Phase shift oscillator

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0.0. 7.3 1

Phase shift oscillator

Example. Let us consider the following phase shift oscillator:

The gain of the amplifier is:

A = −β where

β = R1

R

• The feedback system can be described by the following block scheme:

−V^p -

N.L.

β ^- ⁶^-

VM

V0-

−G(s) ^-

• The nonlinear element (N.L.) is a saturation with a central slope β:

- 6

X1 Vp

VM

V0

−V^M

β X1 = R VM

R1

β = R1

R

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7.3. DESCRIBING FUNCTION: EXAMPLES 7.3 2

• G(s) = ^V_V^p0^(s)(s) is the transfer function of the following electrical network:

6

V0

6

Vp

I-1

C

V1

R I-2

C

V2

R I-3

C

V3

R

• The electrical network is described by the following differential and static equations:

I1 = C d(V0 − V¹)

dt , I2 = C d(V1 − V²)

dt , I3 = C d(V2 − V³) dt V₁ = R(I₁ − I²), V₂ = R(I₂ − I³), V₃ = RI₃

• Using the Laplace transforms, the three differential equations can be expressed as follows:

I1 = C s (V0 − V¹), I2 = C s (V1 − V²), I3 = C s (V2 − V³)

• The mathematical model of the considered physical system can be expressed as follows:

V₀ ^-

C s

?

I1

-

R

6

V₁

-

C s

?

I2

-

R

6

V₂

-

C s

?

I3

-

R

6

V₃

-

Vp

0

• The transfer function G(s) which links the input V⁰(s) to the output V_p(s) can be easily obtained using the Mason formula:

G(s) = V_p(s)

V0(s) = R³C³s³

1 + 5 R C s + 6 R²C²s² + R³C³s³

In fact, the block diagram shows 5 rings with gain − R C s, 6 pairs of rings that do not touch each other, and one set of rings that do not touch three to three. Moreover, the only path starting from V0 and arriving at Vp crosses all the blocks.

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• Nyquist Diagram and Bode Diagrams of function G(s):

−0.2 0 0.2 0.4 0.6 0.8 1 1.2

−0.1 0 0.1 0.2 0.3 0.4 0.5 0.6

Nyquist diagram

10¹ 10² 10³ 10⁴ 10⁵

−150

−100

−50 0 50

rad/sec

db

10¹ 10² 10³ 10⁴ 10⁵

0 50 100 150 200 250 300

rad/sec

gradi

Amplitude

Phase

The Bode diagram intersects the real negative axis at point σ^∗ = −1/K^∗ characterized by frequency ω^∗. The values of parameters K^∗ and ω^∗ can be easily calculated using the Routh criterion. The characteristic equation 1 + K G(s) = 0 of the system is:

R³C³(K + 1)s³ + 6 R²C²s² + 5 R C s + 1 = 0 From the Routh table:

3 R³C³(K + 1) 5 R C

2 6R²C² 1

1 (30 − 1 − K)R³C³

0 1

one obtains that the system is stable for −1 < K < K^∗ = 29.

• A limit cycle appears in the system if:

β > K^∗ dove β = R1

R e K^∗ = 29

• The frequency ω^∗ can be obtained solving the following auxiliary equation:

6 R²C²s² + 1 = 0 → ω^∗ = 1

R C√ 6

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• Simulink block scheme of the considered nonlinear system (Phase Shift Oscillator mdl.mdl):

1 tau.s+1 Transfer Fcn V1

To WS8

V2 To WS7

Vp To WS6

V0 To WS5

t

To WS1

Saturation1 s

1 Int3 s

1 Int2 s

1 Int1

−beta

G7

G6 1/R 1/C G5 G4 1/R

1/C G3 G2 1/R

1/C G1 Clock

• Simulation parameters (Phase Shift Oscillator m):

R=1; % Resistance

C=0.001; % Capacity

beta=60; % Gain

VM=12; % Maximum voltage

V30=0.1; % Initial condition

Q30=C*V30; % Initial condition

wstar=1/(R*C*sqrt(6)); % Oscillation pulse

tau=0.000001; % Amplifier time constant

Tfin=10*2*pi/wstar; % Duration of the simulation sim(’Phase_Shift_Oscillator_mdl’,Tfin) % Simulation

figure(1) % Opening of the figure nr. 1

subplot(211); plot(t,V0); % Plot of voltage V0 subplot(212); plot(t,Vp); % Plot of voltage Vp

• Simulation results (variables V^p and V0):

Let us choose:

R = 1, C = 0.001, β = 60, VM = 12 One obtains:

ω^∗ = 408.2 T = 15.4 ms

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16

-15 -10 -5 0 5 10 15

V0 [V]

Tensione V0

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16

-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8

Vp [V]

Tensione Vp

Time [s]