥Q
B
R
L
Q
o
t
o
1
ω
ω
( )1
11
1
2
2
2
1
2
2
+=
ω==
+
ω=
ω=ω
=
−=
t
o
t
t
t
o
oo
QR
CQ
CRL
C
CRR
L
LR
LC rad/s
ohms
ohms
hertz
LC1
≈
CRoω≈
1
Coω≈
1
12 LQRQ ot ω=≈
t
o
t Qf
LR
CR=
π=
π≈
221
24
TABLE 4 Design Formulas for the Resonant RL||C Circuit
Quantity Exact Expression Units Approximate Expression101 fQ
B
R
L
Q
o
t
o
1
ω
ω
( )1
11
1
1
2
2
2
21
22
+=
ω==
+ω
=
ω=
ω=
−=
t
to
t
t
o
o
t
o
QR
LQCRL
C
LR
CR
CRLCrad/s
ohms
ohms
hertz
LC1
≈
RLoω
≈
Coω≈
1
CQ
RQo
tt ω
=≈ 2
tt
o
CRQf
π=≈
21
Rt
CL
R
25
(c) Tapped Resonant Circuits(i) Tapped Capacitor Circuit
It is usually designed for specified values for R2, R1, resonance frequency fo and bandwidth B, L, C1, and C2 are to be determined.
Using parallel-to-series conversion, R2 and C2 can be converted to the series Rse and Cse as shown.
1
1
CCC*C
Cse
se
+=
26
The following relations can be derived
from Table (4): Bf
Q ot ≅
and12 +
=t
tse
QR
R
from Table (1): 22RCQ op ω= (A)
and12
2
+=
pse Q
RR
Hence;
( )1
1
11
22
22
2
−+=∴
+=
+
ttp
pt
t
RRQQ
QR
QR
for
for
1
10
2
2−=
≥
NQ
Q
Q
tp
t
NQ
Q
Q
tp
t
≅
> 10
where2
2
RR
N t=
Transformer ratio
(B)
27
Design Procedure for the Tapped-Capacitor Circuit
Given Rt , R2 , and B
1-
t
ot
BRC
Bf
Q
π≅
≅
21
proof
from Table (4)
2
2
21
RR
N
CL
t
o
=
ω≅
from Table (4)
tto
o
to
t
o
t
ott
sett
BRRBf
RQ
C
CQ
CQ*Q
RQR
π=
ω≅
ω=∴
ω=
ω=
=
21
1 2
2
28
2- If
21
2
21
10
−≅
≅<
NQ
Q
NQ
Q.e.iQ
tp
tpp
from (A)
( )
CCCC
C
Q
QCC
R
QC
se
se
p
pse
o
p
−=
+=
ω=
1
2
22
22
1
If
>> 1010
NQ
.e.iQ tp
proof
NCCs ≅2
2 R
QC
o
p
ω=
from Table (1) from (B)
1 2
1
2
−=
−=
=
NC
CCCC
C
CC
se
se
se
NC
NRCR
RNQ
C
t
o
t
=
=
ω=
2
22
29
(ii) Tapped-Inductor Circuit
Rt C
L1
L2 R2
A similar design procedure can be derived.
See Table (6) for design equations.
30
Rt
C2
C1
L
R2
TABLE 5 Design Formulas for Tapped-Capacitor Circuit
( )
( )
( )
( ) ( )
( )
column. hand-right in the fomulas thefollow 10N If column. hand-left in the formulas thefollow
and for value thisuse , 10N If 5
4
3
12
211
10For
212
2
,Q
QQ.QNQ
RRN
BfQ
CL
BRC
,BfQ
t
ptpt
t
ot
o
t
ot
<
≥≈
=
≈
ω≈
π≈
≥≈
31
Approximate Formulas10≥pQ Formulas for 10<pQ
( )
( )
( )1
8
7
6
21
2
−=
=
=
NC
C
NCC
NQ
Q tp ( )
( )
( )( )
( )CC
CCC
Q
QCC
R
QC
NQ
Q
se
se
p
pse
o
p
tp
−=
+=
ω=
+=
1
2
22
22
21
2
2
9
18
7
16
32
Rt C
L1
L2 R2
TABLE 6 Design Formulas for Tapped-Circuit
( )
( )
( )
( ) ( )
( )
column. hand-right in the fomulas thefollow 10N If column. hand-left in the formulas thefollow
and for value thisuse , 10N If 5
4
3
12
211
10For
212
2
,Q
QQ.QNQ
RRN
BfQ
CL
BRC
,BfQ
t
ptpt
t
ot
o
t
ot
<
≥≈
=
≈
ω≈
π≈
≥≈
33
Approximate Formulas10≥pQ Formulas for 10<pQ
( )
( )
( ) ( ) 221
2
18
7
6
L:LLNL
NL
C
NQ
Q tp
−=−=
=
= ( )
( )
( )
( ) se
p
pse
po
tp
LLL
Q
QLL
QR
L
NQQ
−=
+=
ω=
−+=
1
2
22
22
21
2
2
9
18
7
116
34
d. TransformersLow frequency equivalent circuit of a perfectly coupled loss less transformer.
22
121222
11
221111
IR
ILLkjILjV
IR
ILLkjILjV
+
ω+ω=
+
ω+ω=
Where R1 and R2 are the resistance of the windings and can be neglected.
L1 is the inductance of winding 1 when I2 = 0 andL2 is the inductance of winding 2 when I1 = 0
The ratio of there self-inductance L1/L2 is equal to the square of the turns ration (N1/N2)2
2
2
1
2
1
=
NN
LL
The “coupling factor” k, and the “mutual inductance” M is defined as:
21LLkM =
I2I1
V1
+
-
L1 L2
V2
+
-
35
For perfect (maximum) coupling k=1, and zero loss (R1=R2=0), V1 and V2 can bewritten as:
121222
221111
and
ILLjILjV
ILLjILjV
ω+ω=
ω+ω=
The ratio of
12
12
12
12
12
22111
12122
1
2
1 II*L
LL
LI
I*LL
ILLILILLIL
VV
+
+=
++
=
recall that2
1
2
1
2
2
1
1
2 and
==
NN
LL
NN
II
This can be modeled by an ideal transformer with a shunt inductor (“magnetizing inductance”)
expected as
12
21
11
1
2
1
2
1
2
LLI
LjV
I
LL
NN
VV
−ω
=
==∴
M2=L1L2(K=1)
I1I2
V1
+
-L1 L2
V2=
+
-N1 N2
Magnetizing inductance=L1 Ideal transformer
2
1
2
2
1
LL
NN
=
36
The imperfectly coupled Transformer
If < 1, the transformer is equivalent to an ideal transformer with one inductor connected in series with one of the windings. This series inductor is known as the “leakage inductance”.knowing that
( ) 222
22121
222211212
1
0
ILkjI`LjI`L`Lj
~R,IRIkLjILLkjV
−ω+ω+ω=
+ω+ω=
122
211 and when LkLLL `` ==
Hence we can model the imperfect transformer as an ideal transformer with turns ratio N1/N2. .
A leakage inductance L2(1-k2) is connected in series with2nd winding.
22
1
2
12
2
1
LkL
LL
nn
`
`==
N1 N2= n1 n2
n1 n2=
I1 I2
+
-V1
+
V2
+
-L1 L2 L1
Leakage inductance ( )22 1 kL −=
Perfectly-Coupledtransformer 2
21
2
2
1
Lk
Lnn
=
Ideal transformer
22
12
2
1
Lk
Lnn
=
Magnetizing Inductance =L1
37
(i) Single-Tuned Transformer
A transformer with one side turned as shown below can be used for impedance matching. The additional advantages of a transformer are; isolation between input and output circuits, and can introduce reversal.
If Rt , R2 , fo , and B are given, a design procedure is required to design the transformer, i.e. L1 , L2 , M and k and the tuning capacitance C.
R1 C L1 R2L2M
38
TABLE 7 Design Formulas for the Single-Tuned Transformer
( )
( )
( ) ( )( )
column. handright theus , 10 If column. handleft in the formulas the
follow , 10 is thisIf k.t coefficien couplingfor valueacceptable
an gives that Qfor valuea Choose 30.-3 Fig. Refer to4
3
12
211
10For
2221
2
2
<
≥
==
ω≈
π≈
≥≈
p
p
p
tt
o
t
ot
Q
Q
RNR;RRN
CL
BRC
,BfQ
R1 C L1 R2L2
M
39
Approximate Formulas10≥pQ Formulas for 10<pQ
( )
( )
( )
N
LL
QR
L
Q
Qk
t
po
t
p
1
1
22
21
LM (8)
(2) step from 7
6
5
=
≈
ω=
≈
( )
( )
( )
( ) ( ) 2121
22
2
1
22
212
8
1
17
6
1
15
LLkM
kQ
QLL
QR
L
Qk
p
pt
po
tp
p
=
−+
+=
ω=
+
+=
40
6. Transmission Lines (TL)
A real wire has distributed capacitance and inductance. When wire length becomes comparable to the wavelength, these distributed capacitances and inductances cannot be ignored. The wires are then transmission lines. Which if designed properly are entirely predictable and can be accounted for in the design.
Fundamentals:
Every increment of the TL contribute series inductance and shunt capacitance. The TL can be modeled by the ladder network shown below.
=Lδ Lδ Lδ Lδ Lδ
Cδ Cδ Cδ Cδ Cδ
41
At RF the series and shunt resistance can be neglected in comparison to the reactances. (Series R is much smaller than , and shunt R is much larger than ).( )Lδω ( )Cδω1
Now consider an L-section of a TL which is loaded by an impedance Zo as shown.
( )CZLxjZZZ
xZCxZCjZ
xLj
xZCjZ
xLjZZ
ooo
o
oo
o
oo
2
2222
2
11
1
−ωδ+≈δ+∴
<<δω+δω−
+δω=
δω++δω=δ+
If input impedance is required to be Zo , i.e. δZ = 0, then L must be equal to C2oZ
orCL
Zo =
This implies that if a TL, which has a “characteristic” impedance , is loaded by a Zo impedance (perfect termination) the input impedance remains to be equal to Zo.
CLZo =
ZZo δ+ 0jZo +
xδ
xLδxCδ
42
Now assume a perfectly terminated TL as shown.
If a voltage V is applied to the input, what is the current?
Sine Zin = Zo , the current is:
oZV
I =
and the voltage drop across the inductor (δV) is:
xLj*ZV
xLj*IVo
δω=δω=δ
Which leads to the following DE:
VLCjVZL
jdxdV
oω−=ω−=
The solution of this DE is:( ) jkx
f eVxV −=
where LCk ω=
oZ
xLδxCδ
tjV ω= eVV δ+
43
Now let us add time dependency to the input voltage, . The voltage signal propagating down the TL I s given by
( ) ( )kxtjf eVt,xV −ω=
tjeV ω=
The propagation of a constant-phase point is governed by ωt-kx = const. Which implies that
phaseVLCkdt
dx==
ω=
1
LC1
has units of velocity and is a measure for “velocity of propagation” denoted as . Some other relations:phaseV
CLCVphase
ε==
1
Were ε is the dielectric constant (relative to vacuum), and C is the speed of light.
and λπ
=ω
=2
phaseVk
where λ is wavelength
44
Impedance Transformation using a TL
The wave propagating down the TL (in the +ve x-direction) is called “incident wave.” If the termination is not perfect , a “reflected wave” will be traveling back to the source (-vd x- direction). This will cause the input impedance to change.
( )oL ZZ ≠
Assume that some source produces an incident wave traveling to the load (e-jkx) and ZLcauses a reflected wave traveling in the opposite direction. At any point X, the voltage on the line is:
jkxeρ
( )
( )o
jkx
o
jkx
jkxjkx
Ze
Ze
xI
eexV
ρ−=
ρ+=−
−
and
45
At X=0
( )
( ) ( )ρ−=
ρ+=
111
oZoI
oV
so the impedance at X=0
( )( ) L
o
ZZ
oZ =ρ−
ρ+= − 1
11
orρ−ρ+
=11
o
L
ZZ
orthe reflectioncoefficient oL
oL
ZZZZ
+−
=ρ What happens if
oL ZZ =
l−=Xat
l−=Xat
( )( ) ( )
( ) ( )
( )( )\\ktanjZZ
\\ktanjZZ*Z
Zeeee
ZZ
Lo
oLo
ojkjk
jkjk
in
ll
l ll
ll
++=
ρ−ρ+
=−= −−−
−−−
46
( )( )l
lktanjZZktanjZZ
*ZZLo
oLoin +
+=
Consider the following cases:
( )
( )impedance. reactive pure a
Z
circuitshort 0 Z-
expected) (as Z
rmination perfect te
in
L
in
lktanjZ
Z
ZZ
o
o
oL
=⇒
=
=⇒
=−
on. so and inductive then and2
for ecapcitativ becomes and
2for inductive is
π<<π
π<
l
l
k
kZin
47
( )
L
l
l
inductive, then and
2for vecapacitati is Z
Z
circit)(open
in
in
π<
=⇒
∞=−
k
ktanjZ
Z
o
L
48
Example:
Find the input impedance and reflection coefficient for the TL shown. It is λ8 long and has a 50Ωcharacteristic impedance.
Solution:
( )( )
( ) ( )( ) Ω−=
++++
=λ
=π=λλπ=
<=+−−+=
+−
=ρ
501004550505045505050
508
4548
2
4634470505050505050
jtanjjtanJJ
*Z
*k
..jj
ZZZZ
o
o
in
o
o
oL
oL
l
Ω= 50oZ
( )8λinZ
5050 jZL +=Ω+= 50100 jZs
osE 010∠=
8λ
== ld 0=d
49
For max. power transfer
( )
cm.
cmf
j
ZZ *ins
753
3010103V
GHz 1At
50100
8
9
10phase
=∴
=×
==λ
Ω+=
λ=
l
( )8λV
( )8λI
( ) Ω−=λ 501008 jZinΩ+= 50100 jZs
osE 0100∠=