# Oscilating System

For a positive feedback system formed by an amplifier network A(s) and feedback network B(s), we have the following transfer function :
H(s) = A(s)/[1 - A(s)B(s)].

Quoting the Barkhaus Criterion, it says that in order generate an oscilator with this system, we must have 1 - A(jwo)B(jwo) = 0
And then he afirms the system will oscilate with frequency wo.That is the same as saying that the poles must be in the imaginary axis for the system to oscilate ( no loss after each loop ).

What i can't understand is why only poles in the imaginary axis introduces no loss into the senoidal after each loop A(s)B(s). In my head, any pole "So" ( be it in imaginary axis or not ) that will satisfy A(So)B(So) = 1 will make the system oscilate, keeping the same amplitude of the original sinusoid and introducing no phase shift ( afterall that's what A(So)B(So) = 1 means ) .

Why only poles in the imaginary axis ( and not any other ) makes the system oscilate indefinitely ? Why do we need poles in the imaginary axis to keep a sinusoidal with the same amplitude and introducing no phase shift after each loop A(s)B(s) ?

Thanks

• The English word is sinusoidal, not senoidal. Oscillating has two l's. The plane to the left of the imaginary axis represents damped oscillations (negative coefficients of exponentials). Stable systems have poles in that plane. The imaginary axis represents a coefficient of zero: oscillations that neither decay nor grow.
– Kaz
Mar 7, 2013 at 8:48
• Yes, that's what i'm asking.Why (mathematically ) do poles in the imaginary axis represent oscillations that neither decay nor grow. Mar 7, 2013 at 10:52

A function is transformed into the s-domain by a convolution against the exponential function. The values at any point s in that domain measure how that function is contributed by $e$ raised to the power of $s$.
The parameter $s$ is complex: $s = \delta + j\omega$, also known as a "complex frequency". Each point in the domain tells us how well the original function correlates with $e^{\delta + j\omega}$.
Now $e^{a+b} = e^ae^b$ and therefore $e^{\delta + j\omega} = e^\delta e^{j\omega}$. Thus point in the s-domain represents a decay factor contributed by the real part $e^\delta$ and a complex sinusoidal part contributed by $e^{j\omega}$. On the imaginary axis, $\delta = 0$, representing the boundary between decay and growth, because $e^\delta = e^0 = 1$.