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Suppose we are dealing with a control problem where the reference trajectory is $0$, then the PID controller is a function,

$$u = K_1 x + K_2 \dot x + K_3 \int_0^t x dt$$ where $x$ is your state.

But this looks almost exactly like state feedback control with integral action,

$$u = K_1 x + K_2 \int_{0}^t \underbrace{Cx}_{y} dt$$

Furthermore, typically $x = (x_1, x_2), x_2 = \dot x_1$. Suppose $K_1 = (K_{11}, K_{12})$ and $C$ selects $x_1$, then this controller looks like, $$u = K_{11} x_1 + K_{12} \dot x_1 + K_2 \int_{0}^t x_1 dt$$ which is exact the form of the PID controller.

Even if there were a reference signal, the structure is remarkably similar and perhaps identical.

Given all these similarities, is it wrong to assume that PID and state feedback (with integral action) are pretty much one and the same? If not, then why is the approach in the standard literature so different when it comes to designing these controllers (Nyquist/Bode plot versus eigenvalue fiddling)?

Your neighbor Todorovich
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1 Answers1

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They are the same under the assumptions you stated, which are plenty:

  • You assume a second order plant
  • You assume $\dot{x}_1=x_2$
  • You assume you measure $x_1$ and $x_2$
  • You assume $r=0$

General state feedback controllers with integral action don't need these assumptions, except the third (access to the full state vector), but even if you don't measure $x$ you might be able to use an observer for the missing states.

As a simple example of a controller where it does make a difference, consider the plant $$ P(s)=\frac{1}{s^3} $$ subject to (ideal) PID control $$ C(s)=k_p+\frac{k_i}{s}+k_ds $$ The closed loop transfer function is $$ G(s)=\frac{P(s)C(s)}{1+P(s)C(s)}=\frac{k_ds^2+k_ps+k_i}{s^4+k_d s^2+k_p s +k_i} $$ The $s^3$ coefficient of the denominator polynomial is zero, so for any $k_p,k_i,k_d$ the system is not asymptotically stable, so PID control cannot stabilize this plant (see also this answer for a related example).

However, $P(s)$ can be written as $$ \begin{align} \dot{x}&=Ax+Bu\\ y&=Cx \end{align} $$ with $$ \begin{align} A&=\begin{pmatrix} 0&0&0\\-1&0&0\\0&-1&0 \end{pmatrix}\\ B&=\begin{pmatrix} -1\\0\\0 \end{pmatrix}\\ C&=\begin{pmatrix} 0&0&-1 \end{pmatrix} \end{align} $$ which is controllable and $$ u=-Kx $$ with $$ K=\begin{pmatrix} -3&3&-1 \end{pmatrix} $$ places all three eigenvalues of $A-BK$ at $-1$, so the closed loop system is exponentially stable.

As you can see, there is quite a difference between PID and state feedback although the plant considered was very simple.

If the plant is more complicated, PID control might even not be uniquely defined anymore. For example, consider a MIMO plant with two inputs, three outputs and three reference signals. How would you distribute the outputs of three PID controllers to only two inputs? There is no single solution to this allocation problem and each method has its own advantages/disadvantages.

So yes, in general it is wrong to assume that PID and state feedback are the same.

SampleTime
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