Lock-in range of classical PLL with impulse signals and proportionally-integrating filter
K. D. Aleksandrov, N.V. Kuznetsov, G. A. Leonov, M. V. Yuldashev, R. V. Yuldashev
LLock-in range of classical PLL with impulse signals andproportionally-integrating filter
K. D. Aleksandrov, N.V. Kuznetsov , G. A. Leonov, M. V. Yuldashev, R. V. Yuldashev
Faculty of Mathematics and Mechanics, Saint-Petersburg State University, RussiaDept. of Mathematical Information Technology, University of Jyv¨askyl¨a, FinlandInstitute of Problems of Mechanical Engineering RAS, Russia
Abstract
In the present work the model of PLL with impulse signals and active PI filter in thesignal’s phase space is described. For the considered PLL the lock-in range is computedanalytically and obtained result are compared with numerical simulations.
Keywords: phase-locked loop, nonlinear analysis, PLL, two-phase PLL, lock-in range,Gardner’s problem on unique lock-in frequency, pull-out frequency
1. Models of classical PLL with impulse signals
Consider a physical model of classical PLL in the signals space (see Fig. 1).
Filter g(t)
VCO ϕ (t)=sign(sin θ (t) cos θ (t)) Input x(0)θ (0)f (θ (t)) =sign(sin θ (t)) f (θ (t)) =sign(cos θ (t)) Figure 1: Model of PLL with impulse signals in the signals space.
This model contains the following blocks: a reference oscillator (Input), a voltage-controlled oscillator (VCO), a filter (Filter), and an analog multiplier as a phase detector(PD). The signals sign (sin θ ( t )) and sign (cos θ ( t )) of the Input and the VCO (here θ (0)is the initial phase of VCO) enter the multiplier block. The resulting impulse signal φ ( t ) = sign (sin θ ( t ) cos θ ( t )) is filtered by low-pass filter Filter (here x (0) is an initialstate of Filter). The filtered signal g ( t ) is used as a control signal for VCO.The equations describing the model of PLL-based circuits in the signals space aredifficult for the study, since that equations are nonautonomous (see, e.g., (Kudrewicz Email address: [email protected] (N.V. Kuznetsov )
Preprint submitted to arXiv May 18, 2019 a r X i v : . [ m a t h . D S ] M a r nd Wasowicz, 2007)). By contrast, the equations of model in the signal’s phase spaceare autonomous (Gardner, 1966; Shakhgil’dyan and Lyakhovkin, 1966; Viterbi, 1966),what simplifies the study of PLL-based circuits. The application of averaging methods(Mitropolsky and Bogolubov, 1961; Samoilenko and Petryshyn, 2004) allows one to reducethe model of PLL-based circuits in the signals space to the model in the signal’s phasespace (see, e.g., (Leonov et al., 2012; Leonov and Kuznetsov, 2014; Leonov et al., 2015a;Kuznetsov et al., 2015b,a; Best et al., 2015). Filterθ (t) G(t)
VCOPD K d φ(θ (t) - θ (t)) Input θ (t) x(0)θ (0) Figure 2: Model of the classical PLL in the signal’s phase space.
The main difference between the physical model (Fig. 1) and the simplified mathemat-ical model in the signal’s phase space (Fig. 2) is the absence of high-frequency componentof the phase detector output. The output of the phase detector in the signal’s phase spaceis called a phase detector characteristic and has the form K d ϕ ( θ ( t ) − θ ( t )) . The maximum absolute value of PD output K d > ϕ ( θ ∆ ( t )) depends on difference θ ( t ) − θ ( t ) (which is called a phase error and denoted by θ ∆ ( t )). The PD characteristicdepends on the design of PLL-based circuit and the signal waveforms f ( θ ) of Input and f ( θ ) of VCO. For PLL with impulse signals the PD characteristic is as follows (see, e.g.,(Viterbi, 1966; Gardner, 1966; Leonov et al., 2012)): K d = 1; ϕ ( θ ∆ ( t )) = π θ ∆ ( t ) , if − π ≤ θ ∆ ( t ) ≤ π , − π θ ∆ ( t ) + 2 , if π ≤ θ ∆ ( t ) ≤ π . (1)Let us describe a model of classical PLL with impulse signals in the signal’s phasespace (see Fig. 2). A reference oscillator and a voltage-controlled oscillator generate thephases θ ( t ) and θ ( t ), respectively. The frequency of reference signal usually assumed tobe constant: ˙ θ ( t ) = ω . (2)The phases θ ( t ) and θ ( t ) enter the inputs of the phase detector. The output of phasedetector is processed by Filter. Further we consider the active PI filter (see, e.g., (Baker,2011)) with transfer function W ( s ) = τ sτ s , τ > τ >
0. The considered filter canbe described as ˙ x ( t ) = K d ϕ ( θ ∆ ( t )) ,G ( t ) = τ x ( t ) + τ τ K d ϕ ( θ ∆ ( t )) , (3)where x ( t ) is the filter state.The output of Filter G ( t ) is used as a control signal for VCO:˙ θ ( t ) = ω free2 + K v G ( t ) , (4)where ω free2 is the VCO free-running frequency and K v > ˙ x = K d ϕ ( θ ∆ ) , ˙ θ ∆ = ω − ω free2 − K v τ ( x + τ K d ϕ ( θ ∆ )) . (5)Denote the difference of the reference frequency and the VCO free-running frequency ω − ω free2 by ω free∆ . By the linear transformation x → K d x we have ˙ x = ϕ ( θ ∆ ) , ˙ θ ∆ = ω free∆ − K τ ( x + τ ϕ ( θ ∆ )) , (6)where K = K v K d is the loop gain. Here (6) describes the model of PLL with the impulsesignals and active PI filter in the signal’s phase space.By the transformation (cid:16) ω free∆ , x, θ ∆ (cid:17) → (cid:16) − ω free∆ , − x, − θ ∆ (cid:17) , (6) with odd PD characteristic (1) is not changed. This property allows one to use theconcept of frequency deviation (cid:12)(cid:12)(cid:12) ω free∆ (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) ω − ω free2 (cid:12)(cid:12)(cid:12) and consider (6) with ω free∆ > ϕ ( θ eq ) = 0 ,ω free∆ − K τ x eq = 0 . Since (6) is 2 π -periodic in θ ∆ , we can consider (6) in a 2 π -interval of θ ∆ , θ ∆ ∈ ( − π, π ].In interval θ ∆ ∈ ( − π, π ] there exist two equilibria: (cid:16) θ seq , x eq ( ω free∆ ) (cid:17) = (0 , ω free∆ τ K ) and (cid:16) θ ueq , x eq ( ω free∆ ) (cid:17) = ( π, ω free∆ τ K ) . As is shown below (see Appendix A) the equilibria (cid:16) θ seq + 2 πk, x eq ( ω free∆ ) (cid:17) = πk, ω free∆ τ K ! (cid:16) θ seq , x eq ( ω free∆ ) (cid:17) . The remaining equilibria (cid:16) θ ueq + 2 πk, x eq ( ω free∆ ) (cid:17) = π + 2 πk, ω free∆ τ K ! are saddle equilibria (see Appendix A).
2. The lock-in range
The model of classical PLL with impulse signals and active PI filter in the signal’s phasespace is globally asymptotically stable (see, e.g., (Gubar’, 1961; Leonov and Aleksandrov,2015)). The PLL achieves locked state for any initial VCO phase θ (0) and filter state x (0). So, there exist no limit cycles of the first kind, heteroclinic trajectories, and limitcycles of the second kind on the phase plane of (6) (see Fig. 3). x θ Δ θ eq τ ω Δ free K possible periodic trajectories: - limit cycle of the first kind- heteroclinic trajectory- limit cycle of the second kind s θ equ θ eq -2 ̟ u θ eq +2 ̟ s θ eq +2 ̟ u Figure 3: Possible periodic trajectories on the phase plane of (6).
However, the phase error θ ∆ may significantly increase during the acquisition process.In order to consider the property of the model to synchronize without undesired growthof the phase error θ ∆ , a lock-in range concept was introduced in (Gardner, 1966): “ If,for some reason, the frequency difference between input and VCO is less than the loopbandwidth, the loop will lock up almost instantaneously without slipping cycles. The max-imum frequency difference for which this fast acquisition is possible is called the lock-infrequency ”. The lock-in range concept is widely used in engineering literature on the PLL-based circuits study (see, e.g., (Stensby, 1997; Kihara et al., 2002; Kroupa, 2003; Gardner,2005; Best, 2007)). It is said that a cycle slipping occurs if (see, e.g., (Ascheid and Meyr,1982; Ershova and Leonov, 1983; Smirnova et al., 2014))lim sup t → + ∞ | θ ∆ (0) − θ ∆ ( t ) | ≥ π. However, in general, even for zero frequency deviation ( ω free∆ = 0) and a sufficiently largeinitial state of filter ( x (0)), cycle slipping may take place, thus in 1979 Gardner wrote: “There is no natural way to define exactly any unique lock-in frequency” and “despite itsvague reality, lock-in range is a useful concept” (Gardner, 1979).To overcome the stated problem, in (Kuznetsov et al., 2015c; Leonov et al., 2015b)the rigorous mathematical definition of a lock-in range is suggested:4 θ Δ θ eq τ ω Δ free K s θ equ θ eq -2 ̟ u θ eq +2 ̟ s θ eq -2 ̟ s - no cycle slipping- cycle slipping Figure 4: The lock-in domain and cycle slipping.
Definition 1. (Kuznetsov et al., 2015c; Leonov et al., 2015b) The lock-in range of model(6) is a range [0 , ω l ) such that for each frequency deviation (cid:12)(cid:12)(cid:12) ω free∆ (cid:12)(cid:12)(cid:12) ∈ [0 , ω l ) the model (6)is globally asymptotically stable and the following domain D lock − in (( − ω l , ω l )) = \ | ω free∆ | <ω l D lock − in ( ω free∆ ) contains all corresponding equilibria (cid:16) θ seq , x eq ( ω free∆ ) (cid:17) . For model (6) each lock-in domain from intersection T | ω free∆ | <ω l D lock − in ( ω free∆ ) is boundedby the separatrices of saddle equilibria (cid:16) θ ueq , x eq ( ω free∆ ) (cid:17) and vertical lines θ ∆ = θ seq ± π .Thus, the behavior of separatrices on the phase plane is the key to the lock-in range study(see Fig. 5).
3. Phase plane analysis for the lock-in range estimation
Consider an approach to the lock-in range computation of (6), based on the phase planeanalysis. To compute the lock-in range of (6) we need to consider the behavior of the lowerseparatrix Q ( θ ∆ , ω free∆ ), which tends to the saddle point (cid:16) θ ueq , x eq ( ω free∆ ) (cid:17) = (cid:18) π, ω free∆ τ K (cid:19) as t → + ∞ (by the symmetry of the lower and the upper half-planes, the consideration of theupper separatrix is also possible). The parameter ω free∆ shifts the phase plane vertically.To check this, we use a linear transformation x → x + ω free∆ τ K . Thus, to compute the lock-inrange of (6), we need to find ω free∆ = ω l (where ω l is called a lock-in frequency) such that(see Fig. 5) x eq ( − ω l ) = Q ( θ seq , ω l ) . (7)By (7), we obtain an exact formula for the lock-in frequency ω l : − ω l K /τ = ω l K /τ + Q ( θ seq , .ω l = − K Q ( θ seq , τ , (8)5 (θ Δ ,ω l ) (θ eq ,x eq (-ω l )) s (θ eq ,x eq (ω l )) x θ Δ ω θ eqs θ eq +2 ̟ s θ eq -2 ̟ s s Figure 5: The lock-in domain of (6) for (cid:12)(cid:12) ω free∆ (cid:12)(cid:12) = ω l . Numerical simulations are used to compute the lock-in range of (6) applying (8). Theseparatrix Q ( θ ∆ ,
0) is numerically integrated and the corresponding ω l is approximated.The obtained numerical results can be illustrated by special diagram (see Fig. 6). Notethat (6) depends on the value of two coefficients K τ and τ . In Fig. 6, choosing X-axisas K τ , we can plot a single curve for every fixed value of τ . The results of numericalsimulations show that for sufficiently large K τ , the value of ω l grows almost proportionallyto K τ . Hence, ω l τ K is almost constant for sufficiently large K τ and in Fig. 6 the Y-axiscan be chosen as ω free∆ τ K .To obtain the lock-in frequency ω l for fixed τ , τ , and K using Fig. 6, we considerthe curve corresponding to the chosen τ . Next, for X-value equal K τ we get the Y-valueof the curve. Finally, we multiply the Y-value by K τ .Consider an analytical approach to the exact lock-in range computation. Main stagesof computation are presented in Subsection 3.1. Consider a system ˙ θ ∆ ( t ) = y ( t ) , ˙ y ( t ) = − K τ τ ˙ ϕ ( θ ∆ ( t )) y ( t ) − K τ ϕ ( θ ∆ ( t )) , (9)where y ( t ) = ω free∆ − K τ ( x ( t ) + τ ϕ ( θ ∆ ( t ))). Relations (9) are equivalent to (6) and allowone to exclude ω free∆ from the computation. Note that equilibria ( θ eq , y eq ) of (9) and thecorresponding equilibria ( θ eq , x eq ) of (6) are of the same type and related as( θ eq , y eq ) = (cid:16) θ eq , ω free∆ − K bx eq (cid:17) . τ =0.9 (p , p ) K τ p = K ω l τ p = Figure 6: Diagram for the lock-in frequency ω l calculation. The separatrix Q ( θ ∆ , ω free∆ ) from (8) corresponds to the upper separatrix S ( θ ∆ ) of thephase plane of (9) (see Fig. 7) and the following relation Q ( θ seq , ω free∆ ) = τ K (cid:16) ω free∆ − S ( θ seq ) (cid:17) is valid. x θ Δ θ eqs x eq (ω ∆ ) free free Q(θ Δ ,ω ∆ ) θ eq +2 ̟ s θ eq -2 ̟ s y θ Δ θ eqs θ eq +2 ̟ s θ eq -2 ̟ s S'(θ Δ ) Figure 7: Phase plane portraits of (6) and (9).
Relation (8) takes the form ω l = 12 S ( θ seq ) . (10)The computation of the separatrix S ( θ ∆ ) is in two steps. Step 1: we integrate theseparatrix S ( θ ∆ ) in the interval (cid:16) π , π (cid:17) (in which the function ϕ ( θ ∆ ) is continuouslydifferentiable) and compute S ( π ). For this purpose, we need to find the eigenvector that7orresponds to separatrix S ( θ ∆ ) on the considered interval. Step 2: we find a generalsolution of (9) on the interval (cid:16) − π , π (cid:17) . Here there exist three cases depending on the typestable equilibrium (cid:16) θ seq , (cid:17) : a stable focus, stable node, and stable degenerated node. Forevery case described above we perform separate computations. Using the computed S ( π )as the initial data of the Cauchy problem, it is possible to obtain an exact expression for S ( θ seq ).The obtained analytical results are illustrated in Fig. 8. The red line in Fig. 8 is usedfor the case of stable focus, and the green line for the case of stable node. The crosses areused for the case of stable degenerated node. K ω l τ K τ τ Figure 8: Diagram for the lock-in frequency ω l calculation. The formulae for three possible cases are given below (redefinitions a = τ τ , b = τ areused to reduce the analytical formulae): A. ( aK ) − bK π > ω l = 1 π c q ( aK ) − bK π (cid:18) − c c (cid:19) − aK q ( aK ) − bK π , (11)where c = π q ( aK ) + 2 bK π q ( aK ) − bK π + 1 , c = π − q ( aK ) + 2 bK π q ( aK ) − bK π . B. ( aK ) − bK π = 0 that corresponds to a stable degenerated node: ω l = 12 c e aK c ! , where c = q ( aK ) + 2 bK π . (12)8 . ( aK ) − bK π < ω l = − aK et Re λ s π ( c cos ( t Im λ s ) + c sin ( t Im λ s )) ++ et Re λ s q bK π − ( aK ) π ( c cos ( t Im λ s ) − c sin ( t Im λ s )) , (13)where t = arctg (cid:18) − c c (cid:19) Im λ s , c = π , c = π q ( aK ) + 4 bK ( π − k )2 q bK π − ( aK ) ,λ s = − aK + i q bK π − ( aK ) π . Rigorous derivation of (11), (12), and (13) is given in Appendix A. The analytical andnumerical results are compared in Fig. 9. τ = 0.5 K ω l τ K τ
23 24 25 26 270.3520.3540.3560.3580.360.3620.364
Figure 9: Comparison of analytical and numerical results on the lock-in computation.
4. Conclusion
In the present work the model of PLL with impulse signals and active PI filter in thesignal’s phase space is described. For the considered PLL the lock-in range is computedanalytically and obtained result are compared with numerical simulations.
Appendix A. The lock-in computation
In this section equations (11), (12), and (13) are rigorously derived. Consider thefollowing relations ˙ θ ∆ = y, ˙ y = − aK ˙ ϕ ( θ ∆ ) y − bK ϕ ( θ ∆ ) . (A.1)Also we consider a normalized 2 π -periodic zigzag function ϕ ( θ ∆ ) = kθ ∆ , if − k ≤ θ ∆ ≤ k ; − kπk − θ ∆ + πkπk − , if k ≤ θ ∆ ≤ π − k (A.2)9or finite k > π in the interval θ ∆ ∈ h − k , π − k (cid:17) . For k = π the function ϕ ( θ ∆ ) is trian-gular and corresponds to (1).From 2 π -periodicity of (A.1) it follows that for each interval the behavior of phasetrajectories on the system phase plane is the same θ ∆ ∈ (cid:18) − k + 2 πj, − k + 2 π ( j + 1) (cid:21) , j ∈ Z . Thus, we can consider a single interval (cid:16) − k , π − k i of the phase plane of (A.1).In the intervals inside (cid:16) − k , π − k i , (A.1) takes the form:I. − k < θ ∆ < k ˙ θ ∆ = y, ˙ y = − aK ky − bK kθ ∆ ; (A.3)II. k < θ ∆ < π − k ˙ θ ∆ = y, ˙ y = aK kπk − y + bK (cid:16) kπk − θ ∆ − πkπk − (cid:17) . (A.4)In each interval there exists only one equilibrium:I. − k < θ ∆ < k y eq = 0 , − aK ky − bK kθ eq = 0; y eq = 0 ,θ eq = 0;II. k < θ ∆ < π − k y eq = 0 , aK kπk − y eq + bK kπk − ( θ eq − π ) = 0 . y eq = 0 ,θ eq = π. To define a type of the equilibria points, we compute the corresponding characteristicpolynomial and eigenvalues. For the first equilibrium ( θ eq , y eq ) = (0 ,
0) the characteristicpolynomial is as follows χ ( λ ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − λ − bK k − aK k − λ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = λ + aK kλ + bK k. The eigenvalues of the equilibrium ( θ eq , y eq ) = (0 ,
0) depend on a sign of ( aK ) − bK k .Here, there exist three cases: A. ( aK ) − bK k > λ s , = − aK k ± q ( aK k ) − bK k , the equilibrium (0 ,
0) is a stable node. B. ( aK ) − bK k = 0: λ s = λ s = − aK k , ,
0) is a stable degenerated node, or stable proper node. C. ( aK ) − bK k < λ s , = − aK k ± i q bK k − ( aK k ) , the equilibrium (0 ,
0) is a stable focus.Denote ( θ s eq , y eq ) = (0 , θ eq , y eq ) = ( π,
0) we have χ ( λ ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − λ bK kπk − aK kπk − − λ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = λ − aK kπk − λ − bK kπk − λ u , = aK kπk − ± r(cid:16) aK kπk − (cid:17) + bK kπk − , which means that ( π,
0) is always an unstable saddle for the considered parameters of thePLL. Denote ( θ u eq , y eq ) = ( π, S ( θ s eq ) from formula (10) for lock-in range is in some stages. First,find two-dimensional eigenvectors X u , X u of saddle point (cid:16) θ u eq , y eq (cid:17) from the interval θ ∆ ∈ (cid:16) k , π − k (cid:17) . Next, compute S ( k ), which is possible due to the continuity of (A.1).Find two-dimensional eigenvectors X s , X s of stable equilibrium (cid:16) θ s eq , y eq (cid:17) in the interval θ ∆ ∈ (cid:16) − k , k (cid:17) . Find a general solution of (A.1) in the interval θ ∆ ∈ (cid:16) − k , k (cid:17) . Using theobtained S ( k ) as the initial data of the Cauchy problem, we can compute S ( θ s eq ).Let us find the eigenvectors X u , X u of a saddle point (cid:16) θ u eq , y eq (cid:17) . First, find theeigenvector X u : − λ u bK kπk − aK kπk − − λ u ! X u = O , − aK kπk − + r(cid:16) aK kπk − (cid:17) + bK kπk − bK kπk − aK kπk − − aK kπk − + r(cid:16) aK kπk − (cid:17) + bK kπk − X u = O , − aK kπk − + r(cid:16) aK kπk − (cid:17) + bK kπk − bK kπk − − aK kπk − − r(cid:16) aK kπk − (cid:17) + bK kπk − X u = O . (A.5)Multiply the second row of (A.5) by aK kπk − + r(cid:16) aK kπk − (cid:17) + bK kπk − bK kπk − − aK kπk − + r(cid:16) aK kπk − (cid:17) + bK kπk − aK kπk − + r(cid:16) aK kπk − (cid:17) + bK kπk − − (cid:18)(cid:16) aK kπk − (cid:17) − ( aK kπk − ) + bK kπk − (cid:19) ( πk − bK k X u = O , − aK kπk − + r(cid:16) aK kπk − (cid:17) + bK kπk − aK kπk − + r(cid:16) aK kπk − (cid:17) + bK kπk − − X u = O . Hence, X u = cc q ( aK k ) + 4 bK k ( πk −
1) + aK k πk − . Let us choose c = q ( aK k ) + 4 bK k ( πk − − aK k bK k . Then X u = q ( aK k ) + 4 bK k ( πk − − aK k bK k ( aK k ) + 4 bK k ( πk − − ( aK k ) bK k ( πk − ,X u = q ( aK k ) + 4 bK k ( πk − − aK k bK k . Next, find the second eigenvector X u in the same way: − λ u bK kπk − aK kπk − − λ u ! X u = O , r(cid:16) aK kπk − (cid:17) + bK kπk − − aK kπk − bK kπk − aK kπk − + r(cid:16) aK kπk − (cid:17) + bK kπk − − aK kπk − X u = O , r(cid:16) aK kπk − (cid:17) + bK kπk − − aK kπk − bK kπk − r(cid:16) aK kπk − (cid:17) + bK kπk − + aK kπk − X u = O . (A.6)12ultiply the second row of (A.6) by r(cid:16) aK kπk − (cid:17) + bK kπk − − aK kπk − bK kπk − r(cid:16) aK kπk − (cid:17) + bK kπk − − aK kπk − r(cid:16) aK kπk − (cid:17) + bK kπk − − aK kπk − (cid:18)(cid:16) aK kπk − (cid:17) + bK kπk − − (cid:16) aK kπk − (cid:17) (cid:19) ( πk − bK k X u = O , r(cid:16) aK kπk − (cid:17) + bK kπk − − aK kπk − r(cid:16) aK kπk − (cid:17) + bK kπk − − aK kπk − X u = O . Hence, X u = − cc r(cid:16) aK kπk − (cid:17) + bK kπk − − aK kπk − . Choose c = q ( aK k ) + 4 bK k ( πk −
1) + aK k bK k : X u = − q ( aK k ) + 4 bK k ( πk −
1) + aK k bK k ( aK k ) + 4 bK k ( πk − − ( aK k ) bK k ( πk − ,X u = − q ( aK k ) + 4 bK k ( πk −
1) + aK k bK k . We can show that the direction of separatrix S ( θ ∆ ) coincides with the direction ofeigenvector X u , which corresponds to eigenvalue λ u . That allows us to find S ( k ). Forthis purpose, we write an equation of straight line, which passes through two points( x , y ) = ( π, , ( x , y ) = π − q ( aK k ) + 4 bK k ( πk −
1) + aK k bK k , . y − − x − π π − q ( aK k ) + 4 bK k ( πk −
1) + aK k bK k − π ,y = 2 bK k q ( aK k ) + 4 bK k ( πk −
1) + aK k ( π − x ) ,y = 2 bK k (cid:16)q ( aK k ) + 4 bK k ( πk − − aK k (cid:17) ( aK k ) + 4 bK k ( πk − − ( aK k ) ( π − x ) ,y = q ( aK k ) + 4 bK k ( πk − − aK k πk −
1) ( π − x ) . Then S ( 1 k ) = q ( aK k ) + 4 bK k ( πk − − aK k πk − (cid:18) π − k (cid:19) == q ( aK ) + 4 bK ( π − k ) − aK . Next, we need to find the eigenvectors of equilibrium ( θ s eq , y eq ) and a general solutionof (A.1) in the interval (cid:16) − k , k (cid:17) . It was shown that for a stable equilibrium ( θ s eq , y eq ) in theinterval (cid:16) − k , k (cid:17) there exist three different cases, which depend on a sign of ( aK ) − bK k .The eigenvectors X s and X s are computed in the case of stable focus only. For othercases the computation of X s , X s is similar to that, considered in Appendix A.1. Appendix A.1. Stable node
This case corresponds to ( aK ) − bK k >
0. Let us find the eigenvectors X s , X s : − λ s − bK k − aK k − λ s ! X s = O , aK k − q ( aK k ) − bK k − bK k − aK k + aK k − q ( aK k ) − bK k X s = O , aK k − q ( aK k ) − bK k − bK k − aK k + q ( aK k ) − bK k X s = O . (A.7)Multiply the second row of (A.7) by aK k − q ( aK k ) − bK k bK k : aK k − q ( aK k ) − bK k − aK k − q ( aK k ) − bK k − ( aK k ) − ( aK k ) + 4 bK k bK k X s = O , aK k − q ( aK k ) − bK k − aK k − q ( aK k ) − bK k − X s = O ,X s = − cc aK k − q ( aK k ) − bK k . Choose c = −
1. Then X s = q ( aK k ) − bK k − aK k . Next, find eigenvector X s : − λ s − bK k − aK k − λ s ! X s = O , aK k + q ( aK k ) − bK k − bK k − aK k + aK k + q ( aK k ) − bK k X s = O , aK k + q ( aK k ) − bK k − bK k q ( aK k ) − bK k − aK k X s = O . (A.8)Multiply the second row of (A.8) by aK k + q ( aK k ) − bK k bK k : aK k + q ( aK k ) − bK k − aK k + q ( aK k ) − bK k aK k ) − bK k − ( aK k ) bK k X s = O , aK k + q ( aK k ) − bK k − aK k + q ( aK k ) − bK k − X s = O ,X s = − cc aK k + q ( aK k ) − bK k . c = −
1. Then X s = − aK k + q ( aK k ) − bK k . In the interval θ ∆ ∈ (cid:16) − k , k (cid:17) for (cid:16) θ s eq , y eq (cid:17) = (0 ,
0) being a node, a general solution of (A.1)has the form: θ ∆ ( t ) = c eλ s t + c eλ s t,y ( t ) = − c aK k − q ( aK k ) − bK k eλ s t − c aK k + q ( aK k ) − bK k eλ s t. (A.9)Let us find coefficients c , c of (A.9) for the solution of the Cauchy problem with initialconditions θ ∆ (0) = k , y (0) = q ( aK ) +4 bK ( π − k ) − aK , which coincide with S ( k ).At moment t = 0 we have k = c + c , q ( aK ) + 4 bK ( π − k ) − aK − c aK k − q ( aK k ) − bK k − c aK k + q ( aK k ) − bK k , c = 1 k − c , q ( aK ) + 4 bK ( π − k ) − aK aK k + q ( aK k ) − bK k k == − c aK k − q ( aK k ) − bK k c aK k + q ( aK k ) − bK k , c = 1 k − c , q ( aK ) + 4 bK ( π − k )2 + q ( aK ) − bK k c k s ( aK ) − bK k , c = 1 k − c ,c = q ( aK ) + 4 bK ( π − k ) q ( aK ) − bK k + 1 : 2 k, c = q ( aK ) + 4 bK ( π − k ) q ( aK ) − bK k + 1 : 2 k,c = − q ( aK ) + 4 bK ( π − k ) q ( aK ) − bK k : 2 k. (A.10)Finally, find y ( t ) under the condition θ ∆ ( t ) = 0. The value of y ( t ) corresponds to S ( θ s eq ). For this purpose, we express y ( t ) in terms of c , c from (A.10). Then c eλ s t + c eλ s t ,y ( t ) = − c aK k − q ( aK k ) − bK k eλ s t − c aK k + q ( aK k ) − bK k eλ s t , − c c = e ( λ s − λ s ) t ,y ( t ) = − c aK k − q ( aK k ) − bK k eλ s t − c aK k + q ( aK k ) − bK k eλ s t , − c c = e − q ( aK k ) − bK k + aK k − q ( aK k ) − bK k − aK k t ,y ( t ) = − c aK k − q ( aK k ) − bK k eλ s t − c aK k + q ( aK k ) − bK k eλ s t , − c c = e (cid:18) − q ( aK k ) − bK k (cid:19) t ,y ( t ) = − c aK k − q ( aK k ) − bK k eλ s t − c aK k + q ( aK k ) − bK k eλ s t , ln (cid:18) − c c (cid:19) = − (cid:16)q ( aK k ) − bK k (cid:17) t ,y ( t ) = − c aK k − q ( aK k ) − bK k eλ s t − c aK k + q ( aK k ) − bK k eλ s t , t = ln (cid:18) − c c (cid:19)q ( aK k ) − bK k ,y ( t ) = − c aK k − q ( aK k ) − bK k eλ s t − c aK k + q ( aK k ) − bK k eλ s t , eλ s t = e ln (cid:18) − c c (cid:19) λ s q ( aK k ) − bK k = e ln (cid:18) − c c (cid:19) q ( aK k ) − bK k − aK k q ( aK k ) − bK k == e ln (cid:18) − c c (cid:19) − aK k q ( aK k ) − bK k = (cid:18) − c c (cid:19) − aK k q ( aK k ) − bK k . Similarly, eλ s t = e ln (cid:18) − c c (cid:19) λ s q ( aK k ) − bK k = e − ln (cid:18) − c c (cid:19) q ( aK k ) − bK k + aK k q ( aK k ) − bK k == e ln (cid:18) − c c (cid:19) −
12 + aK s ( aK ) − bK k = (cid:18) − c c (cid:19) − aK k q ( aK k ) − bK k − . Then y ( t ) = − c aK k − q ( aK k ) − bK k eλ s t − c aK k + q ( aK k ) − bK k eλ s t ,y ( t ) = − c aK k − q ( aK k ) − bK k (cid:18) − c c (cid:19) − aK k q ( aK k ) − bK k −− c aK k + q ( aK k ) − bK k (cid:18) − c c (cid:19) − aK k q ( aK k ) − bK k − ,y ( t ) = − c aK k − q ( aK k ) − bK k (cid:18) − c c (cid:19) − aK k q ( aK k ) − bK k ++ c aK k + q ( aK k ) − bK k (cid:18) − c c (cid:19) − aK k q ( aK k ) − bK k . As a result, for the case ( aK k ) − bK k >
0, when a stable equilibrium (cid:16) θ s eq , y eq (cid:17) isa stable node, S ( θ s eq ) can be found from the following formula S ( θ s eq ) = c q ( aK k ) − bK k (cid:18) − c c (cid:19) − aK k q ( aK k ) − bK k , (A.11)18here c = q ( aK ) + 4 bK ( π − k ) q ( aK ) − bK k + 1 : 2 k, c = − q ( aK ) + 4 bK ( π − k ) q ( aK ) − bK k : 2 k. Appendix A.2. Stable focus
This case corresponds to ( aK ) − bK k <
0. The eigenvectors X s , X s are found inthe same way as in Appendix A.1: X s = − aK k − i q bK k − ( aK k ) , X s = − aK k + i q bK k − ( aK k ) . The eigenvectors X s , X s can be represented as X s , ( t ) = U s , + iV s , , where U s , , V s , are real two-dimensional vectors: U s = − aK k , V s = q bK k − ( aK k ) ,U s = − aK k , V s = − q bK k − ( aK k ) . Consider a solution of (A.1), which corresponds to the eigenvalue λ s : Y ( t ) = e λ s t X s = e (Re λ s + i Im λ s ) t ( U s + iV s ) == et Re λ s (cos ( t Im λ s ) + i sin ( t Im λ s )) ( U s + iV s ) = et Re λ s ( U s cos ( t Im λ s ) − V s sin ( t Im λ s )) ++ iet Re λ s ( U s sin ( t Im λ s ) + V s cos ( t Im λ s )) . A general solution of (A.1) takes the form Y ( t ) = c et Re λ s ( U s cos ( t Im λ s ) − V s sin ( t Im λ s )) ++ c et Re λ s ( U s sin ( t Im λ s ) + V s cos ( t Im λ s )) == et Re λ s U s ( c cos ( t Im λ s ) + c sin ( t Im λ s )) ++ et Re λ s V s ( c cos ( t Im λ s ) − c sin ( t Im λ s )) . In other words, θ ∆ ( t ) = et Re λ s ( c cos ( t Im λ s ) + c sin ( t Im λ s )) ,y ( t ) = − aK k et Re λ s c cos ( t Im λ s ) + c sin ( t Im λ s )) ++ et Re λ s q bK k − ( aK k ) c cos ( t Im λ s ) − c sin ( t Im λ s )) . (A.12)19et us find the coefficients c , c for the solution of the Cauchy problem with the initialdata θ ∆ (0) = k , y (0) = q ( aK ) +4 bK ( π − k ) − aK , similarly to Appendix A.1. k = e λ s ( c cos (0 Im λ s ) + c sin (0 Im λ s )) , q ( aK ) + 4 bK ( π − k ) − aK − aK k e λ s c cos (0 Im λ s ) + c sin (0 Im λ s )) ++ e λ s q bK k − ( aK k ) c cos (0 Im λ s ) − c sin (0 Im λ s )) , k = c , q ( aK ) + 4 bK ( π − k ) − aK − aK k c + q bK k − ( aK k ) c , c = 1 k ,c = q ( aK ) + 4 bK ( π − k ) k s bK k − ( aK ) . Next, let us find t such that θ ∆ ( t ) = 0.0 = et Re λ s ( c cos ( t Im λ s ) + c sin ( t Im λ s )) , − c c = tg ( t Im λ s ) ,t = arctg (cid:18) − c c (cid:19) Im λ s . Finally, all the unknowns for y ( t ) from (A.12) are found and S ( θ s eq ) is as follows S ( θ s eq ) = y ( t ) = − aK k et Re λ s c cos ( t Im λ s ) + c sin ( t Im λ s )) ++ et Re λ s q bK k − ( aK k ) c cos ( t Im λ s ) − c sin ( t Im λ s )) , (A.13)where t = arctg (cid:18) − c c (cid:19) Im λ s , λ s = − aK k + i q bK k − ( aK k ) ,c = 1 k , c = q ( aK ) + 4 bK ( π − k ) k s bK k − ( aK ) . ppendix A.3. Stable degenerated node This case corresponds to ( aK ) − bK k = 0. In this case the eigenvalues λ s and λ s coincide: λ s := − aK k λ s = λ s . A stable equilibrium (cid:16) θ s eq , y eq (cid:17) is a stable degenerated node, or a stable proper node.For the characteristic matrix − λ s − bK k − aK k − λ s ! of (A.1) it is shown that (cid:16) θ s eq , y eq (cid:17) is a stable degenerated node. Find the eigenvector X s corresponding to the eigenvalue λ s of algebraic multiplicity two: − λ s − bK k − aK k − λ s ! X s = O , aK k − bK k − aK k X s = O . Adding the first row, multiplied by aK k aK k aK k ) − bK k X s = O . The eigenvector X s can be written as X s = c − c aK k . Choose c = 1. To find a general solution of (A.1), we need to additionally find the firstassociated vector X s : aK k − bK k − aK k X s = − aK k , aK k aK k ) − bK k X s = ! ,X s = cc − c aK k . Choose c = 1. 21 general solution of (A.1) has the form θ ∆ ( t ) = e − aK k t ! ( c + c ( t + 1)) ,y ( t ) = e − aK k t ! − aK k c + c − aK k t + 1 − aK k !! . (A.14)Similarly to Appendix A.1 and Appendix A.2, let us find the coefficients c and c for the solution of the Cauchy problem with the initial data θ ∆ (0) = k and y (0) = q ( aK ) +4 bK ( π − k ) − aK . In this case we have: k = c + c , q ( aK ) + 4 bK ( π − k ) − aK − aK k c + c − aK k ! , c = 1 k − c , q ( aK ) + 4 bK ( π − k ) − aK − aK k (cid:18) k − c (cid:19) + c − c aK k , c = 1 k − c , q ( aK ) + 4 bK ( π − k )2 = c , c = 1 k − q ( aK ) + 4 bK ( π − k )2 ,c = q ( aK ) + 4 bK ( π − k )2 . Find t such that θ ∆ ( t ) = 0: e − aK k t ! ( c + c ( t + 1)) ,y ( t ) = e − aK k t ! − aK k c + c − aK k t + 1 − aK k !! , e − aK k t ! (( c + c ) + c t ) ,y ( t ) = e − aK k t ! − aK k c + c ) + c − aK k t + 1 !! , k + c t ,y ( t ) = e − aK k t ! − aK c − aK k t + 1 !! , t = − c k ,y ( t ) = e − aK k t ! − aK − aK kc t + c ! . Finally, the expression for S ( θ s eq ) is as follows S ( θ s eq ) = y ( t ) = c e aK c ! , (A.15)where c = q ( aK ) + 4 bK ( π − k )2 . ACKNOWLEDGMENTS
This work was supported by the Russian Scientific Foundation and Saint-PetersburgState University. The authors would like to thank Roland E. Best, the founder of theBest Engineering Company, Oberwil, Switzerland and the author of the bestseller onPLL-based circuits Best (2007) for valuable discussion.
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