You can always add or subtract a multiple of 360°. I think it is easier to see why when drawing the Nyquist diagram.
However in this case the closed-loop will not be stable because the open-loop is unstable (two poles at $s = 4$) and there are no encirclements of the minus one point in the Nyquist diagram. Since one normally only talks about margins when the system is stable, you could say that this system has no margins.
Because of different practical reasons:
- In reality, you can't make $u$ infinitly large, so $k \rightarrow \infty$ which would make $|x| \rightarrow 0$ is not feasible.
- In reality, you also have measurement noise which gets also amplified by large $k$.
- Also, in reality, large $u$ come with a cost (like energy consumption) so large $k$ can be too expensive.
But there are more problems with high gain. In many cases you dont really have a continous controller like in your system, but a digital. For the continous scalar system:
$$
\dot{x} = a x + u \tag{1}
$$
and $u = -k x$ with $k > |a|$ you have $a - k < 0$ so the system is stable, no matter how large you take $k$ as long as its larger than $|a|$.
But if you have a digital controller, you end up with a discrete system. For example $(1)$ has the transfer function
$$
G(s) = \frac{1}{s - a}
$$
which is discretized with sample-and-hold element and sample time $T$:
$$
G(z) = \frac{e^{a T} - 1}{a(z - e^{a T})}
$$
Closed loop with static state feedback and gain $k$:
$$
G_c(z) = \frac{1 - e^{a T}}{k + a e^{a T} - a z - k e^{a T}}
$$
This transfer function has the pole
$$
\frac{k(1 - e^{a T})}{a} + e^{a T}
$$
that is outside the unit circle for large $k$. So high gain feedback can make your system unstable even if the continous system is stable, because of discretization.
Best Answer
No, it doesn't mean that your (closed-loop) system is unstable, quite the opposite. A gain margin of infinty means that no matter how much you increase the gain, the system will always be stable.
And yes, this is true for all integrators: If we have $G(s) = \frac{b}{s}$ with $b > 0$ in negative feedback with a gain $k > 0$ then
$$ L(s) = \frac{k b}{s} $$
so the closed loop function is
$$ T(s) = \frac{k b}{s + k b} $$
which has a pole at $-k b$, which is always negative because $k$ and $b$ are positive. So, no matter how large we take $k$, the system will always be stable and so the gain margin is infinite.
In your case, you have a controller with dynamics included, so your $L(s)$ isn't a (pure) integrator. However, it can still be shown that this structure has also infinite gain margin, if all parameters are positive.
In the general form you have:
$$ L(s) = \frac{k(b_1 s + b_0)}{a_2 s^2 + a_1 s} $$
(in your case: $b_1 = 1800, b_0 =15, a_2 = 1000, a_1 = 16.965$). Closed loop:
$$ T(s) = \frac{k(b_1 s + b_0)}{a_2 s^2 + (a_1 + b_1 k)s + b_0 k} $$
So the characteristic polynomial is
$$ p(s) = a_2 s^2 + (a_1 + b_1 k)s + b_0 k $$
Its Routh table is:
As you can see all entries in the first column are positive (assuming all coefficients are positive). So no sign changes happen and the system is stable, no matter how you choose $k > 0$.
Of course, things get a bit more complicated if some of the coefficients can also be negative. If for example you have a negative $b_1$, i.e. $b_1 = -1800$, then from the second row of the Routh table you can see that
$$ a_1 + b_1 k > 0 $$
is a neccessary stability condition. Insert values:
$$ 16.965 - 1800 k > 0 $$
so we need $0 < k < 377/40000 = 0.009425$. You can confirm this with Matlab:
which will display the same result
This is quite bad: your gain is too large now and has to be reduced by more than $99 \%$ so that the closed loop is stable again.