# Stability of a two poles negative feedback amplifier

let's consider this analysis of a two-pole amplifier with negative feedback (Microelectronic Circuits by Sedra Smith 7th edition):

Before showing you my doubts, let's consider this introduction made at the beginning of the chapter about negative feedback: So, with the term "unstable", this book means "oscillating".

Now, my questions are:

1) Is that definition of stability general? When I studied control theory, I saw a different definition of stability, that is: stable system = system in which bounded input implies bounded output. I do not see any link between this definition and an oscillating behaviour.

2) Now consider again the analysis of the two poles amplifier shown above. The book says that the amplifier is always stable because all the poles of the closed loop system have negative real part. It is true, but they may have also an immaginary part, which means oscillation. It is in contrast with its definition of stability.

• 1. A practical unstable system will reach a limit, which may be either a steady value or oscillatory, depending on the nature of the system. Text books tend to go for the ideal linear definition of stability. 2. Two complex conjugate poles with negative real parts will always be stable since the real parts give rise to exponentially decaying amplitude.
– Chu
Mar 30, 2020 at 15:38
• "A practical unstable system will reach a limit, which may be either a steady value or oscillatory" or also break and completely change its behavior.
– jDAQ
Mar 30, 2020 at 19:22

1) Is that definition of stability general? When I studied control theory, I saw a different definition of stability, that is: stable system = system in which bounded input implies bounded output. I do not see any link between this definition and an oscillating behavior.

The connection is very easy to see. Just write the complete transfer function in terms of the open-loop gain. $$H(s) = \frac{Y(s)}{X(s)} = \frac{A(s)}{1+A(s)\beta}.$$ Clearly, if loop gain $$\L(s) = A(s)\beta = -1\$$, then the output is not bounded. It is only possible if $$\|L(s)| = 1\$$ and $$\\angle L(s) = 180^\circ\$$. For second order systems, such a phase lag is not possible as you know.

2) Now consider again the analysis of the two poles amplifier shown above. The book says that the amplifier is always stable because all the poles of the closed loop system have negative real part. It is true, but they may have also an imaginary part, which means oscillation. It is in contrast with its definition of stability.

Consider a second order system with following transfer function: $$H(s) = \frac{1}{(s-p_1)(s-p_2)}$$ The corresponding time domain response (natural response) will be given by: $$h(t) = A_1e^{p_1t} + A_2e^{p_2t}$$ Here, $$\A_1\$$ and $$\A_2\$$ are constants.
If $$\p_{1,2} = \alpha \pm j\omega\$$, then, the response becomes: $$h(t) = A_1e^{\alpha t}e^{j\omega t} + A_2e^{\alpha t}e^{j\omega t}$$ Thus, if the poles lie in the left half of s-plane meaning $$\\alpha < 0\$$, the response decays over time resulting in decaying oscillations. Thus, depending on the damping $$\\alpha\$$, the system stops oscillating after some time.
If the poles lie in the right half of s-plane meaning $$\\alpha > 0\$$, the response grows over time resulting in increased oscillations.
If the poles lie in the right half of s-plane meaning $$\\alpha = 0\$$, the response is a constant sinusoidal output and the oscillations are sustained over time. This is where the system behaves as oscillator. EDIT
For oscillation to happen the Barkhausen's Criteria needs to be satisfied. In the figure below (assume the disturbance is zero), if you travel around the loop you will have a gain of $$\-A\beta = -L(s) = 1\$$. Another inversion is added due to the difference operation at the input. Thus we have a phase difference of $$\360^\circ\$$ around the loop and the magnitude of gain is unity. This is the condition for sustained oscillation. • About 1), in that case the output is not bounded. But which is the link with oscillations? Mar 30, 2020 at 19:00
• see edit...hopefully you know what is Barkhausen's criteria Mar 30, 2020 at 19:22

To 1: The term "bounded input" has an extreme case: No external input. This is the case for an oscillating system. Hence, the definion (BIBO) is applicable.

To 2: When a pole pair of a closed-loop system has a negative real part, all possible oscillations will die out (due to the term exp(-sigma*t)). Speaking about oscillations, it is necessary to discriminate between

• continuos (steady-state) oscillations with poles on the Im-axis (real part zero)

• rising oscillation amplitudes (poles with positive real part)

• decreasing oscillation amplitudes (poles with negative real part)

• About 1: an oscillator have 0 input and a bounded output. Why should it be considered unstable? Mar 30, 2020 at 17:03
• The explanation for 1 is not that good, "The term "bounded input" has an extreme case: No external input" doesn't help much, since with that input zero, you could not point what would lead it to growing unbounded. If you consider a bounded noise as the input to the circuit, then you would be able to talk about BIBO stability and having those poles in the imaginary axis would lead to an unstable system.
– jDAQ
Mar 30, 2020 at 19:46
• @Kinka-Byo, there is no physical system with "unbounded" input or output. Hence, the term "BIBO" is a theoretical criterion only. Hence, it is more practical to express this criterion in the frequency domain....and we require the poles of a stable system to be in the LHP only. As you know, the poles of a harmonic oscillator are in the RHP (however, rather close to the Im axis) and the rising of the amplitude must be "artificially" limited using a certain non-linearity or the clipping effect caused by the supply rails. Therefore, I think the criterion under discussion is applicable, of course.
– LvW
Mar 31, 2020 at 10:09

Answering only part 1, There are many definitions of stability, and each of them have different purposes. You can have the BIBO stability, that you mentioned, where having a bounded input leads to a bounded output. But in the case you have poles at the imaginary axis, it will mean that with the right choice of $$\u(t)=\sin(\omega t)\$$, a bounded input, you will be able to get an output $$\y(t)\$$ that grows unbounded. But there are different definitions, such as

• Asymptotic stability: as $$\t \rightarrow \infty\$$ the system will go to $$\x \rightarrow 0\$$, no matter the initial conditions.
• Lyapunov stability, or Internal stability: if the system start in some region $$\\|x\|<\epsilon\$$ then it will never grow beyond some $$\\|x\|<\delta\$$ with $$\\epsilon \leq \delta \$$.
• Marginal stability: similar to the Lyapunov one, but usually applied to linear systems. Means the system will not grow unbounded, that is, as $$\t \rightarrow \infty\$$ the system will not go to $$\ \|x\| \rightarrow \infty\$$. Might seem equal to the Asymptotic one, but it does not mean that as $$\t \rightarrow \infty\$$ the system will go to $$\x \rightarrow 0\$$, just that $$\\|x\| being limited to some positive number $$\r\$$.