International Symposium on Nonlinear Theory and its Applications
Xi'an, PRC, October 7{11, 2002

Analysis of Oscillator Circuits Having Multiple Oscillations
Hiroo Yabe, Yoshihiro Yamagami, Yoshifumi Nishio and Akio Ushida

Dept. of E. E. Eng., Tokushima University
2-1 Minami-Josanjima, Tokushima, Japan
Phone:088-656-7470, Fax:088-656-7471
Email:yabe@ee.tokushima-u.ac.jp
1.

Introduction

2.

Basic algorithm

Analysis of oscillator circuits is very important for de- In order to show the ideas of our algorithm, let usa reactance oscillator with a negative resistancesigning communication circuits such as modulators and consider
mixers, where one of the signals is coming from the inter- as given by Fig.1.
nal oscillator and the other from the input signal. Therefore, we need to know the exact oscillator frequency iniorder to calculate the intermodulation. There are many
Lineartype oscillators such as Colpitts, crystal oscillator and soon, each of which has a unique oscillator frequency. On
Reactancev
the other hand, the coupled oscillators with the same
oscillator frequency may have complicated phenomena
Circuitdepending on the initial conditions, some of which have
many di erent phases and the same oscillating frequency[1]-[2]. These phenomena can be analytically explainedin the case of the weakly nonlinear coupled oscillators
Fig.1 A reactance oscillator[1]. Furthermore, it is known that the coupled oscillators
the nonlinear characteristic of the negative remay have quasi-periodic oscillations and/or chaos phe- Assume
is given in the form of power series as follow
nomena [3]. In this paper, we consider the steady-state sistance[12]:
analysis of oscillators with multiple mode oscillations.
i = c0 + c1 v + c2 v2 + 1 1 1(1)
There have been published many papers discussing thesteady-state analysis of oscillator circuits [4]-[8]. The ref- We also assume the waveform in the following form;
erences [5]-[7] are based on the time-domain techniques,
where the initial guess giving rise to the steady-stateK
response is found by the Newton-like shooting method. v = V0 + V1 cos !t + X[V2k02 cos k!t + V2k01 sin k!t]
Especially, W.Ma, L.Trajkovic and L.Mayaram [7] have
k=2
proposed an elegant homotopy method combining with(2)
the shooting method, which can be eciently applied to Observe that the waveform does not contain sin !tthe analysis of high Q oscillators. Of course, the har- component because of the autonomous system. Apply-monic balance method [8] is widely used to the analysis ing the harmonic balance method to the circuit equationof oscillator circuits. The relaxation method in the fre- shown by Fig.1, we have the determining equation as
quency domain [9] is also eciently applied to the weakly follows:
nonlinear large scale oscillators.
9
In this paper, we propose a reactance oscillator which
F0 (V0 ; V1 ; 1 1 1 ; V2K 01 ; ! ) = 0 1 11 DC>>
consists of the Cauer circuit coupled with a negative>
F1 (V0 ; V1 ; 1 1 1 ; V2K 01 ; ! ) = 0 1 11 cos !t>=resistance. We can easily synthesize the Cauer circuit
F
(
V
;
V
;
1
1
1
;
V
;
!
)
=
0
1
11
sin
!t
2
0
1
2
K
0
1
by specifying the resonant and anti-resonant frequen(3)::::::::::::::::::::::::::::::::::::::::::::::::::::::::>
cies. Then, the oscillator may have multiple oscillations>>
F2K 01 (V0 ; V1; 1 1 1 ; V2K 01 ; !) = 0 1 1 1 cos K!t >;around the anti-resonant frequency points [10]. UnfortuF2K (V0 ; V1 ; 1 1 1 ; V2K 01 ; ! ) = 0 11 1 sin K!tnately, since the above mentioned techniques can be only
applied to the analysis with a unique oscillation, theycannot nd out the multiple oscillations. Our method is where ! is the fundamental frequency component to be
based on the harmonic balance method, whose determin- determined and K is the highest component. Of course,
ing equation is described by a set of algeblaic equation. we may have the di erent frequency component dependWe show here an elegant method such that the equation ing on the circuit composition. Thus, the determining
can be replaced by the equivalent coupled cosine and equation is described by a set of algebraic equations consine resistive circuit. Solving the circuit with STC based sisted of (2K + 1)-variables and the same number ofon the Newton homotopy method [11], we can nd the equations. Note that it is not easy to solve the nonmultiple oscillations on the solution curve. The analysis linear equation because it may have many solutions forcan be easily carried out with the transient analysis of the multiple oscillator circuit. For simplicity, set
SPICE. We will show an interesting example in section
f (v) = 0; v 2 R2K +1 ; f (:) : R2K +1 7! R2K +1 (4)
4.

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where v = [V0; V1 ; 1 1 1 ; V2K 01 ; !]T . Applying the New- which satisfy the relation (5). On the other hand,
ton homotopy method [11] to (4), we have the following STC(solution curve tracing circuit) realizes the secrelation:
ond term of (6), where RD is a suciently large
dummy resistance to avoid a L-J cut set. Furthermore,F (v; ) = f (v ) + ( 0 1)f (v0 ) = 0(5) I10 ; I20 ; . . . ; I2K +1;0 in the Fig.2(a) denote the initialThus, we can easily nd the multiple oscillawhere  is an additional variable, and v0 is an initial guesses.
tions
at

= 1 by the use of SPICE.
guess. Thus, the number of variables is 2K + 2. Then,the solution is given by the solution curve in the (2K +2)space. The curve can be traced by the application of the 3. Equivalent circuit models of the determining
equation
arc-length method as follows:9
Oscillator circuits usually consists of many linear
F (v; ) = 0
>
=
2

2
and/ornonlinear reactive and resistive elements. In or2X
K +1 ddvi:(6) der to solve an oscillator with SPICE mentioned in the
+
=
1
>
;
above, the determining equations should be described in
ds
ds
i=1
the functional forms. If the nonlinear characteristics areby the power series function, the Fourier coefThe algebraic-di erential equation can be solve by the described
cientscanbe written in the functional forms. However,backward-di erenece method [13], where the derivative there are manyof nonlinear elements describedj+1of the kth-oder formula at s = s with the step size h by the exponentialkinds
functions,piecewise continuous funcis given bytions, rational functions and so on. In these cases, theFourier coecients cannot be described by the functionaldvi
k0 j +1
jj 0k+1and the above mention technique can no longer be= h vi + Qk;i vi 1 1 1 vi(7) forms,
applied. Fortunately, for the weakly nonlinear elements,
ds s=sj+1
their characteristics can be approximated by the power
series [12], then, we can also apply the above technique.
andIn this section, we propose the cosine and sine cir1
jj +1
j 0k+1
cuits corresponding to the determining equation, which:
Qk;i =
k;1 vi + k;2 + 1 1 1 + k;k vi
his given by the coupled resistive circuits. Each subcircuit
the same structure as the original one except
Thus, (6) is again transformed into a set of nonlinear that thehasreactance
are replaced by the voltageequations, and it can be solved by the Newton-Raphson controlled current elements
sourcesorcurrent-controlled voltage
method, repeatedly. This algorithm is exactly equal sources. Therefore if we use SPICE
simulator, the cirto the transient analysis of SPICE, so that we can use cuit diagram describing the determining
equation will be
SPICE for the curve tracing algorithm.
easily drawn on the computer display with SPICE diagram tool. At rst, we will show the equivalent circuit
models of the nonlinear circuit elements.
v1

f1 (v,ρ)1Ω(ρ−1)Ι 10

VCCS

3.1 Nonlinear inductorI(ρ−1)Ι 20v2

f2 (v,ρ)

VCCS

ρ
I 20

1Ω

Is

Iρ

1(A)

Int

(ρ−1)Ι 2K+1,0
ω

f

(v,ρ)
2K+1

RD

VCCS
I 2K+1,0

1Ω

STC

ρ

Assume that the inductor ux is described by thecurrent-controlled characteristic as follows:^(iL ) = L1 iL + L2 i2L + L3 i3L + 1 1 1 :(10)
Let us consider the K th higher components of the fundamental frequency component !, and assume the waveform as follows:
iL = IL;0 +Σ-1(a) Circuit diagram(b) STC

Fig.2 Circuit diagram of the Newton homotopy method

K
X
k=1

[IL;2k01 cos k!t + IL;2k sin k!t]: (11)Substituting (11) into (10), we haveK
The circuit diagram is shown in Fig.2(a) and (b), where ^(i )  8 + X[8L;2k01 cos k!t + 8L;2k sin k!t];
L = L;0"VCCS" is a voltage-controlled current source, and the
k=1
current sources fi (v; ) are given by
(12)
9
where
8
;
8
;
1
1
1
;
8
are
analytical
functionsoffi(v; ) = Fi (v; ) 0 vi ; i = 1; 2; . . . ; 2K + 2 >
L;
0
L;
1
L;
2
K
=



2X
K +1 
I
L;0 ; IL;1 ; 1 1 1 ; IL;2K . Thus, the inductor voltage is given
dvi 2
d 2
(8) byIs =; I =:>;ds
ds
d^(iL )
i=1v^L (iL ) =dt
Note that if we assume the voltage di erence of 1[ ]KX
resistance is equal to 0vi, then we have
= [0k! 8L;2k01 sin k!t + k!8L;2k cos k!t]: (13)0vi = Fi (v; ) 0 vi ; i = 1; 2; . . . ; 2K + 1: (9)
k=1

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Thus, the sine and cosine components for the kth higher Thus, the k th order higher harmonic of the sine andharmonic component are respectively given bycosine are respectively given by
V 2 = 0k!8 2 01 ; V 2 01 = k!8 2 ; k = 1; 2; . . . ; K
I 2 = I^ 2 ; I 2 01 = I^ 2 01 ; k = 1; 2; . . . ; K (23)(14) Note that the voltage-controlled nonlinear resistors arewhere VL;2k01 , VL;2k are the amplitudes of the voltages described by the voltage-controlled resistors as shown infor the k th frequency components. Thus, the inductor is Fig.3.
replaced by the coupled current-controlled voltage source
as shown in Fig.3.
Cosine component Sine component
L; k

L; k

L; k

L; k

G; k

G; k

3.2 Nonlinear capacitor

Inductor[VC;2k01 cos k!t + VC;2k sin k!t]: (16)k=1

Resistor

VL,2k= − k ω Φ L,2k-1

VL,2k-1 = k ω Φ L,2k

VC,2k-1

I C,2k-1

I C,2k-1 = k ω Q C,2k

VC,2k

I C,2kIC,2k= − k ω Q C,2k-1

^

iR

K
X

I L,2k
VL,2k

iC
vC
^
q =q
(v C )

Substituting (16) into (15), we haveq^(vC ) 
= QC 0 +I L,2k-1
VL,2k-1vL
φ = φ^(i L )Capacitor

vC = VC 0 +

G; k

iL

Assume that the capacitor charge is described by thevoltage-controlled characteristic as follows:q^(vC ) = C1 vC + C2 vC2 + C3 vC3 + 11 1 :(15)
Set the voltage waveform vC as follows:
K
X

G; k

^

I G,2k-1

vR

I G,2k
VG,2k

VG,2k-1[QC;2k01 cos k!t + QC;2k sin k!t];
I
= I
I
=I
i =i^ (v )(17)
Fig.3 Equivalent cosine and sine circuits
where QC 0; QC 1; . . . ; QC;2K are analytical function of
VC0 ; VC 1 ; . . . ; VC;2K . Then, the voltage-current charac4. An illustrative example
teristic of capacitor is given by
Consider a Cauer oscillator with a negative resistor as
shown in Fig.4. Assume that the nonlinear characteristic^iC (vC ) = dq^(vC )dt
is given byK
X
iG = 0C1 vG + C3 vG3 ; C1 = 1: C3 = 1:(24)
= [0k!QC;2k01 sin k!t + k!QC;2k cos k!t]: (18)k=1
rst, we have designed the reactance Cauer such that
Thus, the sine and cosine components for the k th order At
ithas
the following resonant and ant-resonant frequenharmonic are respectively given bycies:
I 2 = 0k!Q 2 01 ; I 2 01 = k!Q 2 ; k = 1; 2; . . . ; K
Anti 0 resonant frequencies : !1 = 1; !3 = 4; !5 = 6 o :(19) Resonant frequencies : !2 = 2; !4 = 5
where 0k!QC;2k01; k!QC;2k are the currents of the kth
Then, we had the following circuit parameters shown infrequency components of the nonlinear capacitors. Thus, thegure. We also assumed small resistances for all thethe capacitor is replaced by the voltage-controlled cur- inductors.
rent sources as shown in Fig.3.
C; k

k=1

C; k

^

C; k

There are two types of the voltage-controlled andcurrent-controlled nonlinear resistors. We consider here
the voltage-controlled resistor as follows:^iG(vG ) = HG0 + HG1vG + HG2vG2 + HG3 vG3 + 1 1 1 : (20)Assume the voltage waveform vG as follows:
K
X[VG;2k01 cos k!t + VG;2k sin k!t]: (21)k=1

Substituting (21) into (20), we have^iG (vG ) = I^G0 +K
X

G,2k-1G^

G,2k-1

G,2k

G,2k

C; k

3.3 Nonlinear resistorvG = VG0 +

G

G

iG

r1

L 1 r2

C1

C2

+
vG

-

L2

r3

C3

L3

Fig.4 Cauer 0scillator
C1 = 0:1; C2 = 0:343; C3 = 0:439; L1 = 0:417
L2 = 0:262; L3 = 1:058; r1 = 1 1 1 = r3 = 0:01

Note that since the nonlinear voltage-current characteristic is symmetrical with respect to the origin, we
can only consider the odd higher harmonic components.
Thus, we assume the waveform vG as follows:[I^G;2k01 cos k!t + I^G;2k sin k!t]:
k=1
(22)

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v (t) = V1 cos !t +
G

K
X

9
=

[V2k cos(2k + 1)!t :;=1+V2k+1 sin(2K + 1)!t]k(25)

Observe that we have neglected the term of sin !t in 5. Conclusions and remarks(25), because of the autonomous system. In the simulation, although we have chosen K = 2, for simplicity, the In this paper, we proposed a simple algorithm forcalculating multiple oscillations of a reactance oscilator
equivalent circuit for K = 1 is given by Fig.5.
which is based on the SPICE simulator, where we neednot to derive the troublesome circuit equation, and tosolve the determining equations. We only need to calculate
the Fourier coecients in the functional forms
cosωt :
for nonlinear elements, where the nonlinear elements
must be described in the form of power series. Then,the equivalent sine and cosine sub-circuits are easily obtained by the use of the SPICE drawing tool. Once the++equivalent circuit is drawn, the solutions will be foundsinωt :
by the transient tool of SPICE simulator.
For the future problem, we need to apply the algorithm to the practical circuits such as crystal oscillators.
r1 I 11C ω L 1I 11S

+
V11C
-

I1C

+
V11S
-

I 1S

+
V31C
-

I3C

r 3 I 13S

+
V13C
-

ω C 3V13S

ω C 2 V12S

r1 I 11S - ω L 1I 11C

r 2 I12C - ω L 2 I12C

- ω C 1V11C V12S

- ω C 2V12C

I31C 3ωL 1I 31S

r1

cos3ωt :+V12C-

ω C 1V11S

r 2 I 12S ωL2 I 12S+
3ω C 1V31S V32C-

ω L 1I 11S

I 13C

r3

- ω C 3 V13C

V13S

r 2 I32C 3ωL2I 32S

r 3 I 33C

+
V33C
-

3ωC3 V33S

3ω C 2V32S- ω L 3I 13C

References

3ωL3 I 33S[1] T.Endo and S.Mori,\Mode analysis of a multimode ladder oscillators," IEEE Trans. Circuits Syst. vol.CAS-23.
no.2,pp.100-113, 1976.
[2] S.Moro, Y.Nishio and S.Mori,\On coupled oscillators+sin3ωt :networks for cellular neural networks,"IEICE Trans.
Fundamentals, vol.E80-A, no.2, pp.214-222, 1997.
[3] Y.Nishio and A.Ushida, \multimode chaos in two couFig.5 Equivalent circuit describing the determining equationpled chaotic oscillators with hard nonlinearities," IEfor K = 1
ICE Trans. Fundamentals, vol.E79-A, no.2, pp.227-232,
Each sub-circuit has the same structure as the original [4] 1996.
K.Kundert, J.White and A.Sangiovanni-Vincentelli,
one as shown in Fig.4. The circuit has been solved by the
Steady-State Methods for Simulating Analog and MiNewton homoptopy method shown by Fig.2. The results
crowave Circuits, Kluwer Academic Publishers, Boston,
are shown in Fig.6(a) and (b) for K = 2, namely, we [5] 1990.
T.J.Aprille,Jr. and T.N.Trick, \A computer algorithm tohave chosen until 5 higher harmonic components. The
determine
the steady-state response of nonlinear oscillasolutions are obtained at  = 1. The solutions are also
tors," IEEE Trans. Circuit Theory, vol.CT-19, pp.354corresponding sin !t-component becoming to zero which
360, 1972.
satis es the condition (25). Thus, we have obtained 6 [6] K.Matsuo, T.Matsuda, Y.Nishio, Y.Yamagami and
A.Ushida, \Steady-state analysis of reactance oscillators
solutions as follows:
having multiple oscillations," Proc. ITC-CSCC 2000,
!1 = 1:0878; !3 = 3:6770; !5 = 5:722 (Stable solution)
vol.1, pp.203-206, 2000.
[7] W.Ma, L.Trajkovic and K.Mayaram, \HOMSSPICE: A!0 = 0; !2 = 2:0001; !4 = 4:999 (Unstable solution)
homotopy-based circuit simulator for periodic steadystate analysis on oscillators," IEEE ISCAS 2002 Phenix,
Arizona, 2002.
[8] A.H.Nayfeh and D.T.Mook, Nonlinear Oscillations,John
Wiley & Sons, 1979.
[9] A.Ushida, T.Adachi and L.O.Chua, \Steady-state analysis of nonlinear circuits based on hybrid method," IEEE
Trans. Circuits Syst.-I: Fundamental Theory and Applications, vol.CAS-39, no.8, pp.649-661, 1992.
[10] M.Kuramitsu and S>Chiba, \Mutual synchronization of
BVP oscillators coupled by reactance," Technical Reportof IEICE, CAS99-47, NLP99-71, pp.23-30, 1999.
Fig.6(a) Solution curves, s < 0, (b) Solution curves, s > 0 [11] A.Ushida,
Y.Yamagami, Y.Nishio, I.Kinouchi and
Y.Inoue, \ An ecient algorithm for nding multiple
DC solutions based on the SPICE-oriented Newton homotopy method," IEEE Trans. on Computer-Aided Design, vol.21, no.3, 2002.
[12] D.D.Weiner and J.E.Spina, Sinusoidal Analysis andModeling of Weakly Nonlinear Circuit: with Application
to Nonlinear Interference E ects, Van Nostrand Reinhold,1980.
Fig.7 Frequency spectrum.
[13] L.O.Chua and P-M Lin, Computer Aided Analysis of(a) ! = 1:0556, (b) ! = 3:6191, (c) ! = 5:68
Electronic Circuit: Algorithm and Computational Techniques, Prentice-Hall, 1975.
I 31S -3ω L1I31C r

r1

+
V31S
-

I 3S

I 32S -3ω L2I32C

2+- 3ω C 2V32C V33S-

- 3ω C 1V31C V32S

25

Fundamental frequency component

sinωt

10

Stable solution
Unstable solution

Stable solution
Unstable solution

ρ

5

15

ω

cosωt

ρ=1
0

sinωt

10

ω

5

ω4

ω2

ω3

ω1

ω0

-5

ρ
cosωt

ρ=1
0

-10

-15

ω5

-5

-20

0

5

10

15

-25

s 25

20

0

5

10

15

20

(a)

25

30

35

s

40

45

(b)

0.9

1

168mHz(ω=1.0556)

0.9-3ω L 3I 33C- 3ωC3V33C

15

Fundamental frequency component

20

-10

I 33S

r3

1.2

576mHz(ω=3.6191)

0.8904mHz(ω=5.68)

1

0.8

0.7

0.7

0.6

0.6

0.8

0.5

0.6

0.5
0.4

0.4

3ω

0.3

0.3

5ω

0.2
0.1

0.1

0

0

0

1

0.4

3ω

0.2

2

3

4

(a)

5

6

7

8
f(Hz)

3ω

0.2

5ω

5ω

0

1

2

3

4

(b)

5

6

7

8
f(Hz)

0

0

1

2

3

4

(c)

5

6

7

8
f(Hz)

These results have small di erences from the results obtained by the FFT analysis to the transient responses,
which is shown in Fig.7. The frequency spectrums still
have higher harmonic components. Therefore, if we want
the exact solutions, we need to consider the more higher
harmonic components.

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