Say we have a linear system with unity feedback, with loop transfer function $L(j\omega)$. The closed-loop transfer function from reference to output is $T(j\omega) = \frac{Y(j\omega)}{R(\omega)}=\frac{L(j\omega)}{1+L(j\omega)}$.

At frequencies for which $L(j\omega)$ approaches $-1$, clearly $|T(j\omega)| \rightarrow \infty$, so the system is unstable - if excited at this frequency, the output is unbounded.

But systems can be unstable even if $L(j\omega)$ never equals $-1$. From the Nyquist plot, we can show that $T(j\omega)$ can have poles in the right half plane if the contour encircles $-1$, even if $L(j\omega)$ never exactly reaches it. (However, I have very little intuition for why this is true - I just see it as a theorem from complex analysis that happens to be useful here).

Alternatively, from the Bode plot, we say a system is unstable if there are any frequencies for which $|L(j\omega)|$ > 1 and $\angle L(j\omega)< -\pi$ (i.e. phase margin is negative, or gain margin is less than unity). However, I'm not sure why these two conditions result in instability, since these don't result in $|T(j\omega)|$ going to $\infty$.

Two questions:

(1) If $|T(j\omega)|$ never goes to infinity (which is the case when $L(j\omega)$ is never exactly $-1$), how can a system possibly be unstable? Is $|T(j\omega)| \rightarrow \infty$ not the right criterion for deciding whether a system is unstable?

(2) Intuitively, why is a system unstable if there are any frequencies for which $|L(j\omega)|$ > 1 and $\angle L(j\omega)< -\pi$? I understand that you can see it from the Nyquist plot because these two conditions tend to result in encirclements of $-1$ in the $L(s)$ plane, but I'm looking for the intuition.

  • $\begingroup$ This video may be useful. $\endgroup$
    – Tendero
    Commented Jan 14, 2017 at 15:45

1 Answer 1


You cannot make conclusions about the stability of a system by only considering its transfer function evaluated on the imaginary axis $s=j\omega$. Replacing $s$ by $j\omega$ in the transfer function only makes sense for a stable system, otherwise you get a function of $\omega$ that does not describe the system, but another (stable) system.

Let me explain this by an example. Assume a causal system with a transfer function


If $a>0$, the system is stable and by substituting $s=j\omega$, we get its frequency response


The corresponding impulse response is


where $u(t)$ is the unit step response. We have

$$H(s)=\int_{-\infty}^{\infty}h(t)e^{-st}dt,\qquad \text{Re}\{s\}>-a\tag{4}$$


$$H(j\omega)=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt\tag{5}$$

Now imagine that we decrease the value of $a$. The system remains stable as long as $a>0$, but as soon as $a$ becomes $0$, $H(j\omega)\rightarrow\infty$ for $\omega=0$, and the system becomes unstable. If we further decrease $a$, the system remains unstable, but the expression $(2)$ does not go to infinity anywhere. However, the point is that the expression $(2)$ is not the frequency response of the system anymore if $a<0$. The reason is that the integral $(5)$ does not exist anymore. So it is pointless to try to deduce stability or instability from the frequency response of a system, because an unstable system has no frequency response.

What you get if you replace $s$ by $j\omega$ in the transfer function of an unstable system is the frequency response of another system with the same expression for the transfer function, but with a different region of convergence (ROC). The ROC is the region in the complex $s$-plane for which the integral $(4)$ converges. For the causal system in our example, the ROC was given by $\text{Re}\{s\}>-a$ (the ROC is always a right half-plane for causal systems). For an unstable system with $a<0$, the ROC does not include the $j\omega$-axis, and, consequently, $H(j\omega)$ does not exist. In that case, the expression $(2)$ is the frequency response of a different system with transfer function $H(s)$ given by $(1)$ but with a different ROC given by $\text{Re}\{s\}<-a$. That system is stable but anti-causal, and its impulse response is given by


which is clearly different from $(3)$.

Concerning your second question, note that $e^{\pm j\pi}=-1$, so a phase shift of $\pi$ for certain frequencies implies a sign change, which turns the negative feedback into a positive feedback. This can be compared to the positive feedback in an audio system where a microphone picks up the loudspeaker signal, which is fed back into the amplifier, sent to the loudspeaker, etc.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.