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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.

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

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

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  • \$\begingroup\$ 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. \$\endgroup\$ – salil87 Feb 21 '15 at 15:05
  • \$\begingroup\$ I have added an additioinal paragraph (time/frequ. domain). \$\endgroup\$ – LvW Feb 21 '15 at 15:38

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