Fundamentals of Linear Control:
A Concise Approach

Maurício C. de Oliveira

Chapter 4: Feedback Analysis

linearcontrol.info/fundamentals

Recall from Chapter 3

Time-invariant linear system in the time-domain

Impulse response

\[\begin{align*} g(t), \quad t \geq 0 \end{align*}\]

Convolution

\[\begin{align*} y(t) &= \int_{0^-}^{\infty} g(t - \tau) \, u(\tau) \,d\tau \end{align*}\]

Time-invariant linear system in the frequency-domain

Transfer-function

\[\begin{align*} G(s) = \mathcal{L} \{ g(t) \} \end{align*}\]

Product

\[\begin{align*} Y(s) &= G(s) \, U(s), & Y(s) &= \mathcal{L} \{ y(t) \}, & U(s) &= \mathcal{L} \{ u(t) \} \end{align*}\]

4.1 Tracking, Sensitivity, and Integral Control

Loop relations

\[\begin{align*} Y(s) &= G(s) \, U(s), & U(s) &= K(s) \, E(s), & E(s) &= \bar{Y}(s) - Y(s) \end{align*}\]

Closed-loop sensitivity transfer-function

\[\begin{align*} E(s) &= S(s) \, \bar{Y}(s), & S(s) &= \frac{1}{1 + G(s) \, K(s)} \end{align*}\]

Short-hand notation

\[\begin{align*} e &= S \, \bar{y}, & S &= \frac{1}{1 + G \, K}, & \text{etc} \end{align*}\]

Example: car model with proportional feedback

Open-loop transfer-function

\[\begin{align*} G(s) = \frac{\frac{p}{m}}{s + \frac{b}{m}} \end{align*}\]

Closed-loop sensitivity transfer-functions

\[\begin{align*} S(s) = \frac{1}{1 + K \frac{\frac{p}{m}}{s + \frac{b}{m}}} = \frac{s + \frac{b}{m}}{s + \frac{b}{m} + K \frac{p}{m}} \end{align*}\]

Closed-loop stability

\[\begin{align*} - \left ( \frac{b}{m} + \frac{p}{m} K \right ) &< 0 & &\implies & K > - \frac{b}{p} \end{align*}\]

Tracking error

Reference input

\[\begin{align*} \bar{y}(t) &= \bar{y} \cos( \omega t + \phi), \quad t \geq 0 \end{align*}\]

Steady-state from frequency response

\[\begin{align*} e_{\mathrm{ss}}(t) &= \bar{y} \, |S(j \omega)| \cos( \omega t + \phi + \angle S(j \omega)) & & \implies & | e_{\mathrm{ss}}(t) | &\leq | S(j \omega) | \, |\bar{y}| \end{align*}\]

e.g. car model with a step reference

\[\begin{align*} | e_{\mathrm{ss}}(t) | &\leq | S(0) | \, |\bar{y}| = \frac{ \frac{b}{m}}{\frac{b}{m} + \frac{p}{m} K} |\bar{y}| = \frac{1}{1 + \frac{p}{b} K} |\bar{y}| \end{align*}\]

Asymptotic tracking

Definition

Error converges to zero for large time \[\begin{align*} \lim_{t \rightarrow \infty} e(t) &= 0 \end{align*}\]

Stability of \(S\) is not enough. It is necessary that \[\begin{align*} E = S \, \bar{Y} \end{align*}\] Poles of \(\bar{Y}\) must be cancelled by zeros of \(S\)!

Lemma 4.1

Let \(S = (1 + G \, K)^{-1}\) be asymptotically stable and \[\begin{align*} \bar{y}(t) = \bar{y} \cos(\omega t + \phi) \end{align*}\] If \(G \, K\) has a pole at \(s = j \omega\) then \(\lim_{t \rightarrow \infty} e(t) = 0\)

Proof: \(S\) has a zero at \(s = j \omega\)

Example: toilet bowl model

Open-loop transfer-function

\[\begin{align} G(s) &= \frac{1}{s A} \end{align}\]

Closed-loop sensitivity transfer-function

\[\begin{align} S(s) &= \frac{1}{1 + G(s) \, K} = \frac{1}{1 + \frac{K}{s A}} = \frac{s}{s + \frac{K}{A}} \end{align}\]

Asymptotically stable for \(K > 0\)

\(S\) has a zero at zero: asymptotic tracking of step (constant) reference inputs

Example: car model with integral feedback

Open-loop transfer-function

\[\begin{align*} G(s) = \frac{\frac{p}{m}}{s + \frac{b}{m}} \end{align*}\]

Integral control

\[\begin{align*} K(s) = \frac{K}{s} \end{align*}\]

Closed-loop sensitivity transfer-functions

\[\begin{align*} S(s) = \frac{1}{1 + \frac{K}{s} \frac{\frac{p}{m}}{s + \frac{b}{m}}} = \frac{s\left (s + \frac{b}{m}\right )}{s^2 + \frac{b}{m} s + K \frac{p}{m}} \end{align*}\]

Asymptotically stable for \(K > 0\)

\(S\) has a zero at zero: asymptotic tracking of step (constant) reference inputs

Example: car model with integral feedback

Controller

\[\begin{align*} U(s) &= \frac{K}{s} E(s) & & \implies & u(t) &= u(0) + K \int_0^t e(\tau) \, d\tau \end{align*}\]

System + Controller block-diagram

Example: car model with integral feedback

Transient response has oscilations!

4.2 Stability and Transient Response

Some facts

  • The zeros of \(1 + G K\) are the poles of \(S\)
  • The poles of the product \(G \, K\) are the zeros of \(S\)
  • When \(G\) and \(K\) are rational the order of \(S\) is the order of the product \(G \, K\)

Example: rational transfer-function with no pole-zero cancellations

\[\begin{align*} G &= \frac{N_G}{D_G}, & K &= \frac{N_K}{D_K}, & S &= \frac{1}{1 + \frac{N_G}{D_G} \frac{N_K}{D_K}} = \frac{D_G D_K}{D_G D_K + N_G N_K} \end{align*}\]

For tracking

  • Add poles to \(K\) so they become zeros of \(S\)
  • More complex \(K\) leads to more complex \(S\)
  • Watch out for stability and oscillations, e.g. car model with integral feedback

Example: car model with proportional plus integral feedback

Proportional plus integral control

\[\begin{align} G(s) &= \frac{\left . {p} \middle / {m} \right .}{s + \left . {b} \middle / {m} \right .}, & K(s) &= K_p + s^{-1} K_i \end{align}\]

Closed-loop sensitivity transfer-functions

\[\begin{align*} S(s) &= \frac{1}{1 + G(s) \, K(s)} %% = \frac{1}{1 + \displaystyle\frac{\frac{p}{m} (K_p %% s + K_i)}{s \left (s + \frac{b}{m} \right )}} = \frac{s \left (s + \frac{b}{m} \right )}{s^2 + \left ( \frac{b}{m} + \frac{p}{m} K_p \right ) s + \frac{p}{m} K_i} \end{align*}\]

No complex-poles (oscillations) if \[\begin{align*} 0 < 4 \frac{p}{m} K_i &\leq \left ( \frac{b}{m} + \frac{p}{m} K_p \right )^2 \end{align*}\]

Example: car model with proportional plus integral feedback

Controller

\[\begin{align*} U(s) &= \left ( K_p + \frac{K_i}{s} \right ) E(s) & & \implies & u(t) &= u(0) + K_p e(t) + K_i \int_0^t e(\tau) \, d\tau \end{align*}\]

System + Controller block-diagram

Example: car model with proportional plus integral feedback

Special choice

\[\begin{align*} K_i &= \frac{b}{m} K_p, & K(s) &= K_p + \frac{K_i}{s} = K_p \frac{s + \frac{b}{m}}{s} \end{align*}\] leads to a pole-zero cancellation \[\begin{align*} G(s) \, K(s) &= \frac{\frac{p}{m}}{s +\frac{b}{m}} \times K_p \frac{s + \frac{b}{m}}{s} = \frac{\frac{p}{m} K_p}{s}, \end{align*}\]

Closed-loop sensitivity transfer-functions

\[\begin{align*} \label{eq:scarpipzc} S(s) &= \frac{1}{1 + G(s) K(s)} = \frac{s}{s + \frac{p}{m} K_p} \end{align*}\]

Zero at the origin: asymptotic tracking

No complex poles: no oscillations

Example: car model with proportional plus integral feedback

Asymptotic tracking, gone are the oscillations!

Beware of control effort: saturation occurs if \[\begin{align*} K_p > \frac{\overline{u}}{\bar{y}} = \frac{3}{60} = 0.05 \end{align*}\]

4.3 Integrator Wind-up

Example: car model with proportional plus integral feedback

Nonlinear input saturation

Output overshoots

Example: car model with proportional plus integral feedback

Output converges slowly

Integrator wind-up

4.4 Feedback with Disturbances

Loop relations

\[\begin{align*} y &= G \, (u + w), & u &= K \, \tilde{e}, & \tilde{e} &= \bar{y} - (y + v) \end{align*}\]

Closed-loop transfer-functions

\[\begin{align*} y &= H \, \bar{y} - H \, v + D \, w, & u &= Q \, \bar{y} - Q \, v - H \, w \end{align*}\]

Tracking error

\[\begin{align*} e &= \bar{y} - y = S \, \bar{y} + H \, v - D \, w \end{align*}\]

Beware \(\tilde{e} \neq e\)!

4.4 Feedback with Disturbances

Closed-loop transfer-functions

\[\begin{align*} y &= H \, \bar{y} - H \, v + D \, w \\ u &= Q \, \bar{y} - Q \, v - H \, w \\ e &= S \, \bar{y} + H \, v - D \, w \end{align*}\]

Gang of four

\[\begin{align*} S &= \frac{1}{1 + G K}, & D &= \frac{G}{1 + G K} & H &= \frac{G K}{1 + G K}, & Q &= \frac{K}{1 + G K} \end{align*}\]

4.4 Feedback with Disturbances

Gang of four

\[\begin{align*} S &= \frac{1}{1 + G K}, & D &= G S, & H &= G K S, & Q &= K S \end{align*}\]

  • \(S\) has as zeros the poles of \(G K\)
  • No pole-zero cancellations in \(G K \, \implies \, S\), \(D\), \(H\), \(Q\) have the same poles!
  • Stabilizing \(S\) stabilizes all other transfer-functions
  • If there are pole-zero cancellations in \(G K\) be careful and see Section 4.7

4.5 Input-Disturbance Rejection

Closed-loop transfer-functions

\[\begin{align*} e &= S \, \bar{y} - D \, w, & D &= G S = \frac{G}{1 + G K} \end{align*}\]

Lemma 4.2

Let \(S = (1 + G \, K)^{-1}\) and \(D = G S\) be asymptotically stable and \[\begin{align*} \bar{y}(t) &= \bar{y} \cos(\omega t + \phi), & w(t) &= \bar{w} \cos(\omega t + \psi) \end{align*}\] If \(K\) has a pole at \(s = j \omega\) then \(\lim_{t \rightarrow \infty} e(t) = 0\)

Proof: \(S\) and \(D\) have zeros at \(s = j \omega\)

Example: car model with proportional feedback

Closed-loop transfer-functions

\[\begin{align*} S(s) &= \frac{s + \frac{b}{m}}{s + \frac{b}{m} + \frac{p}{m} K}, & D(s) &= S(s) \, G(s) = \frac{\frac{p}{m}}{s + \frac{b}{m} + \frac{p}{m} K} \end{align*}\]

No asymptotic tracking, no asymptotic disturbance rejection

Example: car model with integral feedback

Closed-loop transfer-functions

\[\begin{align} S(s) &= \frac{s \left ( s + \frac{b}{m} \right )}{s^2 + \frac{b}{m} s + \frac{p}{m} K}, & D(s) &= S(s) \, G(s) = \frac{\frac{p}{m} s}{s^2 + \frac{b}{m} s + \frac{p}{m} K} \end{align}\]

Asymptotic tracking and asymptotic disturbance rejection, oscillations

Example: car model with integral feedback

Closed-loop transfer-functions

\[\begin{align} S(s) &= \frac{s \left ( s + \frac{b}{m} \right )}{s^2 + \frac{b}{m} s + \frac{p}{m} K}, & D(s) &= S(s) \, G(s) = \frac{\frac{p}{m} s}{s^2 + \frac{b}{m} s + \frac{p}{m} K} \end{align}\]

Asymptotic tracking and asymptotic disturbance rejection, oscillations

Example: car model with proportional plus integral feedback

Closed-loop transfer-functions

\[\begin{align} S(s) &= \frac{s}{s + \frac{p}{m} K_p}, & D(s) &= S(s) \, G(s) = \frac{\frac{p}{m} s}{\left ( s + \frac{b}{m} \right ) \left ( s + \frac{p}{m} K_p \right ) } \end{align}\]

Asymptotic tracking and asymptotic disturbance rejection, no oscillations

4.5 Input-Disturbance Rejection

Example: car model with various feedback controllers

Integral controller is sensitive to disturbances at \(0.1\) rad/s

Watch out for rolling hills!

Example: toilet bowl model

Loop relations

\[\begin{align*} G(s) &= \frac{1}{s A}, & K(s) &= K \end{align*}\]

Closed-loop transfer-functions

\[\begin{align*} S(s) &= \frac{s}{s + \frac{K}{A}}, & D(s) &= \frac{\frac{1}{A}}{s + \frac{K}{A}} \end{align*}\]

Asymptotic tracking but no asymptotic disturbance rejection!

Pole must be at the controller for asymptotic disturbance rejection

\(D\) does not have a zero at the origin

4.6 Measurement noise

Closed-loop transfer-functions

\[\begin{align*} e &= S \, \bar{y} + H \, v, & S &= \frac{1}{1 + G K}, & H &= G K S \end{align*}\]

Fundamental limitation

\[\begin{align*} S + H = 1 \end{align*}\]

At every frequency

\[\begin{align*} S(j \omega) + H(j \omega) = 1 \end{align*}\]

  • Good tracking (\(S(j\omega) \approx 0\)) means bad measurement noise rejection (\(H(j\omega) \approx 1\))

  • Good measurement noise rejection (\(H(j\omega) \approx 0\)) means bad tracking (\(S(j\omega) \approx 1\))

  • Tracking requires good sensors!

  • \(|S(j \omega)| + |H(j \omega)| \neq 1\): both \(S\) and \(H\) can be large at the same time!

4.7 Pole-zero Cancellations and Stability

Open-loop

\[\begin{align*} G &= \frac{N_G}{D_G}, & K &= G^{-1} = \frac{D_G}{N_G}, \end{align*}\]

  • \(G\) cannot be strictly proper or \(K\) will not be proper
  • Even if \(G\) is proper but not strictly proper and \(K = G^{-1}\) \[\begin{align*} y &= G \, K \, \bar{y} + G w = \bar{y} + G w, & u &= K \bar{y} \end{align*}\]
  • Both \(G\) and \(K\) must be stable or \(y\) or \(u\) will be unbounded
  • Stabilization of unstable systems cannot be done in open-loop

4.7 Pole-zero Cancellations and Stability

Closed-loop

Internal stability if

\[\begin{align*} S, \qquad S \, G, \qquad S \, K, \qquad S \, G \, K \end{align*}\] are all asymptotically stable

Lemma 4.3:

Internal stability if and only if \(S\) is asymptotically stable and any pole-zero cancellation in \(G K\) is of stable poles or zeros

Proof (sketch): If \[\begin{align*} G &= \frac{(s - z)}{(s - p)} \tilde{G}, & K &= \frac{(s - p)}{(s - z)} \tilde{K}, & G K &= \tilde{G} \tilde{K}, & S &= \tilde{S} = \frac{1}{1 + \tilde{G} \tilde{K}} \end{align*}\] However \[\begin{align*} S G &= \frac{(s - z)}{(s - p)} \tilde{S} \tilde{G}, & S K &= \frac{(s - p)}{(s - z)} \tilde{K} \tilde{G} \end{align*}\] so that \(z\) and \(p\) must both have negative real part

Example: car model with proportional plus integral feedback

Model and controller

\[\begin{align*} G(s) &= \frac{\left . {p} \middle / {m} \right .}{s + \left . {b} \middle / {m} \right. }, & K(s) &= K_p \frac{s + z}{s}, & z = \frac{K_i}{K_p} = \frac{b}{m} \end{align*}\]

Closed-loop transfer-functions

\[\begin{align*} S(s) &= \frac{s}{s + \frac{p}{m} K_p}, & H(s) &= \frac{\frac{p}{m} K_p}{s + \frac{p}{m} K_p}, & Q(s) &= \frac{K_p \left ( s + \frac{b}{m} \right )}{s + \frac{p}{m} K_p}, & D(s) &= \frac{\frac{p}{m} s}{\left ( s + \frac{b}{m} \right ) \left ( s + \frac{p}{m} K_p \right ) } \end{align*}\]

Canceled pole is a zero of \(Q\) and a pole of \(D\)

Example: car model with proportional plus integral feedback

Imperfect cancellation

\[\begin{align*} K_i &= \gamma \frac{b}{m} K_p \end{align*}\]

More to come in Chapter 6

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