Digital phase-lead and deadbeat compensators

Digital phase-lead and deadbeat controller

The first part is the design of a phase-lead digital controller with small steady state error, small settling time, minimum input action, and minimum overshoot. The second part is a deadbeat controller that reaches zero error very fast at sampling instants in discrete domain systems.

Pre-requisites

  • Digital Control Systems theory.
  • Bode and root-locus analysis theory
  • Jupyter Notebook with Matlab-kernel.
  • Matlab.

Source code

Version PDF/HTML. Matlab and LaTex source code on GitHub.

Digital phase-lead controller

The aim of this task is to design a digital controlled system with some requirements, small steady state error, small settling time, minimum input action, and minimum overshoot.

The plant to be studied is, \begin{align} G_p(s) &= \dfrac{0.04(s+1)}{s^2+0.2s+0.04} \tag{1} \end{align} and the digital controller should have the form, \begin{align} D(z) &= K\dfrac{z-A}{z-B} \tag{2} \end{align}

Matlab script: closed-loop step response and Bode diagram

In [2]:
clear variables;
%% plant Gp 
s = tf('s');
Gp = 0.04*(s+1)/(s^2+0.2*s+0.04); % uncompensated plant

fig = figure(1); 
margin(Gp); % calculates the phase margin and gain margin at their frequencies
[Gm,Pm,Wcg,Wcp] = margin(Gp);
Gcl = feedback(Gp,1); % closed-loop of the uncompensated plant

In [3]:
fig = figure(2);
step(Gcl); % step response to the closed-loop system
stepinfo(Gcl) % system performance values
ans = 

  struct with fields:

        RiseTime: 5.0078
    SettlingTime: 28.6751
     SettlingMin: 0.4631
     SettlingMax: 0.6201
       Overshoot: 24.0167
      Undershoot: 0
            Peak: 0.6201
        PeakTime: 11.1292


The result shows that the plant is very slow with a big overshoot. A phase margin ($PM$) of $60°$ can be the unique requirement. As long as the compensated phase margin is around that value, the settling time and overshoot should be minimized as much as possible.

Phase-Lead compensator

The selected compensator can be written as follows, \begin{align} G_c = K_c~\dfrac{s+z}{s+p},\quad |z|\leq|p| \tag{3} \end{align}

Step 1

Calculate a gain $K$ that satisfies the desired phase margin $PM_d=60°$. Applying the angle condition for a $PM_d$ over $G_p$, \begin{align} \measuredangle G_p(j\omega_c^\prime) &= PM_d - 180° \tag{4} \\ \measuredangle G_p(j\omega_c^\prime) &= 60° - 180° \nonumber \\ \measuredangle G_p(j\omega_c^\prime) &= -120° \nonumber \end{align} where $\omega_c^\prime$ is the new crossover frequency for the desired phase margin $PM_d=60°$.

Using the bode plot of $G_p$, the logarithm gain at $-120°$ is $-8.02~dB$, so the gain $K$ can be calculated as follows, \begin{align} 20\log_{10} K &= |-8.02| \tag{5} \\ K &= 2.52 \nonumber \end{align}

Therefore, the uncompensated $G_p$ that satisfies the desired phase margin is, \begin{align} G_{p1} &= K~G_p \nonumber\\ G_{p1} &= 2.52 \dfrac{0.04(s+1)}{s^2+0.2s+0.04} \tag{6} \end{align}

In [4]:
%% Design requirements
PO = 10; % percentage overshoot
zeta = log(100/PO)/sqrt(pi^2+ (log(100/PO))^2 ); % damping ratio
PM_d = round(100*zeta)+1; % desired PM

%% Obtaining the gain K that meets the desired PM_d
K = 10^(8.03/20); % the gain 8.03 obtained from the bode plot
Gp1 = K*Gp; % new uncompensated plant

fig = figure(11);
margin(Gp1);

The result shows that the new uncompensated plant $G_{p1}$ satisfies the desired phase margin.

Step 2

The digital uncompensated $G_{z1}$ plant of $G_{p1}$ can be calculated using a zero-order holder with a sampling time $T_s=0.01$. \begin{align} G_{z1} &= 1.01 \times 10^{-3}\frac{z-0.99}{z^2-1.99z+0.99} \tag{7} \end{align}

In [5]:
%% Digital uncompensated system with the new gain K
Ts = 0.01; % sampling time
Gz1 = c2d(Gp1,Ts,'zoh'); % convert the continuos time plant Gp2 to the
% discrete time domain with the zero order holder
Gp1_cl = feedback(Gp1,1); % closed-loop system in continuos time
Gz1_cl = feedback(Gz1,1); % closed-loop system in discrete time

fig = figure(3);
step(Gp1_cl,Gz1_cl); % step response of both continuos and discrete systems
stepinfo(Gz1_cl) % system performance values
ans = 

  struct with fields:

        RiseTime: 3.5100
    SettlingTime: 21.4000
     SettlingMin: 0.6452
     SettlingMax: 0.9128
       Overshoot: 27.4986
      Undershoot: 0
            Peak: 0.9128
        PeakTime: 8.0100


It can be seen that the continuous and discrete plant are almost similar. Also, the settling time has been reduced but the steady state error is too big.

Step 3

With the desired phase margin, the value of $\beta$ can be calculated as follows, \begin{align} PM_{act}- & PM_d+\theta = \arctan\dfrac{\beta-1}{2\sqrt{\beta}} \tag{8} \end{align} where $PM_{act}$ is the actual phase margin of the uncompensated plant $G_{p1}$, and $\theta$ is a factor of correction.

After some operations, \begin{align} \beta^2- & \beta\left[2+4\left(\tan(PM_d-PM_{act}+\theta)\right)\right] +1 = 0 \nonumber\\ \end{align} if $\theta=6°$, $PM_d=60°$, and $PM_{act}=61.25°$ obtained from the Bode plot below. The $\beta$ will be, \begin{align} &\qquad \beta = 1.18 \nonumber \end{align}

In [6]:
%% Designing the Phase-lead digital controller 
% introducing a new gain 10 times faster in order to obtain a fast
% response to the step input
Gp2 = K*10*Gp;

fig = figure(4);
margin(Gp2); % checking the desired phase margin PM_d 
[Gm1,PM_act,Wcg1,Wcp1] = margin(Gp2);

% obtaining beta with the actual PM and the desired PM
theta = 6; % correction factor
beta = roots( [1 -(2+4*( tand(PM_d-PM_act+theta) )^2) 1] )
% beta = (1+sind(PM_d-PM_act+theta))/(1-sind(PM_d-PM_act+theta));
beta =

    1.1804
    0.8472


Step 4

Now, the crossover over frequency $\omega_c$ needs to be calculated using the following gain condition formula, \begin{align} \left| G_{p1}(j\omega_c) \right| &= \dfrac{1}{\sqrt{\beta}} \tag{9} \end{align} if we use the peak magnitude $M_{pc}$ relation, \begin{align} M_{pc} &= \dfrac{1}{\sqrt{\beta}} \tag{10} \end{align} and using getGainCrossover command in Matlab, \begin{align} \omega_c& = 1.37 ~rad/sec \nonumber \end{align}

In [7]:
% calculating the new crossover frequency
Mpc = 1/sqrt(beta(1)); % find compensator peak magnitude.
omega_c = getGainCrossover(Gp2,Mpc); % The new gain crossover frequency wc

Step 5

Determining the zero of the controller, \begin{align} \omega_c &= \sqrt{\beta z^2} \tag{11} \\ z &= 1.26 \nonumber \end{align} therefore, the compensator in continuous time can be written as follows, \begin{align} G_c &= \beta\dfrac{s+z}{s+\beta z} \tag{12} \\ G_c &= 1.18~\dfrac{s+1.26}{s+1.49} \nonumber \end{align} and in discrete time, \begin{align} G_cz &= 1.18~\dfrac{z-0.99}{z-0.98} \nonumber \end{align}

The open-loop compensated in continuous and discrete time are, \begin{align} G_{ol} &= G_c~ G_{p1} \tag{13} \\ G_{ol} &= 1.18\quad\dfrac{s+1.26}{s+1.49} \quad 2.52\quad\dfrac{0.04(s+1)}{s^2+0.2s+0.04} \tag{14} \\ G_{olz} &= 0.01\quad \dfrac{z-0.99}{z-0.98}\quad\dfrac{z-0.98}{z^2-1.99+0.99} \tag{15} \end{align}

In [8]:
% determining the zero and the pole of the controller
zc = omega_c/sqrt(beta(1)); % zero of the controller
pc = beta(1)*zc; % pole of the controller

%  controller in continuous and discrete time
Gc = beta(1)*(s+zc)/(s+pc); % phase-lead controller in continuous time
Gcz = c2d(Gc,Ts,'zoh'); % phase-lead controller in discrete time

%% Evaluating the phase-lead controller
Gol = Gc*Gp2; % open-loop compensated in continuous time
Gcl = feedback(Gol,1); % closed-loop in continuous time
Gp2z = c2d(Gp2,Ts,'zoh'); % plant in discrete time
Golz = Gcz*Gp2z; % open-loop compensated in discrete time
Gclz = feedback(Golz,1); % closed-loop in discrete time

fig = figure(5);
margin(Gol);


fig = figure(6);
step(Gcl,Gclz);

The results shows that the compensated system using a Phase-Lead controller achieves with success the desired phase margin of $60°$, and has a small settling time $t_s=4.84~sec$.

Deadbet controller

The deadbeat controller approach finds an input signal in order to bring the output to the steady-state in the smallest number of time steps. It has some characteristics:

  • Zero steady-state error.
  • Minimum rise time.
  • Minimum settling time.
  • Less than 2% overshoot/undershoot.
  • Very high control signal output.
  • Can not be used in continuous time.

As long as the plant $G(z)$ has all zeros and poles inside the unit circle (minimum phase), the plant can be written as, \begin{align*} G(z) &= \dfrac{z^{d} B(z)}{A(z)} = \dfrac{z^{d}(b_o + b_1 z^{-1}+...+b_m z^{-m})}{1+a_1 z^{-1} + ... + a_n z^{-n}}, \quad d=n-m>0 \end{align*} with an step input $r(k)$, the deadbeat response requirement implies, \begin{align*} M(z) &= z^{-d} \end{align*}

The digital deadbeat will have the following transfer function, \begin{align} D(z) &= \dfrac{A(z)}{z^{-d}B(z)}~\dfrac{z^{-d}}{1-z^{-d}} \tag{16} \\ \end{align} and the closed-loop system is, \begin{align*} Y(z) &= z^{-d} U(z) \end{align*} which means, \begin{align*} y(k) &= u(k-d) \end{align*} So when $u(k)$ is a step input, the output will be d-step delay of the same signal.

Let's design a digital deadbeat controller for the plant, \begin{align*} G_p(s) &= \dfrac{0.04(s+1)}{s^2+0.2s+0.04} \\ G_p(z) &= \dfrac{5.47\times 10^{-2}(z-0.34)}{z^2 - 1.78z + 0.82} \end{align*} with sampling time $T_s=1$.

In [9]:
clear variables;
s = tf('s');
Gp = 0.04*(s+1)/(s^2+0.2*s+0.04); % uncompensated plant
Ts = 1; % sampling time
Gz = c2d(Gp,Ts,'zoh'); % System transfer function in discrete time

% Minimum phase case
z=zpk('z',Ts);
Mz = 1/z; % closed-loop because the relative order d = 1
Dz = Mz/(Gz*(1-Mz)); % deadbeat controller
Dz = minreal(Dz); % cancel common factors

fig = figure(1); 
step(Dz*Gz/(1+Dz*Gz)) % step response of closed-loop

In [10]:
fig = figure(2); 
step(Dz/(1+Dz*Gz)) % control signal
title("Control signal");

The closed-loop discrete system results in, \begin{align*} G_{clz} &= \dfrac{(z-1) (z-0.3392)^2 (z^2 - 1.783z + 0.8187)^2} {z (z-0.3392)^2 (z-1) (z^2 - 1.783z + 0.8187)^2} \end{align*} where the controller is, \begin{align*} G_{cz} &= \dfrac{18.288 (z-1) (z-0.3392) (z^2 - 1.783z + 0.8187)^2}{z (z-0.3392)^2 (z-1) (z^2 - 1.783z + 0.8187)} \end{align*}

The output of the deadbeat controller and the closed-loop system response to a unit step input shows clearly that the steady-state error reaches zero value at the first sampled time. In addition, the control signal is very high when the sampling time is increased.

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