Connection between positive feedback at DC and real-axis right-half-plane pole(s)

Question

(Context provided here, question summarized at the end.) Suppose I have an open-loop transfer function with two coincident left half plane poles like

$$ H(j\omega)=\frac{-H_0}{(1+j\omega RC)^2} $$

where $H_0>0$, and so the Bode phase plot of $H(j\omega)$ has a shape like the below (arbitrary pole and gain).

If this system were put into a typical unity-gain negative feedback loop such that we form a closed-loop system

$$ A(j\omega)=\frac{H(j\omega)}{1+H(j\omega)} $$

we would expect based on the phase margin heuristic that the system would be unstable since the phase of the loop gain (here equal to $H(j\omega)$) is below -180° while $|H(j\omega)|>1$. And indeed it turns out that in this case $A(j\omega)$ has a right half plane pole on the real axis at $s=+9000$, so the impulse response is a growing exponential.

In a textbook that discusses the above example (Design of Analog CMOS Integrated Circuits, 2nd edition, p. 610), it states that the system "exhibits positive feedback near zero frequency [...] As a result, it simply 'latches up' rather than oscillates." I can see from the above results that the system does have a real-axis pole instead of a pair of complex conjugate poles and so has a pure exponential growth instead of an oscillatory one. I also see it so happens that the phase approaches -180° near DC instead of some other positive frequency.

My question: Is there some theoretical connection between the fact that the phase approaches -180° at DC and the fact that the system has a real-axis RHP pole instead of being able to achieve a pair of complex conjugate poles? Does -180° phase at DC have special significance in general? Asides from simply plotting the root locus of a given $H(j\omega)$, are there indications/suggestions from a Bode plot that the system might have real-axis RHP poles instead of complex conjugate ones?

Thanks very much for any help.

Answer 1

The formula given as $H(j\omega)$ is the frequency response, but if we are going to discuss poles and zeros for a continuous time system we would need to review the transfer function given as $H(s)$. $H(s)$ is the Laplace Transform of the system's impulse response, and from that we can get the frequency response (and therefore the Bode plot) by restricting $s$ to be $j\omega$.

That said, the OP's transfer function would be given as:

$$H(s) = \frac{-H_o}{(1+sRC)^2}$$

The significance of the phase being 180° at DC is simply the presence of the minus sign. At DC, $s=0$ and therefore $H(s=0) = -H_o$. Asssuming $H_o$ is a positive constant, the result would be 180° at DC.

As far as having complex conjugate poles in the closed loop response, we note that the closed loop transfer function is related to the open loop transfer function as follows:

$$G_{CL}(s) = \frac{G_F(s)}{1+G_{OL}(s)}$$

Where

$G_{CL}(s)$: Closed Loop Transfer Function
$G_F(s)$: Forward Transfer Function from input to output
$G_{OL}(s)$: Open Loop Transfer Function

See my graphic below depicting each of these as a "Gain". Note importantly that the formula as given implies a negative feedback. It is possible that the negative sign the OP used is also accounting for this, in which case the Open Loop gain would just be $\frac{H_o}{(1+sRC)^2}$.

From this we see that the poles of the closed loop system are defined by the denominator given as $1+G_{OL}(s)$. This expression is called the "characteristic equation". To determine if the poles are complex or real, simply solve for the roots of the characteristic equation.