Design a PLL Filter When Only the Zero
Resistor and Capacitor Are Adjustable
By Ken Gentile
Introduction
PLL System Function
As described in the references, a standard procedure can be
used to determine the values of R0, C0, and CP for a secondorder loop filter in a phase-locked loop (PLL). It uses openloop bandwidth (ω0) and phase margin (ϕM ) as design
parameters, and can be extended to third-order loop filters
to determine R2 and C2 (Figure 1). The procedure solves for
CP directly and subsequently derives the remaining values.
The small signal model shown in Figure 2 provides the
means for formulating the PLL response and a template for
analyzing phase variation at the output resulting from a phase
disturbance at the input. Note that the voltage-controlled
oscillator (VCO), being a frequency source, behaves like an
ideal phase integrator, so its gain (KV ) has a 1/s factor (the
Laplace transform equivalent of integration). Hence, the small
signal model of a PLL has frequency dependence (s = σ + jω).
In some cases, CP , R2, and C2 may be fixed-value components
integrated within the PLL, leaving only R0 and C0 available
for controlling the loop response. This nullifies the
aforementioned procedure because CP cannot be adjusted.
This article proposes an alternative procedure that can be
used when the value of CP is fixed, and addresses limitations
imposed by the inability to control the value of CP .
TO
VCO
CHARGE
PUMP
R0
CP
TO
VCO
R2
R0
CHARGE
PUMP
CP
C0
C2
C0
SECOND-ORDER
LOOP FILTER
THIRD-ORDER LOOP FILTER
Figure 1. Typical second-order and third-order passive loop filters.
Assumptions
This loop filter design method relies on two assumptions
that are typically used in third-order passive filter designs
that extend a second-order loop filter design to third-order
by compensating for the presence of R2 and C2 through
adjustment of R0 and C0.
(1)
𝐶𝐶𝑃𝑃
) 𝑇𝑇2 (2)
𝑃𝑃 +𝐶𝐶0
𝐶𝐶𝑃𝑃
𝐶𝐶𝑃𝑃
(𝐶𝐶𝑇𝑇1+𝐶𝐶
)(𝑇𝑇2 )in
(2)
𝑇𝑇1 =transfer
=0function,
𝑇𝑇terms
(2)
The loop filter’s
of T1, T2, and CP ,
2
𝑃𝑃
𝐶𝐶 +𝐶𝐶
0
is important because it plays a significant role in the overall
1
𝑇𝑇
𝑠𝑠𝑇𝑇 +1
response of the PLL:
𝐻𝐻 (𝑠𝑠) = ( ) ( 1 ) ( 2 )
(3)
𝐶𝐶𝑃𝑃
𝑠𝑠(𝑠𝑠𝑇𝑇1 +1)
𝑇𝑇2
𝑇𝑇1 1 𝑠𝑠𝑇𝑇𝑇𝑇21+1 𝑠𝑠𝑇𝑇2 +1
𝐻𝐻𝐿𝐿𝐿𝐿 (𝑠𝑠)𝐻𝐻=𝐿𝐿𝐿𝐿((𝑠𝑠)
) ( ) ()𝑠𝑠(𝑠𝑠𝑇𝑇
)
(3)
(𝑇𝑇1)+1)
(𝑠𝑠(𝑠𝑠𝑇𝑇
)
𝐶𝐶𝑃𝑃 =𝑇𝑇(
2 𝐶𝐶𝑃𝑃
2
1 +1)
1
𝐻𝐻𝐿𝐿𝐿𝐿 (𝑠𝑠)
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑠𝑠) = −𝐾𝐾 (
)
𝑠𝑠𝑠𝑠
𝐻𝐻𝐿𝐿𝐿𝐿 (𝑠𝑠) 𝐻𝐻𝐿𝐿𝐿𝐿 (𝑠𝑠)
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑠𝑠)𝐻𝐻=𝑂𝑂𝑂𝑂−𝐾𝐾
(𝑠𝑠) (= 𝑠𝑠𝑠𝑠
−𝐾𝐾 )( 𝑠𝑠𝑠𝑠 (4)
)
Analog Dialogue 49-02,
February 2015
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑠𝑠)
(4)
(4)
𝑇𝑇1 = (𝐶𝐶
𝐶𝐶𝑃𝑃
VCO
(1)
𝛉OUT
(1)
1/N
(2)
) 𝑇𝑇2
FEEDBACK
𝑃𝑃 +𝐶𝐶0DIVIDER
𝐶𝐶𝑃𝑃
(2)
Figure 2. Small signal
) 𝑇𝑇2
𝑇𝑇1 =PLL(𝐶𝐶model.
𝑃𝑃 +𝐶𝐶0
The closed-loop transfer function
(H
) for𝑠𝑠𝑇𝑇
a PLL
is defined
1
𝑇𝑇1CL
2 +1
as θOUT /θIN . The𝐻𝐻open-loop
(𝐶𝐶 ) (𝑇𝑇function
) (𝑠𝑠(𝑠𝑠𝑇𝑇(H
(3)
OL ),)defined
𝐿𝐿𝐿𝐿 (𝑠𝑠) = transfer
𝑃𝑃
2 transfer
1 +1)
as θFB /θIN , is related to the closed-loop
function. It
1
𝑇𝑇1
𝑠𝑠𝑇𝑇2 +1
( ) ( ) (𝑠𝑠(𝑠𝑠𝑇𝑇
) the (3)
is instructive to 𝐻𝐻
express
because
𝐿𝐿𝐿𝐿 (𝑠𝑠)H=
CL in𝐶𝐶terms𝑇𝑇of HOL
𝑃𝑃
2
1 +1)
open-loop transfer function contains clues about closed-loop
𝐻𝐻𝐿𝐿𝐿𝐿 (𝑠𝑠)
stability:
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑠𝑠) = −𝐾𝐾 (
(4)
)
𝑠𝑠𝑠𝑠
𝐻𝐻𝐿𝐿𝐿𝐿 (𝑠𝑠)
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑠𝑠) = −𝐾𝐾 (
𝐻𝐻
𝑠𝑠𝑠𝑠
)
(𝑠𝑠)
(4)
(5)
)
𝑂𝑂𝑂𝑂 (𝑠𝑠)
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑠𝑠)
(5)
)
(𝑠𝑠)
(𝑗𝑗𝑗𝑗) 𝑁𝑁𝐶𝐶𝑃𝑃
A second-order loop filter has two time constants (T1 and T2)
𝑇𝑇2components:
= 𝑅𝑅0 𝐶𝐶0
(1)
associated with its
𝐿𝐿𝐿𝐿
FB
KV /s
𝐻𝐻𝐶𝐶𝐶𝐶 (𝑠𝑠) = −𝑁𝑁 (1−𝐻𝐻
Transfer Function of a Second-Order Loop Filter
𝑃𝑃
+ 𝑇𝑇2 = 𝑅𝑅 𝐶𝐶
− 𝛉 𝑇𝑇2 = 𝑅𝑅0 𝐶𝐶0
HLF(s)
LOOP
FILTER
K represents the combined gains of the𝑂𝑂𝑂𝑂
phase-frequency
detector (PFD), charge pump, and𝐾𝐾VCO—that
is,
= KD KV ,
𝑇𝑇1
𝑠𝑠𝑇𝑇2K
+1
)
(
)
(
)
(6)
= − (current
𝑂𝑂𝑂𝑂 (𝑠𝑠) pump
where KD is the𝐻𝐻charge
in
amperes
and
2
𝑠𝑠 𝑁𝑁𝐶𝐶𝑃𝑃
𝑠𝑠𝑇𝑇1 +1 KV is
𝑇𝑇2
𝐾𝐾 HL F are
𝑇𝑇1 all𝑠𝑠𝑇𝑇
+1
the VCO gain in Hz/V. HO L , HC L , and
functions
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑠𝑠)
= − ( shows
) (the
) (𝑠𝑠𝑇𝑇2 +1
)
(6)
of s. The negative sign
in Equation𝑠𝑠24𝑁𝑁𝐶𝐶
𝑇𝑇2 phase
𝑃𝑃
1 inversion
implied by the negative feedback to the summing node in
𝐾𝐾
𝑇𝑇
𝑗𝑗𝑗𝑗𝑇𝑇2 +1
in Equation
to1subtraction
Figure 2. Defining
O L as =
− ((𝑗𝑗𝑗𝑗)24 leads
𝐻𝐻𝑂𝑂𝑂𝑂H(𝑗𝑗𝑗𝑗)
)
(
)
(
)
𝑗𝑗𝑗𝑗𝑇𝑇1 +1
𝑁𝑁𝐶𝐶𝑃𝑃
𝑇𝑇2an intuitive
in the denominator of Figure 5, which 𝐾𝐾
provides
𝑇𝑇1
𝑗𝑗𝑗𝑗𝑇𝑇2 +1
explanation of closed-loop
stability.
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗) =
−(
)( )(
)
2
2. T
he load of the series combination of R2 and C2 on the
R0-C0-CP network should be negligible.
𝑇𝑇1 = (𝐶𝐶
KD
PFD AND
CHARGE
0 0
PUMP
𝐻𝐻𝐶𝐶𝐶𝐶 (𝑠𝑠) = −𝑁𝑁 (1−𝐻𝐻𝑂𝑂𝑂𝑂
1. T
he pole frequency resulting from R2 and C2 should
be at least an order of magnitude greater than ω0 (the
desired open-loop unity-gain bandwidth); specifically
f0 ≤ 0.1/(2πR2C2), where f0 = ω0/(2π).
𝑇𝑇2 = 𝑅𝑅𝑇𝑇
(1)
0 𝐶𝐶
2 0= 𝑅𝑅0 𝐶𝐶0
𝛉IN
𝛉ERR
(3)
𝑇𝑇2
𝑗𝑗𝑗𝑗𝑇𝑇1 +1
(7)
(7)
Inspection of Equation 5 reveals a𝐾𝐾potential
stability
𝑇𝑇1loop𝑗𝑗𝑗𝑗𝑇𝑇
2 +1
(𝜔𝜔2 𝑁𝑁𝐶𝐶 of
) (complex
)
(
) (8)
problem. Given𝐻𝐻that
HOL is=
a function
frequency
𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗)
𝑗𝑗𝑗𝑗𝑇𝑇1 +1
𝑇𝑇2
𝑃𝑃
(s = σ + jω), it necessarily has frequency
dependent
magnitude
𝐾𝐾
𝑇𝑇1
𝑗𝑗𝑗𝑗𝑇𝑇
+1
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗)
= (𝜔𝜔2 𝑁𝑁𝐶𝐶
) (simultaneously
) (𝑗𝑗𝑗𝑗𝑇𝑇2 +1) (8)
and phase components.
Therefore,
if H
𝑃𝑃 O L 𝑇𝑇2
1
exhibits unity gain and zero phase shift (or any integer
𝐾𝐾
𝑇𝑇1
multiple of 2π |𝐻𝐻
radians)
for any
s, 1the ) √(1 + 𝜔𝜔
(𝑗𝑗𝑗𝑗)|
= (particular
) (value
) (of
2 𝑁𝑁𝐶𝐶
𝜔𝜔
1+(𝜔𝜔𝑇𝑇
𝑇𝑇
denominator of H𝑂𝑂𝑂𝑂
becomes
zero,
the
closed-loop
gain1 )2
𝑃𝑃
2
CL
𝐾𝐾
𝑇𝑇1
1
becomes undefined,
the =
system
becomes
|𝐻𝐻𝑂𝑂𝑂𝑂and
(𝑗𝑗𝑗𝑗)|
(𝜔𝜔2 𝑁𝑁𝐶𝐶
) (𝑇𝑇 completely
) (1+(𝜔𝜔𝑇𝑇 )2 ) √(1 + 𝜔𝜔
𝑃𝑃
2
unstable. This implies that stability is governed by the 1
frequency-dependent magnitude and phase characteristics of
∠𝐻𝐻𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗) =for𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
2 ) − 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
1)
HOL . In fact, at the frequency
which the magnitude
of HOL
is unity, the phase
far enough from
zero (or
OL must
∠𝐻𝐻of𝑂𝑂𝑂𝑂H(𝑗𝑗𝑗𝑗)
=stay
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
2 ) − 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇1 )
any integer multiple of 2π) to avoid a zero denominator in
Equation 5.
1 = (𝜔𝜔
1=(
𝐾𝐾
2
0 𝑁𝑁𝐶𝐶𝑃𝑃
𝐾𝐾
𝑇𝑇
1
) (𝑇𝑇1 ) (1+(𝜔𝜔
2
𝑇𝑇
2
0 𝑇𝑇1 )
1
) ( 1 ) (1+(𝜔𝜔
𝜔𝜔0 2 𝑁𝑁𝐶𝐶𝑃𝑃
𝑇𝑇2
analog.com/analogdialogue
) √(1 + 𝜔𝜔0 2 𝑇𝑇1 𝑇𝑇2
2
0 𝑇𝑇1 )
) √(1 + 𝜔𝜔0 2 𝑇𝑇1 𝑇𝑇2
1
) 1( 𝑇𝑇)(3)
𝐻𝐻𝐿𝐿𝐿𝐿 (𝑠𝑠) = (𝐶𝐶 ) (𝑇𝑇 ) (𝐻𝐻𝑠𝑠(𝑠𝑠𝑇𝑇
=
(𝑠𝑠(𝑠𝑠𝑇𝑇
)
(3)
1( )𝑠𝑠𝑇𝑇
2 +1
𝐿𝐿𝐿𝐿 (𝑠𝑠)
𝑠𝑠𝑇𝑇21+1
𝑃𝑃
2𝐻𝐻 (𝑠𝑠)
1 +1)
𝐶𝐶1(
+1)
1
=
(
)
)
(3)
𝑃𝑃 )𝑇𝑇(
2
𝐿𝐿𝐿𝐿
𝐻𝐻𝐿𝐿𝐿𝐿 (𝑠𝑠) =𝐶𝐶𝑃𝑃(𝐶𝐶 𝑇𝑇)2( 𝑠𝑠(𝑠𝑠𝑇𝑇
) (𝑠𝑠(𝑠𝑠𝑇𝑇
)
(3)
𝐻𝐻 (𝑠𝑠)
1 +1)
𝑇𝑇2
)
(4)
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑠𝑠) = −𝐾𝐾 ( 𝐿𝐿𝐿𝐿
𝑃𝑃
1 +1)
𝐻𝐻𝐻𝐻𝐿𝐿𝐿𝐿
(𝑠𝑠)
(𝑠𝑠)
𝑠𝑠𝑠𝑠
𝐿𝐿𝐿𝐿
(𝑠𝑠) = −𝐾𝐾 (
))
(4)
𝐻𝐻𝐻𝐻𝑂𝑂𝑂𝑂
𝑂𝑂𝑂𝑂 (𝑠𝑠) = −𝐾𝐾 ( 𝑠𝑠𝑠𝑠
The frequency, ω0, at which the magnitude
great(4)
importance. The phase of HOL at ω0 defines the phase
𝑠𝑠𝑠𝑠holds
𝐻𝐻𝐿𝐿𝐿𝐿 (𝑠𝑠) of HOL is unity,
𝐻𝐻
(𝑠𝑠)
𝐿𝐿𝐿𝐿
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑠𝑠)
= −𝐾𝐾
( 𝑠𝑠𝑠𝑠
)
(4)
margin of the system
ϕM . Both
ω0 and
ϕM can
be
derived
from
H
.
OL
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑠𝑠) = −𝐾𝐾𝐻𝐻𝐿𝐿𝐿𝐿
( 𝐻𝐻(𝑠𝑠)𝑠𝑠𝑠𝑠
)
(4)
(𝑠𝑠)(𝑠𝑠)
𝐿𝐿𝐿𝐿)(𝑠𝑠)
𝐻𝐻𝐻𝐻𝑂𝑂𝑂𝑂
= −𝐾𝐾
( 𝑠𝑠𝑠𝑠
(4)(4)
(𝑠𝑠)
)
𝐻𝐻
=
−𝐾𝐾
(
𝑂𝑂𝑂𝑂
𝑂𝑂𝑂𝑂
)
(5)
= −𝑁𝑁
(01−𝐻𝐻
𝐻𝐻
(𝑠𝑠) 𝑠𝑠𝑠𝑠
Defining R0 and 𝐻𝐻
C0𝐶𝐶𝐶𝐶
in(𝑠𝑠)
Terms
of 𝛚
and𝑂𝑂𝑂𝑂𝛟(𝑠𝑠)
𝐻𝐻𝑂𝑂𝑂𝑂
M
𝑂𝑂𝑂𝑂 (𝑠𝑠) )
(𝑠𝑠)
(5)
=
−𝑁𝑁
(
𝐻𝐻𝐻𝐻𝐶𝐶𝐶𝐶
(𝑠𝑠)
=
−𝑁𝑁
(
) R and
(5)C requires expressions containing those four variables
1−𝐻𝐻
(𝑠𝑠) of
𝑂𝑂𝑂𝑂
Using the design parameters ω0 and ϕ𝐶𝐶𝐶𝐶
to(𝑠𝑠)
determine the
values
1−𝐻𝐻
0
0
𝑂𝑂𝑂𝑂 (𝑠𝑠)
𝐻𝐻M𝑂𝑂𝑂𝑂
𝐻𝐻
(𝑠𝑠)
𝑂𝑂𝑂𝑂
(𝑠𝑠) Start
= −𝑁𝑁
(1−𝐻𝐻
)4,(𝑠𝑠)
(5)
𝐻𝐻𝐶𝐶𝐶𝐶
and other constant
terms.
with
Equation
because
it defines
𝐻𝐻(𝑂𝑂𝑂𝑂 (𝑠𝑠) HOL .)This includes
=
−𝑁𝑁
(5) HLF , which includes R0 and C0 via T1 and T2.
𝐻𝐻
𝐶𝐶𝐶𝐶
(𝑠𝑠)
𝐻𝐻𝑂𝑂𝑂𝑂
(𝑠𝑠)
𝑂𝑂𝑂𝑂 (𝑠𝑠) = −𝑁𝑁 (
1−𝐻𝐻
(𝑠𝑠) can (5)
)
𝐻𝐻stands
𝑂𝑂𝑂𝑂
𝐶𝐶𝐶𝐶𝐻𝐻 𝑇𝑇to
Since HOL has magnitude and phase, it𝐾𝐾
reason
that
ω
and
ϕ
be
incorporated
as well.
(𝑠𝑠)
0
M
)
(5)
=2 +1
−𝑁𝑁
(1−𝐻𝐻
𝑠𝑠𝑇𝑇
1−𝐻𝐻
1
𝑂𝑂𝑂𝑂 (𝑠𝑠)
(𝑠𝑠)+1
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑠𝑠) = − (𝑠𝑠2 ) 𝐶𝐶𝐶𝐶
(
)
(
)
(6)
𝑂𝑂𝑂𝑂
𝐾𝐾
𝑇𝑇
𝑠𝑠𝑇𝑇
1
2
𝑠𝑠𝑇𝑇 +1
𝑁𝑁𝐶𝐶
+1 ) ( 𝑇𝑇1yields
𝑃𝑃 =
1 𝐾𝐾
𝐻𝐻𝐻𝐻𝑂𝑂𝑂𝑂
)
(6)
Substituting Equation 3 into Equation
4(𝑠𝑠)
and
rearranging
6,(6)
which presents HOL in terms of T1 and T2 along
(𝑠𝑠)
=𝑇𝑇2−−((𝑠𝑠𝑠𝑠𝑇𝑇
) (𝑇𝑇𝑇𝑇2 ))((𝑠𝑠𝑇𝑇𝑠𝑠𝑇𝑇12Equation
22𝑁𝑁𝐶𝐶 terms
𝑂𝑂𝑂𝑂
+1 )
𝑃𝑃𝑃𝑃
𝑠𝑠
𝑁𝑁𝐶𝐶
2
1 +1 𝑠𝑠𝑇𝑇 +1
𝐾𝐾
𝑇𝑇
𝑠𝑠𝑇𝑇
+1
with constants K, N, and CP :
1
2
𝐾𝐾
𝑇𝑇
1
2
(𝑇𝑇𝑂𝑂𝑂𝑂)(𝑠𝑠)
(𝑠𝑠𝑇𝑇=+1−) (𝐾𝐾 2 (6)𝑇𝑇)1( 𝑠𝑠𝑇𝑇
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑠𝑠) = − (𝑠𝑠2 𝑁𝑁𝐶𝐶 )𝐻𝐻
) (2 +1
)
(6)
𝐾𝐾) (
𝑃𝑃
2 =
1 (
𝑠𝑠 𝑁𝑁𝐶𝐶
𝑠𝑠𝑇𝑇12)+1 (6)
(𝑠𝑠)
1
𝐻𝐻
𝑃𝑃 )𝑇𝑇(
2
𝑂𝑂𝑂𝑂
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑠𝑠) 𝑇𝑇−
=
(𝑁𝑁𝐶𝐶
)
(
)
(
)
(6)
𝐾𝐾
+1
𝑠𝑠2𝑗𝑗𝑗𝑗𝑇𝑇
𝑠𝑠𝑇𝑇
𝑇𝑇
+1
1 −
2
𝑃𝑃
2
1
2
𝑠𝑠𝑇𝑇1+1
+1
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗) = − ((𝑗𝑗𝑗𝑗)2 𝑁𝑁𝐶𝐶 ) (𝑇𝑇 ) (𝑗𝑗𝑗𝑗𝑇𝑇
) 𝑇𝑇𝑇𝑇1 𝑇𝑇(7)
𝑃𝑃
2 𝑗𝑗𝑗𝑗𝑇𝑇
𝐾𝐾𝐾𝐾𝑠𝑠 𝑁𝑁𝐶𝐶
22 +1
𝑗𝑗𝑗𝑗𝑇𝑇
+1
1
𝑃𝑃
2
1
(𝑗𝑗𝑗𝑗)
𝐻𝐻𝐻𝐻𝑂𝑂𝑂𝑂
==−−of(((𝑗𝑗𝑗𝑗)
))((𝑇𝑇 ))((𝑗𝑗𝑗𝑗𝑇𝑇 +1)) (7)
(7)
𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗)
Evaluation at s = jω, yields the frequency
response
HOL :22𝑁𝑁𝐶𝐶
22
1
𝑗𝑗𝑗𝑗𝑇𝑇
𝐾𝐾
𝑇𝑇1 (𝑗𝑗𝑗𝑗)
𝑗𝑗𝑗𝑗𝑇𝑇2𝑁𝑁𝐶𝐶
+1𝑃𝑃𝑃𝑃 𝐾𝐾 𝑇𝑇
𝑇𝑇1 1 +1𝑗𝑗𝑗𝑗𝑇𝑇2 +1
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗) = − ((𝑗𝑗𝑗𝑗)2 𝑁𝑁𝐶𝐶
)
(
)
(
)
(7)
(𝑗𝑗𝑗𝑗)
𝐾𝐾 2
𝑇𝑇
𝐻𝐻𝑂𝑂𝑂𝑂
) ( 𝑗𝑗𝑗𝑗𝑇𝑇
) (𝑗𝑗𝑗𝑗𝑇𝑇
) (7)
= −1(+1
2 +1
𝑇𝑇=
𝑃𝑃
2 −𝑗𝑗𝑗𝑗𝑇𝑇
(𝑗𝑗𝑗𝑗)𝐾𝐾
𝑁𝑁𝐶𝐶
2)+1 (7)
) (𝑃𝑃𝑇𝑇)1()𝑇𝑇(21𝑗𝑗𝑗𝑗𝑇𝑇
𝐻𝐻 𝐻𝐻(𝑗𝑗𝑗𝑗)
1
2 𝑁𝑁𝐶𝐶
)
(
) (7)
=2(+1
−
((𝑗𝑗𝑗𝑗)
(𝑗𝑗𝑗𝑗)
𝐾𝐾 𝑂𝑂𝑂𝑂
𝑇𝑇1(𝑗𝑗𝑗𝑗)
𝑗𝑗𝑗𝑗𝑇𝑇
+1
𝑂𝑂𝑂𝑂
𝑃𝑃
2
1
2 𝑁𝑁𝐶𝐶
𝑗𝑗𝑗𝑗𝑇𝑇1 +1
(8)
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗) = (𝜔𝜔2 𝑁𝑁𝐶𝐶 ) (𝑇𝑇 ) (𝑗𝑗𝑗𝑗𝑇𝑇
𝑃𝑃 +1 𝑇𝑇2
𝐾𝐾𝐾𝐾 +1) 𝑇𝑇𝑇𝑇
𝑗𝑗𝑗𝑗𝑇𝑇
11
22 +1
𝑗𝑗𝑗𝑗𝑇𝑇
2
2
𝑃𝑃
2–ω
(𝑗𝑗𝑗𝑗)
The (jω) term in the denominator 𝐻𝐻
simplifies
to=
( : 1 ))(( ))((𝑗𝑗𝑗𝑗𝑇𝑇 )) (8)
𝐻𝐻𝑂𝑂𝑂𝑂
(8)
𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗) = (𝜔𝜔𝜔𝜔22𝑁𝑁𝐶𝐶
𝑇𝑇22
1 +1
𝑗𝑗𝑗𝑗𝑇𝑇
𝐾𝐾
𝑇𝑇1
𝑗𝑗𝑗𝑗𝑇𝑇2𝑁𝑁𝐶𝐶
+1𝑃𝑃𝑃𝑃 𝐾𝐾𝑇𝑇
𝑇𝑇1 1 +1𝑗𝑗𝑗𝑗𝑇𝑇2 +1
𝐻𝐻𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗) = (𝜔𝜔2 ) 𝐻𝐻
(
)
(
)
(8)
(𝑗𝑗𝑗𝑗)
𝑇𝑇
=
) ( 𝑗𝑗𝑗𝑗𝑇𝑇
) ( 𝑗𝑗𝑗𝑗𝑇𝑇
) (8)
2 +1
𝑗𝑗𝑗𝑗𝑇𝑇
𝑁𝑁𝐶𝐶𝐻𝐻
𝑇𝑇𝑂𝑂𝑂𝑂
+1(𝐾𝐾𝜔𝜔 2 𝐾𝐾
𝑃𝑃
2
𝑁𝑁𝐶𝐶
2)+1 (8)
) (𝑃𝑃𝑇𝑇)1()𝑇𝑇(21𝑗𝑗𝑗𝑗𝑇𝑇
= (1=
1
2(
(𝑗𝑗𝑗𝑗)
)
(
) (8)
𝐻𝐻(𝑗𝑗𝑗𝑗)
𝐾𝐾𝑂𝑂𝑂𝑂
𝑇𝑇
1
𝜔𝜔
𝑁𝑁𝐶𝐶
+1
2
12
2𝑃𝑃
𝑇𝑇2+1𝜔𝜔𝑗𝑗𝑗𝑗𝑇𝑇
(𝜔𝜔2 𝑁𝑁𝐶𝐶 )𝑂𝑂𝑂𝑂(𝑇𝑇1 ) (1+(𝜔𝜔𝑇𝑇
𝑇𝑇11+1
𝑇𝑇2 )2 + 𝜔𝜔 22(𝑇𝑇2 −2𝑇𝑇1 )2 2 (9)
𝑃𝑃√(1
𝐾𝐾𝐾𝐾 𝜔𝜔 )𝑁𝑁𝐶𝐶
𝑇𝑇𝑇𝑇
𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗)|
The magnitude and|𝐻𝐻
phase
of HOL=are:
2)
11
1
𝑃𝑃
|𝐻𝐻
(𝑗𝑗𝑗𝑗)|
=2 (
)1)((𝑇𝑇 ))((1+(𝜔𝜔𝑇𝑇 )2 ))√(1
+ 𝜔𝜔 2𝑇𝑇1 𝑇𝑇2 ))2++𝜔𝜔𝜔𝜔 2(𝑇𝑇
(𝑇𝑇22−−𝑇𝑇𝑇𝑇11))22 (9)
|𝐻𝐻𝑂𝑂𝑂𝑂
(9)
𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗)| = (𝜔𝜔𝜔𝜔22𝑁𝑁𝐶𝐶
2 √(1 + 𝜔𝜔 𝑇𝑇1 𝑇𝑇2
𝑃𝑃
2
1
)
1+(𝜔𝜔𝑇𝑇
𝑁𝑁𝐶𝐶
𝑇𝑇
𝐾𝐾
𝑇𝑇1
1 𝑃𝑃 𝐾𝐾 2
𝑇𝑇1 2 1
1)2
2
2
2
2
2
(𝑇𝑇
)
|𝐻𝐻𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗)| = ( 2 |𝐻𝐻
) (𝑂𝑂𝑂𝑂
) (1+(𝜔𝜔𝑇𝑇
+
𝜔𝜔 2+2−𝜔𝜔𝑇𝑇1𝑇𝑇1 𝑇𝑇22 ) (9)
𝑇𝑇
11𝑇𝑇2
=1()𝐾𝐾2𝜔𝜔)2√(1
)1+
( 𝑇𝑇𝜔𝜔) (𝑇𝑇1+(𝜔𝜔𝑇𝑇
+2𝜔𝜔 (𝑇𝑇2 − 𝑇𝑇21 )2 (9)
2 ) √(1
𝜔𝜔 𝑁𝑁𝐶𝐶
𝑇𝑇2 (𝑗𝑗𝑗𝑗)|
𝐾𝐾
1)
)√(1
𝑁𝑁𝐶𝐶
)
(𝑇𝑇22(𝑇𝑇
|𝐻𝐻𝑃𝑃𝑂𝑂𝑂𝑂
1
(𝑗𝑗𝑗𝑗)|
=
(
)
(
)
(
+
𝜔𝜔
𝑇𝑇
𝑇𝑇
+
𝜔𝜔
− 𝑇𝑇−
𝑃𝑃
2
1
2
2
1
2
1 )𝑇𝑇 )2 (9) (9)
2
) + 𝜔𝜔
|𝐻𝐻𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗)|
=𝜔𝜔2(𝑁𝑁𝐶𝐶
) ( 1+(𝜔𝜔𝑇𝑇
) (1+(𝜔𝜔𝑇𝑇
𝑇𝑇
2
1
1)
2 ) √(1 + 𝜔𝜔 𝑇𝑇1 𝑇𝑇2
∠𝐻𝐻𝑂𝑂𝑂𝑂 (𝑗𝑗𝑗𝑗) = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
2 ) − 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
𝜔𝜔 2𝑃𝑃𝑁𝑁𝐶𝐶𝑃𝑃 2 1𝑇𝑇)2 (10)
1)
(𝑗𝑗𝑗𝑗)
)
)
∠𝐻𝐻
=
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
−
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
(10)
(𝑗𝑗𝑗𝑗)
)
)
2
1
∠𝐻𝐻𝑂𝑂𝑂𝑂
=
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
−
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
(10)
𝑂𝑂𝑂𝑂
2
1
(𝑗𝑗𝑗𝑗)
)
)
∠𝐻𝐻
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
−
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
(10)
Keep in mind that
T1𝑂𝑂𝑂𝑂
and
T2 are=shorthand
expressions
for
algebraic
combinations
of
R
Evaluating Equation 9 at
(𝑗𝑗𝑗𝑗)
)
) CP.(10)
2
1
∠𝐻𝐻
=
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
−
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
0, C0, and
𝑂𝑂𝑂𝑂
2
(𝑗𝑗𝑗𝑗)
) 1 magnitude
= 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
−
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
(10)(10) of HOL is unity.
𝑂𝑂𝑂𝑂gain
2 )frequency
1the
ω = ω0 and setting |HOL | = 1 defines the∠𝐻𝐻
unity
frequency,
ω
,
as
the
at
which
(𝑗𝑗𝑗𝑗)
)
)
0
∠𝐻𝐻
=
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
−
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔𝑇𝑇
𝑂𝑂𝑂𝑂
2
1
𝐾𝐾
𝑇𝑇
1
) (𝑇𝑇1 ) (1+(𝜔𝜔
√(1 +11𝜔𝜔0 2 𝑇𝑇1 𝑇𝑇2 )2 + 𝜔𝜔022 (𝑇𝑇2 −2 𝑇𝑇1 )2 2 (11) 2
𝐾𝐾𝐾𝐾 𝑇𝑇 )2𝑇𝑇)
1
0)1( 𝑇𝑇1) (
11==2 ((𝜔𝜔 2𝑁𝑁𝐶𝐶
(11)
(𝑇𝑇22−−𝑇𝑇𝑇𝑇11))2
) ( 2 ) (1+(𝜔𝜔
√(1++𝜔𝜔𝜔𝜔00 2𝑇𝑇𝑇𝑇11𝑇𝑇𝑇𝑇22))2++𝜔𝜔𝜔𝜔00 2(𝑇𝑇
(11)
2 ))√(1
00𝑇𝑇𝑇𝑇
11))2
𝜔𝜔00 21
1+(𝜔𝜔
𝑁𝑁𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝑇𝑇𝑇𝑇
2
𝐾𝐾
𝑇𝑇1
𝑇𝑇
1
2
2
2
2
1
Similarly, evaluating
Equation
10
at
ω
=
ω
and
setting
∠H
=
ϕ
defines
the
phase
margin,
ϕ
,
as
the
phase
of
H
at
frequency
2
2
2
2
1 = (𝜔𝜔 2𝑁𝑁𝐶𝐶 ) (𝑇𝑇 ) (1+(𝜔𝜔
𝑇𝑇1 𝑇𝑇2 ) 2+
𝜔𝜔0 (𝑇𝑇
− 𝑇𝑇M)22 ) +(11)
01 = (𝐾𝐾 ) √(1
OL
𝑇𝑇
1
)OL(+ 𝜔𝜔
)M(0 1+(𝜔𝜔
) √(1
+2𝜔𝜔
𝜔𝜔
− 𝑇𝑇21 )
(11)
2
2 0 𝑇𝑇11)𝑇𝑇
2 0 (𝑇𝑇
10 𝑇𝑇)
0
𝑃𝑃
2 1 = ( 0 𝑇𝑇1
)
𝜔𝜔)0 2𝐾𝐾𝑁𝑁𝐶𝐶
) (𝑃𝑃𝑇𝑇)1()𝑇𝑇(211+(𝜔𝜔
+ 𝜔𝜔
−2 𝑇𝑇−
ω0 (the unity gain frequency).
2 𝜔𝜔
0 𝜔𝜔𝑇𝑇12𝑇𝑇𝑇𝑇
2 𝑇𝑇 +
0 𝜔𝜔(𝑇𝑇22(𝑇𝑇
1 )𝑇𝑇 )2 (11)
2 1 √(1
2 𝑁𝑁𝐶𝐶
)
1
=
(
)
(
)
√(1
+
+
(11)
)
𝜔𝜔
𝑇𝑇
0 1 2
0
2
1
0 1(12))2
Φ𝑀𝑀 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔0 𝑇𝑇2 ) −0 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔
𝜔𝜔0 2𝑃𝑃𝑁𝑁𝐶𝐶𝑃𝑃 2 𝑇𝑇02𝑇𝑇1 )1+(𝜔𝜔
0 𝑇𝑇1
)
)
ΦΦ𝑀𝑀𝑀𝑀==𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔00𝑇𝑇𝑇𝑇22 )−−𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔00𝑇𝑇𝑇𝑇11) (12)
(12)
)and
) (12) Equation 01𝑇𝑇for
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔
−
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔
𝑇𝑇1substituting
Φ𝑀𝑀to =
0 𝑇𝑇11
2Φ
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔
(12)
It is a trivial matter
expand
Equation
Equation
12 0by
Equation 2 for T1, which brings
𝑀𝑀 =
0)𝑇𝑇2 ) − 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔
1 ) T2 and
2
)
=
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔
𝑇𝑇
−
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔
𝑇𝑇
(12)
Φ
2
2 +2𝐾𝐾𝐶𝐶
2
𝑀𝑀
0
2
0
1
(Φ
)
𝐾𝐾
𝑁𝑁𝜔𝜔0 2 ) K, N,
𝐾𝐾𝐾𝐾√1−cos
have
succeeded
in
relating
ω
and
ϕ
to
the
variables
R
and
with𝑃𝑃constants
R0 and C0 into the equations. Hence,𝜔𝜔we
)
)
(12)
𝑇𝑇22 0− 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔
𝑇𝑇1
Φ𝑀𝑀 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝜔𝜔
𝑃𝑃
0 Ccos(Φ
𝑀𝑀 )+(𝐶𝐶
0
𝑀𝑀
M
0𝑁𝑁𝜔𝜔
0 along
0
0
2
2 22
2
𝑅𝑅
=
𝐶𝐶
=
−
(
) )+(𝐶𝐶
2
2
𝐾𝐾𝐾𝐾 +2𝐾𝐾𝐶𝐶
cos(Φ𝑀𝑀
𝜔𝜔𝜔𝜔00𝐾𝐾𝐾𝐾√1−cos
0𝐴𝐴
2
2 2 2(Φ
2
2 +𝐾𝐾 cos(Φ
𝑃𝑃𝑃𝑃𝑁𝑁𝜔𝜔
𝑃𝑃𝑃𝑃𝑁𝑁𝜔𝜔
and CP .
(Φ𝑀𝑀𝑀𝑀)) 0𝐴𝐴
𝑁𝑁𝜔𝜔0𝑀𝑀
𝑁𝑁𝜔𝜔00 2))
𝐾𝐾𝐾𝐾√1−cos
𝑁𝑁𝜔𝜔=
𝑁𝑁𝜔𝜔=
𝐾𝐾2 +2𝐾𝐾𝐶𝐶
0 ))cos(Φ
𝑀𝑀 )+(𝐶𝐶
0 cos(Φ𝑀𝑀 )+(𝐶𝐶
𝑃𝑃 𝑁𝑁𝜔𝜔0 )
0 (𝐶𝐶
𝑃𝑃 𝑁𝑁𝜔𝜔
0 +2𝐾𝐾𝐶𝐶
𝑅𝑅𝑅𝑅𝑃𝑃0𝐴𝐴
𝐶𝐶
−
(
))
𝐶𝐶0𝐴𝐴
(
2
2 2
2
222+𝐾𝐾 cos(Φ ))
0𝐴𝐴 =𝐾𝐾𝐾𝐾22+2𝐾𝐾𝐶𝐶
0𝐴𝐴 = −
2 cos(Φ 𝑁𝑁𝜔𝜔
20)
2 +2𝐾𝐾𝐶𝐶
𝑀𝑀𝑀𝑀)+(𝐶𝐶
00 2))2
0022(𝐶𝐶
𝑃𝑃𝑃𝑃𝑁𝑁𝜔𝜔
𝑀𝑀𝑀𝑀 ))
2 (Φ𝑃𝑃𝑃𝑃𝑁𝑁𝜔𝜔
)+(𝐶𝐶𝑃𝑃𝑃𝑃𝑁𝑁𝜔𝜔
(𝐶𝐶𝑁𝑁𝜔𝜔
+2𝐾𝐾𝐶𝐶
𝑁𝑁𝜔𝜔
cos(Φ
𝑁𝑁𝜔𝜔
𝑁𝑁𝜔𝜔
𝑁𝑁𝜔𝜔
+𝐾𝐾
cos(Φ
2
)+(𝐶𝐶
) 00 2cos(Φ
𝐾𝐾
𝑁𝑁𝜔𝜔
2
0
𝜔𝜔
𝐾𝐾𝐾𝐾√1−cos
2
®
𝑃𝑃
0
𝑀𝑀
𝑃𝑃
0
0
𝑀𝑀
)+(𝐶𝐶
(Φ
)
𝐾𝐾)+(𝐶𝐶
+2𝐾𝐾𝐶𝐶
2
2 0 cos(Φ𝑀𝑀
𝑃𝑃 𝑁𝑁𝜔𝜔
𝑃𝑃 𝑁𝑁𝜔𝜔
𝑀𝑀The symbolic
Simultaneously solving
equations
for R0 2and
C𝜔𝜔
no trivial
task.
processor
in 𝑃𝑃MathCad
2
2𝑀𝑀
(Φ
)0 0is𝐾𝐾𝐾𝐾√1−cos
𝑁𝑁𝜔𝜔0 2 cos(Φ
𝑁𝑁𝜔𝜔
𝜔𝜔 𝑅𝑅𝐾𝐾𝐾𝐾√1−cos
2 (Φ
𝑅𝑅0𝐴𝐴 =the2resulting
𝐶𝐶0𝐴𝐴
) 2 𝑀𝑀
02 )
)+(𝐶𝐶𝑃𝑃 𝑁𝑁𝜔𝜔
) = − ( 2𝐾𝐾2 +2𝐾𝐾𝐶𝐶
𝐾𝐾cos(Φ
𝑁𝑁𝜔𝜔
cos(Φ
𝜔𝜔0𝑀𝑀𝐾𝐾𝐾𝐾√1−cos
=
=2available
−
(+2𝐾𝐾𝐶𝐶
22
2 0
2 )2 2
2𝑃𝑃 𝐶𝐶𝑁𝑁𝜔𝜔
𝑃𝑃))
0 )
2 +2𝐾𝐾𝐶𝐶
0𝐴𝐴 arccos
2𝑀𝑀
2+𝐾𝐾
22cos(Φ
22)𝑃𝑃
2 +2𝐾𝐾𝐶𝐶
20𝑁𝑁𝜔𝜔
2 +𝐾𝐾𝑀𝑀
𝑁𝑁𝜔𝜔
22(Φ𝐶𝐶
( equations,
)
=
−
(
)
𝑅𝑅0𝐵𝐵 =𝐾𝐾 −+2𝐾𝐾𝐶𝐶
)+(𝐶𝐶
(Φ
)𝑃𝑃 𝑁𝑁𝜔𝜔
𝐾𝐾
cos(Φ
𝑁𝑁𝜔𝜔
𝜔𝜔
𝐾𝐾𝐾𝐾√1−cos
𝑃𝑃 𝑁𝑁𝜔𝜔𝑅𝑅
0 cos(Φ
𝑀𝑀 )+(𝐶𝐶
𝑃𝑃 𝑁𝑁𝜔𝜔
0𝑃𝑃
0𝐶𝐶(𝐶𝐶
𝑃𝑃 0𝐴𝐴
0
𝑀𝑀
can solve the two simultaneous
but
must
be
substituted
for
arctan.
This
transformation
enables
the
symbolic
)+(𝐶𝐶
)
(𝐶𝐶
))
2
)+(𝐶𝐶
)
𝑁𝑁𝜔𝜔
cos(Φ
𝑁𝑁𝜔𝜔
𝑁𝑁𝜔𝜔
cos(Φ
𝐾𝐾
𝐾𝐾
+2𝐾𝐾𝐶𝐶
𝑁𝑁𝜔𝜔
𝑁𝑁𝜔𝜔
𝜔𝜔
𝐾𝐾𝐾𝐾√1−cos
𝑃𝑃
0
0𝐵𝐵
0
𝑀𝑀
=
=
−
(
0
𝑀𝑀
0
0
𝑃𝑃
0
𝑀𝑀
2
2
2
2
2
2
𝑃𝑃
0
𝑀𝑀
𝑃𝑃
0
0
𝑀𝑀
)+(𝐶𝐶
(Φ
)
𝐾𝐾0 −
+2𝐾𝐾𝐶𝐶
𝜔𝜔0 𝐾𝐾𝐾𝐾√1−cos
0𝐴𝐴
0𝐴𝐴(𝐶𝐶
+2𝐾𝐾𝐶𝐶=
𝑁𝑁𝜔𝜔
𝑁𝑁𝜔𝜔
+𝐾𝐾(cos(Φ
𝐾𝐾 𝑅𝑅
2 0 )
2
𝑃𝑃 𝑁𝑁𝜔𝜔2𝑀𝑀
0 ))cos(Φ
𝑀𝑀
𝑃𝑃 𝑁𝑁𝜔𝜔0 )
𝑀𝑀 𝑁𝑁𝜔𝜔 2 )2 ) 𝐶𝐶 𝑁𝑁𝜔𝜔0
𝑃𝑃
0(𝐾𝐾
𝑀𝑀 )+(𝐶𝐶
𝑃𝑃 𝑁𝑁𝜔𝜔
𝑅𝑅−−
=22+2𝐾𝐾𝐶𝐶
𝐶𝐶−−
𝑁𝑁𝜔𝜔
=
((0D=
0𝐴𝐴
0𝐴𝐴
𝑃𝑃 𝑁𝑁𝜔𝜔𝑃𝑃
0 2 cos(Φ
𝑀𝑀 )+(𝐶𝐶
𝑃𝑃 R 0,2C2 2;))R
0 (𝐶𝐶
𝑃𝑃
0 +𝐾𝐾2cos(Φ𝑀𝑀 )) ))
2 +2𝐾𝐾𝐶𝐶
20C𝐶𝐶
2 cos(Φ
2𝑁𝑁𝜔𝜔
(𝐾𝐾cos(Φ
𝑅𝑅0𝐵𝐵
the
following
solution
sets
(R
,𝑀𝑀C)+(𝐶𝐶
,0𝐵𝐵
C
and
R
, C0D𝑁𝑁𝜔𝜔
). See
Appendix
processor to solve for R0 and C0, yielding
22(𝐶𝐶the
222+𝐾𝐾
(𝐶𝐶
))
𝑁𝑁𝜔𝜔
𝑁𝑁𝜔𝜔
𝑁𝑁𝜔𝜔
+𝐾𝐾
cos(Φ
𝐾𝐾
0A
0A ;𝑁𝑁𝜔𝜔
0B𝑁𝑁𝜔𝜔
0B
0C ; =
0𝐵𝐵 =
0𝐵𝐵
)+(𝐶𝐶
)
))
+2𝐾𝐾𝐶𝐶
𝑁𝑁𝜔𝜔
cos(Φ
𝑁𝑁𝜔𝜔
cos(Φ
2
2
2
2
𝑃𝑃
0
𝑃𝑃
0
0
𝑃𝑃
0
𝑀𝑀
2
2
2
𝑀𝑀𝑀𝑀 )+(𝐶𝐶
𝑃𝑃𝑃𝑃2𝑁𝑁𝜔𝜔
00 )𝐾𝐾 +2𝐾𝐾𝐶𝐶 𝑁𝑁𝜔𝜔 cos(Φ 𝑁𝑁𝜔𝜔
0 (𝐶𝐶
𝑃𝑃𝑃𝑃 𝑁𝑁𝜔𝜔
0)
𝑀𝑀𝑀𝑀 ))
2 (Φ𝑃𝑃𝑃𝑃 𝑁𝑁𝜔𝜔
cos(Φ
𝐾𝐾 +2𝐾𝐾𝐶𝐶
2 cos(Φ
) 00 𝜔𝜔
0 +𝐾𝐾
𝜔𝜔0to𝐾𝐾𝐾𝐾√1−cos
𝑃𝑃
0
𝑀𝑀 )+(𝐶𝐶
𝑃𝑃 𝑁𝑁𝜔𝜔
0𝑃𝑃 𝑁𝑁𝜔𝜔
𝑀𝑀function.
(Φ𝑀𝑀
)
𝐾𝐾02 +2𝐾𝐾𝐶𝐶
for details on transforming
use the arccos
0 cos(Φ𝑀𝑀 )+(𝐶𝐶𝑃𝑃 𝑁𝑁𝜔𝜔
0 𝐾𝐾𝐾𝐾√1−cos
2
2
2
2
( 2 +2𝐾𝐾𝐶𝐶12𝑁𝑁𝜔𝜔
)
𝐶𝐶
=
−
(
)
𝑅𝑅0𝐵𝐵 = −Equation
)+(𝐶𝐶
𝑁𝑁𝜔𝜔20 cos(Φ
𝜔𝜔0 𝐾𝐾𝐾𝐾√1−cos
= − ( 2 +2𝐾𝐾𝐶𝐶
) 𝐶𝐶
−𝐾𝐾 (+2𝐾𝐾𝐶𝐶
𝑅𝑅 2 cos(Φ
2 )2 2 0𝐵𝐵(Φ2𝑀𝑀 )
2 +𝐾𝐾
𝑃𝑃))
𝑀𝑀
𝑃𝑃 𝑁𝑁𝜔𝜔0 )
2 cos(Φ
0𝐵𝐵 0=
2
𝐾𝐾
)+(𝐶𝐶
𝐾𝐾2 +2𝐾𝐾𝐶𝐶
𝑁𝑁𝜔𝜔
𝜔𝜔
𝑃𝑃
𝑃𝑃 𝑁𝑁𝜔𝜔
𝑀𝑀
)+(𝐶𝐶
(𝐶𝐶𝑃𝑃0𝑁𝑁𝜔𝜔
))
𝑁𝑁𝜔𝜔
𝑁𝑁𝜔𝜔
𝐾𝐾 𝑃𝑃 𝑁𝑁𝜔𝜔0𝑃𝑃
𝑃𝑃𝑀𝑀
0 𝐾𝐾𝐾𝐾√1−cos
𝑀𝑀 )𝑃𝑃 𝑁𝑁𝜔𝜔0𝑁𝑁𝜔𝜔
= − ( 𝑀𝑀2()+(𝐶𝐶
)2 )20)𝐶𝐶2 (𝐶𝐶
=−
(−cos(Φ
𝑅𝑅0𝐵𝐵𝑅𝑅00𝐵𝐵
0 cos(Φ𝑀𝑀(Φ
0𝑃𝑃 𝑁𝑁𝜔𝜔
0 +𝐾𝐾𝑀𝑀cos(Φ
0𝐵𝐵𝐶𝐶
2 cos(Φ
2 )2
2 (𝐶𝐶
2 +𝐾𝐾
=
(
)+(𝐶𝐶
𝑁𝑁𝜔𝜔
𝑁𝑁𝜔𝜔
𝑁𝑁𝜔𝜔
𝑁𝑁𝜔𝜔
cos(Φ
𝐾𝐾 +2𝐾𝐾𝐶𝐶
0𝐵𝐵 = −
0𝐵𝐵
𝑃𝑃
0
𝑀𝑀
𝑃𝑃
0
0
𝑃𝑃
0
𝑀𝑀 )) ))
2 2 (𝐶𝐶 𝑁𝑁𝜔𝜔 2 +𝐾𝐾 cos(Φ
2 +2𝐾𝐾𝐶𝐶 𝑁𝑁𝜔𝜔 2 cos(Φ )+(𝐶𝐶
2 )2 2
𝑁𝑁𝜔𝜔
𝑁𝑁𝜔𝜔
𝐾𝐾
2
2
2
𝑃𝑃
0
𝑀𝑀𝐾𝐾𝐾𝐾 + 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝑃𝑃 𝑃𝑃𝑃𝑃0𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 cos( Φ𝑀𝑀𝑀𝑀) + �𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 � 0
𝑃𝑃
0
𝑀𝑀
𝜔𝜔𝜔𝜔0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1− cos ( Φ𝑀𝑀𝑀𝑀)
�
𝐶𝐶𝐶𝐶0𝐴𝐴𝐴𝐴 = − �
𝑅𝑅𝑅𝑅0𝐴𝐴𝐴𝐴 = 2
2
2 2
2
2
1 = (𝜔𝜔
2
0 𝑁𝑁𝐶𝐶𝑃𝑃
𝐾𝐾𝐾𝐾 + 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 cos( Φ𝑀𝑀𝑀𝑀) + ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 )
𝜔𝜔𝜔𝜔0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−cos2( Φ𝑀𝑀𝑀𝑀)
�
2
𝐾𝐾𝐾𝐾 + 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 cos( Φ𝑀𝑀𝑀𝑀) + ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 ) 2
𝑅𝑅𝑅𝑅0𝐵𝐵𝐵𝐵 = − �
𝑅𝑅𝑅𝑅 0𝐶𝐶𝐶𝐶 =
𝜔𝜔𝜔𝜔0 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1− cos2( Φ𝑀𝑀𝑀𝑀)
2
𝐾𝐾𝐾𝐾 − 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 cos( Φ𝑀𝑀𝑀𝑀) + ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 ) 2
𝑅𝑅𝑅𝑅0𝐷𝐷𝐷𝐷 = − �
𝜔𝜔𝜔𝜔0 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1− cos2( Φ𝑀𝑀𝑀𝑀)
�
𝐾𝐾𝐾𝐾2− 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 cos( Φ𝑀𝑀𝑀𝑀) + ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 ) 2
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 + 𝐾𝐾𝐾𝐾 cos( Φ𝑀𝑀𝑀𝑀))
𝐾𝐾𝐾𝐾2 + 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 cos( Φ𝑀𝑀𝑀𝑀) + �𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 2�
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 + 𝐾𝐾𝐾𝐾 cos( Φ𝑀𝑀𝑀𝑀))
𝐶𝐶𝐶𝐶0𝐵𝐵𝐵𝐵 = − �
𝐾𝐾𝐾𝐾2 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 2 cos( Φ𝑀𝑀𝑀𝑀) + �𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02�
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 2( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 − 𝐾𝐾𝐾𝐾 cos( Φ𝑀𝑀𝑀𝑀))
𝐶𝐶𝐶𝐶0𝐶𝐶𝐶𝐶 = − �
2
𝐾𝐾𝐾𝐾2 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 2 cos( Φ𝑀𝑀𝑀𝑀) + �𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02�
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 2( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 − 𝐾𝐾𝐾𝐾 cos( Φ𝑀𝑀𝑀𝑀))
𝐶𝐶𝐶𝐶0𝐷𝐷𝐷𝐷 = − �
2
�
�
2
�
𝐾𝐾𝐾𝐾 2 − 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔0 2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐( Φ𝑀𝑀𝑀𝑀 ) + (𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔0 2) 2 (13)
2
b 2 = a 2 + c 2 – (2ac)cos (β) (14)
𝑅𝑅𝑅𝑅0 =
2
𝜔𝜔𝜔𝜔0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1− cos2( Φ𝑀𝑀𝑀𝑀)
2
2 2
(15)
Analog Dialogue 49-02, February 2015
𝑅𝑅𝑅𝑅 =
b 2 = a 2 + c 2 – (2ac)cos (β) (14)
𝜔𝜔𝜔𝜔0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1− cos2( Φ𝑀𝑀𝑀𝑀)
(15)
0
2 in Figure
2 2
( Φ𝑀𝑀𝑀𝑀) +3,( 𝐶𝐶𝐶𝐶in
This is depicted
graphically
which
𝐾𝐾𝐾𝐾2−2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 cos
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
0 ) the left and
right sides of Equation 17 are each equated2(to y)(blue and
𝜔𝜔𝜔𝜔0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1− cos Φ𝑀𝑀𝑀𝑀
green curves)𝑅𝑅𝑅𝑅
with
horizontal axis sharing ω and ϕ 2.
(15)
0 =the
2−2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔 2 cos( Φ ) + ( 𝐶𝐶𝐶𝐶0 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔 2)M
𝐾𝐾𝐾𝐾
𝑃𝑃𝑃𝑃
0 marks 𝑀𝑀𝑀𝑀
𝑃𝑃𝑃𝑃
0
2
The intersection
of the two2curves
the
boundary
2
2
𝐾𝐾𝐾𝐾 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 cos( Φ𝑀𝑀𝑀𝑀)2+ �𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 �
2 + 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
and
.+
The
under which Equation
condition
for 0ω20cos
𝐶𝐶𝐶𝐶0𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
=
(16)
( Φϕ𝑀𝑀𝑀𝑀
�𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃condition
𝐾𝐾𝐾𝐾
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0) 2�
2M()𝐾𝐾𝐾𝐾cos
2
(Φ
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
− 𝐶𝐶𝐶𝐶portion
0
𝑀𝑀𝑀𝑀The
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 ) of the horizontal
�
�
𝐶𝐶𝐶𝐶
=
−
𝑅𝑅𝑅𝑅0𝐴𝐴𝐴𝐴 = 2
17
is
true
appears
as
the
red
arc.
0𝐴𝐴𝐴𝐴
2 cos( Φ ) + ( 𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔 2 ) 2
2 ( 𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔 2 + 𝐾𝐾𝐾𝐾 cos( Φ ))
𝐾𝐾𝐾𝐾
+
2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
2
0
𝑀𝑀𝑀𝑀 modeling
𝑃𝑃𝑃𝑃
0a PLL, all of the variables
0
𝑃𝑃𝑃𝑃
0
𝑀𝑀𝑀𝑀
2 range
2 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
Note that in𝑃𝑃𝑃𝑃the context
of
( Φ𝑀𝑀𝑀𝑀of
) 2+ϕ2�M𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃and
axis beneath
the red
defines
𝐾𝐾𝐾𝐾arc
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0ω20�that
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔the
0 cos
2 cos
2 + 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
2( Φ )
(
)
�
�
𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
cos
𝐾𝐾𝐾𝐾
Φ
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
+
𝐶𝐶𝐶𝐶
2
𝐶𝐶𝐶𝐶
=
0
𝑀𝑀𝑀𝑀
0
𝑀𝑀𝑀𝑀
𝑃𝑃𝑃𝑃 the
0 horizontal
in the above equations
possess
values, including
2 0cos𝑃𝑃𝑃𝑃( Φ Note
2 on
2 + 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶�
ensures
is 0positive.
point
2(positive
2𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
( 𝐾𝐾𝐾𝐾cos
� 02 ) axis (16)
𝐾𝐾𝐾𝐾
Φ𝑀𝑀𝑀𝑀)
+ �𝐶𝐶𝐶𝐶0the
𝑅𝑅𝑅𝑅�0𝐴𝐴𝐴𝐴 = 2 𝜔𝜔𝜔𝜔0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−cos
𝐶𝐶𝐶𝐶�0𝐴𝐴𝐴𝐴
= − C𝑃𝑃𝑃𝑃0𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝑀𝑀𝑀𝑀) 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝑃𝑃𝑃𝑃
0 �( Φ𝑀𝑀𝑀𝑀) − 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
2
2
2
2
2
(
(
)
(
)
(
))
2
�
�
constrained
between
0 and π/2.
𝐾𝐾𝐾𝐾 +ϕ2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 cos Φto
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔
+ 𝐾𝐾𝐾𝐾
+ 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝐶𝐶𝐶𝐶
= − M ) because
= − directly𝐾𝐾𝐾𝐾below
𝑅𝑅𝑅𝑅0𝐵𝐵𝐵𝐵 cos(ϕ
𝐶𝐶𝐶𝐶0𝐵𝐵𝐵𝐵
𝑀𝑀𝑀𝑀 values
0
0)
𝑀𝑀𝑀𝑀 and green curves
the
intersection
ofcos
the Φ
blue
M is 𝑃𝑃𝑃𝑃
Φ
(17)
>
2 ( 𝐶𝐶𝐶𝐶 (𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
2
𝑀𝑀𝑀𝑀
𝑃𝑃𝑃𝑃 (0Φ𝑀𝑀𝑀𝑀0
))
𝐾𝐾𝐾𝐾2 + 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 cos( Φ𝑀𝑀𝑀𝑀) + ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 ) 2
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0cos
𝑃𝑃𝑃𝑃
0 + 𝐾𝐾𝐾𝐾 cos
As a result,
C0A and R0B are clearly negative quantities.
,
the
maximum
value
of ϕM 2to ensure C0
establishes
ϕ
M
_M
A
X
2( Φ )
𝜔𝜔𝜔𝜔0C𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−cos
𝐾𝐾𝐾𝐾2 +2 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 cos( Φ𝑀𝑀𝑀𝑀)2 +2 �𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 2�
𝑀𝑀𝑀𝑀
Therefore,
R0A
2(, Φ
is2positive.
0A)and R0B , C0B are immediately
(
)
�
�
� 2setscos
�
�
�
𝜔𝜔𝜔𝜔−
𝐾𝐾𝐾𝐾
−2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
cos
Φ
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
+
𝐶𝐶𝐶𝐶
=
−
𝑅𝑅𝑅𝑅0𝐵𝐵𝐵𝐵 =solution
𝐶𝐶𝐶𝐶
0 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
𝑀𝑀𝑀𝑀
𝑃𝑃𝑃𝑃
0
𝑀𝑀𝑀𝑀
𝑃𝑃𝑃𝑃
0
2
0𝐵𝐵𝐵𝐵
2 2
2
2 +𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
cos
𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔
(�Φ
) + ( 𝐶𝐶𝐶𝐶are
+ 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 component
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 cos( Φ𝑀𝑀𝑀𝑀
𝐾𝐾𝐾𝐾𝑃𝑃𝑃𝑃cos
𝑅𝑅𝑅𝑅0𝐶𝐶𝐶𝐶 =
ruled2 out because𝐾𝐾𝐾𝐾2negative
values
not
𝑀𝑀𝑀𝑀 )0>
0𝑀𝑀𝑀𝑀)) (17)
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
0 ) 𝐶𝐶𝐶𝐶0𝐶𝐶𝐶𝐶 = − �
0 ((𝐶𝐶𝐶𝐶Φ
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾 − 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 cos( Φ𝑀𝑀𝑀𝑀) + ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 ) 2
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 2( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 − 𝐾𝐾𝐾𝐾 cos( Φ𝑀𝑀𝑀𝑀))
𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾02
acceptable. The R0C , C0C and R0D , C0D results require further
2
Φ𝑀𝑀𝑀𝑀_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀
=𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
� +cos
�Φradians
(18)
2𝜔𝜔𝜔𝜔0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
( Φ𝑀𝑀𝑀𝑀) 𝐾𝐾𝐾𝐾
�𝐶𝐶𝐶𝐶2𝑃𝑃𝑃𝑃(𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝑀𝑀𝑀𝑀0)2�
𝜔𝜔𝜔𝜔0 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
cos2( Φ𝑀𝑀𝑀𝑀)
𝐾𝐾𝐾𝐾2 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝑃𝑃𝑃𝑃(
0) cos
2 cos
2�2
2 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
analysis,
however.
2( Φ
)
�𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝑅𝑅𝑅𝑅
=
𝐶𝐶𝐶𝐶0𝐴𝐴𝐴𝐴
𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
cos
𝐾𝐾𝐾𝐾
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
Φ
+
�
�
=
𝐶𝐶𝐶𝐶
=
−
𝑅𝑅𝑅𝑅
0
𝑀𝑀𝑀𝑀
𝑃𝑃𝑃𝑃
0
𝑀𝑀𝑀𝑀
0
0𝐴𝐴𝐴𝐴
2
2
2
2
(Φ
)+
( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃2𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 )
20− 𝐾𝐾𝐾𝐾
𝐾𝐾𝐾𝐾 +2(2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
cos
� 2 )2
�𝐶𝐶𝐶𝐶𝑀𝑀𝑀𝑀
( Φ ) + ( 𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
(Φ
))
𝑅𝑅𝑅𝑅0𝐷𝐷𝐷𝐷 = − �0𝐶𝐶𝐶𝐶2 𝐾𝐾𝐾𝐾2 − 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶2𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔(02 cos
𝐶𝐶𝐶𝐶0𝐷𝐷𝐷𝐷 = − �0𝐶𝐶𝐶𝐶
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝑀𝑀𝑀𝑀
2cos
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
Equation
than
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾
0 K in order
(0Φ
( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃requires
))0 be less
P Nω
𝐾𝐾𝐾𝐾 − 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 cos Φ𝑀𝑀𝑀𝑀) + ( 𝐶𝐶𝐶𝐶𝑀𝑀𝑀𝑀𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02𝑃𝑃𝑃𝑃) 2 0
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 218
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔002 − 𝐾𝐾𝐾𝐾that
cosC
𝑀𝑀𝑀𝑀
Φ𝑀𝑀𝑀𝑀_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀 =of𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
� function
� radians (18)
Note that the four equations involving R0C , C0C and R0D , C0D
to satisfy the
constraints
the
arccos
𝐾𝐾𝐾𝐾2( 2 2 )for ϕM_MAX
(Φ
𝜔𝜔𝜔𝜔0 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1− cos2( Φ𝑀𝑀𝑀𝑀)
𝐾𝐾𝐾𝐾2 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 2 cos𝜔𝜔𝜔𝜔
0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−cos
𝑀𝑀𝑀𝑀) + �𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0Φ�𝑀𝑀𝑀𝑀
possess
the
common
factor:
�
�
=
−
𝑅𝑅𝑅𝑅
𝐶𝐶𝐶𝐶0𝐵𝐵𝐵𝐵
�
𝐾𝐾𝐾𝐾 establishes
between
0 and π/2.2This
ω0_MAX , the �upper 2limit
= − �0𝐵𝐵𝐵𝐵
𝑅𝑅𝑅𝑅0𝐷𝐷𝐷𝐷 = − � 2
𝐶𝐶𝐶𝐶0𝐷𝐷𝐷𝐷
2 cos
2
(2𝑀𝑀𝑀𝑀
) + ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 ) 2
𝐾𝐾𝐾𝐾
2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
Φ))
(Φ
(2𝐶𝐶𝐶𝐶+
𝐾𝐾𝐾𝐾 − 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 cos( Φ𝑀𝑀𝑀𝑀) + ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 ) 2
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾 0cos
𝜔𝜔𝜔𝜔
=
radians/s
(19)
𝑃𝑃𝑃𝑃−
𝑀𝑀𝑀𝑀
�
0is
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
0
2
0_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀
2
2
2
on
ω
to
ensure
C
positive.
2
0+2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
0cos
𝐾𝐾𝐾𝐾))+
𝜔𝜔𝜔𝜔0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−2cos
cos2((Φ
2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 cos
Φ
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02��
Φ𝑀𝑀𝑀𝑀))
+��𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
((𝐶𝐶𝐶𝐶
𝑃𝑃𝑃𝑃𝑀𝑀𝑀𝑀
𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾2 +
Φ
2
2 2
This result is problematic because the goal was to solve for R0
and C0 given ω0 and ϕM , but this indicates four possible R0, C0
pairs instead of a unique R0, C0 pair. However, closer
inspection of the four results leads to a single solution set as
follows.
𝜔𝜔𝜔𝜔0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1− cos2( Φ𝑀𝑀𝑀𝑀)
𝑀𝑀𝑀𝑀
=𝐾𝐾𝐾𝐾 22 − 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶0𝑃𝑃𝑃𝑃 𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔
0 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐( Φ𝑀𝑀𝑀𝑀 ) + (𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔0 )
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅0𝐴𝐴𝐴𝐴
0𝐴𝐴𝐴𝐴 =
𝐾𝐾𝐾𝐾 + 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔 22 cos(( Φ )) +(( 𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔 22))22
(13)
=−
−��
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶0𝐴𝐴𝐴𝐴
0𝐴𝐴𝐴𝐴 =
��
𝜔𝜔𝜔𝜔cos
))cos2( Φ𝑀𝑀𝑀𝑀)
0 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0022((𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0022+
+𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾
cos
Φ
((Φ
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝑀𝑀𝑀𝑀
𝑀𝑀𝑀𝑀))
𝐶𝐶𝐶𝐶0𝐶𝐶𝐶𝐶
𝑅𝑅𝑅𝑅
0𝐶𝐶𝐶𝐶 =𝜔𝜔𝜔𝜔𝐾𝐾𝐾𝐾0_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀
radians/s
2
2 )2
2 − 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶=𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
�0𝐶𝐶𝐶𝐶 cos
( Φ𝑀𝑀𝑀𝑀22) + ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0(19)
𝑃𝑃𝑃𝑃
𝐾𝐾𝐾𝐾
2
2
2
2
Closer inspection
reveals
that
Expression
13
has
the
form
𝑃𝑃𝑃𝑃
2
2
2
2
𝐾𝐾𝐾𝐾 −
2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
(13)𝐾𝐾𝐾𝐾2 + 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝜔𝜔𝜔𝜔0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−cos
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−cos
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔002 cos
cos((Φ
Φ𝑀𝑀𝑀𝑀
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02��
Φ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
+��𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝜔𝜔𝜔𝜔
𝑃𝑃𝑃𝑃 𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔20((Φ
𝑀𝑀𝑀𝑀)) ( Φ𝑀𝑀𝑀𝑀 ) + (𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔0 )
𝑀𝑀𝑀𝑀))+
𝑀𝑀𝑀𝑀
2
�
=
−
=−
−��𝐾𝐾𝐾𝐾 + 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
–b(2ac)cos(β)
+c c22.0–Equating
this(β)
with
the arbitrary
quantity,
��𝐾𝐾𝐾𝐾a22 2++
�� 2(Φ𝑀𝑀𝑀𝑀)
ΔΦ
–𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
arctan
(ω
R
C2))))) 0 (20)
−
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅0𝐵𝐵𝐵𝐵
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶0𝐵𝐵𝐵𝐵
0𝜔𝜔𝜔𝜔
2𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
2=
0𝐵𝐵𝐵𝐵a =
0𝐵𝐵𝐵𝐵 =
cos
2 cos(( Φ
2))22�
2=
2 + 𝐾𝐾𝐾𝐾
(2ac)cos
(14)
0
2
2
2
2
(
)
(
(
2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
cos
Φ
+
𝐶𝐶𝐶𝐶
𝐶𝐶𝐶𝐶
(
)
(
(
𝐾𝐾𝐾𝐾 + 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔00 cos Φ𝑀𝑀𝑀𝑀
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔�00 + 𝐾𝐾𝐾𝐾 cos Φ𝑀𝑀𝑀𝑀
𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃−
𝑀𝑀𝑀𝑀 + 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔00
𝑀𝑀𝑀𝑀
00 =
�
𝑅𝑅𝑅𝑅y 0𝐷𝐷𝐷𝐷
𝐶𝐶𝐶𝐶0𝐷𝐷𝐷𝐷
b2, yields:
2
2 )2
2
(
) (
𝐾𝐾𝐾𝐾 − 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 cos
2 Φ𝑀𝑀𝑀𝑀 + 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0
2
2
2
2
2
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
cos22((Φ
𝐾𝐾𝐾𝐾2 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0ΔΦ
cos
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔002R�� C ) (20)
+��𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃(ω
))
))+
𝜔𝜔𝜔𝜔
cos
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
02 cos
=((ΦΦ
–𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀
arctan
0 2 ��2
b𝜔𝜔𝜔𝜔020𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
= 2a 2 +
c –Φ𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀
(2ac)cos
(β) (14) 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶0𝐶𝐶𝐶𝐶
= 22
=−
−��𝐾𝐾𝐾𝐾 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃2𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅0𝐶𝐶𝐶𝐶
0𝐶𝐶𝐶𝐶 =
0𝐶𝐶𝐶𝐶 =
y = CPN𝛚02
2))22
2(( 𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
2 − 𝐾𝐾𝐾𝐾–cos
2
2
2
(
(
)
(
))Φ
𝐾𝐾𝐾𝐾
−
2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
cos
Φ
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
Φ
+
𝐶𝐶𝐶𝐶
(
(
)
(
=
Φ
ΔΦ
=
(21)
Φ
2
𝐾𝐾𝐾𝐾 − 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔00𝜔𝜔𝜔𝜔cos
Φ𝑀𝑀𝑀𝑀
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔M_
+ 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃(𝑃𝑃𝑃𝑃Φ
𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔00 −M𝐾𝐾𝐾𝐾 cos Φ𝑀𝑀𝑀𝑀
𝑀𝑀𝑀𝑀 cos
𝑀𝑀𝑀𝑀))
M + arctan(ω0 R2 C2 )
)0
00 MAX
0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
𝑀𝑀𝑀𝑀0
(15)
𝑅𝑅𝑅𝑅0 = 2
K
2
2
2
2
2
2
2 cos
( Φ2𝑀𝑀𝑀𝑀
2(()Φ
𝐾𝐾𝐾𝐾 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
cos
+𝑀𝑀𝑀𝑀)()𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
(13)
Equation 14,
the Law
of02Cosines,
a, 0b,2)and
c as the
�2 ( Φ𝑀𝑀𝑀𝑀 ) + (𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔0 )
𝜔𝜔𝜔𝜔𝑃𝑃𝑃𝑃00𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
cos
𝐾𝐾𝐾𝐾22 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔002𝐾𝐾𝐾𝐾
Φ2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
+�𝑃𝑃𝑃𝑃�𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔
00022�𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
((Φ
𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
cos
cos−
Φrelates
𝑀𝑀𝑀𝑀))+
𝑀𝑀𝑀𝑀
𝑀𝑀𝑀𝑀
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
�𝐾𝐾𝐾𝐾 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
� ) the interior
�
=−
−��of22the three
=
−
𝐶𝐶𝐶𝐶
�
�
�
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅0𝐷𝐷𝐷𝐷
=
=
−
𝐶𝐶𝐶𝐶
22(βΦ
0𝐷𝐷𝐷𝐷lengths
0𝐷𝐷𝐷𝐷
sides
of
a
triangle
with
being
0𝐷𝐷𝐷𝐷
=
Φ
–
ΔΦ
=
Φ
+
arctan
(ω
R
C
)
Φ
𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
cos
2
2
2
2
M_ 0MAX
M
M
0 2 2
2 −𝐾𝐾𝐾𝐾
0𝑀𝑀𝑀𝑀))+
))
−2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
cos((Φ
Φ𝑀𝑀𝑀𝑀
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔002 ))2 𝑀𝑀𝑀𝑀
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔002((𝐶𝐶𝐶𝐶2𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾cos
cos
Φ𝑀𝑀𝑀𝑀
+((𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
))
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾 −
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
2 ((Φ
2 2
0 −
𝑅𝑅𝑅𝑅
=𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0202𝜔𝜔𝜔𝜔cos
(𝑀𝑀𝑀𝑀Φ+
( Φ𝑀𝑀𝑀𝑀
+ 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
+ �𝐶𝐶𝐶𝐶=
0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
𝑃𝑃𝑃𝑃=
0M_cos
𝑀𝑀𝑀𝑀)ΔΦ
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
0 �
angle of the vertex
side
b.cos
b2)is the square
of the (15)
arccos
(ω0 2 NCP/K) – arc
Φ
2Since
2 2
M_𝐾𝐾𝐾𝐾
MAX_NEW
MAX
2Φ
𝐾𝐾𝐾𝐾 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
�
K�
(𝚽
)
>
C
N𝛚
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 cos( Φ𝑀𝑀𝑀𝑀) + ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 )
𝑅𝑅𝑅𝑅0𝐴𝐴𝐴𝐴 = 02 opposite
𝐶𝐶𝐶𝐶
=
−
COS
M
P 2 0
0𝐴𝐴𝐴𝐴
2 2 2
2 it must
2
( Φ𝑀𝑀𝑀𝑀)be
length of one
side
a triangle,
𝐾𝐾𝐾𝐾 +of2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
cos( Φ𝑀𝑀𝑀𝑀))
+ (a𝐶𝐶𝐶𝐶positive
2=
𝑃𝑃𝑃𝑃2𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 cos
𝑃𝑃𝑃𝑃2𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 ) quantity,
0 (a𝐶𝐶𝐶𝐶2𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾2 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
2
b𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
+ c02 +– 𝐾𝐾𝐾𝐾(2ac)cos
(β)
(14)
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 cos( Φ𝑀𝑀𝑀𝑀) + �𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 �
KCOS (𝚽M) =
CPN𝛚
0
which
the 2right side of Equation2 14 must also
be
𝐶𝐶𝐶𝐶0 implies
=
(16)
2
2
2
2
2
(
(
)
)
2 cos
2�=
2Φ
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾cos
Φ
−
𝐶𝐶𝐶𝐶
2
2
2
2
2
=
Φ
+
ΔΦ
arccos
(ω0 2 NCP/K)
M_
MAX_NEW
M_
MAX
0
𝑀𝑀𝑀𝑀
𝑃𝑃𝑃𝑃
0
(
(
)
)
�
(
)
(
)
𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−cos
𝐾𝐾𝐾𝐾
+
2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
Φ
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
Φ
+
𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾 −
− 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶�𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔
𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔00 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
Φ𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀)be
𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔00quantity,
(13)
+a(positive
) 2�(13)
𝑃𝑃𝑃𝑃
0
𝑀𝑀𝑀𝑀
𝑃𝑃𝑃𝑃
0
𝐾𝐾𝐾𝐾
Φ
𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔
+
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑀𝑀𝑀𝑀
positive.
130(must
�
�
= −Expression
=
−
𝑅𝑅𝑅𝑅0𝐵𝐵𝐵𝐵Thus,
𝐶𝐶𝐶𝐶
2
2
2
0𝐵𝐵𝐵𝐵
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0of
cos((Φ
Φpositive.
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
2 �2
2 +−2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
02 cos
𝑀𝑀𝑀𝑀))++(�𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾
2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 + 𝐾𝐾𝐾𝐾 cos( Φ𝑀𝑀𝑀𝑀))
which means𝐶𝐶𝐶𝐶the=denominator
R
The
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
00 )
0D is𝑀𝑀𝑀𝑀
(16)
0
2 ( 𝐾𝐾𝐾𝐾cos
2 ) be
𝜔𝜔𝜔𝜔0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1− cos2( Φ𝑀𝑀𝑀𝑀)
(
)
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
Φ
−
𝐶𝐶𝐶𝐶
numerator of R0D is also𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
positive,
therefore
R
must
0
𝑀𝑀𝑀𝑀
𝑃𝑃𝑃𝑃 0D 0
𝑅𝑅𝑅𝑅
=
(15)
2 cos( Φ ) + �𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔 2�2
2 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
2( Φ )
2
0
𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
cos
𝐾𝐾𝐾𝐾
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
2 𝑃𝑃𝑃𝑃( Φ 0) + ( 𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔 2) 2
2−2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
0 the R0D , C
𝑀𝑀𝑀𝑀
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾0
set. This leaves
negative,
out
𝐾𝐾𝐾𝐾bbcos
Φ
𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔
(17)
>
𝑃𝑃𝑃𝑃 𝑀𝑀𝑀𝑀 0 cos
𝑀𝑀𝑀𝑀
0D solution
0
� 𝑃𝑃𝑃𝑃) 0
=(which
a22𝑀𝑀𝑀𝑀+
+)2rules
c22𝐶𝐶𝐶𝐶
–𝑃𝑃𝑃𝑃(2ac)cos
(2ac)cos
(β)
(14)
𝐶𝐶𝐶𝐶0𝐶𝐶𝐶𝐶 = − �
𝑅𝑅𝑅𝑅220𝐶𝐶𝐶𝐶
=
a=
c
–
(β)
(14)
2
2
2
2
2
y
=
K
(𝚽
(
(
)
(
)
(
))
𝐾𝐾𝐾𝐾 − 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔 cos Φ + 𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔 𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔 − 𝐾𝐾𝐾𝐾 cos Φ
𝐾𝐾𝐾𝐾 + 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔00 cos Φ𝑀𝑀𝑀𝑀
𝑀𝑀𝑀𝑀 + 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔00
only the R0C , C0C pair as𝑃𝑃𝑃𝑃 a contender
0
𝑀𝑀𝑀𝑀for the𝑃𝑃𝑃𝑃 simultaneous
0
solution of Equation
11
and
Equation
12.
2
(
)
𝐾𝐾𝐾𝐾 cos Φ > 𝐶𝐶𝐶𝐶 𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔 2 (17)
𝑀𝑀𝑀𝑀 𝜔𝜔𝜔𝜔 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
𝑃𝑃𝑃𝑃
0 (Φ )
cos
0
𝑀𝑀𝑀𝑀
�
𝑅𝑅𝑅𝑅0𝐷𝐷𝐷𝐷 = − � 2𝜔𝜔𝜔𝜔 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾20222(
2 2
cos
Φradians
(𝑀𝑀𝑀𝑀
(�Φ
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔−00𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1−
2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
Φ))𝑀𝑀𝑀𝑀) + ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
cos
𝑀𝑀𝑀𝑀
Φ𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑀𝑀𝑀𝑀_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀
=
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
�
(18)
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
0 cos
0 )
=
(15)
=
(15)
𝐾𝐾𝐾𝐾
0
0
22
22)22
22
(
) (
−2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔00 cos
cos( Φ
Φ𝑀𝑀𝑀𝑀
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔00 )
+( 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝑀𝑀𝑀𝑀) +
𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾02
Φ𝑀𝑀𝑀𝑀_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 �
� radians (18)
(𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔0 2) 2
𝐾𝐾𝐾𝐾 2 − 2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔0 2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐(𝐾𝐾𝐾𝐾Φ𝑀𝑀𝑀𝑀 ) +
𝐾𝐾𝐾𝐾
22
2
2
2
2
2
2
(
)
�
�
𝐾𝐾𝐾𝐾 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔radians/s
cos( Φ
Φ𝑀𝑀𝑀𝑀
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔00 �
+�(19)
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝑀𝑀𝑀𝑀) +
𝜔𝜔𝜔𝜔𝐶𝐶𝐶𝐶0_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀
�𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
00 cos
= 𝐾𝐾𝐾𝐾 =−2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
(16)
𝐾𝐾𝐾𝐾
𝐶𝐶𝐶𝐶00 =
(16)
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0𝑃𝑃𝑃𝑃022((𝐾𝐾𝐾𝐾cos
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0022))
𝐾𝐾𝐾𝐾cos((Φ
Φ𝑀𝑀𝑀𝑀
−𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
𝑀𝑀𝑀𝑀))−
2 = a 2=+ c 2𝐾𝐾𝐾𝐾– (2ac)cos
(14)
𝜔𝜔𝜔𝜔b0_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀
�𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾 radians/s(β)(19)
(13)
𝑀𝑀𝑀𝑀
𝑃𝑃𝑃𝑃 Equation
0
Although Equation
15 and
16 are contenders for
2( Φ )
𝜔𝜔𝜔𝜔
cosand
0𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾�1− 11
𝑀𝑀𝑀𝑀
the simultaneous
solution
of
Equation
Equation 12,
𝑅𝑅𝑅𝑅
=
0 = –𝐾𝐾𝐾𝐾arctan
ΔΦ
(ω
R
C
)
(20)
2
2
0
2
2
)2
−2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
cos( Φ𝑀𝑀𝑀𝑀)values
+ ( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
they are only valid if they
result
for0 2both
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔in
0 positive
22 arctan(ω R C )
𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾
=
Φ
–
ΔΦ
=
Φ
+
Φ
M_
MAX
M
M
0
2
2
𝐶𝐶𝐶𝐶
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾
R0 and C0. Close inspection of𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃R0 shows
it to be positive—its
00
Φ𝑀𝑀𝑀𝑀_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀
= 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ��
radians
(18)
Φ
=
��radians
(18)
𝑀𝑀𝑀𝑀_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀
numerator
is positive, because 𝐾𝐾𝐾𝐾
the
𝐾𝐾𝐾𝐾 range of cos2(x) is 0 to 1—
0
𝑀𝑀𝑀𝑀
0
𝑃𝑃𝑃𝑃
𝑃𝑃𝑃𝑃
0
0
𝑀𝑀𝑀𝑀
COS
M
2
𝐾𝐾𝐾𝐾2 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 2 cos( Φ𝑀𝑀𝑀𝑀) + �𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02�
2 −2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔 2 cos( Φ ) + �𝐶𝐶𝐶𝐶 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔 2�2
𝐾𝐾𝐾𝐾
0 ( Φ ))𝑀𝑀𝑀𝑀
𝑃𝑃𝑃𝑃
0
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0 2( 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02𝑃𝑃𝑃𝑃− 𝐾𝐾𝐾𝐾 cos
𝑀𝑀𝑀𝑀
0
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 ( 𝐾𝐾𝐾𝐾cos( Φ𝑀𝑀𝑀𝑀) − 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔02 )
𝐶𝐶𝐶𝐶0𝐷𝐷𝐷𝐷 = − �
0
�
𝐶𝐶𝐶𝐶 =
𝐾𝐾𝐾𝐾 cos( Φ𝑀𝑀𝑀𝑀 ) > 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔0 2
(17)
Φ𝑀𝑀𝑀𝑀_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀
= =𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
�
𝛚
[K/(C N)]
� radians
𝚽M = 𝛑/2
2
𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾0
𝚽M_MAX = arccos (CPN𝛚02/K)
𝐶𝐶𝐶𝐶
𝑃𝑃𝑃𝑃
Constraints on R0 and C0 2
ΔΦ
= –((arctan
2 C20)2 (20)
𝐾𝐾𝐾𝐾cos
cos
Φ𝑀𝑀𝑀𝑀))>
𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔
(17)
>(ω
𝐾𝐾𝐾𝐾
Φ
𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔
(17)
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶0𝑃𝑃𝑃𝑃R
𝑃𝑃𝑃𝑃
O_MAX
P
1/2
Figure 3. Constraint on C0 denominator.
𝐾𝐾𝐾𝐾
(16)
𝛚0, 𝚽M
(18)
(15)
Compensating for R2 and C𝐾𝐾𝐾𝐾2 (Third-Order Loop Filter)
(21)
𝜔𝜔𝜔𝜔0_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀 = �
radians/s (19)
𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 filter,
𝐾𝐾𝐾𝐾
In the case of a third-order loop
components R2 and C2
2 was
introduce
additional
phase
shift,
Δϕ,
relative
to the secondand its denominator
is
the
same
as
Expression
13,
which
2
2
2
ΦM𝑃𝑃𝑃𝑃–𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔
ΔΦ
= (Φ
(21)
ΦM_ MAX𝐾𝐾𝐾𝐾 =−2𝐾𝐾𝐾𝐾𝐶𝐶𝐶𝐶
�𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0(ω
� 0 R2 C2 )
ΦM
+ arctan
0 cos
𝑀𝑀𝑀𝑀)+
order loop filter:
previously
shown
to
be
positive.
The
numerator
of
C
is
also
𝐶𝐶𝐶𝐶
=
(16)
2
0
0
(ω 2 NCP/K) – arctan(ω0 R2 C2 )
(22)
ΦM_MAX_NEW = ΦM_MAX
2 (ΔΦ =
( Φarccos
𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔0+
𝑀𝑀𝑀𝑀) − 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝜔𝜔𝜔𝜔00 )
the same as Expression
13,
so C𝐾𝐾𝐾𝐾cos
0 is positive as long as its
ΔΦ = – arctan(ω0 R2 C2 ) (20)
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾
denominator
following condition:
𝜔𝜔𝜔𝜔0_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀
=�
radians/s
(19)
� theradians/s
𝜔𝜔𝜔𝜔
=
(19)
0_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀satisfies
𝐾𝐾𝐾𝐾
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝐾𝐾𝐾𝐾
= ΦM_MAX + ΔΦ = arccos (ω0 2 NCP/K) – arctan(ω0 R2 C2 )
(22)
ΦM_MAX_NEW
To deal with this excess phase shift, subtract it from the
2
(
)
𝐾𝐾𝐾𝐾 cos Φ𝑀𝑀𝑀𝑀 > 𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝑁𝑁𝑁𝑁𝜔𝜔𝜔𝜔0
(17)
desired value of ϕM :
ΦM_ MAX = ΦM – ΔΦ = ΦM + arctan(ω0 R2 C2 )
ΔΦ =
= ––arctan
arctan(ω
(ω00RR22CC22)) (20)
(20)
ΔΦ
Φ𝑀𝑀𝑀𝑀_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 �
𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾02
𝐾𝐾𝐾𝐾
� radians
=Φ
ΦMM–– ΔΦ
ΔΦ =
=Φ
ΦMM+
+ arctan
arctan(ω
(ω00RR22CC22))
ΦM_
M_ MAX
MAX =
Φ
𝐾𝐾𝐾𝐾
(18)
(21)
(21)
ΦM_MAX_NEW = ΦM_MAX + ΔΦ = arccos (ω0 2 NCP/K
𝜔𝜔𝜔𝜔0_𝑀𝑀𝑀𝑀𝐴𝐴𝐴𝐴𝑀𝑀𝑀𝑀 = �
radians/s (19)
𝐶𝐶𝐶𝐶𝑃𝑃𝑃𝑃+
=Φ
ΦM_
+𝐾𝐾𝐾𝐾ΔΦ
ΔΦ =
= arccos
arccos(ω
(ω0022NC
NCPP//KK)) –– arctan
arctan(ω
(ω00RR22CC22))
ΦM_
M_MAX_NEW
MAX_NEW=
M_MAX
MAX
Φ
Analog Dialogue 49-02, February 2015
(22)
(22)
3
ΔΦ = – arctan(ω0 R2 C2 ) (20)
ΔΦ = – arctan(ω0 R2 C2 ) (20)
ΦM_ MAX = ΦM – ΔΦ = ΦM + arctan (ω0R2C2)
(21)
Applying ϕM_NEW to Φ
Equation
15 Φ
and
Equation
16Mresults
in different
0 and C0 than for the second-order solution, with
ΔΦ = Φ
+ arctan
(ω0 R2values
C2) for R(21)
M_ MAX =
M–
the new values compensating for the excess phase shift introduced by R2 and C2 . The presence of R2 and C2 also affects ϕM_MAX , the
= ΦM_MAX
+ of
ΔΦ
(ω0 2 NCP/K) – arctan(ω0 R2 C2 )
(22)
M_MAX_NEW
maximum allowable value of ϕMΦ
. The
new maximum
value
ϕM=(ϕarccos
M_MAX_NEW ) is
ΦM_MAX_NEW = ΦM_MAX + ΔΦ = arccos (ω02NC P /K ) – arctan (ω0R2C2)
Conclusion
(22)
This article demonstrates using open-loop unity-gain bandwidth (ω0) and phase margin (ϕM ) as design parameters for secondorder or third-order loop filters when only components R0 and C0 are adjustable. Simulation of a PLL with a second-order loop
filter using R0 and C0 yields an exact match to the theoretical frequency response of HOL and the resulting phase margin, thereby
validating the equations. The parameters ω0 and ϕM have upper bounds for a second-order loop filter per Equation 19 and
Equation 18, respectively.
The procedure for determining R0 and C0 assumed a second-order loop filter, but is extendible to third-order loop filter designs by
adjusting the desired phase margin (ϕM ) to a new value (ϕM_NEW ) per Equation 21, yielding a new upper bound (ϕM_MAX_NEW ) per
Equation 22.
Although simulations using a second-order loop filter validated Equation 15 and Equation 16, validating the equations that
extend the design procedure to third-order loop filter designs requires a redefinition of the loop filter response, HLF (s), to include
R2 and C2 as follows:
𝐻𝐻𝐿𝐿𝐿𝐿 (𝑠𝑠) =
𝑠𝑠(𝑠𝑠 2 𝑅𝑅0 𝑅𝑅2 𝐶𝐶0 𝐶𝐶2 𝐶𝐶𝑃𝑃
𝑠𝑠𝑅𝑅0 𝐶𝐶0 + 1
+ 𝑠𝑠𝑅𝑅2 𝐶𝐶0 𝐶𝐶2 + 𝑠𝑠𝑅𝑅0 𝐶𝐶0 𝐶𝐶𝑃𝑃 + 𝑠𝑠𝑅𝑅2 𝐶𝐶2 𝐶𝐶𝑃𝑃 + 𝑠𝑠𝑅𝑅0 𝐶𝐶0 𝐶𝐶2 + 𝐶𝐶0 + 𝐶𝐶2 + 𝐶𝐶𝑃𝑃 )
𝑦𝑦 CP = 1.5 nF
The incorporation of this form of HLF into the𝑥𝑥HOL and
= 𝜃𝜃2simulations
− 𝜃𝜃1 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
( ) −loop
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ( )
enables
of third-order
HCL equationsΦ
1
1 R2 = 165 kΩ
filter designs using R0 and C0. Such simulations reveal the
calculated values of R0 and C0 deviate from the theoretical
C2 = 337 pF
frequency response and phase margin associated with
2
2
= using
𝑎𝑎2 +a𝑏𝑏third-order
+ 2𝑎𝑎𝑎𝑎 cos(𝜃𝜃)
is the
opposite
D = 30 μA side c)
HOL for a PLL𝑐𝑐when
loop filter.(θ
This
is angle K
predominantly due to the effect of R2 and C2 on HOL in a thirdKV = 3072 (25 ppm/V at 122.88 MHz)
order loop filter.
N = 100
2
2
2 ) +a (√1
Recall that the(𝑥𝑥
formulas
and
C0 𝑥𝑥
assume
second-order
= R(0√
1+
+ 𝑦𝑦 2 ) − 2√1 + 𝑥𝑥 2 √1 + 𝑦𝑦 2 cos Φ
− 𝑦𝑦)2for
Simulation 1 and Simulation 2 use ω0 = 100 Hz, which is near
loop filter, but R2 and C2 do not exist in a second-order filter,
the calculated upper limit of 124.8 Hz (ω0_MAX ). As such,
so including them as part of the loop filter constitutes a source
Simulation 1 and Simulation 2 deviate from the design
of error in spite of adjusting R0 and C0 to compensate for the
parameter
values (ω0 and ϕM ) by nearly 10%. On the other
phase shift introduced by R2 and C2. Even
in the
1+
𝑥𝑥𝑥𝑥 presence of
hand,
Simulation
3 and Simulation 4 use ω0 = 35 Hz, which is
Φ = simulation
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 (indicates that using the )
this error, however,
2 )(1
approximately ¼ the upper limit. As expected, Simulation 3
+ 𝑦𝑦of2 ω) 0 to a
+ 𝑥𝑥the
adjusted values of R0 and C0, but√(1
limiting
choice
and Simulation 4 hold much closer to the design parameters
maximum of ¼ of the value dictated by Equation 19 yields
(ω
0 and ϕM ), yielding an error of only about 1%.
acceptable results. In fact, the simulated open-loop bandwidth
and phase margin results deviate only slightly from
the design
1 + 𝜔𝜔2 𝑇𝑇1loop
𝑇𝑇2 filter.
parameters (ω0 and ϕM ) for a PLL using a third-order
Φ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 (
Simulation Results
√[1 + (𝜔𝜔𝑇𝑇2 )2 ][1 + (𝜔𝜔𝑇𝑇1 )2 ]
The following is the result of running four simulations of a
PLL with a third-order loop filter. The simulations all have the
following fixed-loop filter components and PLL parameters:
4
Table 1 summarizes the simulation results and also includes
)the calculated values of R0 , C0, ω0_MAX , and ϕM_MAX for the given
design parameters, ω0 and ϕM . Note that for the purpose of
comparison it would be preferable for both Simulation 1 and
Simulation 3 to use ϕM = 80°, but Simulation 1 must satisfy the
constraint imposed by Equation 22 of ϕM < 48° (hence the
choice of 42°).
Analog Dialogue 49-02, February 2015
Table 1: Simulation Summary
Simulation 1
Parameter
Simulation 2
ϕM
ω0
ω0
Simulation 3
ϕM
ω0
Simulation 4
ϕM
ω0
ϕM
Design
100 Hz
42°
100 Hz
30°
35 Hz
80°
35 Hz
30°
Simulation
93.1 Hz
38.7°
92.5 Hz
27.1°
34.9 Hz
79.0°
34.7 Hz
29.3°
R0
969.6k kΩ
1118 kΩ
240.1 kΩ
139.9 kΩ
C0
14.85 nF
3.670 nF
225.5 nF
21.24 nF
𝛚0_MAX
124.8 Hz
124.8 Hz
124.8 Hz
124.8 Hz
𝛟M_MAX
48.0°
48.0°
84.8°
84.8°
50
80
90
60
80
40
40
70
30
20
60
0
50
−20
40
−40
30
−60
20
−80
10
1
10
100
1k
10k
100k
1M
0
FREQUENCY (Hz)
SIM 1 GAIN
SIM 1 PHASE
SIM 2 GAIN
SIM 2 PHASE
SIM 3 GAIN
SIM 3 PHASE
SIM 4 GAIN
SIM 4 PHASE
MAGNITUDE (dB)
20
10
0
−10
−20
PEAKING
50
MAGNITUDE (dB)
−100
0.1
PHASE (DEGREES)
MAGNITUDE (dB)
Figure 4 and Figure 5 show the open- and closed-loop response for each simulation.
−30
45
40
35
0.1
1
−40
−50
0.1
1
10
100
FREQUENCY (Hz)
10
1k
100
1k
10k
FREQUENCY (Hz)
SIM 1 GAIN
Figure 4. Open-loop gain and phase.
𝐻𝐻𝐿𝐿𝐿𝐿 (𝑠𝑠) =
SIM 2 GAIN
Figure 5. Closed-loop gain.
𝑠𝑠𝑅𝑅0 𝐶𝐶0
SIM 3 GAIN
SIM 4 GAIN
+1
𝑅𝑅0 𝑅𝑅2 𝐶𝐶Function
𝑠𝑠𝑅𝑅the
𝑠𝑠(𝑠𝑠 2Arctan
Appendix—Converting the Discontinuous
Continuous
Arccos
0 𝐶𝐶2 𝐶𝐶𝑃𝑃 + to
2 𝐶𝐶0 𝐶𝐶
2 + 𝑠𝑠𝑅𝑅0 𝐶𝐶0 𝐶𝐶
𝑃𝑃 + 𝑠𝑠𝑅𝑅Function
2 𝐶𝐶2 𝐶𝐶𝑃𝑃 + 𝑠𝑠𝑅𝑅0 𝐶𝐶0 𝐶𝐶2 + 𝐶𝐶0 + 𝐶𝐶2
𝑠𝑠𝑅𝑅
𝐶𝐶
0
0 + 1 ) and
Equation 10 demonstrates that the angle ϕ𝐻𝐻is the
difference between angle𝑠𝑠𝑅𝑅
θ2 and
angle θ , where θ2 = arctan(ωT
2
0 𝐶𝐶0 + 1 1
𝐿𝐿𝐿𝐿 (𝑠𝑠) =
2
𝑠𝑠𝑅𝑅
𝐶𝐶
+
1
0 2y/1:
0 0 𝐶𝐶2 𝐶𝐶𝑃𝑃 + 𝑠𝑠𝑅𝑅2 𝐶𝐶0 𝐶𝐶2 + 𝑠𝑠𝑅𝑅0 𝐶𝐶0 𝐶𝐶𝑃𝑃 + 𝑠𝑠𝑅𝑅2 𝐶𝐶2 𝐶𝐶𝑃𝑃 + 𝑠𝑠𝑅𝑅0 𝐶𝐶0 𝐶𝐶2 + 𝐶𝐶0
θ1 = arctan(ωT1). Furthermore,
ωT2 is expressible as x/1 𝑠𝑠(𝑠𝑠
and ωT
(𝑠𝑠)
𝐻𝐻
=
𝑅𝑅
𝑅𝑅
𝐶𝐶
1 as
𝐿𝐿𝐿𝐿
0
2 𝑅𝑅 𝑅𝑅 𝐶𝐶 𝐶𝐶 𝐶𝐶 + 𝑠𝑠𝑅𝑅 𝐶𝐶 𝐶𝐶 + 𝑠𝑠𝑅𝑅 𝐶𝐶 𝐶𝐶 + 𝑠𝑠𝑅𝑅 𝐶𝐶 𝐶𝐶 + 𝑠𝑠𝑅𝑅 𝐶𝐶 𝐶𝐶 + 𝐶𝐶 + 𝐶𝐶 + 𝐶𝐶 )
𝐻𝐻𝐿𝐿𝐿𝐿 (𝑠𝑠) =
𝑥𝑥0 0𝐶𝐶 𝑃𝑃𝐶𝐶 + 𝑠𝑠𝑅𝑅
0 2 𝑠𝑠𝑅𝑅
0 2 𝑃𝑃 + 𝑠𝑠𝑅𝑅
2 𝐶𝐶
0 𝐶𝐶
2
2 𝑦𝑦2
𝑃𝑃
0 0 2
0
2
𝑃𝑃
𝑠𝑠(𝑠𝑠 2 𝑅𝑅0 𝑅𝑅2 𝐶𝐶𝑠𝑠(𝑠𝑠
0 𝐶𝐶2 𝐶𝐶𝑃𝑃 +
0 𝑃𝑃 +(𝑠𝑠𝑅𝑅
𝑃𝑃
Φ =2𝜃𝜃𝐶𝐶20 𝐶𝐶
−2 𝜃𝜃1 = 0𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
) 2− 2𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
(0 𝐶𝐶)0 𝐶𝐶2 + 𝐶𝐶0 + 𝐶𝐶2 + 𝐶𝐶𝑃𝑃 )
1
1
𝑥𝑥
This implies the geometric relationship shown in Figure
with θ1 and θ2𝑦𝑦defined by the triangles 𝑦𝑦
𝑥𝑥𝜃𝜃26,−
Φ
=
𝜃𝜃1ϕ=
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
( )between
− 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
( ofθ).Figure 6 (b) and (a),
𝑥𝑥
𝑦𝑦
and
respectively. Figure 6 (c)
combines
these
two
triangles
to
show
as
the
difference
θ
Φ
=
𝜃𝜃
−
𝜃𝜃
=
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
(
)
−
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
(
)
1
1
12
2
1 (2 ) − 2
Φ = 𝜃𝜃2 − 𝜃𝜃1 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
) cos(𝜃𝜃)
1 (θ is the angle opposite side c)
𝑐𝑐 1 = 𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
+ 𝑏𝑏12 +(12𝑎𝑎𝑎𝑎
The Law of Cosines relates an interior angle (θ) of a triangle to the lengths of the three sides of the triangle (a, b, and c) as follows:
𝑐𝑐 2 = 𝑎𝑎2 + 𝑏𝑏 2 + 2𝑎𝑎𝑎𝑎 cos(𝜃𝜃) (θ is the angle opposite side c)
2
2
2
𝑐𝑐
=
𝑎𝑎
+
𝑏𝑏
+
2𝑎𝑎𝑎𝑎
(θ is the2 angleside
opposite
𝑐𝑐 2 = 𝑎𝑎2 + 𝑏𝑏 2 + 2𝑎𝑎𝑎𝑎 cos(𝜃𝜃)
(θcos(𝜃𝜃)
c) 2 side c)
2is the√angle opposite
(𝑥𝑥 − 𝑦𝑦)
= ( 1 + 𝑥𝑥 2 ) + (√1 + 𝑦𝑦 2 ) − 2√1 + 𝑥𝑥 2 √1 + 𝑦𝑦 2 cos Φ
Applying the Law of Cosines to angle ϕ in Figure 6 (c) yields:
2
2
2 𝑦𝑦)2 = (√1 +2 𝑥𝑥 2 ) + (√1 + 𝑦𝑦 2 ) − 2√1 + 𝑥𝑥 2 √1 + 𝑦𝑦 2 cos Φ
(𝑥𝑥
−
2√
2
2
2 ) + (√1
2 √1 + 𝑦𝑦 2 cos Φ
(𝑥𝑥
−
𝑦𝑦)
=
(
1
+
𝑥𝑥
+√𝑦𝑦12 )+ 𝑥𝑥−2 √1
2√1++𝑦𝑦 2𝑥𝑥cos
2
2
(𝑥𝑥 − 𝑦𝑦) = (√1 + 𝑥𝑥 ) + (√1 + 𝑦𝑦 2 ) − 2
Φ
1 + 𝑥𝑥𝑥𝑥
)
Φ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 (
√(1 + 𝑥𝑥 2 )(1 + 𝑦𝑦 2 )
1 + 𝑥𝑥𝑥𝑥
1+
)
Φ
= 𝑥𝑥𝑥𝑥
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 (
1
+
𝑥𝑥𝑥𝑥
2 )(1 + 𝑦𝑦 2 )
)
Φ
=
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
(
√(1
+
𝑥𝑥
Φ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 (
2)
2
√(1++𝑦𝑦𝑥𝑥
2 ) )(1 + 𝑦𝑦 ) 1 + 𝜔𝜔 2 𝑇𝑇 𝑇𝑇
√(1 + 𝑥𝑥 2 )(1
1 2
Φ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 (
)
2
2
√[1 + (𝜔𝜔𝑇𝑇2 ) ][1 + (𝜔𝜔𝑇𝑇
2 1) ]
1 + 𝜔𝜔 𝑇𝑇1 𝑇𝑇2
𝜔𝜔2 𝑇𝑇1 𝑇𝑇(2
=+𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
)
2 Φ1
1
+
𝜔𝜔
𝑇𝑇
𝑇𝑇
1 2
2 ][1 + (𝜔𝜔𝑇𝑇 )2 ]
Φ
=
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
(
)
)
(𝜔𝜔𝑇𝑇
√[1
+
2
1
Φ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 (
)
2
2
2 + (𝜔𝜔𝑇𝑇2 ) ][1
2 + (𝜔𝜔𝑇𝑇1 ) ]
√[1 + (𝜔𝜔𝑇𝑇√[1
2 ) ][1 + (𝜔𝜔𝑇𝑇1 ) ]
Analog Dialogue 49-02, February 2015
5
Φ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 (
𝐻𝐻𝐿𝐿𝐿𝐿 (𝑠𝑠) =
𝛉2
2
(1
+
x
√(1 + 𝑥𝑥 2 )(1 + 𝑦𝑦 2 )
)
But, x/1 = ωT2 and y/1 = ωT1, allowing ϕ to be expressed in
terms of T1 and T2.
½
)
X
𝑠𝑠(𝑠𝑠 2 𝑅𝑅0 𝑅𝑅2 𝐶𝐶0 𝐶𝐶2 𝐶𝐶𝑃𝑃
1
𝑠𝑠𝑅𝑅0 𝐶𝐶0 + 1
1 + 𝜔𝜔2 𝑇𝑇1 𝑇𝑇2
+ 𝑠𝑠𝑅𝑅2 𝐶𝐶0 𝐶𝐶2+ +
𝑠𝑠𝑅𝑅
𝐶𝐶
𝐶𝐶
+
𝑠𝑠𝑅𝑅
𝐶𝐶
𝐶𝐶
+
𝑠𝑠𝑅𝑅
𝐶𝐶
𝐶𝐶
+
𝐶𝐶
+
𝐶𝐶2 + 𝐶𝐶𝑃𝑃 )
Φ
=
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
(
)
)
0
0
𝑃𝑃
2
2
𝑃𝑃
0
0
2
0
y
(1
√[1 + (𝜔𝜔𝑇𝑇2 )2 ][1 + (𝜔𝜔𝑇𝑇1 )2 ]
y
2½
𝛉1
1
𝑥𝑥
𝑦𝑦
(B)
Φ = 𝜃𝜃2 − 𝜃𝜃1 = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ( ) − 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
( )
1
1
(A)
2
+
(1
x
½
)
x− y
x
)
𝑐𝑐 2 = 𝑎𝑎2 + 𝑏𝑏 2 + 2𝑎𝑎𝑎𝑎
cos(𝜃𝜃)
(θ is the angle
(1 +y
2½
𝛉2
φ
y
𝛉1
1
2
(𝑥𝑥 − 𝑦𝑦)2 = (√1 +(C)𝑥𝑥 2 ) + (√1 + 𝑦𝑦 2 )
Figure 6. Geometric representation of Equation 10.
Solving for ϕ:
Φ = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 (
1 + 𝑥𝑥𝑥𝑥
√(1 + 𝑥𝑥 2 )(1 + 𝑦𝑦 2 )
2
References
Brennan, Paul V. Phase-Locked Loops: Principles and Practice.
McGraw-Hill, 1996.
Keese, William O. AN-1001, National Semiconductor
Application Note, An Analysis and Performance Evaluation of
opposite
side Filter
c) Design Technique for Charge Pump Phase-Locked
a Passive
Loops. May 1996.
MT-086: Fundamentals of Phase Locked Loops (PLLs).
with
Integrated
2 cos
− 2√1 + PLLs/PLLs
𝑥𝑥 2 √1 + 𝑦𝑦
Φ VCOs.
)
1 + joined
𝜔𝜔2 𝑇𝑇1Analog
𝑇𝑇2 Devices in 1998 as a system design
Ken Gentile [ken.gentile@analog.com]
Φ
=
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
(
)
engineer with the Clock and Signal Synthesis
product
line in
2 ][1 + (𝜔𝜔𝑇𝑇
2 ]Greensboro, NC, where
(𝜔𝜔𝑇𝑇
)
)
√[1
+
2 analog filter design,
1
he specializes in direct digital synthesis,
and writing GUI-based
engineering tools in MATLAB. Ken holds 10 patents. He has published 14 articles in
various industry trade journals and over a dozen ADI application notes, as well as having
presented at ADI’s annual General Technical Conference (GTC) in 2001, 2005, and 2006.
He graduated with honors in 1996 with a B.S.E.E. from North Carolina State University. In
his spare time, Ken enjoys reading, mathematical puzzles, and most anything related to
science, engineering, and “backyard” astronomy.
6
1 + 𝑥𝑥𝑥𝑥
Ken Gentile
Also by this Author:
Reconstruct a DAC
Transfer Function
from Its Harmonic
Spectral Content
Volume 43, Number 1
Analog Dialogue 49-02, February 2015