Feedback Criterion for Oscillations [closed]

In the Feedback equation A/(1+AB) the criterion to be fulfilled is |AB|=1 and phase shift = 180 deg. In this case the poles of 1+AB would be on img axis at $w_0$. But in practical cases |AB| is made slightly higher than 1. It is said that it makes the poles go in RHP. I know it should.. but it is difficult to visualize it. How can it be proved that poles do go to RHP?. Thanks.

closed as unclear what you're asking by Olin Lathrop, Leon Heller, Vladimir Cravero, Daniel Grillo, Scott SeidmanFeb 23 '15 at 13:20

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• "A/(1+AB)" is no equation. – Olin Lathrop Feb 21 '15 at 14:27

The correct equation for a finite gain amplifier having frequency-dependent feedback is

A(s)=Ao/[(1-AoB(s)]

with Ao=finite gain and feedback function B(s); the product AoB(s) is the loop gain Aloop(s)=AoB(s).

• Oscillation condition: Aloop(s)=1

• Interpretation: Because the loop gain must be positive ( |Aloop|=1; phase=360deg) we have two options: (a) Ao>0 and B(s) with 360deg (0deg) phase shift at f=fo or (b) Ao<0 and B(s) with 180deg phase shift at f=fo.

• Pole distribution: Solving the oscillation condition for the nominal (ideal) case Aloop(s)=1 results in a pole pair directly on the imag. axis of the s-plane. Because this condition cannot be exactly fulfilled (tolerances!) and to ensure a safe self-start of oscillations we design the circuit for Aloop(jw=jwo)>1. Now - for calculation of the pole distribution we have to solve the oscillation condition for Aloop(s)>1. This results in a pole pair with a positive real part sigma (right half of the s-plan).

• Time domain: In the time domain, the pos. real part of the poles is equivalent to a positive sigma value in the expression that determines the amplitude : exp(sigma*t). Thus, the amplitude rises with time and must be limited using a kind of non-linearity within the circuit. As a result, the loop gain will be reduced for large amplitudes approaching the case Aloop(jw=jwo)=1. Therefore, the poles are shifted back (automatically) in direction to the imag. axis.

• Relation between time and frequency domain: The denominator D(s) of a transfer function T(s) for an active circuit with feedback (frequency domain) is identical to the "characteristic polynominal P(s)" which results from the differential equation (time domain). That means: The solutions of the charact. equation P(s)=0 are identical to the zeros of D(s) - identical to the poles of the transfer function T(s). Hence, if the real part "sigma" of the time domain exponential solution [exp(sigma*t)] is positive, we have instability with rising amplitudes - equivalent to a positive real part of the zeros of D(s) being the poles from T(s).

• Thanks LvW. Just a query. How can we see Aloop(s)>1 has polepair in RHP. Do we have to solve it to prove that or is it obvious from the equation in some other way. – salil87 Feb 21 '15 at 15:05
• I have added an additioinal paragraph (time/frequ. domain). – LvW Feb 21 '15 at 15:38